RAR.Y
U N VERS ITY
or 1LL1 NOIS
L
B
I
OF THE
I
<V 510.7
\S59
Digitized by the Internet Archive
in
2012 with funding from
University of
Illinois
Urbana-Champaign
http://www.archive.org/details/highschoolmathem07auniv
HIGH SCHOOL MATHEMATICS
Unit 7.
MATHEMATICAL INDUCTION
UNIVERSITY OF ILLINOIS COMMITTEE ON SCHOOL MATHEMATICS
MAX BEBERMAN, Director
HERBERT E. VAUGHAN, Editor
UNIVERSITY OF ILLINOIS PRESS
URBANA,
1961
HIGH SCHOOL MATHEMATICS
Unit 7.
MATHEMATICAL INDUCTION
UNIVERSITY OF ILLINOIS COMMITTEE ON SCHOOL MATHEMATICS
MAX BEBERMAN,
HERBERT
UNIVERSITY OF ILLINOIS PRESS
Director
E.
VAUGHAN,
•
URBANA,
Ed/for
1961
©
1961 by the Board of Trustees of the University of Illinois.
Manufactured in the United States of America.
[7-iii]
PREFACE
In Unit 2 you formulated some of your knowledge about the real
numbers in ten basic principles. [These principles are repeated on
page 7-17 of the present unit.] You found, as you derived theorems
from these basic principles, that the latter contained, at least implicitly,
what you knew about the real numbers and, perhaps, some things
you hadn't known. In this unit you will, for one thing, discover a few
additional basic principles which you can use, as in Unit 2, to organize
more of your knowledge about real numbers and to learn more about them.
For example, you will discover simple basic principles which you can
use to derive the transformation principles for inequations [see page 7-36]
that you learned to use in Unit 3. Knowledge of these new basic principles
will also help you to discover new things about inequations, and, in par-
most
of
ticular, about the positive
numbers.
The fact just mentioned- -that some of the new basic principles will
imply new theorems about positive real numbers- -should remind you
that the basic principles and theorems of Unit 2 formulate only properties
which hold of all real numbers [or, in the case of division, of all nonzero
Special kinds of real numbers --for two examples, the
real numbers].
positive numbers and the positive integers - -have special properties
which do not hold of all real numbers. For instance, although it is the
case that between any two real numbers there is a third real number,
no positive integer between, say, the positive integers 5 and 6.
Among the new basic principles which you will discover in this unit there
are some which will take account of the basic differences between the
+
of positive integers.
set of all real numbers and the set, I
Using these
you will be able to discover interesting properties peculiar to the positive
integers. In particular, you will learn a special method--proof by mathematical induction- -for proving generalizations about positive integers.
You will find this method of proof very useful, not only here, but in all
your later work in mathematics. [Incidentally, you will find many
opportunities in this, as well as in later units, to practice the techniques
you learned in Unit 6 for writing column proofs and paragraph proofs.]
there
is
,
Just as the theorems you proved in Units 2 and 6 justified techniques
which were helpful in applying your knowledge of real numbers and
geometric figures to the solutions of various sorts of problems, so the
theorems you learn in this unit will help in solving other kinds of problems. However, as you should have learned by now, mere knowledge
of the theorem which justifies a technique does not guarantee that one is
adept at using the technique. Such techniques --simplification of algebraic
expressions is an example- -must be practiced, both by themselves and
as steps in the solutions of problems, if one is to learn to make good
use of them. The Miscellaneous Exercises throughout this unit and the
Review Exercises at the end furnish opportunity for the practice of both
old and new techniques.
[7-v]
TABLE OF CONTENTS
Preface
[7-iii]
Introduction
7.01
[7-1]
Some problems
[7-1]
The real numbers
[7-3]
Addition
[7-3]
The addition table
[7-4]
Algorithm
[7-4]
A
proof of the theorem
'2+2=4*
[7-6]
Multiplication
[7-7]
The multiplication table
A
proof of the theorem
4
[7-7]
2» 4 =
8'
[7-9]
Miscellaneous Exercises
[7-10]
Division-with-remainder algorithm
[7-12]
Using the division-with-remainder algorithm to find
the value of a function for a given
argument
[7-14]
Other computing facts
The
[7-16]
10 basic principles for real
numbers used
[7-17]
in Unit 2
1^0
The eleventh basic principle:
The notations '=^
>
*
[7-18]
for conditionals and '<==>
,
for
biconditionais
[7-19]
Miscellaneous Exercises
[7-20]
7.02 The positive numbers
Basic principles (Px ) and (P2
[7-22]
)
for the set of positive
numbers
[7-22]
Closure principle, (P3 )> for addition
of positive
numbers
[7-23]
Closure principle, (P4 ), for multiplication of
positive
numbers
[7-24]
[CONTENTS]
[7-vi]
A
principle, (N), which defines the negative
7-25]
Miscellaneous Exercises
7-27]
Exploration Exercises --
7-29]
Investigating the relation {(x,
7.
numbers
y)
y
:
-
x
P}
€
03 Inequations
7-30]
Basic principle (G) for the relation >
7-30]
Inequality theorems
Two
7-29]
proofs
7-32]
of the
theorem:
VVV
[x + z >
x y z
l
y3 + z
—q=5
The inference scheme:
—M->
=>
x >
y]
7J
+*
not p
The transformation principles for inequations
7-34]
7-35]
7-36]
Addition; multiplication; factoring
7-36]
Proving part
7-37]
of the
factoring transformation principle
Inequality theorems on the squaring function
7-38]
A
7-39]
geometric application
VV x2
of 'VX
+ y2 > 2xy*
The reciprocating function
7-39]
Inequality theorems on the reciprocating function
7-40]
Monotonicity
7-41]
functions
of
Graphing inequations on a number line picture
Systems
two variables
of linear equations in
7-41]
7-42]
Miscellaneous Exercises
7-43]
Exploration Exercises --
7-45]
Discovering whether a given set
Proofs that
2,
is
an inductive set
and 4 belong to I*
3,
7.04 The positive integers
of all
Basic principles,
7-46]
7-47]
Assumptions, used in Unit
Defining the set
7-45]
+
I-
,
4,
about positive integers
positive integers
l*
t
I* tor the positive integer
7-47]
7-48]
7-49]
Exercises on inductive sets
7-50]
Proofs by mathematical induction
7-51]
V
[CONTENTS]
A
[7-vii]
3 +
of 'V
proof
r
n
The principle
of
n
€ I
+'
by mathematical induction
mathematical induction [PMI]
[7-51]
[7-53]
Inductive proof: initial step, inductive step
(using the inductive hypothesis), final step
A
[7-53]
short cut in writing inductive proofs
Closure
of
+
I
[7-54]
with respect to + and X
Two methods
[7-56]
for proving generalizations
Combining both methods
to
prove
m Vn m + n e
*
[7-56]
+*
I
Miscellaneous Exercises
[7-57]
[7-59]
Reflection of a point with respect to a point
Recursive definitions
[7-60]
[7-61]
Computing triangular numbers
[7-61]
Recursion--recursive definition
[7-62]
The function T
[7-62]
Explicit definition
[7-62]
Square numbers --S
[7-62]
Finding an explicit definition for T
[7-63]
Proofs
Tn
of 'V
n
=
SjE + iT
Column
[7-64]
Paragraph
[7-65]
Tree-diagram
[7-65]
Even and odd positive integers
Finding that S
is a function of
[7-66]
T
[7-66]
Practice with recursive definitions
[7-68]
Compound
interest formulas
[7-68]
Graphing recursively defined functions
[7-69]
Oblong numbers
[7-69]
Pentagonal numbers --polygonal numbers
[7-70]
A
pattern for recursive definitions of polygonal
number functions --the notation: P
n
The number of p-membered subsets of an n-membered
set--the notations:
C(n,
and
p)
x
tr '
:
C
n p
[7-70]J
*•
[7-71]
l
J
Finding a recursive procedure for computing values
of
l
C(n, p)'
Applications
[7-72]
[7-74]
[CONTENTS]
[7-viii]
"The arithmetic
of the positive
Recursive definitions,
+
I
+
and
,
4
integers
I
,
5
[7-76]
for addition and
multiplication, respectively, of positive integers
*
Derivation of the apa for positive integers from
I
4
[7-78]
Derivation of the cpa for positive integers from
I
4
[7-79]
Miscellaneous Exercises
[7-80]
7.05 The relation greater than for the positive integers
For each positive integer there
Vm Vn
Lower bounds and
4
Definitions of
The
7.06
[7-76]
least
is
[7-84]
a next one:
[n>m=>n>m+l]
~
L
[7-84]J
L
J
members
least
[7-87]
a lower bound' and of *a least
number theorem
[for
member'
+
I
[7-87]
[7-88]
]
The cofinality principle
[7-89]
Exercises on bounds
[7-90]
Misuses
[7-90]
of
induction
Miscellaneous Exercises
[7-91]
The integers
[7-94]
A
definition
(I)
for the set of integers
Closure theorems for integers
:
[7-94]
opposition, addition,
subtraction, and multiplication
[7-95]
Exploration Exercises--
[7-96]
Investigating the translating function,
each integer k onto k
-
ti,
which maps
[7-96]
j
Investigating the reflecting function, r:, which
each integer onto
its
The least number theorem
maps
opposite
[7-97]
for integers
[7-98]
k>j}
Induction over {k:
[7-100]
Induction over the set of all integers
[7-100]
Miscellaneous Exercises
[7-101]
The greatest integer function
[7-102]
The notation
'[
.
.
.
]
*
for 'the integral part of
*
.
.
.
[7-102]
[CONTENTS)
[7-ix]
{(x, y)
[xj}
y =
:
is called
4
the greatest integer
[7-104]
function'
A
proof
of
,
Vx Vk
[x] <=> k < x
[k =
< k +
[7-104]
1]'
Division -with- remainder
[7-106]
The
[7-106]
least integer function
The notation
{(x, y)
y =
:
'{{... J' for 'the fractional part of
x }}
{[
is
.
.
.'
[7-107]
called 'the fractional part
[7-107]
function'
^Discovering a technique for "splitting'* each positive
integer into multiples of powers of a given
integer >
[7-107]
1
Partial quotient and remainder
[7-111]
Miscellaneous Exercises
[7-112]
Divisibility
[7-115]
A theorem
that
The
|
is
stating that
j
is transitive,
and one stating
antisymmetric
[7-115]
divisibility relation
[7-116]
Similarities and differences between
Examining the way
in
which
J
and <
orders the positive
|
integers
[7-117]
Prime numbers
"^Highest
[7-116]
common
[7-118]
factor
[7-119]
The symbols 'HCF' and 'GCD'
[7-120]
Integral linear combinations
[7-121]
Euclid's algorithm for finding greatest
common
divisors
[7-123]
Exploration Exercises--
Discovering whether a graph
[7-125]
of a linear
equation
contains a point with integral coordinates
[7-125]
Finding pairs of integers which satisfy a linear
equation
Finding
all
[7-125]
integral solutions of '6x + lOy = 0'
Diophantine problems
Finding
[7-125]
[7-127]
all the integral solutions of a linear
equation in two variables
[7-128]
.
[CONTENTS]
[7-x]
Relatively prime integers
[7-129]
Solving linear Diophantine equations
[7-130]
Review Exercises
[7-133]
Basic Principles and Theorems
[7-145]
Basic principles for the real numbers
[7-145]
Theorems from Unit
[7-145]
2
Basic principles for P--basic principle for >
[7-149]
Theorems about >
[7-149]
Basic principles for
Theorems about
+
[7-151]
I
positive integers
[7-151]
Cofinality principle and basic principle for
I
Theorems about integers
[7-152]
[7-152]
Definitions of the greatest integer function and the
fractional part function
Theorems involving
*|[
.
.
.
[7-153]
J'
and '|
.
.
}'
[7-153]
Definition of the divisibility relation
[7-153]
Theorems about
[7-153]
|
Table
of
Trigonometric Ratios
[7-155]
Table
of
Squares and Square Roots
[7-156]
[Introduction]
[7-1]
Some problems --Try
.
I.
Al Moore
problems:
to solve these
starts at the rate of $4000 per year.
The
first offer
an annual increase of $400.
If
he plans on staying with one
him more money?
position for five years, which one will pay
II.
Stan
Brown also has a choice between two
The
for 30 days.
with
1
first job
is to last
cent the first day, 2 cents the second day, 4 cents the third
Which job pays more? By how much?
Study the following sentences and complete the last two.
Vl +
3
= 2
Vl +
3 + 5
= 3
Vl +
3 + 5 + 7
= 4
Vl +
3 + 5 + 7 + 9
= 5
•
IV.
Each job
jobs.
pays $100 per day. The second job starts
day, etc. doubling each day.
III.
promises a
The second assures him
salary increase of $100 each six months.
of
Each position
decide between two job offers.
is trying to
V1 +
3 + 5 + ...
V1 +
3 + 5 + ...
•
•
+87 =
+999 =
Study these sentences and complete the last one.
2<-L+J-+JL+_L<3
VT
JL
+
VT
18 <
Use
V3
Vz
•
VT
+
*
vr
-L
—
—
VT
vr
-L
•
+
4<JL
-2_
VT
8 <
< 7
•
the facts that
—i—
»
< 19
= 1.414,
V3
= 1.7 32,
+
..
#
+
-L
-2<
< 9
+ ...
V25
+
+
•n
and VT
-L«
V?
i
fio
=
-i.+ J__ + J_ + JL+J-.< zJl
VT
VI
V3
V4
VT
<
i
2.236 to check
that
2/5
+
V2
-L
VT
Vioo
VI
-L
JL
VI
VT
VTe
+ ...
+
•
•
+
.
1
[Introduction]
[7-2]
V.
You can easily check by computing
and that (1.0
l)
4
Do
> 1.04.
that (1.0
1)
2
Do you think
so.
> 1.02,
that
(
1.
(
l)
3
>
1.
I)
100
> 2?
1
.
That (1.05) 100 > 6?
VI.
You can probably expand
3
(x + y) =
(x + y) 4 =
2
'
much
without
thinking:
x2 + 2xy + y 2
this are below.
Complete the last two.
(x + y)
Other equations
'(x + y)
like
2
=
x 3 + 3x 2 y + 3xy2 + y 3
x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4
(x + y) 5 =
(x + y)
VII.
6
=
Whatever route Milton takes
in going
burg High School he has to walk
from
his
home
at least nine blocks.
to
Zabranch-
How many
routes are there which are just nine blocks long?
r —
i
—
l/
HOME
L
1
i
r
_
__
.j
i
1
r
_
—
j
1
|
l
i
!
i
-\
J
Z.H.S.
You could certainly solve each
enough computation.
of
the foregoing problems by doing
But, as you will learn in this unit and the next
one, there are easier ways.
Also, each problem suggests a generalization which you couldn't
prove by any amount
of
computation.
ditional basic principles for real
unit.
In fact, the proofs
numbers which you
depend upon
will study in this
ad-
03,
[7-3]
[7.01]
The real numbers
7.01
tells
.
--If
you were to write a proof
you that the expanded form
find yourself
making use
(x + 2) 2
'
is *x 2 +
of basic principles,
and computing facts such as that
of
of
4
4x +
theorem
that
you would
4',
previously proved theorems,
and that 2*2
2 + 2 = 4
of a
The theorems,
= 4.
course, are consequences of the basic principles, but where do the
computing facts come from?
[There
is
a list of theorems on real
num-
bers starting on page 7- 145. The first 78 were proved in Unit 2.]
ADDITION
Statements
quences
4
*1
10' is
computing facts
of the basic principles
What you need
for
of
4
+ T,
to
3* is
know
is that
first studied place value,
Now,
let's
'9
kind just referred to are conse-
together with some obvious definitions.
the numeral
an abbreviation for
an abbreviation for
*3« 10 2 + 4« 10 +
of the
'2
+
'2* is
1',
'4'
merely an abbreviation
for
4
3 + l\
example,
is
an abbreviation for
7\
prove the theorem
2 +
'2 + 2 = 4'.
(l+l)
= (2 +
1)
+
[definition]
[apa]
1
=
3+1
[definition]
=
4
[definition]
same way, you can prove, in succession, *2 +
6',
and 2 + 8 = 10'. Do so. If you try, in
In the
2 + 4 =
prove
'2
4
.
.
.
and
And, as you have known since you
+ 1*.
'347', for
2 + 2 =
4
. . ,
.
,
3 = 5*,
this
way, to
+ 9 = 11', you need an extra step, and a definition relating to
place value:
2 + 9 =
2 +
(8+1)
= (2 + 8) +
A
A
definition
apa
Which step
is justified
1
=
A
10+1
theorem
=
1-10+1
=
11
A
A
theorem
definition
by a place value definition?
[7-4]
[7.01]
same step-by-step manner you could establish any computing
concerning sums of positive integers. For example, to prove
In this
fact
*54 + 368 = 422' you would prove a sequence of theorems.
would prove '54 +
use
it
in proving
4
Once you had proved
= 55'.
1
54 +
First, you
theorem, you would
this
Continuing in this way, you would
2 = 56'.
eventually have proved *54 + 367 = 421', and would use this theorem in
proving
4
54 + 368
Each
= 422'.
your proofs would make use only of
of
definitions [including place value],
proof] of the preceding
theorem
of the apa,
and [except for the first
[instead of proving
sequence,
in the
368 theorems, you could get by with proving 54 theorems
in grade school
you learned an easier way to establish this kind
of fact of arithmetic --that is,
the
sums
that is,
of all pairs of the
You
to add positive integers.
numbers
0,
you learned the addition table
this in adding larger
you used an
Which one?]
additional one of our basic principles for real numbers.
Back
if
1,
2,
4,
5,
6,
7,
8,
and 9~~
Then, you learned how to use
.
For example,
numbers.
3,
learned
first
to
add 675 and 237, you
placed the second numeral under the first and, after adding corresponding
digit
numbers and "carrying'', you ended up with something
/
like:
i
675
237
912
sum
This procedure for finding the
an example of an algorithm
is
though we shall not do so here, the steps you take
in
.
Al-
using this algorithm
can be justified on the basis of place value definitions, the addition table,
the cpa, the apa, and the dpma.
There
is
a similar algorithm for simplifying certain types of alge-
Study the following examples of
braic expressions.
Example
1
Simplify:
.
Solution
.
3x
(3x
+ 2x
+ 2x
-
7
x
-
9
5x + 5x + 3x
-
16
2
2x + 5x +
3
3
[Explain
3
how
2
- 7)
+ (2x
3
its
use.
+ x + 5x
2
- 9)
the principle for subtraction is used here.]
[7-5]
[7.01]
Example
2
Simplify:
.
3b
(a +
- c)
-
+ (a
(b - 3a
-
2c)
v
-
3c)
;
^
Solution.
Explain.
5a + 2b
EXERCISES
A.
Simplify.
1.
3a + 4b) + (7a + 9b)
2.
(5x
3.
-6t +
4.
(-9c + 7d) + (3c
5.
8x
6.
y + 7z) + (5y
7.
5a
8.
8x
9.
7y
-
2
+ (-9y
- z)
6b + 8c) + (3a
3x +
-
Ilk +
11.
5t
12.
7m
-
4b
-
5y + 8y)
- (2t 2
-
12)
- 5)
- 5)
+ (-2x + 3) + (2
13.
3a
-
14.
3x + 7
15.
8t
16.
lis 4
17.
3a
18.
y + 2y
19.
3x 3
-
2x +
20.
8a 3
-
3a b + 7ab 2
2
5y)
2k + 4n)
-
3x
+ 8t
- x)
7b) + (-2a + 5b) + (-a + 8b) + (4a
-
3t
2
5x)
+
7)
3s
-
-
-
2
-
+
2
- 5t
+ 9) + (5s
2ab + 3b 2
- 7)
2
4x
(9 +
(9t
)
+ (2y
1)
2
2
2
-
3
+
-
-
b
3x3
3
)
-
6b)
2
)
2
+ (3t - 5t + 9)
t)
4s 4 + 11)
4y +
-
+ (5x
5x
-
-
(-a 2
-
7d)
2
+ 7)
2
-
3c)
-
- 3t
3
-
6y) + (7x + 3y)
3z)
+ 9x
(8m
3n) +
-
2
7) + (5x
+ 6y) + (2
- 3
10.
2
5s)
(-2x + 3y) + (-6x +
3y) +
2
-
+ (-4t
s)
-
-
(3s 4
-
7
-
2s
2
)
5ab + 3b 2 )
3) -
(-8y
-
10 + 3y
3
3) - (6x - 2) -
+ (9b
3
-
3a 2 b + 7ab 2
2
)
-
2
+
1
-
3a 3 )
-
(a
(7x
2x)
3
-
b3 )
[7-6]
[7.01]
Sample
B.
.
Solution
3x
=
If f(x)
2
-
5x +
f(3a) = 3(3a)
1.
If
f(x) =
4x 2
2.
If
g(x) =
2x3
3.
If
g(x) =
x3
If f(x) =
5.
If f(x)
6.
Suppose
=
2
-
7
5(3a) + 7
=
27a 2
-
15a +
f(a) + f(3a) =
30a 2
-
20a + 14
8x
-
-
x2 +
x2 +
-
g(-x)
-
g(3x)
7,
,
,
2
and f(10)
f(3) =
9,
2
for each
f(x) =
x,
3g(x) =
[3o g](x) =
3
8x + 7x + 5x + 6 and g(x)
that,
=
g(3x) =
7,
7x + 2x +
7
f(2a) + f(a) =
- 3,
[3g](x) =
4.
5a +
f(a) = 3a* -
.
f(a) + f(3a) =
7,
the line containing the points
x
=
4x 2 + 3x +
-
2x
(2, f(2))
-
and
3.
=
f(10) + g(10) =
9,
What
(2 + h,
is the
f(2+h)),
slope of
where
h ^ 0?
C_.
Here
is a
ginal
comments.
column proof
of the
V x
x
(1)'
x
=
theorem
x
2+2=2+2
(2)
2 =
(3)
1
+
1
2 + 2 = 2 + (1 + 1)
(4)
V V X
= (x + y) +
1
(6)
2 + (1 + 1) = (2 + 1) +
1
(7)
2 + 2 = (2 + 1) +
1
(5)
(8)
(9)
x y
+ (y +
3=2+1
2+2=3+1
4 =
(10)
(11)
1)
2 +
3+1
2=4
'2 + 2 = 4'.
Fill in the
mar-
[7-7]
[7.01]
MULTIPLICATION
Just as statements of the computing facts of addition are really
theorems derivable from the basic principles and definitions
numerals and
•2«3
so are statements of the computing facts
of place value,
As an example, here
of multiplication.
a proof of the
is
=
2(1 +
=
2
•
1
+
1
+ 2
•
[definition]
1)
1
+ 2
•
[idpma]
1
= 2 + 2 + 2
[pml]
=
[addition table]
4 + 2
= 6
[addition table]
we had previously proved
[Explain.
proof.
rows
2*2
= 4,
we could have given a shorter
of the addition table
V V x
*
is a
that
]
In filling out the
which
+ (y +
1)
=
is
you could use the theorem:
(x + y) +
1,
y
consequence of the apa.
In filling out a
cation table, the first entry is given by the
it
theorem
6\
=
2*3
If
of certain
pml
row
in the multipli -
and, to find the others
convenient to use a theorem:
V V x(y
+1)
x y w
=
xy7 + x
[What basic principles do you need to refer to
theorem?]
order to prove this
For example,
2-4
Now, prove 24
As
in
in the
=
2(3 +
=
2*3
[definition]
1)
[theorem]
+ 2
= 6 + 2
[theorem]
=
[addition table]
8
5 = 10',
'2-6
= 12',
...,
and
l
2-9
=
18\
case of addition, the multiplication table gives us theorems
which are used
in
which you learned
carrying out the steps
in
grade school.
in the multiplication
algorithm
)
)
[7.01]
[7-8]
Again, there
a similar algorithm for simplifying certain types of
is
Study the following examples of
algebraic expressions.
Example
1
Simplify:
.
(3x
Solution.
2
+
2x
its
use.
+ l)(7x + 3)
3x' + 2x +
1
7x +
3
9x 2 + 6x +
3
2
Zlx 3 + 14x + 7x
21x 3 + 23x 2 + 13x +
Example
2
Simplify:
.
(5y
-
J
3y + 2)(2y
3y +
2
-
7
21y
-
14
3
2
41y + 4y + 21y
-
14
Solution.
5y
2
2/
35y*
-
10y
5
3
+
-
- 7)
2
6y + 4y
-
lOy5
3
EXERCISES
A.
The procedure used
3 -digit
plying a
in
Example
Example
in
number by
2 is like
1
above
a 2 -digit
multiplying a
(
is like that
number by a
-digit
in
multi-
The procedure used
number.
?
used
?
(
-digit
number.
B^.
Simplify.
2
-
(2t
3.
(8y
5.
(6b
7.
(8k + 2k
9.
(5 -
2
3
(7x
1)
4.
(2a
5)
6.
(7r
)
8.
(lis
+ 3x 4 )(7 + 5x)
10.
2
+ y
-
-
3)(5y
-
3b + 7)(4b +
2x
2
2
2.
3t + 5)(5t + 3)
1.
3
-
3)(5
-
7k 2
(8y
+ 3x
2
3
3
+ 2r
-
-
5)(-x + 7)
3a + 7)(5a + 4a
-
2
-
s
2
+ r)(3r
-
3)(2s
2
2
2
-
2
-
1)
2)
+ 3)
2y + 7y )(4y
2
-
y +
3)
[7-9]
[7.01]
<3.
D.
2
3x + l)(x
11.
(2x 4
13.
(-a 3
15.
[5(q
16.
[3x(2x
-
a
-
2
- l) 3
2
-
+ 2a)(a
-
2(q
x +
Complete each
12.
(3y
-
14.
(~a 4
2
- l)
1) -
a)
2
+ 7(q
2.
V
(
x
)(x + 2) =
3.
V
(
)(4x +
l)(2y
|a 3
-
[2<q
x 3 + 2x 2 + 3x +
1) =
- l)
1) -
+ a
2
-
2
-
2
3(q
2y
- 1)
l)(fa
- 1)
3
-
-U2
+
1)
+ 7]
3x(2 -x)]
column proof
of the
theorem
(1)
V x
=
x
(2)
2*4
=
2-4
X
3 +
(3)
2-4
(4)
V V
x y
6
20x 3 + 17x2 + 19x + 4
comments.
(5)
-
2
6x + lOx
)(2x) =
is a
y
of the following.
(
Here
- 8]
2
V
x
- 1)
-
5x (3x-2)][4x(x-
1.
x
2
- 4)
1
= 2(3 + 1)
= xy
x(y
+ x
7 +1)
}
2-3 +
(6)
2(3+1)
(7)
2-4
(8)
2-3=6
(9)
2-4
=
6+2
(10)
6 + 2
=
8
(11)
2-4
=
8
=
= 2
•
3
2
+ 2
'2
•
4 = 8'.
Fill in the
marginal
[7-10]
[7.01]
MISCELLANEOUS EXERCISES
A.
If f(x)
=
3x
2
1.
2.
If f(x)
=
5x
2
and
3.
C.
5x +
-
1
7
and
2x 2 + 3x
g(x) =
and g(x)
3x
-
7
and g(x)
=
2x +
=
then [fg](x)
- 5
then [fg](x)
=
[gf](x) =
If f(x)
=
and [g°
B_.
-
4x 2 + 2x
- 3
1
then
[f° g](x) =
f](x) =
Simplify.
1.
(x + 2)(x
2.
(y - 5)(y - 7) + (y - 3)(y + 5)
3.
(2y
4.
(a
5.
(3x
-
y)(2x + 5y)
6.
(7x
-
4)(15x
7.
(5y
-
7)(y+
8.
(2x + 3y)(2x + 7y)
9.
(3x
-
4y)(x
10.
(5a
-
2b)(7a
-
3) + (x + 7)(x - 5)
-
4)(y + 8) + (3y
2b)(3a + b) + (7a
-
-
9) -
3) +
-
2y)
-
-
-
5)
b)(5a + 3b)
(6x + y)(3x + y)
(8x + 5)(13x
(5y
-
-
7)(2y +
-
-
7)(y
-
3)
(2x + 3y)(x
(4y
3b) + (3b
-
4y)
3x)(3x + 5y)
-
-
7a)(5a
If
1
-
2b)
the perimeter of this rectangle is
34,
5x +
- 7)
what
is
the
area-measure?
2
If
the
area-measure
of this
rectangle
i
is 33,
what
is
the perimeter?
3x + 5
3.
Find the measures
cises
1
and
2.
of the
diagonals of the rectangles in Exer-
[7-11]
[7.01]
D.
E.
Simplify.
- 3)
2.
(y - 3)(y + 8)(y
3.
(a + l)(a + 2)(a + 3)(a + 4)
4.
(b + 2)(b - 2)(b + 3)(b - 3)
5.
(a - 6)(a + 7)(a - 8)(a + 9)
6.
(2x
1.
Suppose that
2.
-
4)(x
f(l),
7)(x + 2)(x
f(x) =
4
g(x) = 5x
-
5x
2
-
2x +
8.
-
7)(5x
Compute
f(2),
l)(3x
-
3)(4x + 7)
f(-2),
-
3x + x
-
x +
1.
Compute
g(6),
g(~5),
g(9).
Replace the letters by
1.
3x 3
-
-
f(-|), and £(0.05).
f(0),
Suppose that
and
G.
2)(y + 6)
(x
£(j),
F_.
-
1.
digits.
1ABCDE
* 3
SEND
+ MORE
ABCDE1
MONEY
2.
Solve these problems.
1.
The weight
of a circular disc varies jointly as the
radius and the thickness.
discs
is 9 to 8
and
if
If
square of the
the ratio of the thicknesses of two
the weight of the first disc is twice that of
the second, what is the ratio of their radii?
2.
Two numbers are
in the ratio 5 to 37.
added to each so that the sums are
What number should be
in the ratio
1
3.
Find three consecutive even numbers whose sum
4.
How many
pounds of a 12%
84 pounds of salt
5.
A
is
204.
are required to yield
?
printer used 1890 digits in numbering the pages of a certain
book.
6.
salt solution
to 3 ?
How many
Prove that none
pages are
of the
in the
numbers
the square of an integer.
11,
book?
111,
1111,
11111,
... is
:
[7-12]
[7.01]
DIVISION-WITH-REMAINDER ALGORITHM
The
subtraction, and multiplication algorithms are pro-
addition,
cedures for simplifying indicated sums, differences, and products
positive integers
We can
named by decimal numerals.
algorithms to apply to
all
of
extend these
integers by taking account of various theorems
about opposites.
The
Subtraction, which in
situation with division is not so simple.
some ways
is
analogous to division, always "comes out even".
you know, division doesn't.
In earlier grades you learned an algorithm
for finding the quotient of a first
a factor of the first.
Take
But, as
number by
example
the
a second
when the second
is
simplifying '867 t 17'.
of
51
17/867
1
50-17
=
850
I
17
1
•
17
=
17
I
867 = 51-17
1
The content of the dashed box suggests what
algorithm shows that 867
51
•
-
50
•
17
-
1
•
behind the algorithm.
867 = 50
So,
17 = 0.
•
17 +
1
•
The
17 =
17.
As you know,
this
algorithm can
still
information in cases in which the quotient
this
is
be used to yield important
is not
an integer.
Look
at
example
51
17/868
r
50-17
=
850
18
The result
is
of
1
•
17
=
868 = 51
•
17
+
17
L
--n
868
applying the algorithm in this example shows us that —r=
not an integer.
integers 51 and 52,
More
specifically,
Further,
it
it
tells us that
tells us that
86 8
.
All this information is conveyed by the equation
^
4
868
.
_
differs
868
=
is
between the
from
51-17 +
1
51 by y=1*.
[7-13]
[7.01]
In fact, the
So, the
purpose
algorithm
of the
algorithm
arrive at just such an equation
is to
called the division - with - remainder algorithm .
is
This algorithm has an analogue used for transforming certain alge-
For example, consider the expression:
braic expressions.
20x 3 + 17x 2 + 19x +
and the problem
of
transforming
is a
way
to an expression of the form:
)(4x+
(
Here
it
of finding
6
+(__.)
1)
expressions to
fill
the blanks:
5x 2 +
4x +
/20x 3
1
5x 2 (4x +
3x
+ 17x
1
+ 19x + 6
2x 2 +
12x 2 +
1) =
3x(4x +
2
5x2
20x 3 +
1) =!
4
+
1
9x
3x
I6x + 6
1) =
4(4x +
20x 3 + 17x 2 + 19x +
2
6 = (5x +
3x+4)(4x
I6x + 4
"2
+ 1) +
Let's do another problem:
V 5x 4
2x 2 + 3x +
+
5
=
(_?_)(x
5x 3 + 15x 2 + 47x
-
3/5x'
5x 4
+
-
15x
-
3) +
(_?_)
+ 144
2x 2 +
3x +
:
15x J +
15x 3
-
2x^
45x
2
47 x
2
47x
,:
+
3x
-
141x
144x +
144x
-
5
432
437
So,
V 5x 4
+
2x 2 + 3x +
5 =
(5x 3 +
1
5x 2 + 47 x + 144)(x
-
3)
+ 437.
_n
[7-14]
[7.01]
EXERCISES
A.
Use the division-with-remainder algorithm
complete the following
to
sentences.
-15+
1.
379
3.
V 2x 3
4.
V
5.
V 6x 4
+ 17x
6.
V 3x 4
-
7.
V x4
8.
V 27x3
+
64
9.
VJ^ x 3
8
-
10.
=
+ x2 +
x + 16
15x 4 + 5x 3
-
+ 7x + 25 =
7x +
-
-
35 =
=
2 =
6x 2 + 12x
=
2
-
+
3x +
-
7 =
V 2x 4
3x + 7
-
=
5x 4 + 2x2 + 3x +
)(x + 4) +
(
5)
f
is
way
is to
use
vO%
*"l
5 =
that
5x 4 + 2x2 + 3x +
this
it
is
(
_)
_
)
(
™ ™
™"
)
)
*«,
*v
told us that, for each x,
f(x) =
-
3)
+ 437.
5x 4 + 2x 2 + 3x +
5,
and sup-
to do this is to substitute *3*
2x 2 + 3x +
5'
and then simplify. Another
Do you see how?
we wish
with-remainder algorithm
From
+
(
_)
)
One way
f(3).
expression '5x 4
Now, suppose
+
+
(5x 3 + 15x2 + 47x + 144)(x
a function such that
(*).
" ™ " ~
'
(
pose that we wish to find
x' in the
+
2)
""
The second example on page 7-13
4
-
(
-
5)
+ (_
)
+
)(x
(
28 +
)
'
vU
for
(
(
)(x
(
5)
'
'
Suppose that
+
-
•
_)
2x +
-
2x
+
7)
•"l"
(*)
2x2
2
)(x
(
=
)(3x+l)
(
)(
)(3x+4)
(
=
(
)(x
(
J^
11.
)(x + 2) + (_
(
'
2x 4
V
8x 3
2x 2
-
2
=
3x2 + llx + 4
-
5032
2.
5 =
to
show
(5x 3
We
to find f(-7).
-
that, for
35x 2 + 247x
easy to see that f(-7)
o^
,,x
O,
^x
each
-
12087.
=
vi.
,
>
t
can use the divisionx,
1726)(x +
7)
+ 12087
[7-15]
[7.01]
B.
1.
Suppose that
6x 4 + 2x3
f(x) =
with-remainder algorithm
2.
(a)
Suppose that
x4
g(x) =
I6x2
-
lOx +
-
compute
to
-
and
4
3.
Show
4
that
x
- 5' is
14x 3 + 71x2
-
a factor of
4
x3
-
9x2 + 23x
-
Do
what would the remainder be?]
Show
5.
Which
(a)
(c)
(e)
C.
1.
If
x2 + x
4
that
of the
(x + 5)(x
x
x 2 + 4x
X2 + x
X3
-
2' is a
factor of
If
-
-
2)
21
12
-
of
cube
is
3x
-
2
volume
is a
is a factor,
7x
-
2'.
(d)
x2 + lOx - 21
x2 - 7x + 12
-
x3 + 3x2
2x
-
-
3
25x + 21
(
is
cubic inches.
3x + 12)(2x +
8)
square inches,
volume?
is
long
-
x2 + 7x - 18
x - 2
is
the
volume
of the
solid is 2x 3 + x 2
the
- 5*
inches long, then the surface area
the surface area of a cube is
is its
x
(b)
(f)
If
If
x 4 + 9x2
-•
3.
4.
-
x2
6x 2
7x + 60
x*
3x - 18
-
an edge
what
x5
4
[If
following fractions can be reduced to lower terms?
square inches and the volume
2.
4
You could
15'.
-
problem a different way.
the
2
-
g(3),
= 0'.
154x + 120
use the division-with-remainder algorithm.
4.
g(2),
g(5).
Solve the equation x 4
(b)
Use the
154x + 120.
division-with-remainder algorithm to compute
g(4),
and f(2/3)
f(-7),
f(6),
f(2),
14x 3 + 71x2
-
Use the division-
1.
of a
cube
-
pictured rectangular
8x + 21 cubic feet, what
the area of the shaded face?
is
x3
diagonal of a face?
-
1
5x 2 + 75x
How
-
125 cubic inches,
how
long is a diagonal of the cube?
1
[7-16]
[7.01]
OTHER COMPUTING FACTS
In
5
j-
•
5
2 =
Here
order
we use
-_-,
is a
to establish a
computing fact such
example, that
as, for
the principle of quotients and the division
theorem.
column proof:
(1)
V V V xyz
=
x(yz)
(2)
|*
3
=
| (2
2*3
=
6
[theorem]
7^
=
X
f
6/0
[
Xyz
(3)
V V
(4)
y^0
x
2
"
(5)
|
(6)
-2
(7)
|
(8)
VV
V
xy/Oz
/
rt
'
[apml
Lr-j
3)
[(!)]
[(4)
"
5
I<
3
=
—
zv = x then z =
if
(11)
3
=
5
|-2
=
|
2
•
[
then
|
•
]
and
2 >'
(5)]
a ^d(6)]
3 >'
<
[theoreml
y
•
|-
?
5
3/0
if
^
-6 =
(9)
(10)
p
2 =
|-
?
]
[(8)
and
(9)]
[(7)
and
(10)]
[apm and pq] and
division theorem], we
In the foregoing proof, in addition to basic principles
theorems derived from them
used some inequations
['2
•
3 = 6'
and the
/ 0' and '3 / 0'].
['6
Similarly, in Unit
2,
in
proving the important theorems:
Vx
we used
4
1
/.
0'
-r-
and — 1 /
'
=
x
V
and:
x
1
0'.
Now,
it
is
—
-r
easy
=
-x,
to see that
inequations can be derived from our ten basic principles.
do
is
to imagine there are no real
another
name
for 0.
Nov/,
numbers except
0,
none
of
these
All you need
and that
'1' is
look at each of the ten basic principles:
just
y
[7-17]
[7.01]
Commutative principles
Associative principles
VV
x + y =
x y
Vx V V x
'
Vx V xy7
=
y
y + x
y
+
z
y+z
V V V xyz
xyz
yx
7
7
'
=
x + (y+z)
7
= x(yz)
7
Distributive principle
V V V
xyz (x + y)z
7
=
and
Principles for
V x
+
x
V x
•
X
=
1
x
x
=
1
xz + yz
7
Principle of opposites
Principle for subtraction
Vx x
VV
x-y = x+
x y
-x
+
=
7
—'
Principle of quotients
VVy^O„
(x t y)
/
•
y = x
:
Since
+
+
=
and 0*0
certainly true for the
If
'0,
the
and 0*0
=
commutative principles are
number system you are imagining.
you also imagine, as definitions
+
that
=
and multiplication,
of addition
then you can check at once that the associative
=
principles and the distributive principle also hold.
Remembering
that *1' is, for the
you see that the principles for
and
moment,
1
example!
-0
[If
numbers except
0,
= 0/
you
the principle of
you don't believe this, try to find a counter-
if
we could prove from
the ten basic principles that, for example,
then, since the basic principles are true for your
system,
0-0
= 0, and:
]
Now,
1^0
true, too.
is
for 0,
opposites and the principle for subtraction hold.
since there aren't any
Finally,
quotients
of
name
also hold.
Defining oppositing and subtracting by:
see that the principle
just another
4
1
/
0'
would also have
to be true of this
imagined number
system.
But,
it
isn't.
:
[7-18]
So,
[7.01]
we want
if
theorems
of
How
4
-l /
0'
have enough basic principles to reaLly prove
Unit
we should
2,
[We have
principles.
do this.
to
Do
it
4
add, at least,
1
all the
/ 0' to our list
of
basic
space below the pml on page 7-17 so you can
left
now.]
For example, do we need
about the other inequations?
The answer
as a basic principle?
to take
to this question is 'no';
we
can prove
Theorem
V if x
X
[From
this
79.
and the basic principle
(1)
-a
=
(2)
VX x
=
a +
(4)
Vx x
(5)
=
a +
(6)
V x
(7)
0' it
follows that
-1^0. Why?]
x
[Identity]
-a
[(2)1
-a
[(1)
x
= a
[(5)]
-x
=
[po]
a +
-a
=
r
a =
(9)
and
'.7)]
(6),
[(9);
*(D]
if
-a
(11)
if
a
fi
then -a £
[(10)]
if
x £
then -x /
[(H)]
V
(12)
then a
=
You can also derive Theorem 80 ['-0
principles.
Do
Notice that
this
]
now.
Theorem
79
Vx
and that Theorem 80
1
=
is
if
is
equivalent to:
-x
=
then x
= 0,
equivalent to:
V
x
if
x
=
then -x
=
and
[(4),
(10)
=
(3)]
[paO]
+
x
(8)
fi
:
= a +
+
'1
[assumption] *
a + -a = a +
(3)
then -x ^
i
(8)]
from our eleven basic
[7-19]
[7.01]
'-0=0' can be deduced from
[To see this last equivalence, note that
then -x = 0' and the principle of identity. And, on the other
if x =
x
hand, notice that from the assumption 'a = 0' and the theorem -0 = 0'
'V
4
= 0'.
one can infer '-a
from -0=0'.]
4
Complete the derivation
Theorem
So,
79 and
Theorem
V
4
of
x
if
then -x
=
combined
80 can be
= 0'
into
a single statement:
(*)
V
•
[x =
if
and only
'i^
In writing conditional
and biconditional sentences we shall find
convenient to abbreviate the connectives
only
by
'
if
Theorem
==>
4
and
'
.
.
.
'
<^^>
4
.
.
.
[x
/
L
^
V
L
[x =
<=> -x
The brackets indicate the scope
necessary when we use the
.
.
if
and
Thus,
-x /
0]
/J
4
...
if
as 'for each x,
written with the
if
i
only
is not
x
is
^=^>
if
\
not
it is
= 0]J
then
Theorem
s
2^0,
as
4
is
for each
and
1
to justify the first 80 theois not
it
possible to
[You can show this impossibility
argument we used
no real numbers other than
:
it
If it
it
"J
However,
derived from the ten basic principles
Hint
is not 0'.
to read
read
still
OxN
O,
"I
7- 149.
prove, from these alone, that
(
x
you
79,
the opposite of x is not 0'.
rems on pages 7-145 through
multiplication,
of
sometimes convenient
Our eleven basic principles are enough
of
aren't brackets
notation?
'
then the opposite
Ox
"C
by the same kind
Why
of the quantifier.
Of course, whichever way you write
way.
.
can be written:
(*)
x
and
'
respectively.
',
V
x
x,
then
...
'if
it
4
79 can be written:
x
and
= 0]
Ox
*V
Ox
T
Ox
-x
if
to
of
show
Unit
and that
that
1
p 0' cannot be
Imagine that there are
2.
-f
4
1
.
Then, define addition,
opposition, subtraction, and division in a "natural"
Let
1
+
1=0.) Now, show
that all eleven basic principles
are true statements.]
Since, for the real numbers,
basic principles.
Z
f-
In the next section
,
we
it
follows that
shall introduce additional basic
principles which will enable us to prove not only
but will also give us a basis for proving
we need more
'2
£ 0',
'3
fi
0',
etc.,
theorems about the relation >.
[7.01]
[7-20]
MISCELLANEOUS EXERCISES
A.
B.
C.
Expand and simplify.
1.
(5a + b)(5a-b)
2.
(x + 7)(x + 8)
4.
(y + 9)(y
5.
(3x
7.
(3z + 4) 2
8.
(7a+2)(5a-6)
10.
[(a
12.
5pq(p
14.
(2a
-
-
-
9)
b) + c][(a - b)
2
pq
-
-
3b)(5a 2
q
-
4)
2
c]
1 1
2
)
2ab + 7b 2
-
-
)
[x
.
3.
(y + 9)(y + 9)
6.
(3z
9.
(ax + b)(ax
y + z][x
-
13.
7x2 y 3 (x 4
15.
(3x 2
4)(3z +
-
-
-
y
x2 y + y 4
-
2x + l)(5x 2
-
-
4)
c)
z]
)
7x
-
-
2)
Simplify.
1.
81-53+81-47
2.
93-67 + 67-7
4.
(7-33-j) -(9-3)
5.
12
87
•
i
+
12j
•
12
3.
8-19*12i
6.
(
16 |)
•
1 1
•
(5
•
6)
Simplify.
x
-
3
2_
x+8x
7
1U
*x-4
x +
Z
*x-5
7
x +
1
'x+4
-
8
8
7a 3 b
8xy4
3ab2
„
L1
(x- 3)(x +5) y
(x + 4)(x- 5)
x 2 - x - 12
x 2 + 3x- 10
IS
x2
17
*
-
x
x+5
x+5
5x 2 y 3
*
x
3
x
+
5_
-
^
4.
x + 4
5
(x- 3)(x-
2)
2x
3
~
*5x+l
3m 3 n 3
"
2p
(x.- 7)(x
2
q3
-
5)
(x- 5)(x +
5)
x2 + 2x - 15
x^ + 7x + 12
x +
x -
1
x
3
x-
-
x
x
+
8
-
3
-
5
3x
~
x
,
-
x
4
*x+4
7
q
5x-l
8pg 4
* 12mn4
**'
7y +
*
3
+
y-2
x2
x2 - 5
- 5x +
5
3y
3y
-
2
-
6
8(a - b) 3 v 3(a + b) 4
9(a + b)*
2(a - bp
x2
6
12x + 35
-
x2 + 2x
6x 2 + 7x - 5
16x 2 - 26x + 3
R
-
x+5
x 2 + 2x- 15
x* - x - 20
b
2
1
'yy+2
.,
4
-
y
"
-
15
32x 2 - 12x+ 1
12x 2 + 17x-5
2x+
1
x
3
-
5x x -
3
2
[7-21]
[7.01]
ia
19
'
Z1
3x2 + x
x 2 + 5x
-
23.
1
"
~x~^8~
2x2 + 7x- 15
8x +
x +
-
xz +
"
4x +
1
(x .
24
1
"
2x-7
+
~ZTT
3
4x+8
5
8x- 12
*+*)-<*£!.
X
a
2a- -i
a -.11
-
x
- 3
x2 + 4xx* +
(
5
x2
2
llx
-
9
"
x*
-
3x - 7
10x + 9
x2 4 3x - 10
x2 + 9x+20
21
+28
-^l + (7-^)
1
-
8
x + 2
1
•
x +
3
Solve each system of equations.
1.
4.
E.
3x
24.
£.
D.
-
??
""
y)
x + y
5x+7
.
CK)
1.
x
-
y
^
= 2
2x + y =
5
2x
-
y =
5
^
2x
-
3y
= 5
J
2.
J
5.
4x + 3y
= 10
^
2x
-
2y = 5
J
3x
-
2y =
5
^
= 9
J
6x-4y
x + 3y
3.
^
= 5
2x + 4y = 7
6.
2x
-
5x +
y = -2
1
=
-2y
Find two consecutive integers the difference between whose squares
is 23.
2.
If
A
how
3.
can do a certain job in 6 days, and
long will
take
Mr. Allen drives
per hour.
trip,
4.
it
how
In
to
B can do
in 8 days,
it
them working together?
Zabranchburg--a 100-mile trip- -at 45 miles
order to average 55 miles per hour for the entire
fast should he drive during the return trip?
To bring
Bill has an average grade of 74 after the first two tests.
his average up to 80, what
5.
if
A
must he get on the third test?
car travels north at 45 miles per hour, and 2 hours later another
car leaves the same place and travels north at 55 miles per hour.
How
6.
long will
Rectangle
A
it
take the second car to pass the first?
is 14 feet
longer than
it is
wide.
Rectangle B
feet smaller in width but 16 feet longer in length.
same area, what
tangles have the
7.
Given a circle
of
radius 10.
circle a function of
there a function
f
its
Is the
i = f
4
the rec-
is it?
distance, d,
such that
If
is
«d?
measure,
i,
of a
chord
from the center- -that
If
so,
give such an
f.
of this
is,
is
[7.02]
[7-22]
The positive numbers --We concluded the
7.0 2
numbers are
out that our eleven basic principles for real
among
to prove,
other things, that 2/^0.
Now, by
some by
trying to prove that
definition, this
amounts
to proving that
by the principle
only
-1 =
if
what we need to prove
1,
One way
to see that -1 and
and, of course, that
positive
= *1.
of the things
-1/^1
that in proving that
if
we need
consist of the opposites
(Px
(P
€
By
P
2^0]
how
"1
recall that -1 =
1
is
a
-1^1.
is,
paragraph suggest
will be useful to
have
=>
)
V
not both x
^
P
N
of
negative numbers to
What we need
P.
of
numbers
of positive
to
express
the operation oppositing and the property
The following two basic principles do
[x
)
it
the set
members
V
either x
€
P
P and -x
e
or
e
-x
€
this.
P]
P
says that, no matter what nonzero real number you
),
and (P2
-1^1.
or -1
(P2
)
e
What does basic principle
positive.
is
),
and the basic principle
For, since
1^0,
P. But, this being so,
this is not the case.
Since -1 £
is,
[Explain.]
say?
)
prove that
1,
it
follows
-
if
the
1
=
1
j£
0',
it
from (Px
then both
1
)
e
is
easy to
that either
P and -1
€
P.
-1^1.
Hence,
from
follows
it
'1
0-sum theorem
that
1
+ 1/^0.
That
2^0.
However, even with
it
is
or its opposite
it
Using (Px
1
of the
)
Basic principle (Px
(P2
in the last
define the set
being positive are related.
pick, either
and
if
deal with positive and negative numbers.
to,
new basic principles
in the
we said
we may as well concentrate on
we can,
=
1
a negative number, and
is
[and, so, that
some basic principles which
since
-1
So,
is to
Since,
+
1
and -1 are not the same number --that
1
Now, several
Actually,
-1^1.
2^0.
1^0.
Hence, since the same number cannot be both positive
number.
and negative,
1
is that
are different
1
+
1
0-sum theorem,
opposites and the
of
not sufficient
we need more
Evidently,
Let's find
basic principles.
of
last section by pointing
.
seems
likely that
(Pjl)
we are on
and (P2
),
that
the right track, and that what
more basic
principles about the set P.
that 3 ^
we could show
if
we cannot prove
that 3
e
P and
3^0.
But,
we need are
For example, we could prove
^ P.
Now, we can prove
that
-
[7-23]
[7.02]
This follows from (P 2
i P.
-0=0.
)
and the theorem [Theorem 80] that
we have:
So,
Theorem
4 P
how we might show
Let's see
By
2+1.
3 =
definition,
two positive numbers
6
(P3
V V
)
[(x €
1
e
P.
[Explain.]
)
number.
that 2 +
P and
y
So,
we replace
4
P' by *N\
about the set
of
from (P^, (P2
we could,
in
we
each
to
of the
e
of
another
that 2
P and
e
P]
2 e
1
proving:
is that of
82.
prove this?
We can
principles (P^, (P2
see that we
)»
if
we could derive
and (P3
)»
and (N 3 ),
P'
'1 e
and (P3 ), and our eleven other basic principles, then
same way, derive
in just the
can't do this
P by showing
P we shall take:
Now,
and our eleven other basic principles.
we
this as
get true statements, (N x ), (N 2 ),
negative numbers.
)»
we adopt
e
1
our only problem
Do we know enough basic principles
if,
If
P) ==> x + y
e
showing that
Theorem
1 6 P
don't by noticing that
sum
P and that 6 P.
you can prove that 2 e P once you have proved
let's tackle the job of
Evidently, using (P3
that
P.
So, as a third basic principle for
P.
Now,
a positive
is
e
Also, as you probably recall, the
we can show
basic principle then
1
that 3
81.
4
1
e
from (N 1 ), (N 2 ), and (N 3 ),
N'
Since
1
is not a
negative number,
.
Our trouble
is that
none
of the
basic principles which
we have
accepted really distinguishes between positive numbers and negative
numbers.
which
is
We need
a basic principle
not also a property of N.
most obvious omission, so
far, is
which states some property
of
What haven't we considered?
any consideration
of
The
multiplication-
P and
set N is
and, with respect to multiplication there is a difference between
N.
not.
The
So,
set
P
is
P
closed with respect to multiplication, but the
let's take the following as a fourth basic principle for
P:
[7.02]
[7-24]
(P4
Vx V
)
Now we can prove Theorem
Since 1/^0,
pose that -1
the
1
6
it
pml, -1
P
or -
eP
1
1*1
=
So,
= 1.
easy to prove that
(P 2 ) to show that -1
But, 5
then
Since
3 = 0.
We
that either
•
-1
-1
€
that
2 e P,
P, 4
3
2
/
0,
it
P, etc.
€
e
P.
if
-
Sup-
P.
Theorem
23 and
Since either
1
P
e
then
[Incidentally,
'3
Theorem
'4
0',
^
'5
Theorem
fi
0',
For,
2'.
£
(3 + 2)
30,
P,
6
1
81,
it
we can use
we have a
etc.
-
5-2
= 2
then
2 = 3.
So,
if 5
if 5 =
'2
like
*
^
-=-
The key theorem here
.
1.
3^0,
follows that
it
0]
?
if
j
proving the theorem
In
Vx Vy x
***
Prove
2.
B.
is
54:
then
=
But,
2 = 0.
2^0.
2
So,
j
^ 0.
EXERCISES
A.
= 0.
follows that 5/^2.
VV
,[^ = 0=>x =
x y t o y
Since
1
e
[using definitions and (P 3 )]
In addition, using
and, by
-
or -1
But, by
P then
Now
P.
e
3 e
P.]
f(
1
can also prove theorems
Theorem
P.
e
P
e
1
P.
e
also prove theorems like
2 = (3 + 2)
-
1
2^0, and can prove
that
We can
if
a proof:
is
and since, as we have just shown,
we have proved
new proof
)
Then, by (P4 ), -1
follows that, in any case,
So,
is
-1
•
Here
82.
from (P1
follows
it
P.
e
P andyeP) => xyeP]
[(xe
y
t
(*)
1.
4
4.
No
5.
P
e
5
=
2'
y=^
we used
^
of the
-
y
x
= 0]
= 0*.
following.
2.
positive
-
the theorem:
of (*).
P
is not
fi
by first proving 'V x
Prove the converse
Prove each
4
number
is its
If
3
own opposite.
closed with respect to subtraction.
3.
-j
£
1
2
[7-25]
[7.02]
C.
Prove the following theorems.
Sample
1
Vx V
.
Solution
P and
[(x e
y
Suppose that a
.
y
P and
€
So, by (P 4 ), (a + b)
2
Hence, a 2
2
+ 2ab + b
then a 2 + 2ab + b 2
V V
x
y
[(x
e
P and
V V [(x-yeP and
y
e
P)
=> x
2.
V V
P and
y
e
P)
=>
Sample
-
y
e
Vx V
2.
Solution
Suppose that a
.
(Px
),
either a
Theorem
either a
Vx V
y
3.
Vx V
4.
V V V
[(x
5.
Vx V V
[x
6.
Vx V Vz
L
y
not both x
y
z
y
-
-
[(z €
y
y'
e
e
yJ
-
-
[x ^ y
e
P =>
y7
e
it
e
e
Look
.
e
is
If
we are
to
2
P and
€
P.
+ 2ab + b
b
e
2
.
P
Then
b
P
a
-
or -{a
= b
a
e
P].
Theorem
at
32.
]
P]
or
e
a
= a
2
+ 2xy + y
x-yeP
- b)
x
if
2
-
£
a.
-
y-xeP]
[Why?] and, by
fi
b) e
But, by
P.
Hence,
if
a ^ b then
Consequently,
P,
x-yePory
-xeP].
P
-
(y
w
+
vl'»^
z) €
-
P]J
yz
7
e
PiJ
k",
-V
defined the negative numbers to be the opposites
numbers.
numbers
=> x2
y
P) =£> xz
«.•-
itive
-
So,
b)
a + b
y-zeP)=>x-zeP]
*.,%
We have
yJ
But, (a +
P]
Consequently,
P or b •
=> either
(x + z)
-
P.
6
P and
P and
P and x
-(a
33,
b
b
-
€
2
-
b.
fi
P.
e
By (P3 ),
P.
P] [ Hint
e
x2
e
e
P)
e
[x ^ y ==> either
y
b
P.
e
y
1.
[(x
=> x2 + 2xy + y2
P)
e
prove theorems about the set
N
of
of
the pos-
negative
convenient to incorporate this definition in a basic principle
[7.02]
[7-26]
V [-xeN <=> x
(N)
We
P]
€
How-
shall not include (N) in our "official" set of basic principles.
ever, you will be asked to prove
next set
some theorems about
the set
N
in the
exercises, and for these proofs you will need to use (N) as a
of
basic principle.
o^
vl^,x
D.
Prove each
1.
of the following.
Vx [xe N <=> -xe
L
Sample
.
V
Solution
V
(1)
V
(4)
P]J
=>
[x £
either x
e
P
or x
N]
e
.
[x £
=>
either x
€
P
or
-xe
P]
[basic principle]
a /
=>
either a
e
P
or
-a
P
[(1)]
(2)
(3)
o^
,,v
^,<s
xeN <=>
-x
ae N <=>
-aeP
P
e
^=> either a
a £
(5)
€
[theorem--Ex.
[(3)]
e
P
or a
N
e
[(2)
Vx [x/0=> eitherxePorxeN]
(6)
from step
[Note that step (5) follows
(4)
and
(4)]
[(1) - (5)]
and step
(2)
by virtue
This principle
substitution rule for biconditional sentences.
logic tells us that, given a biconditional sentence [(4)] and a
sentence
[(5)] is
e
4
-a
e
of the biconditional
second
[in
compon-
[(4)
and
(2)].]
paragraph proof for the same theorem:
Since, by (P L
),
and only
a
if
of
second sentence, the new sentence so obtained
N'] in the
is a
of the
sentence
P'] is replaced by its other
a consequence of the given sentences
Here
if
component
component
this case, the
ent ['a
either
if
[(2)],
1]
if
e
Consequently, V
a
N,
[x
-f
it
£
then either a
follows that
=>
if
either x
e
a
e
P
fi
P
or
-a
e
P, and, since -a
then either a
or x
e
N].
e
P
or a
e
e
N.
P
[7-27]
[7.02]
P and x
N
2.
Vx
not both x
3.
V
[x £
4.
Vx
not both x
5.
V V
[(x e
N and
6.
V V
[(x €
N and y£N)=> xy
7.
V
f
[x
=>
e
either x
€
N
e
N and -x
e
=> x 2
y
N]
€
N
e
=> x
N)
e
-x
or
yeN]
+
P]
6
P]
e
MISCELLANEOUS EXERCISES
A.
x2
3x
1.
f
g(x) =
2.
f
f(x) = (x
3.
f
h(x) =
4.
f
f((x, y)) =
5.
i
g(x, y) =
6.
f
2
h(x, y) = (5x )
7.
f
f(a, b, c) =
8.
f
g(a, b, c) = 18
9.
f
h(m,
10.
£
f(x) = 3(x
11.
f
f(x) = 8
12.
£
13.
f
-
7)(x + 4) then
-
x2 + x
{
^-
-
-
-
+
1
a
g(6) =
f(27) =
,
and
f(6) =
h(-6) =
and
f(3, 2) =
m
2
xy 2 then h(3,
x~^T
-
4(x 2
,
then
g(x) = 9
have the same value?
-
5)
f(2) =
Si
-
1
)
=
.
=
f(2, 3) =
and g(2,
2) =
n then h(3, -6)
and h(35)
,
2)
3)
3)
and
=
=
.
h(2, 3) =
=
6c + 2ab then g(-4, -12,
-
f(l) =
and
and
2
13
x2 +
,
,
,
then g(3,
g(27) =
+ 5a c then f(-4, -12,
2)(x + 2)
then
17) =
f(
+ x) 2 +
- (5
-3ab
g ^ X * = x~^2 +
f(x) =
2
y then
-
-^
mn 2
,
30 then h(15) =
-
6x2
n) =
X
28 then g(17) =
-
3)
=
=
+ (x
-
,
and
»
g( 3 )
8)(x +
1)
then f(-5) =
f(3) =
and
»
1
g(
j)
=
x2 then for what argument do
f
and
g
[7-28]
B.
[7.02]
Tell which expressions are equivalent to the given one.
1.
2.
mx
+
(A)
(m
my
nx + ny
+
+ n)(x + y)
T
T(l +
Kk
_D.
(B)
y(x +
s)
(B)
s(l + T)
z)
(B) y(l + x)(
(B)
K(k
(B)
(t
(B)
z(x
2.
8
5.
2-3 2
+ 3 3 = 35
2.
3
-3 3 = 9 5
5.
3
k(K
tx
-
(t
xy
-
-
1)
ty + sx
-
x(m
+ y)
(C)
x(y +
2) +
x
+ xz
(C) x(y + z + x)
(C) (T + 1)(1 + s)
1
+ x2 )
(C)
yx +
(C)
k(K
y(
1
+ x) 3
-
-
1)
-
1)
+ k
s)(x + y)
-
z)
-3 2
3
2
4.
(2
4
3
)
+ s)(x
-
y)
y) +
xy
(
C)
(t
+ x)(s
-
+ yz
-
-x
(C)
-y(
3.
(-3)
6.
5
3.
3
6.
(3
-
y)
-
z)
-r
2
U
9
-f
8
7
2
-
(-2)
+ 2- 3 Z
True-or -false ?
1.
3
4.
3
2
2
K
sy
Compute.
1.
-
xz + yz
(A) x(y
C.
+
K
-
(A)
8.
m(y + n)
2
2
y(l + x) + y(l + x) x
(A)
7.
C)
Ts
+
(A) y(l + x) 3
6.
(
xy + xz + yz
(A)
5.
+ (x + y)n
(B) x(y + 3)
(A) x(y + z) + yz
4.
m(x + y)
xy + 3x
(A) y(x + 3)
3.
(B)
2
8
+ 33 = 6 5
~
3
2
= 34
2
•
3
4 2
)
3
= 36
=3 6
;
xz
2
[7-29]
[7.01]
E.
Solve these equations.
1
F.
5x
.
2 + 7( 3
-
3.
x + 2(5x
1.
Draw
-
x) =
-
-x
=
1)
1
-
a square and
3
mark
7x
2.
2
(1
-
x) +
4.
6 + (x
-
3)
= (x +
plane.
Draw
a point in
-
its
2
= 5
3(x
-
l)
2
-
6)
2
-
a line
through this point such that the line cuts the square region into
regions
equal area-measure.
of
for a regular hexagon.
2.
Repeat Exercise
3.
Given an equilateral triangle
1
cular arc with center at
Draw
4.
a square
ABCD
A
AABC. Find
the radius of the cir-
which bisects the triangular region.
and draw perpendicular lines
that & intersects each of a pair of opposite sides
each
of i
determined by the intersection points
segment
Two
5.
6.
m
of
is
and
m
and
m
such
intersects
segment
that the
congruent to the
determined by m's intersection points.
players at a circular table take turns in putting pennies on
the table.
is
Show
other pair of opposite sides.
of the
Jt
The pennies may touch but not overlap.
the player
who
can't find a place to put
down
The loser
The
a penny.
How?
first player
can always win.
Remove two
opposite corner squares from a chessboard.
Can
the remainder of the board be covered without overlapping by
•
rectangular cards, each card covering exactly two squares?
EXPLORATION EXERCISES
1.
Graph the sentence
2.
What name
is
l
y
-
x
e
P'
customarily given
on a picture
of
the
number plane
to the relation {(x, y)
which you pictured in Exercise 1?
:
y
-
x
e
P}
[7.03]
[7-30]
Inequations
7.0 3
P
of positive
numbers are also basic
the relation >.
we defined
4
--The principles (P1
.
In fact,
shall
now accept
7
> 9'.
'5
that 5
know
-
that
7 >
4
7 >
.
.
<
How
When we
of the
[Since 7
about
'as merely another way
form
'
.
.
>
.
greater than
.
inequality such as
'.
.
x
to derive
y
[y >
>
'
.
'
a >
(2)
(3)
Vx x
(4)
a
(5)
Vx
number, we
[x
L
of
>,
is just
it
we
occa-
form
'
>
*
.
.
83
e
a
-
x
-
numbers which are
We
state this as:
P]
from our basic principles?
Try
€
P]
[basic principle]
e
P
[ID]
[theorem]
x
-
[(3)]
P
> o <^=>
a
>
xe P]
e
.
an abbreviation for
column-proof which follows.
<^=>
=
saying
]
<^ —> x
x >
=> y
x
> 2*
83.
Theorem
to do so before reading the
x
u
of
However, we shall use
tells us is that the real
V
V V
way
writing one of the form
.
'7
saying that
of
2 is a positive
-
are precisely the positive numbers.
Do you see how
way
merely the converse
is
=
.
.
Theorem
(6)
of
> 9'?]
'5
of
form
or
Another thing that (G)
(1)
in Unit 2 that
do, you should think of a sentence of the
[Similarly, a sentence of the
one
theorems about
> 9' is another
'5
could get along without the symbol '<'.
'.
for the set
)
principle.
Similarly,
Of course, since the relation <
sionally.
a ^d (P4
xeP]
-
just another
2' is
number.
true.
2' is
y
meant by statements
number.
9 is a positive
4
p3 )>
from your work
recall
new basic
(G) as a
The statement
2 is a positive
-
(
for establishing
Vx V [y> x <=>
y
(G) tells us what is
and
),
>' by:
(G)
We
may
you
(P2
),
[(2),
[(1)
(4)]
-(5)]
[7-31]
[7.03]
In
paragraph form we could write:
By
(G), a >
and only
if
rem 43, a - = a. So,
V [x > <=> x € Pi.
Theorem
of the
form
a >
if
83 is useful in that
*.
.
.
can restate the theorem
be restated as
'0
/
4
€
1
€
form
4
P* as
But, by Theo-
P.
and only
1
Also, since
0'.
-
if
a
e
P.
Hence,
enables us to replace any sentence
it
P' by one of the
£
a
if
*.
.
.
> 0'.
So, for example,
we
Theorem 81--'0 i P'--can
Theorem 81 is equivalent to the
> 0'.
statement:
V
x
=>
[x =
x 4 P]
and, so, to the statement
Vx [xe
P
l
Theorem
/o],
83 can be used to translate
V
Other convenient ways
x
c
P
of
=> x
>
[x
Theorem
81 to:
£ 0]
saying the very
same
thing are:
\>o^°
and:
r
EXERCISES
A.
1.
Use Theorem
(P3 ), (P4
The
2-.
),
83 to translate our basic principles (P.^, (P ),
2
and (G), into statements which do not contain *P*.
last of these five
statements we shall call 'Theorem 84'.
Can you use Theorem 83
obtained in Exercise
1
to translate the five statements you
back into the basic principles
(P 2 ), (P3 ), (P 4 ), and (G)?
O,
Theorem
V V
*U
s.K
84.
[y >
J^
m.h
x
<=> y-x
O,
> 0]
(Pj^),
]
[7.03]
[7-32]
B.
One useful corollary
of
Theorem
Theorem
Vx
Prove
Part
C.
1.
this
D
and
-x >
<r-^>
this
0]
theorem with Exercise
look at the theorems in Sample 2 and Exercises
6 of
Part C on page 7-25.
we can
about >.
d,
<
[x
l
1
of
on page 7-26.]
Now
83
85.
[Compare
theorem.
84 is:
With the help
3,
4,
5,
and Theorem
of (G)
translate these five theorems into useful theorems
Do
[These five new theorems are parts
so,
and e respectively,
could be taken as a set
of
Theorem
of
basic principles for >.
These
86.
five
a,
c,
b^,
theorems
See Part
^D
on page 7-35.
2.
Give a column-proof to show that Theorem 86b
and the theorem
of (G)
of
Exercise
is
a consequence
on page 7-25.
3
[See the
solution of the Sample on page 7-26.]
INEQUALITY THEOREMS
In Part
C
This theorem
of the
is
preceding exercises you discovered Theorem 86.
an immediate consequence
of
earlier theorems on
positive numbers, and basic principle (G).
Theorem
86.
y => either x > y or y > x]
a.
Vx V
[x
b.
Vx V
and y7 > x
not both x > y
'
c.
Vx V V
d.
Vx V V
fx > y
e.
Vx V V
[(z >
y
y
y
y
y
z
z
z
/£
[(x
L
> y7 and y7 >
=>
z)
=^ x
x + z > v' +
and x >
y)
=>
>
z]J
zlJ
xz > yz]
:
[7-33]
[7.03]
Other theorems on inequations can be proved in a simiiar manner, but
theorems from Theorem 86 and our
often easier to deduce such
it is
first eleven basic principles.
For example, consider:
Theorem
87.
Vx x f x
As
it
happens, there
Theorem
be used in proving
which theorem this
much simpler
Do so now.
86.
is
Theorem
a previously proved theorem about
is
Hint
.
it
proving Theorem 87.] However,
87
Recall that
*x
from one
f
x' is
of the
parts of
it
Theorem
short for 'not x > x'.]
87 is equivalent to:
Vx V
=> x ii*
f y]
= y
[x
L
J
y
This, and another consequence of
[Explain why.]
Vx V
(*)
are very useful.
x=> x^
[y >
y
We can combine them
Theorem
Vx Vy
[Explain.
in
Theorem
to derive
[
[Before reading further, try to guess
87.
and use
is,
P which can
Theorem
86b:
y]
into a single
theorem:
88a.
[y
L/
> x => x f
—
i
y]
i i
]
To see how Theorem 88a
is
used, consider the following theorem
which you probably recognize as the addition transformation principle
for inequations
:
Theorem
89.
w
V W
V
V W
x y
The
all
if
I
z
L
\
=>
x+z>y+z<=
-part of this theorem
we need do
is
prove
its
i
i
^-
'
is
Theorem
only-if-part
86d.
^
x > y]i
So, to prove
Theorem
89,
.
[7.03]
[7-34]
[x+z>y+z=>x>y]
VVV
x y z
(**)
This can be proved in a considerable number
simplest
of
of
ways.
Probably the
(a + c)
-
these ways is the following:
since (a +
(b + c) = a
-
c)
Then, by
c > b + c.
Suppose that a +
However, for purposes
-
a
b,
-
b
€
(G),
(b + c) €
P
and,
So, by (G) again, a > b.
P.
consider the following proof
of illustration,
of (**):
Suppose that a +
Theorem 86a
bora>
Suppose that
b > a.
Theorem 86d,
But, by Theorem
b > a then, by
b + c > a + c.
Hence,
it
Since a = b or a / b,
c.
that either a =
a > b or b > a.
if
> b +
c
if
b
^
So, a > b.
VVV
x y
z
^
Therefore,
+ z > y + z
[x
L
One advantage
(***)
if
=> x
method
VVV
Jt
from Theorem 86e.
b = a then b + c = a + c,
88a, ifb +
c>a
So,
if
b
>_
y
Do
of
a then
Since, by hypothesis, a + c > b +
•;
a.
Consequently,
> y]
7J
proof
is that it
can be used to derive:
and xz > yz) => x >
[(z >
Z
this
and
+ c then a + c f b + c.
a + c > b + c then a > b.
'
of this
If
But, as shown above, either a > b or b
a.
from
b or b > a- -that is, that either
b + c > a + c.
a then a + c p b + c.
follows that b
follows
it
y]
now.
EXERCISES
A.
On page 7-33 we mentioned
(*)
that the
theorem:
VV[y>x=>x/y]
x
y
is
an immediate consequence
Vx V
86b:
sounds as though anyone who accepted Theo
it
rem 86b would agree
Theorem
not both x > y7 and y7 > x
y
By common sense,
of
to (*).
inference of the form:
That
is,
it
looks as though each
c,
[7-35]
[7.03]
not [p and q]
q => not p
is a
Let's use the rules of reasoning
valid inference.
Unit 6 to
show
in
that this is the case:
*
t
e
a
p and s
p
we studied
=>
and
[p
*
not [p and q]
q]
not p
q =^ not p
1.
You can use
is
argument
this kind of
show
to
a consequence of the single premiss:
that:
b > a => a
^ b
not both a > b and b > a
What assumptions would you use and discharge during the proof?
2.
As shown
Give the rules
ences.
B.
1.
in the tree -diagram, the proof
Prove the converse
second sentence
2.
C.
V x
2.
V V V V
3.
Vx V V
y
4.
1
z
second proof
of the
of (**)
86a.
[See
on page 7-34.]
(**).
of the following.
1.
+
reasoning which justify the inferences.
Theorem 88a from Theorem
of
Find a third proof for
Prove each
of
would contain four infer-
> x
[(x
> y and
[(x > y
u>v)=>x + u>y +
and y >
=> x
z)
>
z]
Vx V [(x>yandy>x) r=> x = y]
Use Theorem 86a on page 7-29.]
[ Hint
v]
[Theorem
90]
[Theorem
91]
[Theorem
92]
[Theorem
93]
[Theorem
94]
.
5.
^"JD.
Vx V [-x
Show
> -y
<=>
y > x]
that the five parts of
Theorem
86 constitute an adequate set of
basic principles for > by deriving from them the five statements you
obtained in solving Exercise
1
of
Part
A
on page 7-31.
[7.03]
[7-36]
THE TRANSFORMATION PRINCIPLES FOR INEQUATIONS
In Unit 3
you studied theorems which justified the various steps taken
As you learned there,
and inequations.
in solving equations
the trans-
formation principles for equations are derivable from the first ten
basic principles.
Theorems
Theorems 6 and 7, Theorems
You are now ready to consider
[See
56 and 15.]
of
our
and 48, and
11
the proofs of the
transformation principles for inequations:
The addition transformation principle
[x+z>y+z <==>
VVV
x y z
J
for inequations
x > y]
J
The multiplication transformation principle
_
V V
a.
Vxyz>0
—
> yz
[xz
L
7
<=>
x >
VVV.,,
x y z <
< yz
[xz
L
J
<=>
x > y]
' J
.
b.
—
for inequations
y]
7J
The factoring transformation principle for inequations
a.
V V
[xy >
<=>
([x >
and y >
0] or [x <
and y <
0])]
b.
V V
[xy <
<=>
([x >
and y <
0] or [x <
and y >
0])
x y
[These principles are Theorems
95,
89,
and 96.]
The addition transformation principle
of
Theorem 86d and
Part
quence
a_
of
of the
(**) on
an immediate consequence
page 7-34.
86e and (***) on page 7-34.
Theorems 86a and 88a are used
It is
is
multiplication transformation principle
Theorem
easy to derive part
principle
from part
Theorem
94.
Here
a_
By Theorem
in deriving (***)
of the
b_
[Recall
now how
from Theorem 86e.]
Theorem
-c >
0.
20,
and
So, by part a of
the multiplication transformation principle,
-c
<=>
a > b.
-(ac) > -(be)
<=>
a > b.
•
a conse-
proof:
85, for c < 0,
a
is
multiplication transformation
by using Theorem 85,
is a
]
-c > b
•
Hence, by Theorem 20,
[7-37]
[7.03]
So, by
Theorem 94 [Exercise
ac < be <=>
Consequently, V V
Parts a and b
will be
proved
[xz < yz <==>
_
<
a > b.
x >
y].
factoring transformation principle for inequations
of the
in
V
Part C on page 7-35],
5 of
Part
A
exercises.
of the following
EXERCISES
A.
1.
Prove the
if
Do
ciple.
-part of part a of the factoring transformation prin-
this by first proving:
if
(a >
and b >
0)
then ab >
and b <
0)
then ab >
and then proving:
if
2.
Now,
let's
(a <
prove the only-if-part
of
part a of the factoring trans-
formation principle.
Suppose that ab >
By Theorem
0.
principle for multiplying by
Suppose that b >
mation principle,
it
By
0.
0,
it
0,
it
only
if
a > 0.
if
a > 0.
follows that a > 0.
and b >
both a >
Next, suppose that b < 0.
we have
So, for b > 0,
0.
[Show that in this case
it
follows that
and b < 0.]
Since either b >
(a >
Hence,
part a of the multiplication transfor-
follows that, for b > 0,
ab >
both a <
Hence, by Theorem 86a,
follows that
ab > 0b only
Since ab >
b ^ 0.
So, by the
or b < 0.
either b >
Since 0b =
0,
81, ab £ 0.
if
ab >
or b < 0,
and
then
it
follows that either
and b <
b > 0) or (a <
[(a >
and b >
0)
or
0).
(a <
Consequently, we have proved the only-if-part
of
and b <
0)].
part a of the
[7.03]
[7-38]
This together with Exercise
factoring transformation principle.
1
3.
gives us part
Prove part b
As
in
a_.
factoring transformation principle.
of the
proving part
t>
of the
principle on page 7-36, you
85;
[
Hint
.
multiplication transformation
may
find
Theorem
helpful to use
it
then use part a of the factoring transformation principle.]
B.
1.
Prove
maps each
that the squaring function
nonzero real number on a positive number.
Interpret this graphically.
2.
[Theorem 97a]
Suppose that a and b are arguments
of the
squaring function for which the corres-
ponding values are the same.
know about a and b?
3.
Prove
What do you
it.
Suppose that a and b are nonnegative argu-
ments
of the
squaring function for which
the corresponding values are the
What do you know about a and b?
Suppose that a and b are arguments
of the
squaring function such
that the corresponding values are different.
In particular, sup-
pose that b 2 > a 2
a?
that b
2
Can you conclude
.
that b >
If
> a 2 and b > 0, can you conclude that b > a?
Can a be
5.
Prove
[Theorem 98a]
it.
4.
same.
Suppose that a and b are arguments
Does
Can a be -b?
-b?
less than
that b > a.
you suppose
it
of the
follow that b 2 > a 2
?
squaring function such
Would additional knowl-
edge about b allow you to conclude that b 2 > a 2 ?
How
about
more
knowledge about a?
6.
theorems you discovered
(a)
State the
(b)
Prove them.
in
Exercises 4 and
5.
[Theorems 98b and
c]
[7-39]
[7.03]
Suppose that quadrilateral
C.
ABCD
a
rhombus whose diagonals are 2a
units
1.
and 2b units long.
Compute
the
area-measure
the square and the
2.
area-measure
Is the
of the
use of the rhombus?
3.
State and prove a
that
the
if
square
square ever less than the area-meas-
Are they ever equal?
is not
a square then the
find
it
helpful to use the
The figure
't
graph
Q(
T'
of the
[Theorem 97b]
theorem *V
/
x^
at the left
of the
you
tells
area-measure
greater than that of the rhombus.
You may
[Hint.
D.
is
of
rhombus.
theorem about real numbers which
rhombus
n
x 2 > 0'.]
shows the
function
2)
«x,
1.
y),
x > 0: y = i}.
What are the side -measures
What
is the
of
Q?
the rectangle with vertex
area-measure
of
this rectangle?
2.
What
is
the
area-measure
the rectangle with vertex
What
4.
Must the two rectangles shown have
5.
Could you choose a point on the curve such that perimeter
perimeter
of the
corresponding rectangle
is
rectangle with vertex
the
of
P?
3.
is the
Q?
same perimeter?
greater than 100?
If
of the
there are such
points, give the coordinates of one of them.
6.
a
CDEF
square and that quadrilateral
is
is
Among the rectangles we have been considering, which
minimum perimeter? Which a maximum?
has a
.
[7.03]
[7-40]
E.
1.
Graph the reciprocating function
—
=
r(x)
r
defined by:
for x ^
,
[Use cross -section paper and a unit length
2.
On
(a)
'y =
(b)
drew
the figure you
for Exercise
where
—
x +
p(x) =
draw
to
for x >
,
3.
Does p have a minimum value?
4.
State a
5.
about
1
inch.]
graph the equation
1,
x\
Use these two graphs
cise
of
theorem which
3.
Compare
[V
L
theorem
the
in
J
Exercise 4 with Theorem 97b.
*,
o,
T
Exercise
in
T»
of
1
v^
T
Part E, the reciprocating function
has positive values just for positive arguments.
V
x
/
<=>
>
[x
n
L
f-
Use the
[Theorem 97c]
former.
latter in proving the
As you have seen
["addition of ordinates"]
.
what you have guessed in doing Exer-
>
— —
x
x +
_
.
x >
tells
the graph of the function p
x >
That
is,
0].
J
Let's prove this
By Theorem
97a, for a f 0, a 2 > 0.
So, by the multiplication
transformation principle for inequations, for a f
I
a
So,
since
•
<=> -
>
a
a2 =
I
a
Consequently,
^
'
«a 2 >
-a 2
<=>
>
V
x
/
f-
_
,
0,
—
x >
01.
J
and since, for a ^
a > 0.
f—
x
L
*!,
•v
>
t
«JU
<=>
O^
-v
0,
•
a 2 = a,
[7-41]
[7.03]
F.
Prove each
1.
VVV
xyz/=0n
2
Vx V
«
3'
zzi
L
<=> xz
>
[2L
,
[Theorem 99a]
> yz]
y^0 [~>0 <=> xy>0]
<*>
V
(b)
V
V
[yz >
z ^ Q
y ^ Q
V
V
n
x>0y^0z^0
'
.
.
.
/
/
->
(y > z
>
[yz
L7
=>
<^> I
> I)]
> z
(y
7
<=> z
V.
n V/ n V._[y>z->~>-]
z
x>0y/=0z>0
y
(c)
G.
—
following.
of the
l
_
/
[— >
L
f-
x
if
and only
'
if
x >
y
[Theorem
L7
The theorem Vx
> -)]J
100]
that the
you
7
tells
Ol'
J
reciprocating function has positive values just for positive argu-
What does
ments.
about the values
H. Suppose that
real
f
is
theorem
the
of the
is
3(a) of
Part F
tell
you
reciprocating function?
domain
is
the set of
such that
V V [x>
(*)
x y
y
Prove each
of the
1.
V V
[£(x)
>
f(y)
=>
x>
y]
2.
Vx. V
[f(x) =
f(y)
=> x=
y]
y
Exercise
a real-valued function whose
numbers and
x
in
=>
f(x) > f(y)].
following.
y
I_.
Graph each
1.
5x
-
x- 5r
5.
x(x
~
> 7
3
3.
-
of the
-
5x
2.
2(x +
>
x
- + -4
4.
1
3)
>
6.
(2x
.
T
number
following sentences on a picture of the
<
> 4 + 6x
3)
5x
—3—I6
-
-
2
l)(x +
-
3)
,
>
line
.
[7.03]
[7-42]
J.
7.
x(x
9.
x
> x(x
- 2)
10.
x+ i
x
13.
(x2 + 5x)
15.
<*
-
>
1.
y > 3x
3.
x
2
< 36
4>
14.
*16.
>
-
<
ordered pairs
2x + y
2.
4
3.
4x + y > 4 and 2x
4.
There
-
of
and x
4 >
1
-
x
x
-
6 <
-
<
-
6x
1
1
- 5
> 7
(x + 5)(x
JLLL
-
l)(x
-
4)
>
> 2
1
sentences.
5
1
x2
Vx* +
of the following
2y +
all
12.
%
2
+ 5
x»><V
Graph each
K, Find
8.
3)
> 10
11.
-
-
2.
2x + y
4.
y
-
-
3
>
3 <
integers which satisfy the given sentence
-
2y + 3 <
and y
4 <
-
y < 4x < 3y < 9
is
-
y <
and x + 2y < 6
a triangle whose sides are x inches, y inches, and 5
inches long and y = 2x.
JL.
1.
If
three cans of peaches and four cans of pears cost less than
$1.60, and four cans of peaches and four cans of pears cost
more
than $1.80, can you buy a dozen cans of peaches for $2.40?
2.
In a certain
bubblegum.
candy store, a lollipop costs less than a piece
If
you have just enough money to buy a piece
blegum and someone gives you
2 cents
have enough to buy three lollipops.
But,
have more than you need to buy a piece
lollipops.
of
What
bubblegum?
is the
more, you
of
if
still
of
of
bub-
won't
you have 8 cents, you
bubblegum and two
cost of a lollipop and the cost of a piece
1
1
[7-43]
[7.03]
MISCELLANEOUS EXERCISES
A.
Simplify.
1.
7
11
4.
7x + y
x+y
7.
10.
13.
16.
B^.
x -y
2(x+y)
x
2x
2y
-
b+
x 2 - 16
x+5
-
9
y
11.
1
x
4x +
- x2
-
4
xy
3
x
of the
4.
y = 4
|x|
7
y =
1.
-
3x+15
-
Which
I
a
-
4
x + x
=
4x - 8
x + 4
1
b
c
-
3
x
-
2x
2y
x 2 -16
x+5
1
a
17.
....*?.„
9 - 2
-
6.
k
14.
9.
y
3x+15
x
+
-
12.
4
2
V
+
b
-
!
t -
15.
18.
3
a - b
a + b
a + b
4
a
x2
x
5
b
b
-
4y
2y
-
-
9mb
b+1
2
5x
m
(x + 2y) 2
9m
.
b+1
k2
k
1
k2
- 1
t*
9
-
a
-
2
k
-
.
t
*
t -
3
following sentences.
2.y
|
x~hr
of the
i
k
8.
9m
9mb
b+1
x2
5.
5a + b
3.
12
11
k
k
y = |3x
-
ID.
2.
Graph each
1.
C.
2
3
"
=
x+|x|
3.
y = 4x
x|
6.
y = |4x
r^w
9-
y = x2 +
5.
y = |4
8
y =
-
-
x2
-
-
x
xTtt
following equations have no roots?
x
2.
3x - 12
x + 4
_
x+i=i
xx
3
*
3.
x +
x +
2
3
5
x
x(x
-
1)
x(x
-
1)
Solve.
4) = (x +
1.
(x + 5)(x
3.
a2
5.
k(k- l)(k-3)
7.
x(x +
-
-
5a + 6 = 2a 2
7)
=
-
5)(2x
- 9)
6a
k(2k + l)(k-
= (x + 7)(x + 3)
1)
-
2.
y(y
4.
a2
6.
(y +
8.
^-i
3
-
3)
=
y(2y +
5a + 6
=
a2
2)(y-4)(y +
=
3a + 7
3a
5)
-
3a
5) =
40(y
-
4)
[7.03]
[7-44]
E.
1.
Find a point on the y-axis which
from (-1,
2.
3.
A
0)
as
from
(6,
0)
and
8)?
moves so
point
coordinates
A
(4,
the graph of 'y = x' is equidistant
Find an equation
4.
twice as far from
0).
What point on
(6,
is
point
that
is
equidistant
of its path.
Then,
(x, y).
moves so
distance from
it
.
[
.
.
Hint
4).
(0,
Suppose that the point has
from
Find an equation
0).
and
0)
(4,
]
that its distance
(3,
.
from
0) is
(0,
twice
its
"What is the
of the path.
path?
5.
Suppose that P
center
(0,
0)
is
a
moving point whose path
and radius
of
6.
7.
A
is
the point
the circle with
of the
0).
(8,
path
M,
of
Sketch the path
M.
By how much does
0.
Find an equation
2.
the midpoint of PA, where
is
0.
142857 differ from
3333 differ from
j
highway.
of the trip is
If
the
By how much does
?
Mr. Johnson travels by car between
One tenth
-^ ?
Zilchville and Zabranchburg.
through towns and the rest
speed limit
in
towns
is
is
on the open
15 miles per hour and on
the open highway 50 miles per hour, can he average 40 miles per
hour for the entire trip without exceeding the speed limits?
8.
Which
9.
Mr. Jones starts from Zabranchburg
is
the larger,
miles away.
accident.
proceeds
the road.
is his
or
£-=-,
or are they equal?
to drive to
Frampton, 24
Fifteen minutes after starting he meets with an
After wasting
at half
5
minutes trying to
speed to a service station
Three quarters
and by driving
to get to
-rj-
12 miles an
Frampton
of
he
miles farther along
an hour later he starts out again,
hour faster than
50 minutes later than he
usual driving speed?
2
fix the car,
at first,
he manages
had planned.
What
[7-45]
[7.03]
EXPLORATION EXERCISES
A.
Consider the eleven sets listed below.
the set of all real
(b)
the set of all rational
(c)
the set of all irrational
(d)
the set of all negative
(e)
the set of all nonnegative
(f)
the set of all integers
(g)
the set of all integers greater than 5
(h)
the set
+
I
{1,
(J)
{-y>
(k)
{5,
1.
For each
0,
the
2.
5-j,
4,
5,
6,
y,
1,
l~,
6,
For which
4.
Name
1
numbers
7,
10}
9,
8,
2,
.
.
.
7-j,
'
.
.
.
mean?]
8y, ...}
8,
eleven sets, put a check
mark
column
in
(1) if
belongs to the set.
check mark in column
S
S a set:
V
and
1
x
c
[x
L
S and
1.
Name some members
2.
How many
3.
Do you
1
(2) if, for
each
belongs to this set.
check marks
in both
columns?
such that
five other sets S
e
4
2
2
7,
[What does
}
sets do you have
1
each
numbers
of the sets, put a
3.
{S,
y
numbers
x belongs to the set then x +
if
Eet?7\=
2
6-j,
of the
number
For each
numbers
of all positive integers
3,
2,
(i)
(2)
numbers
(a)
2
B_.
(1)
S
e
V
=> x
[x € S
+
1
e
Si.
J
=> x
+
of 777
which
1
e
S]}.
of77"\.
sets belong to/71?
think that there is a
member
of 777?
member
is a
subset of
x,
[7.03]
[7-46]
C.
In
A
Part
+
you noticed that
e#(--that
I
that
is,
+
1
(i)
€ I
and that
V
(ii)
Here
is
a proof, using
(i)
=^ x
I
and
+
e I*].
1
+
that 2
(ii),
e I
2=1+1
[xel =»x+le
(1)
+
V
(2)
+
[x e
=>
+
(3)
e I
1
1
+
:
[definition]
+
I
[(ii)]
]
+
€ I
1
[(2)]
+
(4)
(5)
+
1
(6)
1.
In
A
Part
1
e I
[(i)]
1
€l +
[(3),
(4)]
2el +
[(1),
(5)]
you also noticed that the set
How
belongs toTTT.
R
of all rational
numbers
could you easily obtain a column proof, in
which you used no other fact about R, for the theorem '2eR'?
2.
Suppose that
is that
3.
Does
S eTf\.
Prove that
theorem
3 € I
'2 e
you are told about some set S
all
+
I
+
[
.
'.
Repeat Exercise
5.
Repeat Exercises
6.
Do you think
view
of
Hint
this that 2
You may use
.
e
S?
[Why?]
proved
the previously
with
2
4
3 e S'
and 4 with
3
in place of
'4' in
4
2 e S'.
place of
that there is a positive integer
prove belongs to
In
from
follow
]
4.
7.
it
numbers
of real
+
I
by using
(i)
and
(ii)?
'3'.
which you could not
[Given lots
of
time!]
your answer for Exercise 6 [and your answers for the
preceding exercises], what are some
of the
numbers which you
believe belong to each set S which belongs toT^T?
8.
Consider the following statement:
(iii)
Vg
[( 1
€
Do you believe what
S and
this
Vx
[x e S
says?
=>
x +
1
e S])
==>
+
I
C
S]
[7-47]
[7.04]
7.04 The positive integers --In an earlier section you proved that
that 2 € P, that 3 € P,
and shown that
the real
5 e
numbers
we have proved
One reason
and that 4
P, that 6
2,
1,
P,
of
P.
we could have continued
Clearly,
we
until
. • •
got tired.
them really
is
*
it is
nothing about the positive integers.
P
is that
of
Another
it.
impossible at this stage to prove that each
'positive'* integers belongs to
is 'no*.
number
for this is that a generalization with an infinite
reason why
Could
The answer
positive?
instances cannot be proved just by proving instances of
called
As you know,
are called the positive integers.
...
3,
each
that
€
€
P»
€
1
.
of the
so-
our basic principles say
theorems about
In order to prove
+
I
--the set of positive integers --we need additional basic principles.
Back
in Unit 4 you did
to do so you
prove some theorems about
made certain assumptions. They were
(1)
The positive integers belong to P, that
(2)
I
(3)
For
+
is
is,
I*,
and
in order
the following:
C, P.
I*
closed with respect to addition and multiplication.
all positive integers
m
and n,
if
n >
m
then n
-
m
is
a
positive integer.
is the only positive integer
(4)
1
(5)
Each nonempty set
(6)
For each real number
k < x < k +
(7)
between
of positive
and 2
member.
integers has a least
x, there is
an integer k such that
1.
Each positive integer other than
1
has just one prime factori-
zation.
So, in effect, in Unit 4 you adopted (l)-(7) as additional basic principles.
Now, certainly,
this is a
heterogeneous
lot
and there
reason to believe that they are adequate for proving
may want
to
prove about
I*.
Let's try to find
ciples which [together with the sixteen
We
even any
is not
all the
theorems we
some simpler
we already have]
will do the job.
said that the positive integers are the real numbers
Our problem
is that of
explaining the
*
...
*
making
this statement
basic prin-
1,
2,
3,
more precise- -actually,
.
someone who doesn't know asks you what the
positive integers are. You might begin by saying that the positive
Let's suppose that
.
.
.
of
.
[7.04]
[7-48]
integers are
2,
1,
To
and so on.
3,
test his understanding of your
You would
answer, he might then ask, "Is 4 a positive integer?'
probably
tell
him
that, yes,
And you might add
on.
tive integer.
that, for
each positive integer
(a)
1
(b)
Vx
e
This last
,+
x +
€ I
1
try to restate
real
information
it
K
e
real
+
I
If
a real
using
(a)
+
number
and
4
by
*I
's
So,
if
S satisfies
in it
In other
C^ S.
S's,
if
you say that
(*)
On
v
K
+
be the set
by using
I
x +
1
e
+
(a)
I
those
of all
and
Let's
.
(b).
Then,
K
integers
all the positive
-
suppose that S
V
x
e
x +
c
s
1
e
-amounts
is a set
S
then, by the definition of K, there is a proof
+
e I
.
Consider such a proof.
you then have a derivation
then each
member
of
K
is a
you replace
If
of 'c
e
S'
member
[(1 e
+
I
S and
C K
V
from
of S,
x +
1
e S)
=> K C
x
£
c
o
x +
1
e S)
=>
+
I
leK, you are implying
that
+
I
CK.
C
—
that is,
Si.
S].
Since
1
e
all
(1).
then you are implying that
VQ [(leSandV
o
c
the other hand, suppose that you assert (*).
V..X+
To
words,
VQ
So,
K
(1)
€
K
this entails,
and
that c
(b)
way?"
that
S
e
c e
the
K
To see what
numbers such
1
x
K contains
-that
C_K.
(1)
V
of
"
positive integers there are.
Let
to be in
and
Your answer, above-
+
I
question:
a third basic principle about
more usable form.
in a
1
of
is
all the
number
out an unlimited
positive integers which can't be gotten this
numbers which can be proved
to saying that
things about
+
more important
But, he still has one
bit of
posi-
e I
you would say, "No, those are
this
is a
1
T
Knowing these, he has a method for picking
"Are there any
x +
x,
and so
is 5,
him two important
So, so far, you have told
positive integers.
and so
4 is a positive integer
K and
:
[7-49]
[7.04]
+
So, (*) says precisely that
I
C
each positive integer can be gotten from
Let's use
P
Since
if
(
€
1
Now, by Theorem
V
),
P and V
82,
1
p x
e
P.
+
1
principle for
V
(c)
Using
In
s
(a)
+
I
Vx
S and
(b) it is
more
=> x +
[x e S
e I+
l
l
l
and
(*)
S' as
(c)
'p',
'n',
S' or
V q e S'. In
theorem 'V
4
for the trivial
and
l
q'
particular,
t+
+
I
are equivalent.
of later
as variables
*
2
)
Vn
n +
»*3
)
Vs
[( 1
(I
Note that when
+
(I
2
)
is
Also, one can rewrite
1
el
1
e I
theorems,
x
+
'
e I
Vm
l
T
m
+
x
+
e I
'
e
whose domain
we
S',
is
is I
+
.
shall write
an abbreviation
Using these conventions about
variables we can restate our three basic principles for
(I\)
simpler
in a
a variable whose values are sets of real
So, for example, in place of the sentences 'V
e
P.
^Vx£l+ xeS]
e S])
order to state our three basic principles for
shall (as above] use
m m
C
useful to take as our third basic
possible to show that
numbers, and use m',
'V
+
convenient for proving
(*) is
form, and to simplify the statements and proofs
we
I
that
(*)
By Theorem 82 and the basic closure
+
P. So [by modus ponens], I C_P.
e
will be
P) then
e
1
from
the following:
[(l e
and
it
us that
(b).
follows
it
x +
Although, as you have just seen,
certain theorems,
and
(a)
(*) tells
on page 7-47:
(1)
a set of real numbers,
is
principle (P3
prove
(*) to
words,
In other
K.
+
I
as follows:
+
+
e
S and
Vn
[n e S
=»
n 1
1
€ S])
=> V n
n
written out fully, we have *VX [x
Theorem
101 as 'V
n
ne
P'.
+
e I
e
S]
==> x +
1
e I*]
1
[7.04]
[7-50]
EXERCISES
A. In each of the following exercises you are given a set
is to
Your job
S.
prove or disprove:
V
n
Sample
S = {m:
.
Solution.
[n €
{m:
can simplify
[3 +
L
n
1
e
S]
}
to
3 +
prove [or disprove]
m
e
T} => n
immediately
this
€
=> n +
+
e I
What we want
Vn
T =>
3 + (n
+
+
1
e
following:
is the
{m
3 +
:
m
familiar
way by supposing
3 + (p +
Proof
T]J
1) €
.
4
3 + p e
I
+'
and then deriving
eT\
1)
Suppose that
(3 + p) +
i
Therefore,
That
3 + p e
Then, by
T.
B.
is,
L
S =
{m:
m
> 1}
2.
S =
{m:
m
-
3.
S =
{m:
m
4.
S = {m:
m
< 4}
{m:
m
< 0}
Prove
that 3
5 e I*}
>
+
e I
if
p
e
3
+ p
+
e I
.
m}
So,
then
S then p +
J
1.
2
if
= 3 + (p + 1).
[neS=>n+leS].
Vn
S =
}]
to:
+
(I
2
),
(3 + p) +
But, by the associative principle for addition,
*5.
+
e I
order to prove this generalization, we proceed in the
In
4
m
3 +
V
We
[n € S
1
e
3
+ (p +
3 + (p +
S.
T.
1) e
+
1) e
I
.
Consequently,
+
1
e I
.
[7-51]
[7.04]
PROOFS BY MATHEMATICAL INDUCTION
The
positive integers in an earlier unit.
+
proved by using (l*^,
(I
and
2 ),
+
(I
[To prove
3 ).
These theorems
basic principle.]
first of
The other six assumptions can also be
which we have already proved.
more
made about the
these is Theorem 101
stated certain assumptions you
On page 7-47 we
will be
(6)
we
proved
shall need one
in this
and later
units
As an instructive example
of
how such theorems are proved,
let's
prove:
Vn
(*)
[Notice that since 3
+
e I
3 +
n
+
e I
this is an instance of a
,
theorem, V V m + n€
m n
+
4
I
-
+
which says that
To prove
that
when
I
(*) is
it is
(*) is
to
prove that each positive integer has the property
added to
to saying that {x:
words,
closed with respect to addition.]
is
3 +
x
the
3,
+
e I
}
1
6
(I
is
This amounts
a positive integer.
contains all the positive integers.
In other
equivalent to:
Vn n
Looking at
sum
+
we see
3 ),
{x: 3 +
x
+
€ I }
that
(i)
3 +
(ii)
Vn
if
1
{x:
l
3 +
+
x
€ I
we can prove
and V
And, we can prove this
€
[n e {x:
3 +
}
J
this
x
e
if
we can prove:
T} => n
+
1
e
{x: 3 + x
we can prove:
e I*
and:
Now,
let's start
[3 +
proving.
n
+
€ I
=>
3 + (n
+
1) e I
+
]
[Explain.
]
e
T}]
',
:
:
[7.04]
[7-52]
Part
(i)
Vn
(1)
n +
(2)
3
(3)
Part
e
T
(i*
3e
r
theorem]
€
r
(1), (2)]
1
+
1
2
(ii)
3 + q €
(4)
+
Vn n
+
1
6 I*
(6)
(3 + q) +
1
£ I
V V x+
x
(7)
y
assumption]*
I
(5)
(y + 1) =
+
u 2 )]
+
(4), (5)]
(x+y) +
1
theorem]
1
(7)]
(8)
3 + (q +
1)
= (3 + q) +
(9)
3
+ (q +
1)
€ I*
3 + q £
(10)
V
(11)
Part
)]
(iii)
[3 +
n
+
I
+
£ I
3
+ (q +
1) e I
=>
3
+ (n +
1) £ I
{x: 3 +
x£
(13)
Vn [n€
{x: 3 +
x
(14)
VJdeSandVn [n
(15)
*(4)]
(9);
+
(4)
]
-(10)]
:
1
r
+
=>
£
(12)
(6), (8)]
(1 €
Vn [n
£
+
I
(3)]
}
+
€ I
}
£
=> n
S
+
{x: 3 + x £
I
{x: 3 + x £
I
}
+
=> n
}=>
{x:
3 +
x£ T}
(16)
V n
£
{x:
3 +
x
(17)
V
+ n
£ I
+
n +
£
3
€
1
{x:
£ S])
1
3 +
x£
+
I
=> V n
n
(11)]
}]
£ S]
and
V n
n
+
+
£ I }
1
£
{x:
3 +
x
€ I
+
(14)]
}])
J
(12), (13), (15)]
+
(16)]
[7-53]
[7.04]
This proof
a typical proof by mathematical, induction [or
is
+
The basic principle
tive proof].
(I
3
)
some
[or
called the principle of mathematical induction
an inductive proof
of a
sentence
Vn
of
indue
:
equivalent statement]
-
is
As illustrated above,
.
a generalization about positive integers--that is,
form:
of the
[V
1
n
F(n)
3 +
n
c
T]J
consists of three parts.
Part
[steps (1) -(3)] is a proof of the generalization's
(i)
instance"
:
F(l)
[3 +
[The whole of part
Part
n
of
part
assumption [step
Part
(iii)
(I
(11)] is a
proof
(ii) is
(4)] is
of
[V
n
el + ]
initial step of the proof.]
a sentence of the form:
[3 +
n
+
e I
=>
3 + (n +
+
1) € I ]]
and the
step;
called the inductive hypothesis .]
This part
).
1
sometimes called the inductive
[steps (12) -(17)], which is
+
3
-
sometimes called the
=> F(n +1)]
V^ [F(n)
[The whole
is
(i)
[steps (4)
(ii)
brings in
"first
much
is
the
sometimes called
same
the final step
in all inductive proofs.
It
consists of six steps of the form:
€
1
{x: F(x)}
(
)
(
)
V
[n € {x:
(
)
V
[(1 €
n
g
C
(
)
I
(
Vn
[n
€
V
R
[n
(x: F(x)}
e
{x: F(x)
)
V n
e
{x:
^
)
V
n
n
F(n)
1
S=>
F(x)}=> n
{x:
n
6
+
e
{x: F(x)}]
n +
1
€ S])
=> V n
n
}
J
F(x)}J
6
S]
[(I
">
and
V n
V.
(
(1 €
F(x)}=> n
S and
]
[
+
1
e
{x: F(x)}])
J
[
,
+
3
)]
,
[7.04]
[7-54]
The marginal comments for the
proof, to the conclusions of parts
comment
ginal
fifth
first
and
(i)
two steps refer, in each such
The mar-
respectively.
(ii),
That for the
for the fourth step refers to the third step.
step refers to the first, second, and fourth steps.
Finally,
the
marginal comment for the last step refers to the preceding step.
Obviously, once you have completed parts
proof, writing part
is just
(iii)
and
by printing this form after part
comments on the right. Instead
we have completed parts (i) and
(
numerals
doing even this, let's agree that once
of
(ii)
Vn
)
we can skip
step
makes use
(i)
and
+
(I
3
),
(ii)
that is,
principle of mathematical induction.
consists
(ii),
of
and a
(iii),
the proof of
3 +
n el
+
[(3),
(1).
of
part
(ii)
As usual, paragraph proofs are shorter.
(i)
(ii)
(*)
3 e
+
I
,
it
Suppose that
3
But, (3 + q) +
(iii)
since this is the
(5),
would be
Here
4
(1), (4) -(9)'.]
paragraph
is a
:
Since
Vn
PMI]
(11),
Doing so, the succeeding steps would be renumbered
and the comment for the conclusion
proof for
(i)
final step:
[We could also shorten the proof by leaving out step
as step
theorem
make up parts
twelve steps --the eleven steps which
Vn
(12)
same
of
PMI]
,
,
will refer to the conclusions of parts
of this
form:
of the
[
With this agreement for shortening part
and
the first five steps given
F(n)
and indicate that the justification
(*)
and the marginal
at the left
and finish the proof by writing a single step
(iii),
The marginal comment
of the
finish any inductive proof
writing the appropriate sentences in
(ii),
the nine blanks, and filling in the
in part
an inductive
like the one above, but leave blanks
Then you could
'F(n)'.
of
(ii)
hack work. You could even have a rubber
stamp made which would print a form
in place of 'F(x)'
and
(i)
[3
From
+
+ n eI
(i)
and
follows by
+ q
1
e
I\
(I
It
+
2
)
(ii)
3
+ (n +
it
3
follows by
= 3 + (q + 1).
=>
that
So,
+
(I
1
e I
+
2
)
+
.
that (3 + q) +
3 + (q + 1) €
+
I
.
+
6 I
1
Hence,
4
1) € I ].
follows, by the PMI, that
Vn
3 +
n
+
e I
.
.
[7-55]
[7.04]
Before writing an inductive proof, you will find
the
theorem you wish
to
prove
and to write the sentences
*•
J
the forms:
of
V
and:
F(l)
=> F(n
[F(n)
of parts
which are to be the conclusions
in seeing
F(x)
{x:
e
helpful to restate
form:
in the
Vn n
it
(i)
and
(ii)
what these last two sentences are, you
+
If
.
may
1)]
you have trouble
find
it
write a sentence -form with a blank in place of the variable.
you were to prove
helpful to
Thus,
if
(*):
Vn
3 +
n
€
T
you would, before beginning the proof, write:
V ne
n
3 +
(i)
3 +
{x:
v
.
.
3 +
.
e
I
xc T}
+
+
1
e I
J
(ii)
V
[if
3 + n
e I*
then 3 + (n +
+
1) € I
]
EXERCISES
A_.
1.
2.
Give a column proof
of
Theorem
101 [*V n
e
—
mathematical
induction.
[Before stating the proof, carry out the suggestion
made
preceding paragraph.]
Write
in the
all six
steps of part
(iii)
of the
inductive proof of
101.
B.
P'] by
1.
Give a column proof
r
2.
Give a paragraph proof
of 'V
n
3n
of the
e I
+ *.
same theorem
Theorem
[7.04]
[7-56]
CLOSURE OF T WITH RESPECT TO
Two
of the
AND
+
X
assumptions about positive integers which you made
in
an earlier unit amount to the following theorems:
Theorem
102.
m
Vm Vn
+ n
€ I
+
and:
Theorem
103.
m Vn m
V
Theorem
and
its
on page 7-51
(*)
is
n
•
the "third instance" of
Theorem
proof will suggest how to prove
We now have
instances
To use
of the
Theorem
The
r.jd-ke
up a testing pattern for the
The second method involves
generalization.
first
depends on the test-pattern
in earlier units,
method, you
this
102,
102.
two methods for proving generalizations.
method, which you used
principle.
+
e I
the
PMI
[and applies only to generalizations about positive integers].
In Unit 2
you learned that you could often find a test-pattern for a
generalization by first proving an instance
example,
in
order to find a proof for
by trying to prove, say,
H
(2)
V X-
+
12
X
1
:
1
:
(3)
3
(4)
a + oi
(5)
x(y +
(6)
(7)
U(l
EJ +
z)
1)
[I]
2
(8)
(9)
+
'
E]
=
+ ,3
-
Q] +
of
-
:=
==
==
Here
m
--
=
is
x
•
2'
you might begin
such aproof:
[pml]
[(2)]
(II-
xy
1
GOa
+
+ QJ.
1
xz
+
[H-i
3
=
For
[Identity]
X
i
the generalization.
x + x
-®
=-
=
V
3*2'.
'3 + 3 =
(1)
4
+
+
i
-2
[(1), (3)]
[idpma]
GOi)
1
[(5)]
[(4), (6)]
[definition]
[(7), (8)]
[7-57]
[7.04]
Now,
is
in
order to obtain a proof
replace each boxed
4
3'
by, say, an 'a',
V x
(10)
of *V
m
we begin by trying
€ I
suggests that a proof
V
(*)
may
indicate
n
how
3 +
to write
e
of
3 +
1
p +
For,
if
6 I
we can do
(
-
[(1)
Theorem
(9)]
102:
form:
r
one such sentence, say:
e
I
(*)
on page 7-52.
V
n
This
is
an inductive
to prove:
+
[3
L
n
+
e 1
=>
3 + (n +
+
1)
e I
]
J
to do is to get test-patterns ending with:
+
1
and add one more step:
such a test-pattern.
and:
T,
What we now want
you need do
all
,
main problem was
e
2',
+
n
Let's look at the proof of
proof, and the
•
to find a test-pattern for sentences of the
p + n
n r
it
x
+
+ n
V
And,
=
x-2
+ x =
This suggests that, in order to prove
Vm Vn
x + x
V
and:
,
+
I
=>
p + (n +
+
1)
e I
[
,
]
we can conclude:
this,
Vn
)
[p + n e
p + n
e
T
Then, we shall have a test-pattern for sentences
V
+ n
n p
of the
,
and PMI]
form:
+
e I
,
and, by the test-pattern principle, can complete the proof of
Theorem
102 by writing
(
)
m Vn m
V
+
nel +
[(1)
l\ /
-
(
\
)]
/j
[7.04]
[7-58]
EXERCISES
A.
Theorem
1.
Write a column proof
2.
Write a paragraph proof
of
of
vUN
"l
If
Part
Vn
n +
p +
(3)
Parts
VL, V_
n
(i)
m
and
(ii),
,
+ n €
+
I
make up
(iii)
1
Part
of
at
A
part
(i)
probably
e
I
[(!*)]
p
e
T
[theorem]
1
€
I
+
We can somewhat
our convention that the domain
is
a trivial
theorem — step
(3) is
of *p' is
comment
plete part
(5)
and
for (15).
(i)
(13),
numeral
a
for
substitution, step
+
I
.
Because
x»*
e I
+ '.
some
(2) will
'pel +
of this,
»
(1).
>.**
"V
'i %
of
and
Theorem
fill
in (14)
After you have seen
and part
r + n
n p
'V
an immediate consequence of step
Complete the following proof
(i)
form
simplify the proof by taking account
K^
V
First look at
'p'
for
making such a
So, after
(2)]
a test-pattern for instances of
we substitute
of
Part
[(1),
*
positive integer.
be a theorem.
+
1
'--that is, for sentences of the
test such an instance
1.
*v
(i):
(2)
B_.
o„
*t*
*(*
:
(1)
To
102.
proof on page 7-52, your answer to Exercise
begins like this
m
Theorem
you followed the discussion on test-patterns, and looked
of the
*
102.
10 3.
[
Important
:
and the marginal
how
to do this,
com-
(ii).]
:
+
(1)
V n
e
I
(2)
P
e
I
[trivial
theorem]
+
[(1)]
(3)
[basic principle]
(4)
[
(5)
pi
e
T
]
[(2),
(4)]
[7-59]
[7.04]
Part
(ii)
:
(6)
V V
(7)
x(y +
+
pq
e
l
[inductive hypothesis ]*
1)
=
xy + x
[theorem]
(8)
[
(9)
[
pq + p
(10)
e
r
[
]
(9)]
,
(11)
[
(12)
[(H); *(6)]
Vn [pnel + => p(n+
(13)
Part
<iii)
l)e
+
I ]
]
-
[
]
:
(14)
[
Vm Vn mn
(15)
1
'
2.
]
,
,
PMI]
+
€ I
Write a paragraph proof
of
Theorem
103.
MISCELLANEOUS EXERCISES
1.
If
the area of a rectangular enclosure is 48 square yards and the
perimeter
2.
Mr. Jones
is
is
28 yards, what are the length and the width?
24 years older than Bill.
twice as old as Bill.
3.
A
do
4.
can do a certain task in
it
in 4 days.
One leg
of
hypotenuse
5.
How
How
0)
days.
long would
it
is 7 feet
now?
B works along with A, they can
take B to do it alone?
If
longer than the other leg.
If
the
13 feet long, what is the area of the triangle?
Suppose that the vertices
A(-5,
8
a right triangle
is
old are they
Eight years ago he was
and B(5,
0).
of the
acute angles
of a right
triangle are
Write a sentence whose graph contains just the
points which can serve as the third vertex of this triangle.
[7.04]
[7-60]
6.
The average velocity
about
7.
A
X
5
10 4
of
unit.
It is
long.
If
room temperature
of air at
About how many miles per hour
cm/sec.
commonly used
unit
molecules
in
measuring atomic particles
is
is
this?
is the
Angstrom
8
a unit of length, and is defined to be 10~ centimeters
o
the average atom
placed side by side would
8.
5A
is 3.
take to span a
it
how many such atoms
in diameter,
1
-inch distance?
Physicists in the 18th century discovered that the force of attraction
between two particles with opposite electrical changes
inversely
is
proportional to the square of the distance between them, and jointly
proportional to the magnitude
of the
How
charges.
attraction between the charged particles changed
between them
(c) if
9.
Q
R
if
rilateral
Let
doubled?
(b) if
(a) if
the force of
the distance
the charge of one particle is doubled?
the charges of both particles are doubled?
[A point
point
is
is
P2
is
P
said to be a reflection of a point
and only
ABCD
if
is a
R
midpoint
is the
of
PQ.
Px
be the reflection of
Suppose that quad-
]
parallelogram and that P
with respect to a
plane.
is a point in its
with respect to A,
P3
be the reflection
P2 with respect to B, P4 be the reflection of P3 with respect
and P5 be the reflection of P4 with respect to D. Prove that P5
of
10.
(a)
If
V
f(p) =
P
(b)
If
V
p
11.
f(p) =
*P + 1 then
^P " 3
V
f(p +
and V
=
1)
p
andV
f(p + 1) =
(c)
If
Vn
g(n) = n(n + l)(n + 2) then
V
(d)
If
V
2
g(p) = p
g(p +
(e)
If
Vn
S(n) = n 2 then
A
-
(p
- l)
Vn
2
then V
S(n +
1)
student looked at the statement
thought
it
was false because
it
=
f(p-l)
=
Px
=
,
P
P
p(p-l) then V
- 1)
f(p
to C,
n
g(n +
1)
1)
and V g(n
=
n
and V g(p
=
and V S(2n
=
n
*2X(3
looked as
+ 4) = 2 +
if
+
1)
- 1)
- 1)
=
=
=
(3X4)' and
someone were trying
to
use a commutative principle on the multiplication and addition signs.
But, of course, the statement is true.
and
z
such that x X (y +
z)
=
x + (y X
z).
Find other integers
x,
y,
[7-61]
[7.04]
RECURSIVE DEFINITIONS
One
of the
discovery
of
mathematical pastimes
ancient Greeks was the
of the
interesting generalizations concerning whole numbers.
For example, consider the following diagram:
\*
V
V
•\
V
A
•
#
,\.
.
\
If
we count
•
•v
••V
•
•
*
•
•
*
•
•
•
•
y
•
•
v
•
\
\
the dots in each of these "triangles" and in those
get by continuing the obvious construction,
we would
we obtain a sequence
of
num-
bers which begins
28,
21,
15,
10,
6,
3,
1,
...
.
The Greeks called these whole numbers triangular numbers.
purposes
will be
it
more convenient
For our
regard the corresponding positive
to
integers as triangular numbers.
Suppose that we use
T. = 1.
1
Also, T2
'
= 3,'
'T
T,3
'
= 6,'
to
name
T,4
Suppose you are told that T
to find
T X1
10,
and T 5
=
What
15.
Can you use
is 55.
this
number.
is T, ?
6
Then
T_
7
?
information
?
Given that T 25
As you
=
the first triangular
= 325,
find T 26
.
see,
T1
r
(1)
These formulas give
IV
V n
a
way
T
,
n+
of
=
1,
and
=
T
+ (n + 1).
n
l
computing triangular numbers. For example,
T6
=
T5
=
(T 4 +
=
(T 3 + 4) +
5 + 6
=
(T2 +
4 +
=
(T
=
1
1
+
+ 6
5)
+ 6
3) +
5 + 6
+ 2) + 3 + 4 + 5 + 6
2+3
+
4+5
+ 6 = 21.
[7.04]
[7-62]
The process
compute T
to
using T 5 to compute T
of
is
Evidently,
That
is,
(1)
What
whose domain
of the
is the
+
4
,
it
is
Use recursion
.
T 10 instead
'
compute T6
positive integers and
in
in
T are
T with
(1,
1),
8 as first
(2,
3),
(3,
component?
to
whose
and
6),
(4,
10).
With 9 as first
of 'T(IO)' .]
of
formulas
(1) is
Here are recursive
Vn
+
l=
(n+1
called a recursive definition
f
>
>
f
Vn
n
n +1 =
working with functions
to a different
way
of defining
consider the sequence,
S,
of the
func-
definitions of two other functions:
r
n+
of
>
><
<
£
-
=
l
1
|r
\
n
+
r
/2
n
n
in earlier units,
functions
l =
Complete
the first four values of each function.
Vn f n + 2 -<
you became accustomed
-explicit definition.
For example,
square numbers.
•
An
.
modify the usual functional notation and write, say,
r
In
to
[Notice that in the case of a function like T whose domain is
The pair
Compute
and T
,
computing any triangular number.
of
is the set of
ordered pairs
customary
tion T.
way
,
triangular numbers.
ordered pair
component?
1
recursion
compute T 5
to
gives us a procedure for computing the successive values of
is the set of
Some
of
gives us a
(1)
a function, T,
range
an example
T4
,
fe
•
•
•
•
•
•
•
•
•
•
•
•
explicit definition of S is
n
With this kind
of definition,
S
n
=
n2
you can find any value
of S
directly without
previously computing other values.
Can we discover an
explicit definition for
play with this problem before reading further.
T?
You might want
to
[7-63]
[7.04]
following figure shows that T 6 + T
X. The
Let's reconsider
=
6*7,
fe
\
and suggests that
Vn Tn
(2)
We can check
and
that, for
=
n n +
(
recursive definition by noting that
this guess against the
each
*)
2
1(1 + 1)
=
+!)+!]
=
j
n,
(n + l)[(n
n(n +
1)
+ (n +
x)
#
[Verify this last generalization by transforming the right side of the
equation into the
left.
]
This checking procedure shows that there
--which satisfies the recursive definition.
the one defined by
(2)
we want to do now
is to
show
We
is to
(2)
V
prove:
n
shall use the principle of
What
that there is only one such function.
we can accomplish by deriving
Our job
function- -namely,
is a
from
Let's do so.
(1).
n (n +
T
This
1)
n
mathematical induction to prove
generalization about positive integers.
this
Restating the generalization,
we have
w
(
-r
V ne, {m:
T
n
m
As a help
in
=
Ti =
+
2
=
1)
»
-}
formulating the proof we write:
...(... +
T
<i)
m(m
—
Ki
+
D
(ii)
T
V
n
n
=
n n +
(
2
*)
1)
==>
T
n+
_
l
(n-H)[(n-H)+l]
[7.04]
[7-64]
Part
(i)
T 1=
(1)
Kl
(2)
+
.
*
_
T
=
definition]
[theorem]
,
2
(3)
Part
1)
[recursive
1
(ii):
(4)
V
(5)
n
T
n+
,
Id
^
q(q+
1)
[(1), (2)]
[inductive
1)
hypothesis]*
[recursive
T
+ (n +
1)
T
+ (q
VM +
1)'
n
l
+
definition]
(6)
T
(7)
W-aafiUu +
(8)
q+L
(12)
ais^Ll) + (q +
i)
=
la +
_
(q + l)[(q + 1) + 1]
2
Part
t
=
(t
=
q(q +
q
n
[(4),
<*+'>[<*+l>
"
V
i)
=
q+
7
[(5)]
1)
(10)
(in
q
+ (x +
V i<i + i>
{9)
=
l
n
l)
2
n(n +
2
*>
1
=> t
q+
=> T
n+
=
+ l]
[theorem]
m +ILLU
(q j
D[(q
[(8)]
=
(n + l)
l
Hn
UJLIJJ
+
2
l)
(9)]
[(7),
2
l
(6)]
+
Ll
[(10);
[(4)
*<4)]
-(H)]
(iii)
Vn T
(13)
In the future
n
we
sive definition]'.
=
n( n +
*)
[(3),(12),PMI]
shall omit steps like (5) and label steps like (6) '[recur-
[7-65]
[7.04]
one could replace steps
In practice,
q(q +
1)
(8),
+ 2 (q +
and
(9),
(10)
by
1)
2
[algebra]
/
<q + D[(q +
You might want
Here
1)
_
knew what you were doing.
T
,
'
T
nition,
Kl
=
q+
l
T
q(q +
pi
_
Here
1)
is
it
n(n +
1)
aia +
i) =
then T
'
(iii)
-.
?
Since, by the recursive defi-
follows that T
q+
q
q+1
w fT
n
=
+ (q + 1),
But, stfL+Ji + (q +
_
n+i
From
(i)
and
+ 2 'q +
U
(q+l>[(q+l)
=
T
2
and since
= 1,
+ 1)
2
l
Suppose that T
(ii)
=
steps than these to convince
Since, by the recursive definition of T, T
(i)
T
1]
a paragraph proof of (2):
is
1(1 +
2
more manipulation
to put in
a reader that you
1)
+
(ii) it
U
-
=
,
l
'
+ (q+
M
2
<q +
+ l]
q^ q *
l
^
+ z>
1).
'
s ».
.
Consequently,
(n + l)[(n + 1) + 1]
2
"|
'
J
follows, by the PMI, that V^ T^
n n
=
"^ ^
a tree-diagram of the proof on page 7-64.
(5)
(8)
(4)
(9)
(6)
(7)
(10)
(2)
(1)
(3)
(11)
[PMI]
(12)
(13)
It
shows clearly
that (13) is a
consequence
the recursive definition consisting of
(1)
of the
and
(5),
if
theorems
(2)
and the PMI.
and
(8),
.
[7.04]
[7-66]
Use either column proofs or paragraph proofs
for the following
exercises
EXERCISES
A.
1.
Consider the following recursive definition
domain
is
of a
function
E whose
+
I
:
E1
E
V
n
(a)
Compute E p
(b)
Complete
E
,
=
n+
E
,
= 2
E
i
+ 2
n
and E
,
.
the following
E
V
n
n
=
i
to obtain an explicit definition of E.
(c)
2.
By mathematical induction, derive
E from its recursive definition.
the explicit definition of
Give a recursive definition for the function
O whose
values are
the successive odd positive integers, and repeat Exercise
using your recursive definition
of
O
in place of the
1,
recursive
definition of E.
3.
Here
is
a recursive definition of a function S
V
S
n
Repeat Exercise
B_.
1.
1
sum
of
16
=
S
n+
,
l
of
n
+ (2n +
1)
triangular numbers.
15
10
9
1
for the function S.
Consider the sequence
4
s 1=
25
21
28
T
tP
Notice that the
T
/p +
i
36
each two consecutive triangular numbers appears to be
[7-67]
[7.04]
a square number.
which begins
This suggests, as a theorem, a generalization
:
V
S
n
=
n+
,
1
Complete the generalization, and prove
2.
The
A
explicit definition of S
it
by mathematical induction.
which you obtained
in
Exercise
3 of
Part
is:
Vn
Use
(a)
this definition to
Vn
(*)
'
v
Use
(b)
this result
numbers
to
S
n
complete the following generalization:
S
=
,
2n +
l
and the explicit definition
may
(*)
of the
triangular
discover a relation between the odd square numbers
and the triangular numbers.
side of
n2
=
[
Hint
Transforming the right
.
Express your discovery as a generali-
help.]
zation beginning with:
V
n
S
=
,
2n +
l
Give a second proof- -this time by mathematical induction- -of
(c)
the generalization you discovered in (b).
3.
Consider these three se qu ences
:
1
2
3
4
1
3
6
10
P
.
.
.
T
P
1
4
16 ...
9
S
P
Discover how
to find a
corresponding terms
this
term
of the
of the first
second sequence from the
and third sequences.
discovery in a generalization and prove
it
Express
by mathematical
induction.
4.
Prove
of
that, for
each
n,
the nth even positive integer is the average
the 1st and the 2nth odd positive integers.
.
^
.
[7-68]
C.
~~*
1.
t
Here
of a function
recursive definition
is a
f
f
Compute
2.
sum
the
r ± = 3,
and V
r
If
f
3
= 7,
and V
4.
If
i
1
= 9,
and V
5.
If gj.
= 4,
and v
c,1 = 1,
and V
gl =
and V
6.
If
7.
If
8.
If
f
9.
If
t
,
1
n
n
1,
c
f, =
1
11.
If
f,
1
=
and V
=
f
t
and V
1,
_
=
f
t
n
_
=
(2n+
—rr>
nn+1
•
"
nn+1
•
,
,
(b)
Find the smallest
m
m
1Q
.
"
then c 35
1),
then g 4
=
.
lo
1
10 =
.
3 =
and
f,
then
f
4 =
and
f
in
m
=
§8
=
f
such that k
:
.
16 =
and §9
1),
f
r
then
such that k
+
I
f
- f
1T
-i, then
2
is
.
then §5
and that for each
Find the smallest
± =
f
then
m -2,
3
= f
+
_ ±
=
n
f
=
f
l
nn+i
1
(a)
is
=
,
>
i
then
+ 10,
04 ]
n
2
+ (3n + 3n +
n
gn = gn
±
,
-p,
m
§n + n
= c
and V f
nn+i
1,
f
l
m m+
Suppose that k 1
Here
=
,
n>
n
-
y^
then compute
+ 5 then
f
x
j_
n+
1= i,andVn>3
If
=
f
n gn +
= 2,
10.
p
m m+
n+1
^p,
+
±
n+i
=
"first ten" values of
= r
p +
£
f
n
of the
p
3.
12.
D.
If
1
and V
= 3,
whose domain
f
7
n,
k
-
>
1.11.
>
-^r-.
k
x
—
1
+
9
a recursive definition for a function
A
-whose
domain
is the
set of nonnegative integers.
V
n
A
1.
Compute
2.
How much
=
100
=
1.03A
n
- i
.
does $100 amount to
for 4 years at
^3.
AQ
An
if it is
deposited in a savings bank
3% interest compounded annually?
Guess an explicit definition for A.
[7-69]
[7.04]
E.
[Plot at least 5 ordered pairs.]
Graph each function.
Sample,
v
f
and V
= 3,
.
i
f
=
,
n n+
f
l
n
+ (5n +
1)
f
n
100
Solution,
f
3'
l =
f
4
-
80 -
f
2 = 9
= 36,
f
3 = 20
f
»
= 57,
5
f
6
= 83
60
40 -
[Note that the graph is
20 -
a set of discrete dots.]
n
f
Vn
f
n+
f
V
f
n
n+
=
f
i =
°
=
f
i=
1
l
,
5.
f
°
i =
i
n
n
f
+
Vn
1
f
V
f
n
=
.
n+
l
f
+ n
i =
n+
i
,
6.
n+
£
7.
r
£,
vn
F.
1.
f
n+
= 2
=
i
l^lli4
2f
i
V
n
n
Each function defined
f
n+
,
+
explicitly below has
I
l
=
f
1
,
i
=
+4
n
=
!=
f
Vnn+i
n
l
f
l
,
n
2
+ n
n
1
n
f
n+1
n
= 2
= 2f
n
as domain.
Give a
recursive definition for each and then use mathematical induction
from
to derive the given explicit definition
(a)
2.
f
n
=
5n +
(b)
1
f
-3n
n
+
the recursive definition.
(c)
1
f
n
=
3n 2 +
1
The oblong numbers are defined recursively by
B1
V„
Use
this
B
n +1
= Z
=
B
n
+2 n+1
and the recursive definition
<
>
of the triangular
of the
as the basis of an inductive proof
r
theorem: V
n
numbers
B = 2T_n
n
[7.04]
[7-70]
G.
1.
Besides triangular and square numbers, the Greeks studied
pentagonal numbers, as well as polygonal numbers of
still
Here are pictures showing how triangular,
higher orders.
square, and pentagonal numbers are generated geometrically.
Study the pictures and the recursive definitions
of
triangular
and square numbers, and figure out a recursive definition for
Pn
pentagonal numbers,
.
•
•! •
•
•
7 n
#
\
i
pl
Pn +
V
n
^
•
P
=
n+i
n
=
i
Use mathematical induction
V
3.
•
\"
r
2.
\
y
V
' \*
«
Pn
+
to prove:
=
T
n
+ S
n+i
Notice that the explicit definitions for T and S can be written:
n
T
=
n( n +
_
n
*)
Vn
2
S
=
n(2n +
"
n
^
Guess an explicit definition for the function P
and use mathematical induction
definition of Exercise
4.
°>
2
to derive
it
of
from
Exercise
the recursive
1.
The polygonal numbers
of
order
3
are the triangular numbers,
those of order 4 are the square numbers, those of order
the pentagonal
for T,
is the
4
numbers,
p( 4 )' for
l
S,
1,
P (5 ^'
Let's use 'P' 3 '' as a
etc.
for P, etc.
fourth hexagonal number.
are
new name
So, for example,
From
5
P^
(
your work on Exercise
you should see how to write a pattern for recursive definitions
">f
polygonal number functions
v
n
of all
_,(m+2)
_
p(m+2)
n+ l
=
orders.
p (m+2)
n
1
[7-71]
[7.04]
^5.
Guess a pattern for an explicit definition
of
polygonal number
functions of all orders:
p (m+2)
n
V
n
and derive
Exercise
H.
1.
it
from
the recursive definition you found in solving
4.
in a certain football league,
If,
=
each two teams play just one
how many league games are played? For
example, how many games are played if the league has 4 mem-
game
with each other,
bers?
2.
5
Suppose
members?
that,
1
member?
for each n,
G
is the
number
of
games played by
an n-teamed league in which each two teams play together just
Find a recursive definition for G.
once.
many
additional league
is
Gx ?
How
games must be played when a new team
joins a league which formerly
3.
[What
Use the recursive definition
had n members?]
compute some values
to
of G.
Guess
an explicit definition for G:
Vn
Gn
=
and prove your guess correct by deriving
from
this explicit definition
the recursive definition.
In solving
v»,
v"^
«V
*i
o^
<v
s
Part H, above, you found the number
of
ways
in
which
2
teams can be chosen from a league with n members.
More generally,
you learned how to find out, for any n € I + how many 2-membered subsets an n-membered set has. It is customary to abbreviate 'the number
of 2-membered subsets of an n-membered set' by 'C(n, 2)' [or, sometimes, by n C 2 ']. So, the result of Exercise 3 of Part H could be written:
,
'
V
n
In the next
C(n,
3),
of
C(n,
2)
=
n(l
y
"
exercise you will work on the problem
3-membered subsets
of
vO
of finding the
an
n-membered
si,
si.
set.
number,
[7.04]
[7-72]
Mark
many
I.
many
whose vertices
triangles are there each of
points?
is to
How
points A, B, C, D, E, and F, no 3 of which are collinear.
6
[This
is C(6,
ask how many
do not.
Why?] One way
3).
have
of the triangles
The sum
of
A
of
one
is
A
these six
attacking this problem
as one vertex and
these two numbers is C(6,
Each triangle which has
of
3).
how
[Why?]
as one vertex corresponds with
some
2-membered subset of the 5-membered set {B, C, D, E, F}. So,
the number of such triangles is C(5, 2)--that is, 10.
The number of triangles which do not have A as vertex is the
number of 3-membered subsets of {B, C, D, E, F}. [Explain.]
So,
This number is C(5, 3).
C(6,
C(5,
3)
+ C(5,
= C(5,
3)
+ 10.
3) =
2)
Continue this recursive procedure until you have computed
C(6,
3).
«^
*!*
*-,>.
Now,
let's find a
Suppose that S
particular
member
-
«J.
^.v
recursive procedure for computing values
'C(n, p)'.
(n
*,•«.
of S.
is a set
Then S
with n
= S
Q
w
l)-membered subset
of
S [in fact,
Now, S Q has C(n-
p)
p-membered
1,
members, and
{e
SQ =
Q },
of
that e Q is
where S
is
some
a particular
{e Q }].
subsets, and each of these
is
a
p-membered subset of S.
What other p-membered subsets does S have? Obviously, each of
the other p-membered subsets of S contains e and, in addition, contains
Q
Since S Q has n - 1 members, S has exactly
p - 1 members of S Q
C(n - 1, p - 1) p-membered subsets which contain e Q
So, the total number of p-membered subsets of S is
.
.
C(n-
1,
p) +
C(n
- 1,
p
-
1).
Hence,
Vn V
p
C(n, p) =
C(nj^
1,
-j-
p) +
o-
C(n
- 1,
p-
1).
[7-73]
[7.04]
J.
Here
is
The entry
a table for values of 'C(n, p)\
\
p
2
1
4
3
8
7
6
5
in the
6-row and
9
1
2
3
4
5
20
6
7
8
9
3-column
Note
*p'.
is C(6,
that,
which, as you discovered in Part E,
for the table
What does
5-row and
3)
4
C(5, 0)'
,
we have allowed
mean?
What
is 20.
as a value of
4
n*
and
the proper entry for the
is
-column?
What does
'C(0,
5)'
mean?
What
is the
proper entry for the
0-row and 5-column?
What
proper entry for the 0-row and 0-column?
is the
The recursion formula:
V V
n
p
should help us to
in the
6-row.
C(n, p)
fill
=
C(n
- 1,
in the table.
p) +
Try
- 1,
to use
it
In that case, the instance of the
V
C(6, p) =
C(5, p) + C(5,
So, to find an entry in the 6-row, you
To
C(n
p-
- 1)
to fill in the entries
formula
is:
1)
need entries
find these, you need entries in the 4-row.
Complete the table.
p
Etc
in the
5-row.
[7.04]
[7-74]
K.
Let's recall problem VII of the Introduction, page 7-2:
Whatever route Milton takes
from
in going
home
his
to
Zabranch'
How many
burg High School, he has to walk at least nine blocks.
routes are there which are just nine blocks long?
r —
.Home
—
L
N
1
1
r
p
i
—
1
1
r
-E
i
-
_i
S
"1
ZHS
During each
each
of
of his 9
-block walks to school, Milton walks south on
notebook on which he keeps track
is
his
record
The page
routes he takes.
route pictured above looks like this
of the
1st
^
2d
b
W
(a)
3d
^
w
S
I
4th
^
5th
^
w
y
6th
>
7th
>
8th
w
i
)
columns [and 'W's
(b)
Mark
(a)
In
(b)
How many
in the
and
:
9th
w
I
I
s
Fill in the next line in Milton's table by writing
of the
2.
of the
a special page of his
ruled into 9 columns headed '1st block', '2nd block', etc.,
s
1.
He has
four blocks and west on five.
remaining
4
S's in four
five].
the route you have described on the figure.
how many ways could you have answered Exercise
9-block routes can Milton take?
1(a)?
[7-75]
[7.04]
^3.
How many
10-biock routes can Milton take from his
home
to
Z.H.S.
?
"^4.
How many
11
-block routes can Milton take from his
home
to
Z.H.S.
?
*5.
How many
11
-block routes are there in which no block
is
walked
over more than once?
L. Here
is
another way to solve Milton's problem.
During each
Mil-
of
ton's 9-block walks to school he has to walk south for one block at
each
He
times.
of 4 different
distributes these 4 times of southerly
[For the route marked in the
walking among 6 north-south streets.
figure for Part G, none of his southerly walks took place on the 1st
north-south street,
4th,
1
took place on the 2nd, none on the 3rd,
on the 5th, and none on the 6th.
2
solved by finding the number
kind can be put into
1.
6
of
boxes.
ways
the
out
[
n-
1
Use your answer for Exercise
3.
In
how many ways can you
[right
1
.
of the
same
will help you do this.
of distributing p things of
Think
putting
of the p things as laid
them
n boxes amounts
in
partitions.]
2.
pockets
which 4 things
ways
of
Hint
same kind in n boxes.
in a row on your desk. Then,
to erecting
in
Exercise
Find a formula for the number
problem can be
So, Milton's
]
on the
1
and
left
1
to solve Milton's
distribute six pennies
coat pockets,
problem.
among
four
right and left trouser
pockets]?
4.
In
how many ways can you
pockets so that there
5.
How many
distribute six pennies
is at least
among
four
one penny in each pocket?
selections of six coins can be
made
if
each coin
is
each coin
is
either a penny, a nickel, a dime, or a quarter?
6.
How many
selections of six coins can be
made
if
either a penny, a nickel, a dime, or a quarter, and each selection contains at least one coin of each kind?
[7.04]
[7-76]
^THE ARITHMETIC OF THE POSITIVE INTEGERS
In
proving Theorem 102 we made use
+
of (1*^,
(I
2
+
),
(I
3)
and a
theorem about real numbers:
Vx V x
y
which
+
+ (y
'
[We used
a consequence of the apa.
is
= (x + y) +
1)
deriving the addition table in section 7.01.]
we could have used
with positive integers,
(I*
\
of this
4
m
Vm V
)
/
n
theorem.
+ (n +
=
1)
same theorem when
Since we are dealing only
this
a special case:
(m
+ n) +
we were unacquainted with
If
1
1
the other real
numbers,
and with our basic principles for real numbers, we might adopt
a recursive definition
+
(I
is
+
)
-(I 4
alone,
),
of
addition of positive integers.
+
(I
4
)
as
In fact, using
possible to prove that addition of positive integers
it is
commutative and associative.
Similarly, in proving
Theorem
of the
102, (1^),
+
(I
),
2
Theorem
U
idpma and the pml.
+
3
)>
103 on page 7-58
the
pml, and a consequence
[We used the
multiplication table in section 7.01.]
m
Vm
of
[step (7)]
two when deriving the
last
Again, since we are dealing only
we could have replaced
with positive integers,
we made use
1
=
m
1)
=
mn
the last two by:
(i*.)
V
m Vn
If
we were
adopt
(I
+
5
)
+
+
m
to disregard our basic principles for real
numbers, we might
as a recursive definition of multiplication of positive integers.
Using {l\)
integers
m(n
is
to addition.
-
+
U5
alone,
)»
we could prove
that multiplication of positive
commutative and associative, and
whole arithmetic
In fact the
derived from (1^
)
- (I
+
5
)
with respect
of the positive integers
can be
together with one other pair of principles:
m m
Vm Vn if m
V
is distributive
+
1
i
+
1
=
1
n +
1
then
m
=
n
[7-77]
[7.04]
^EXERCISES
In
answering the following exercises, use:
l
+
(I
2
Vn n
)
+
(I
3
Vs
)
+
tf
el +
+
[(1 €
r
S and
m
Vm Vn
4>
€
1
V
+ (n +
1)
^1.
Theorem
(m
=
m
and the logical principle 'V
use
+
1
(i)
1
+ n) +
=
m
V n
then
e S])
r
1
Of course, you
+ 1'.
of 'V^
n
1
+ n =
may
also
n + 1\
:
logical principle]
(1)
(2)
]
1+1
(3)
Part
€ S]
[Why?].
102
Complete the following proof
Part
=> n +
[n € S
=
1
+
1
1
.
(ii)
inductive hypothesis]
(4)
(5)
]
(6)
]
(7)
+ (p +
1
1)
= (1 + p) +
1
(8)
(5),
(6)]
(4),
(7)]
(9)
]
(10)
V [l+n
Part
(iii)
=
n+l=>
1
+ (n+
1)
= (n + 1) + 1]
(9)]
(ID
^2.
Complete
the following
proof
tt>r
[
Hint
.
of the
Up
m
V V
mnp
of 'V
+ (n + p) =
(m
+ n) + p*.
to the last step, the proof is a test-pattern for sentences
form V
4
in + (n + p) =
(m
+ n) + p'
.
In
completing the proof,
[7-78]
[7.04]
you will find
and
Look
(16).
comment on
Part
(i)
at
(12).
(
Then,
14) again.
fill
in
Finally, complete part
(
1
3)
(i),
and
(
fill
in (15)
except for the
12),
and part
(ii).
:
+
[U J]
(1)
(2)
Part
Then
helpful to look, first, at (2) and (14).
it
m
+ (n +
1)
=
(m
+ n) +
m
+ (n + q)
=
(m
+ n) + q
n + q
e
I
1)
=
[m
+ (n + q)] +
1
1])
=
[m
+ (n + q)] +
1
1)
=
[(m +
1
[
1
(ii)
(3)
(4)
(5)
(6)
m
+ ([n + q] +
m
+ (n + [q +
+
(7)
(8)
(9)
(10)
(m
(11)
+ n) + (q +
n) + q] +
1
(12)
(13)
[(12);
*
}
V [m
(14)
Part
(iii)
m
+ (n + p)
=
+ (n + [p +
(m
+ n) + p
1])
=
(
m
=>
+ n) + [p +
l]]
[(1),
:
(15)
[
(16)
[(1)
.
-
(15)]
]
[7-79]
[7.04]
3.
Complete the following proof
proof
fill
Part
is
4
of
V V
+ n = n + m'.
a straightforward inductive proof.
in (17),
(16),
(15),
(14),
and
(13),
First,
except for the
fill
[
Hint
The
.
in (1).
Then,
comment on
(i):
[theorem]
(1)
Part
m
(ii)
(2)
Vn qn
+ n = n + q
n
q + p
(3)
=
p + q
[
]
(4)
[theorem]
(5)
[
(6)
(7)
q +
(l + p)
1
=(q+
+ p = p +
I)
1
+p
I
]
[
(8)
I
(9)
[(4)
1
•
]
)
.
(5)]
(10)
(11)
(12)
(13)
[
1
[(4)
[
(5)]
.
(14)
[(1)
(15)
[(14); *
(16)
[(1)
Part
(iii)
(17)
]
-(13)]
-(15)1
]
(13).]
[7.04]
[7-80]
MISCELLANEOUS EXERCISES
1
.
2.
Vx V x6
•
y
6
y'
=
xy 12
12
36
(A)
(xy) 6
If it
takes x minutes to walk along two sides of a rectangular field
(B)
(C)
(xy)
(D)
(xy)
how many minutes can be saved by walking
30 feet by 40 feet,
diagonally across the field?
3.
Find two positive integers whose sum
is
less than 29 and
whose
difference is less than 7.
4.
Solve this system of equations.
3a
4b +
-
a + 2b
5.
Ten
is
water.
total
The rest
70% disinfectant.
are drawn off and replaced by
If
2
quarts
of this liquid
disinfectant, what is the resulting ratio of disinfectant to
mixture?
Simplify.
(a)
Sx 3
--^
lbx*
(b)
4
Solve the equation
8.
A
(c)
(a + b)
/IT
/—
= b
•
-y^-a
make
How many pounds
cylindrical bar
of
-
b
for 'x'.
grocer mixes coffee worth 88 cents a pound with coffee worth
cents a pound to
A
ab'-J^
a b
7.
9.
3 =
pints of a certain liquid contain
3 pints of
6.
-
=
1
6 8
20 pounds of blend worth 75 cents a pound.
the 88 cent coffee should he use?
made
in radius.
Another bar
in radius.
If
of a
certain metal
of the
same metal
is 5 feet
is 4 feet
long and 2 inches
long but 3 inches
the longer bar weighs 50 pounds, what is the weight of
the shorter one?
[7-81]
[7.04]
10.
If
A
B
=
A
varies inversely as the square of B, and
when
11.
Graph
12.
If
1
when B
= 9,
then
= 4.
lj
|x + 2| <
{(x, y):
f^}
r\
{(x, y)
|x + 2| + y <
:
0}.
x years old y years ago, how old will he be z years
man was
a
A
=
from now?
13.
Which
14.
3
m\
(B)
l
IA\
(A)
numbers
of the given
j
is
^5
(C\
< c
io
the largest?
m
{D)
-TT
)
(a)
(5a 2 + ab + c) + (4c
(b)
(7k2 + k
(c)
(-5mn
T
2m 2
-
+ n3 +
-
+ (5k
)
m
2
)
-
2a z
-
-
, E
(E)
.
"oTT"
4/5
2
ab)
3m 2
(2m 2
-
2k 2
)
n 3 + 5mn)
-
^S
«_
r
What
is the
ures
of the
ZS,
16.
What
17.
A
is the
distance between the points
nut mixture is
The remainder
is
How many pounds
of filberts
$. 85 a
of
of the
meas-
angles ZP, ZQ, ZR,
andZT?
(1,
9)
and
(6,
pecans, and there are five pounds
the mixture weigh?
pounds
sum
-3)?
l/3 peanuts, 3/l0 almonds, and l/5 cashews.
many pounds does
to 10
19.
2
Simplify.
15.
18.
(0.5)
,
pecans worth
of
these.
How
worth $1.03 a pound should be added
$. 76
a pound to
make
pound?
Solve this system of equations.
\
6x +
3x
-
y+3
2y +
=
9 =
a mixture worth
[7.04]
[7-82]
ABCD
quadrilateral
Given:
20.
is a
parallelogram,
AE
EB, DF
=
the ratio of the
Find:
of the
A
22.
=
given parallelogram
^
V.
21.
t
area-measure
shaded region to that
of the
V ioo
FC
=
100(V-Vo
Vy ioo - Vo
Derive a formula for
in
terms
t
A, V, and
of
Vr
)
J
number
circular regions, each with an area
What
is the
of 13
square inches, which can be stamped from a rectangular
greatest
of
sheet of steel 9 inches by 13 inches?
23.
A
motorist buys 14 gallons
full to
24.
90%
What
full.
is
of gasoline to
bring his tank from 20%
the capacity of the tank?
Simplify,
10
(a)
25.
m
m+
- 1
A merchant who
(b)
4 -t
1
5
had a certain number
through thirteen gates
of
(c)
-t
of
thirteen cities.
bags
charge at each gate was his friend.
each officer gave him one bag as a
had exactly the same number
of
gift.
+
moment.
y
y
-
2x
passed
at
The
each
officer
After the levy was paid,
To
his surprise, he still
bags after he had completed his
How many bags did he have?
Magazine Nov. - Dec, 1959]
journey.
of flour
-
The curious levy
gate was one half of what he possessed at the
in
2x
[Chih-yi Wang, Mathematics
,
26.
When successive discounts
price
of
of
30% and 10% are applied
an article, the selling price comes to $44. 10.
the list price?
to the list
What
is
[7-83]
[7.04]
27.
A
gardener uses 25 pounds
How many
28.
of
seed
seed for a lawn 50 feet by 90 feet.
per square yard?
is this
Simplify.
<
29.
pounds
of
a > <! +
Bill and
and
+
!><t
t>
<
Emily are traveling on their bicycles
miles per hour, respectively.
5
"-h&th -rrr>
b>
miles per hour
start at the
same direction.
turns back, how far from their
and time, and travel
hours, Bill
They
at 6
in the
same place
after traveling 3
If
starting point will they
meet?
30.
A
rectangle and a circle have the
mula for the area-measure, K,
sions,
and w,
S.
31.
Complete:
^
32.
Solve
33.
Simplify.
.
(a)
:
x
V
of the
—^—
x + 62.008
1.5749
2
„
0.8tJ5
circle in
terms
dimen
of the
-
I+i
4
3
"
v
(b)
*"'
~
1
34.
of the
Derive a for-
rectangle.
= =-
=
same perimeter.
-
,
11
1
-rr'
v
~
12
,
(c)
vw
74
1
1%
(d)
v ~'
125
375
T*4
Solve this system of equations.
-14y
-2x
+ 4x = 28
+ 7y =
-14
how much do Z\ dozen cost?
35.
If
tangerines sell
36.
If
a radius of a circle is x inches long
the
same area
is
to the given side?
5 for 16£,
y inches long,
and a side
of a triangle
how many inches long
is
with
the altitude
[7.05]
[7-84]
7.05
The relation greater than for the positive integers. --From our
basic principles for real
and
numbers
possible to prove that
it is
no least positive number,
(1)
there
(2)
between any two positive numbers there
is
is
a third.
[Given a positive number, how can you find a smaller one?
positive numbers,
Do
(1)
and
(2)
how can you
remain true
'number'
there a positive integer less than 1?
say, 5 and 6?
between them?]
find one
if
Given two
Is
replaced by 'integer'?
is
Is
there a positive integer between,
Using our basic principles we can show that the answer
to both of these questions is 'no'.
The
theorem:
first question is settled by the following
This theorem can be proved easily by mathematical induction.
You
will
be asked to do so in the next set of exercises.
The second question
Vm Vn
(*)
For, according to
if
(*),
settled by:
is
[n >
L
How
n >
m
+
1]
a positive integer is greater than 5 then
greater than or equal to 6- -that
than 6.
m
is,
greater than
if it is
5,
it is
it
not less
So, 6 is the next positive integer after 5.
can we prove (*)?
V
m
Vn
We
>
[n
L
note that
(*) is
m =>
m—
n
-
:
equivalent to:
1]J
And, we could use Theorem 104 to prove this once we have proved:
Theorem
V
m
[Explain.]
105.
Vn
[Theorem 105
is
>
[n
L
m
n
assumption
-
(3)
m
+
e I
]
on page 7-47.]
is
[7-85]
[7.05]
Theorem 105. A
we must prove:
So, let's prove
To do
the job.
so,
Vn
(i)
[n >
=>
1
n
-
1
el
+
n
-
straight -forward induction will do
]
and:
Vm
(ii)
'
Vn
Consider
(i)
Vn
(a)
1=>
[n/*
And, we could prove this
V
(b)
Now,
(b) is
m
(n >
[n =
1=>
+
n
n
el
1
-
Also, since (p +
= 1,
1
-
1)
€ I
1
p >
=
1
)]
then p f
1
[Explain.]
1.
So,
+
]
[Explain.]
]
Here's how.
+
or
follows that
it
,
e
1
+
e I
or
1
-
1
(p +
[(n =
n
or
1
- 1
=>
+
e
I
)
-
1)
(n +
Thus, by mathematical induction,
we have established
SO;,
Suppose that Vn
q > p +
Vn
+
1) € I
1
€
+
I
.
[by conditionalizing and generalizing],
V
1.
-
+
follows that
it
p and p
=
1
p +
(q
(m
+
-
1=1
q >
=>
we could prove:
if
or n
1
-
T)
e
easy to prove by induction.
Clearly, since
Hence
m
By Theorem 87, if
if we could prove:
(i).
we could prove
m =>
[V
(n >
L
n
1.
So,
Then, q
from
+
1)
-
p
[n > p +
So,
e I
1
from
.
=>
(i)
-
But,
-
and
q
-
I
(ii),
+
and,
-
p = q
e I
(q - 1)
(p +
1)
+
p
€ I
Since p +
> p.
1
-
€
I*].
or
(n + 1)
- 1
e
T)]
1
(p
to (ii).
Further, suppose that
].
from
-
1
(b)
Now, on
.
[n > p '=> n
(i),
n
(i)
=
1
> p
>L
1
,
it
also follows that
the inductive hypothesis,
+
1).
So, q
Consequently,
-
(p + 1) e
T.
Hence,
(ii).
by mathematical induction,
Theorem
105.
[7.05]
[7-86]
EXERCISES
A.
B.
Prove.
1.
Show
that (a) and (b) on page 7-8 5 are equivalent by deriving (b)
from
(a)
104
and
2.
Prove: V
n
3.
Show
that
if
n 4
is
1
from
(a)
'
1
>
then n —
Use Exercise
2.
Prove the converse
of (*)
converse give us
its
Theorem
rems prove
2.
IS.
r
F_.
Prove each
1.
Vn
2.
V
n/t
m Vn
Prove
(1)
and Theorem 104.1J
<
[n
l
and
2.
on page 7-84
(*)
T
«A.
«A.
+,<*
*-,v
*X
:
>
[n
L
—
prove
that, for
VV
n x
Prove:
(a)
.
m
+
*^
o,v
to
prove
to
'5 e I
each
x < n +
1
<=>
and
*
x,
1
n > mlJ
*•>
*i n
*i x
'i
You know how
Use
[Hint.
L
2
106.
Vm Vn
1.
1
2.]
Use Theorems 104 and 105
and
1
the only positive integer between
1.
(*)
+
(b).
J,
V
D^.
>
Theorem
[Hint.
C.
Vn n
1.
if 5
*6 e
I
'.
Accepting these theo
< x < 6 then x
"=> x
tf
1*
.
+
fi'
*
I
]
J
of the following.
[Theorem 107a]
1
[n <
m
and
(2)
L
+
1
<=>
< mlJ
n —
on page 7-84.
[Theorem 107b]
[7-87]
[7.05]
LOWER BOUNDS AND LEAST MEMBERS
Consider the set P
that
P has no smallest member—that
which
that, for
each y
How many
another.
a lower bound of
that
Consider the set K
it
P, x
y.
<_
no positive number
is
But, there is a real
Name
one such real number.
Each such real number
are there?
P does
have many lower bounds.
called
is
{x:
not have a least
member
but that
it
does
Does P have a greatest lower bound?
of all
positive integral multiples of
Does K
3.
Does K have a lower bound?
member?
Consider
€
noted earlier
P
You have seen
have a least
there
is,
than every other positive number.
is less
number x such
Name
We have
numbers.
of all positive
x > 6}.
member?
Does
this set
have a least
Does
this set
have a lower bound?
Does
have a lower bound?
Consider
have a least
{n:
n > 6}.
member?
Consider the set
Do you
think
it
Does
it
[Recall what the domain of n' is.]
4
of
Does
negative integers.
Do you think
has a lower bound?
integers has an upper bound?
it
have a least
member?
the set of positive
[To settle these last two questions, we'll
need another basic principle.]
By now you probably have
bound
of
a set of real numbers' and 'a least
numbers'.
meant by
a good idea of what is
member
lower
of a set of real
But, for definiteness, here are formal definitions:
V..
S
Vx
VoQ Vx
[x
is a
L
lower bound
of S
<==>
V
_
y
member
[x is a least
of S
€
S
x <
—
yl
7i
<^=>
x is a lower bound of S and x
Use
*a
of a set is
S]
number which
the first definition to prove that each
some lower bound
e
also a lower bound
is less
than
of the set.
Using both definitions and Theorem 93, you can prove that a set
cannot have two least
Suppose that S
members.
is a set of
must S have a lower bound?
least
Do
so.
real numbers.
If
If
S has a least
S has a lower bound,
must
it
member,
have a
member?
Suppose that T and S are sets
S has a lower bound, does
T?
has a lower bound, does S?
If
If
of
real
numbers such
a least
T
C_ S.
member, does T?
member, does S?
S has a least
T has
that
If
If
T
[7.05]
[7-88]
THE LEAST NUMBER THEOREM
A
member
set which has a least
certainly not empty.
many nonempty
you have seen, there are
not have least
is
members.
One
of the
sets of real
But, as
numbers which do
+
most useful theorems about
I
says
that none of its subsets is of this kind:
Theorem
108
.
Each nonempty set
formulation
is often
is:
Each
positive integers
member.
has a least
This theorem
of
number theorem
called the least
[An equivalent
.
numbers which contains
set of real
a positive inte-
ger contains a least positive integer.]
We
set of positive integers
Suppose that S
Since
ber.
number theorem by showing
shall prove the least
integers,
1
is
1
is
member
which has no least
is a set of
that the only
empty
is the
set.
mem-
positive integers which has no least
the least positive integer, and since S is a set of positive
a lower bound of S.
member,
Since S has no least
1
i
S.
So,
VjneS=>
n
(i)
Suppose now that V
1<
n].
[n e S
=>
p < n].
By Theorem
106,
it
follows [using
the substitution rule for biconditional sentences] that
Vn
--that is, that p +
member,
it
1
is a
[n
S
e
=>
lower bound
follows that p +
1
< n]
So, since S has no least
of S.
Hence, V
( S.
1
p +
[n e S ~=> p +
1
< n].
Consequently,
(ii)
From
(i)
V
[~V
mLn
and
(ii)
m
(n e S =s>
it
n
V n
g
S.
The
V
[n e S ==>
n
=> V
n
(n
e
S
=>
m
+
1
< nfl.
J
by mathematical induction, that
follows,
Vm V
In particular,
< n)
"
[n
L
n]
e
.
S
'
m
Since
< n].
V x
i-
x,
it
follows that
--that is, that no positive integer belongs to S.
least
interesting to
number theorem is assumption (5) on page 7-47. It is
recall how we used this theorem back in Unit 4. We were
[7-89]
[7.05]
number whose square
trying to show that there is no positive rational
8.
We argued
that
if
there were such a rational number, V8, then there
would be positive integers p and q such that v8 q
•
+
{q: V"8 *q €
I
}
is
From
not empty.
= p--that is,
this supposition
All
shown
not
we know
there
ber.
In a later unit
make
it
we
+
it
Then
.
is
now
not a rational
it is
is 8.
num-
shall adopt a final basic principle which will
possible to prove that each positive real
square root.
€ I
number whose square
such a number,
is
•
[Notice that up to
is 8.
that there is any positive
is that, if
V8 q
and so concluded that there
to derive a contradiction,
no positive rational number whose square
we have
that
and Theorem 108
follows that there is a least positive integer q such that
we proceeded
is
number has a
positive
]
THE COFINALITY PRINCIPLE
A
few paragraphs back we raised the question
lower bound for the set
equivalent to asking
integers.
if
of
negative integers.
there
is
number which
is
That
to say 'no'.
is,
up
of the
we believe
greater than or equal to each
positive integer which is greater than
so,
By Theorem
but
to
it
now.
is
a
94, this is
it.
real numbers,
we
that there is no real
of the positive
other words, we believe that, for each real
still
whether there
an upper bound for the set of positive
And, from our intuitive knowledge
would be inclined
of
number
As a matter
,
integers.
there
is
In
a
of fact, this is
cannot be deduced from the basic principles we have set down
So, let us
now take
this as a basic principle.
Cofinality Principle
Vx 3 n
When,
in a later unit,
[How do you read
n > x
we adopt our
'3
'?]
final basic principle for the real
numbers, we shall see that the cofinality principle then becomes a
theorem.
From
the cofinality principle
it
follows that
V 3 n>
-x.
[Explain.]
Vx 3 n x> -n. Hence, Vx 3 n x^-n. But, this is just to say
real number is a lower bound of the set of negative integers.
So,
that no
.
[7.05]
[7-90]
EXERCISES
A.
Give an example
both an upper bound and a lower bound;
2.
a lower bound but no upper bound;
3.
an upper bound but no lower bound;
4.
neither an upper nor a lower bound;
5.
a greatest
6.
an upper bound which belongs to
7.
a least upper bound which does not belong to S;
member
but no upper bound;
1.
Prove: V .>
x
2.
that
Suppose
«*r
n
3
+
&, =
-
x >
n
n
and
I
(a)
Does & f have an upper bound?
(b)
Does
(ft-
have a greatest
with
4.
Repeat Exercise
2
with
1.
What does Theorem 108
[
1.
Hint
.
n
(b)
{q:
that, for
Prove: V„[f
n n
L
In (a)
zation
each
= g
&n
4
Vn f n
member?
n
*
n
-
1
'
replaced by
tell
about (n: 3
you
}
m xn
n,
v
m
=
f
=>
(ii)
f
4
n
(b)
=
m)
if
J
f
n+
of
-
x
is
an irrational number?
that there is a p such that
q =
= g '. Is
n
A least member?
nil,
w(-1)
*
number?
you gave part
n
lower bound?
*
n
3
—
=
n
—
Show
> n.
Consider
Suppose
(a)
m
A
replaced by
1
a positive rational
Suppose that
f
'
1
2
(a)
n,
'
i
i
—
Repeat Exercise
•
D.
each
that, for
f
3.
*2.
S;
a lower bound but not a greatest lower bound.
»
C.
—
numbers which has
1.
**&.
B.
—
real
of a set S of
<
m
-
np <
np}.]
n(n +
l
= e
g
1)
n+
and g
i
= (n +
the result you obtained in (a)?
2
.
]
J
an inductive proof
= g ?
4
-=-)
of the
generali-
Does your answer contradict
Explain.
n.
[7-91]
[7.05]
2.
Consider the following generalization:
Vm Vn
(*)
x
'
^^
— m
<
[n
L
n
=
m]J
(a)
Give a counter example for
(b)
Criticize the following purported proof of
From Theorems
(i)
104 and 93
Vn
(ii)
(#).
=>
Suppose that V
[n < p
that q < p +
Then, q
1.
-
inductive hypothesis, q-
Hence, Vn
Vm Vn
(iii)
So,
(c)
— m=>
<
[n
L
from
Vm Vn
Compare
[n < p +
(i)
and
m =>
<
fn
L
—
n
1
=
=>
=
1
£
p.
Further, suppose
n = p +
from
So,
and q
= p,
J
1.
Consequently,
1],
—m
[n <
L
p +
=
the
+
l=>
n =
m
+
l]J
by mathematical induction,
m].
J
with the proof
(b)
1
= l]<
J
= p].
ml => Vn
(ii),
n
n
follows that
it
< 1=> n
—
[n
L
(*)
of
Theorem
105 on page 7-85.
MISCELLANEOUS EXERCISES
1.
For each
(a)
n,
Cn
is
number
of
subsets of an
Find a recursive definition for C.
n-membered
(n +
e
(b)
the
Q ?
Show
set
and that
l)-membered
How many
that
set, S.
eQ
^ SQ
.
[
Hint
.
n-membered
Suppose that S Q
Then, S Q
How many
w
subsets
of
{e Q
} is
is
an
an
S do not contain
do contain e Q ?]
each nonempty set has the same number
membored
set.
subsets as
it
has even-membered ones.
of
odd-
[Check
this
result in the table you completed on page 7-73.]
(c)
A
ladder which reaches just to the roof of a building has 10 rungs
By how many
routes can a beetle climb along the ladder, from
the ground to the roof, supposing that, on each route, he never
retraces his path?
[7.05]
[7-92]
2.
+
Each function defined below has
(a)
f
(d)
N
f
'
3.
7n +
=
= 6
n
Each
-
from the recursive
1
(b)
f
5n
(e)
f
of the following
Give a recursive
mathematical induction to derive
definition for each and then use
the given explicit definition
as domain.
I
n
definition.
=
7n + 9
(c)
f
=
n2
(f)
f
n
generalizations is false
R
n
=
-n
=
2n2
- 5
Give a counter-
.
example for each.
'^Vo^
(c)
4.
If
Vo
= 5
=k
lZk + k
the length-
and width-measures
(b)
v*^-*-
(d »
V
x/3 -IFTT
=
1
~2
rectangle are doubled, by
of a
what per cent does the area-measure increase?
5.
The surface area-measure
second cube.
If
cube
of a solid
made
the cubes are
16
is
times that
of
a
same substance and
of the
the
larger cube weighs 80 pounds, what does the smaller one weigh?
6.
7.
Simplify.
2r 2 + 7s 2 + (2r 2
(a)
8rs
(b)
7x(2x 2
(c)
(3a
-
-
-
3x +
2)
2b)(a + b)
How much
-
-
3rs
-
5x(4x
-
-
it
)
3x 2 +
4(a + 2b)(a
longer does
2
s
-
-
1)
3b)
(r
2
- s
+ (x
-
5(a
-
2
)
2)
-
2
b)(a + b)
take a train to travel
1
mile at 55 miles
per hour than at 60 miles per hour?
8.
9.
AABC
and
eter.
If
AADC
AB
= 1,
are inscribed in a circle for which
AC
Find an ordered pair
= 5,
and
(x, y)
AD
= 3,
such that x
find
-
AC
is a
diam-
CB and CD.
y = 8 and 5x + 2y
-
5 = 0.
[7-93]
[7.05]
10.
hn
+ -=-mv 2
E
=
11.
imv
=
terms
/
.
eV
Derive a formula for V in
)
2
Solve the equation 'm
Suppose that
which
13.
=
A
A and B
=
3
of
15.
X
you
added
7
16.
-^-)
a pipe
If
27
(B)
you subtract
If
3
3
—
4
n
value of
is the
A
for an
argument for
the midpoint of the segment whose endpoints
For each positive integer
27
n'.
-
-
have the same value?
Find the coordinates
(A)
l
for
'
r-
What
c"B.
have coordinates {y,
14.
e.
J
1
12.
E, and
n,
of h,
3
^-
and
x,
3
2x
y)
(-5-,
x
x
•
9
(C)
3
=
3x
from a number and
to the
(D)
get
1
-=-,
3x
(E)
9
if
number?
inches in diameter can drain a tank in x minutes then a
minutes.
If it
takes 24 days for 20 people to consume a certain stock of food,
how
long would
it
x%(800)
=
18.
Solve:
19.
Simplify.
,
.
(a)
20.
2x
what would you get
pipe 4 inches in diameter can drain the tank in
17.
3
If (0,
a 2 + 7a + 12
+ 2a - 3
^
and
3)
pair of
f
(1. 5,
take 25 people to
consume
it?
32
a 2 + 4a
^
X
"im
0)
-
5
,.
<
.
b>
—sT~ ~3i~
rs + r
belong to a linear function
has equal components?
2
.
T
f,
r s + r
2
which ordered
[7.06]
[7-94]
The integers. --Up
7.06
introduced
to
now, in our study
numbers, we have
of real
basic principles to single out the positive real numbers
some
We
and some other basic principles to single out the positive integers.
also adopted the basic principle (G) in order to introduce the relation >,
and the cofinality principle to help specify how the positive integers occur
among the real numbers.
The next step in our program
positive, negative,
are, and
and
we formulate
V
(I)
We
0.
this
[x
<=>
number
So, to say that a real
integers --
of
I,
already know what real numbers these
knowledge
e I
study the set,
is to
in a basic principle:
+
(x e
an integer
is
+
or -x
or x =
I
is to
e I )]
say that
a positive
is
it
integer, or is 0, or its opposite is a positive integer.
Our
theorem gives us another way
first
Read
integers.
carefully and see
it
Theorem
V
you believe what
m3
<= ->
I
3
This theorem tells us that a real number
prove that this
In the first place,
m
m
105, n
n
-
m€
-
Hence, by
it
so, by
= 0;
+
I
(I),
(I),
But, n
.
m
-
n
m
105,
-
m
m =
either a
a +
+
€ I
.
m
n
-
-n]
an integer
e I*
or a =
or -a
I
;
so,
Finally,
n)
-
Since, by (1^),
I
+
1
e I
,
(1)
.
So,
by
(I),
m-nel.
if it is
easy
is
it
.
a is a difference of positive integers.
m
If
n
=
n >
-
Suppose that a
and since a
If
m then, by Theorem
+
n) e
if n > m then -(m
m> norm = norn> m,
if
positive integers belongs to
+
e
and only
if
and n are positive integers.
+
€
e I.
-(m
of
m
second place, suppose that a real number a
(I),
1
is
x =
Since, in any case,
€ I.
follows that each difference
In the
n
-
n
a proof.
is
suppose that
Theorem
> n then, by
then
Here
is so.
says.
it
Using the basic principle
a difference of positive integers.
to
of
109.
[x
e
L
x
if
characterizing the set
of
= (a +
1)
Suppose that a
.
-
.
I.
an integer. By
is
e 1*
I
+
Then, by
1,
= 0.
it
(I
2
),
follows that
Since
+
1
e I
[7-95]
[7.06]
and
1
-
= 0,
1
a
suppose that -a
-
1
(-a +
+
e I
Then, as before, -a +
.
+
e I
1
+
1
,
€
I
Finally,
and a
,
So, again, a is a difference of positive integers.
1).
each integer
again, a difference of positive integers.
is,
=
Hence,
a difference of positive integers.
is
Just as, for the purpose of stating generalizations about positive
integers,
4
it
was convenient
whose domain
q']
is
+
I
shall choose the letters
Among such
[
m',
4
n',
4
p',
so, in stating generalizations about integers
,
will be convenient to use special variables
we
4
to introduce special variables
4
4
i',
and
j',
4
whose domain
it
For these
is I.
k'.
generalizations are those which say that
I
is
closed
with respect to opposition, addition, subtraction, and multiplication.
[Is I
closed with respect to division?]
Th eorem
a.
V.
110.
-j e
I
J
b.
V.
J
c
V.
J
d.
V.
J
\k+
\ k je
\ kj
e I
j
-
I
e I
Let's prove the first two parts of this theorem.
Suppose that
positive integers
-j = n
of
m.
-
j
an integer.
is
m
Then, by Theorem 109, there are
and n such that
j
=
m
Hence, by Theorem 109, -j
-
So [by
n.
Theorem
33],
This completes the proof
€ I.
part a.
Suppose that
j
and k are integers.
are positive integers m,
k
So, k +
j
=
(m
-
n,
m
=
n) + (p
-
and q such that
p,
and
n
q) =
(m +
respect to addition [Theorem 102],
Theorem
109, k +
j
e I.
Now, prove parts
c
Then, by Theorem 109, there
p)
m
-
j
= p
(n + q).
+ p e
+
I
-
Since
and n
This completes the proof
and
d.
q
+
I
+ q
of
closed with
is
+
e I
.
part b.
Hence, by
[7.06]
[7-96]
Here are other theorems about integers which you can prove:
Theorem
V
V.
112.
[k +
1
>
i
<= =>
k >
j]
Theorem 111 involves the same techniques used in proving
Theorem 110. To prove Theorem 112 use Theorems 111 and 104, and
The proof
of
theorems on >
.
EXPLORATION EXERCISES
A.
1.
Suppose that S
=
subtracting -5 from each
2.
Let
t
Describe the
k > -7}.
{k:
member
k
(k) =
-
-5.
Complete
this picture to
-8-7-6-5-4-3-2-10
<e-|
,
1
1
1
,
1
1
\
,
obtained by
of S.
whose domain
be the function [mapping],
t
S
set,
show
is
I,
such that
the
mapping
t_ g
6
7
8
9
1
1
1
1
2
3
4
5
1
,
,
,
1
1
2
3
4
.
,_>
I*-!
-8-7-6-5-4-3-2-10
3.
What
is the
set does t_
5
4.
range
map
of t_
{k:
For a given integer
integer k,
(a)
What
(b)
If i
t.
is
(k) =
k
j,
?
Does
t
H
h
5
6
have an inverse?
4->
7
8
Onto what
k > -5}?
let
t.
be the function such that, for each
- j.
the range of t.?
> k, what can you say about
t.
J
(i)
and
t.
J
(k)?
9
[7-97]
[7.06]
(c)
Look
at
Part
H
The results there hold
on page 7-41.
well for functions whose domain is
and
1
5.
(a)
t_ 10 to
each
Describe the set obtained by applying t 27 to each
of {k: k > 27}.
(c)
Describe the set obtained by applying
k >
r to
Does
r.
each
member
is
I,
such that
r
A
to
r(j) = -j.
show the
have an inverse?
members.
of its
> -1}
{j:
j
(c)
{j:
-5 <
(e)
{j:
j
at
member
set listed below, describe the set obtained by applying
(a)
Look
to each
t.
a picture like that in Exercise 2 of Part
For each
member
j}.
Let r be the function, whose domain
mapping
3.
you about t.?
(b)
Draw
2.
tell
So, what do Exercises
k > ~10}„
of {k:
1.
H
Describe the set obtained by applying
of {k:
B.
Part
2 of
I.
just as
j
< -4}
< -8}
Theorem
V
94.
V. [r(i)
Use
>
it
to
< 5}
(b)
{j:
j
(d)
{j:
-6 <
(0
0:
j
j
< 5}
< k}
complete the following statement
r(j)
]
J
4.
(a)
Pick a real number.
By
that there is an integer
ber.
the cofinality principle, you
which
is
Can you prove that there
than this real
number?
know
greater than this real
is
an integer which
is
num-
less
[You can do this by using the cofi-
nality principle to get an integer which is greater than the
opposite of the chosen real number.
(b)
]
Suppose that a set S has a lower bound.
integral lower bound.
Prove
that
it
has an
[7.06]
[7-98]
THE LEAST NUMBER THEOREM FOR INTEGERS
Consider a nonempty set S
integers.
of
If
S contains only positive
we have proved [Theorem 108], S has a least member.
other hand, we know that if S contains negative integers, it
integers then, as
On
may
the
not have a least
an example
member.
Give an example
some
a set of integers,
of
have a least member.
of
113
which are negative, which does
.
Each nonempty
set of integers which
has a lower bound has a least
For, suppose that S
is a
nonempty
member.
set of integers which has a lower
Then, by the cofinality principle, there
member
less than each
Now,
is
an integer
such that
I,
t-(k) =
S
>c
S'
C^
f
be the set obtained by applying
+
So, by
[Explain],
I
Since
me
Theorem
S*, there is a k
6
t.
to
108, S
each
,s
S such that
has
t
k
is
a lower bound of S [Why?].
The proof
of
Theorem
bound
of S*,
is
k-
j.
J
J
Let
which
j
[Explain.]
of S.
be the function, whose domain is
let t.
give
Can you guess a theorem?
Theorem
bound.
Now
such a set.
of
•
member of S. Then,
a least member m.
(k) =
m.
Since
m
So, S has a least
113 illustrates a general
is a
lower
member, k
method for deducing
theorems about integers from analogous theorems about positive integers.
In
Part
A
of the
preceding Exploration Exercises you discovered certain
properties of the functions
erty of
t-
is that
it
maps
For
t;.
way
lower bounds
of {k:
the
image
bound
of S.
of the
subset S
And, by Part
image
j,
the essential prop-
the set of integers greater than
positive integers in such a
of a
a given integer
of S is the
H
j
that order is preserved.
k > j} are
mapped
on page 7-41,
image
of
it
on the set of
So, for example,
onto lower bounds of
follows that each lower
some lower bound
of S.
In general,
[7-99]
[7.06]
because
ti
has this property, anything you can say about the positive
integers which has only to do with their order, you can also say about
the set of integers greater than
j.
For example, the principle
method
for proving
mathematical induction gives us a
of
theorems concerning
erties of the function
t.
for any
,
Using the prop-
k > 0}.
{k:
we can transform
j,
the principle
J
of
mathematical induction [see page 7-49] into a principle which enables
us to prove theorems concerning {k:
k >
j
Theorem 112 tells us that, for each k, k
So, the PMI tells us that, for each set S,
(the least integer > 0)
[k
k>
e
—>
S
in
to
see that the validity
k >
j
-
1}
of this
j}.
_>
integer > k.
result depends only on the
l
j
k >
and
we have
-
j
<=^> k >
1
j
-
may
=
1)
members
replace the
of
4
0's in
j,
j,
(the least integer > k) =
k +
1,
the following theorem:
Theorem
114.
(] e
S and
Vk
We can use
this
for proving
theorems about
theorem
specified integer
instance
Since the
way
Since
l's.
-
(the least integer >
such
k
'
are ordered in the same way, we
the statement above by
r
{k:
(the least integer > k) e S]
k>o keS
easy
In
is the least
1
which the positive integers are ordered.
{k:
--that is,
}
S
€
V
then
is
1
and
if
It
+
-
of
j.
Theorem
Part B
of the
that, for
each
[k
>
±)
to give us
all
is
S => k +
V1S .keS
6 S])
1
mathematical induction principles
integers greater than or equal to
For example,
114
e
in
case
j
equivalent to the
=
1
,
PMI
some
the corresponding
itself.
Exploration Exercises you worked with the function
j,
r(j) =
-j
.
This function maps
I
onto itself in such
:
[7.06]
[7-100]
a
way
of
So, for example, lower bounds of a set S
that order is reversed.
integers are
mapped by
r onto
Similarly, upper bounds of S are
this
upper bounds
mapped
of the
image S
of S.
onto lower bounds of S*.
From
and Theorem 113 we have:
Theorem
115
.
Each nonempty
bound
set of integers which has an upper
has a greatest member.
Theorem
Also, since, as
we can do mathematical
114 tells us,
"upward" from any given integer, so we can do mathematical
induction "downward" from any given integer. Using the order-revers
ing property of r we can obtain from Theorem 114:
induction
Theorem
116.
(jcSand Vk£j [keS=> k-1 eS])=> Vk <. keS
From Theorems
114 and 116
we
obtain a theorem of mathematical
induction for the set of all integers
Theorem
117.
(0 €
S and
[k
V,
€
S
=>
(k +
1
€
S and k
- 1 e
S)])
=> V k
e
S
EXERCISES
1.
Complete the following recursive definition
of the
numbers, D
,
of
diagonals of an n -sided polygon:
Vn .
>^
[
2.
Hint
.
Draw
-
3
Dn +
Discover an explicit definition
Use
i
a diagonal joining a vertex to a next but one.
n>
3.
3
of
]
D:
n
the appropriate instance of
definition
=Dn +
,
Theorem
from the recursive one.
114,
and derive the explicit
[7-101]
[7.06]
MISCELLANEOUS EXERCISES
A.
1.
Here
a recursive definition for a function
is
=
nn+i
f
n
n+ 2
(a)
Compute
(b)
Guess an explicit definition for
f
2,
f
3
f
,
4,
and
f
5
f:
.
f,
and try to derive
from
it
the recursive definition by mathematical induction.
n +
—
JTT'
1
If,
for each n, g
=
3.
If,
for each n, g n
=
4.
Complete these generalizations about the positive integers.
/
[
*
n
n n +l
}
u
1.
n+ 2
n n+ 3
"
=
n+2
n+
3
1
(n+ 3)(n
n+ 3
n n+4
+ 4)
n+4
n+ d
,
Complete.
VA. x 3
If
—
r
6 for
3.
I
(n+2)(n+3)
j
w
,
(a)
2.
+
w
.
= g
n +
1
n n+ 2
'
=
then, for each n, g
1)
(n+l)(n+2)
n+1
V
(b)
.
n(n +
^
»
K
B.
then, for each n, g
2.
If
of
=
—
s
r
+
1
+
—
= (x +
1)(
find the value of
t
and 4 for
t
(b)
)
s
fill
X
-
= (x -
1
1)(
)
which corresponds
with the value
r
.
a cubic centimeter of water weighs
water will
Vv x 3
1
gram, how many kilograms
a rectangular container 45
cm. by
35
cm. by
8
i
4.
If
the square of the
sum
of
two numbers
their difference is 4, find their product.
is
16
and the square
of
cm.?
[7.06]
[7-102]
THE GREATEST INTEGER FUNCTION
One consequence
number, there
x,
{k:
k
<.
Since the nonempty set {k: k
Theorem
follows from
it
For each
is that,
an integer less than or equal to
is
x} f 0.
[Explain.],
principle
of the cofinality
of the sets listed
115 that
below, give
it
its
That
it.
<^
given a real
is, for
x} has an upper bound
member.
greatest member.
has a greatest
(1)
{k:
k < 4.3}
(2)
{k:
k <
(3)
{k:
k < 13/3}
(4)
{k:
k <
(5)
{k:
k < 8}
(6)
{k:
k < -1.3}
(7)
{k:
k<?r/2}
(8)
{k:
k<2/3}
(9)
{k:
k<
(10)
{k:
k<2223/llll}
-0.5}
each
2y}
tt}
Instead of writing:
the greatest
it
customary
ple, instead of asking:
we can ask: What
Now complete
is
[[7.
(14)
[7T+ 1J
Here
is
I]
x' or as 'the integral
What
[4.
3
=
is the
greatest
part of x'.
member
So, for
of {k:
exam-
k < 4.3}?
=
(12)
[[0I|
(15)
I-6]|
=
(13)
E-5.2D
(16)
[[0.
62]
=
=
a list of the integers.
-6,
-5, -4,
-3, -2, -1, 0,
Where would you draw
ir
than or equal to
(17)
and
1,
2,
3,
4,
5,
6,
...
a line to separate the integers which are less
from those which aren't?
Where would you draw
Repeat
x}
<^
J?
=
than or equal to
(19)
k
the following sentences:
(H)
...,
(1.8)
of {k:
to write:
Read 'HxJ' as 'brackets
(17)
member
([ -n
J
a line to separate the integers
which are less
from those which aren't?
(18) for the real
numbers /I,
>/l,
V4, and
-/IT.
[7-103]
[7.06]
Questions (17) -(19) suggest the following theorem which follows
immediately from the definition
v
\
[
k
1
tt
x H <==> k
of '[[...
]]':
[Theorem 118a]
xl
l
EXERCISES
A.
1.
What
is the
member
greatest
2.
3.
Which
greatest
member
x <
of {x:
(b)
(d)
19+2.7])
(e)9
(g)
12.9.6]
(a)
Use Theorem 118a
Hint
(h)
But,
.
.
c if
.
V
(d)
Use part
(e)
Prove or disprove: V
(f)
Which
(")
V V
x
to prove:
1 1
2
>
k
V
prove:
x] < x
[[
c ] is less
is less
if it
i|[9.6]
(j)
than or
than or equal to
tt
x li-
x
(b)
y{[x]]
H
]]
V V
(ii)
V V [xj
[[
c J.
< 2x
prove
to
:
V 2[[xj£[[2x]]
A
yx
£ Hyx]j
theorem?
kx B
y>Q ylx]< [yxl
(i)
*
[[
y{[x]] <
Tell which of the following
x
V
{[
8a and part
the following is a
vx vk
Ij'9.61
(i)
]
Use Theorem
(i)
[ 7. 5 ] + |[7. 5 ]
(f)|[9] + 2.7
[2.7j
and only
(c)
of
(c)
118a, the integer
Use part
(a) to
+
to prove:
(b)
(a)
7.5]
2[9.6]
By Theorem
.
equal to
(h)
3}?
4.
[[7.5 +
[[7.5]]
(g)
is the
of the following do not stand for integers?
(a)
[
What
k < 4.3}?
{k:
of
+ k <
k Jx]]
+ y <
ffx +
kj
Ix
y]
+
y
Prove or disprove:
is
VV
x y
a
JxJ
theorem and prove
+
dyJ^Ix
+
yJ
it.
[7.06]
[7-104]
B.
The
set of ordered pairs,
y = [x]},
{(x, y):
is a
maps each
function which
number onto
real
not greater than the real number.
greatest integer function
Make a sketch
1.
_4
is
sometimes called the
.
12
-1
1
1
it
greatest integer mapping.
of the
_2
-.3
1
So,
the greatest integer
j
1
4
3
1
,
1
1
5
6
,
,
-4-3-2-10123456
H
4
Graph y
2.
h
1
=
[xj\
Graph y
4
3.
*'-
T-
You probably discovered
in
T
Vx j[x+
I]]
[x
+
1
J'.
*J*
<A,
Part
=
N
"l
A
that
[x]|+
=
1.
Let's try to prove this.
In
which
Exercise
tells
3(g) of
A
Part
Theorem
93,
all
we have
prove a theorem
Evidently, to prove this,
it
+
1
< |[x+ lj.
to prove is that
Vjx]
rem 118a
to
you that
Vx [xj
So, by
Theorem 118a
you used
+
1
>
Ix+
1].
would be helpful to have a theorem
like
Theo-
but which would start with:
Vx Vk [k> Ix] «=>
Theorem 118a, with the help
number c and any integer k,
k >
(*)
of
Tiieorem
[cj <=> k
This tells us about > rather than
>.
88, tells us that,
for any real
> c.
[Explain.]
But, > -sentences about integers
can be translated into > -sentences with the help
fact, for integers
(**)
k and
j,
k >
j
<=> k
+
1
>
j.
of
Theorem
112.
In
[7-105]
[7.06]
By
(*),
since k +
an integer,
is
1
k+
[cj
So, since
is
1
[cj <=> k
>
an integer,
k > [cj
1
from
follows
it
+
<=> k
+
>
[Explain.]
c.
and
(*)
(**) that
> c.
1
Consequently,
Vx V,k
Now we can prove
[[
c J +
He
1
1
[Theorem 118c]
> x].
J +
<=>
1
<=>
So, since each real
II
[x+
>
1
1]
1]
|[c +
Theorem 118c
1)
[[c] +
>
1
+
an integer,
is
<==> ([c J +
+
1
> c +
1
\ Why?
> c
Theorem
cj > Ic].
number
is
|[
c
]}
is
1
18c,
an integer
greater than or equal to
ffcj +
Consequently,
+
:
Vx Ix]
Since, for any c,
<=> k
[k > |[x J
1
>
Ic
itself,
+ I]].
V [x]+l>[[x+l]|.
Notice that Theorems 118a and 118c can, with the help of
93, be
combined
into:
Theorem
Vx Vk
118e.
[k
=
JxJ <=> k
This theorem gives us a quick way
integral part of a given real
T*
Use Theorem 118e
Hint
.
fl> +
Since, for any real
an integer,
k+
f)
proving that a given integer
is
the
o^
"V
T
«J,
to prove:
VV V.
[
of
< x <
number.
O-
C.
Theorem
all
jB
=
number
[xj
c
+
[Theorem
j
and any integer
you need, by Theorem 118e,
j,
is to
[[
c
]]
+
show
Ic] + j<c + j<(|[c]| + j) + l.
Use Theorem 1 1 8e to show that [c]l<c<|[c]|+l.]
119]
j
is
that
:
[7.06]
[7-106]
D.
Exercise
Part
3(h) of
two real numbers
of the
E.
1.
A
tells
Can
sum
of
sum of the integral parts
greater? By how much? Prove this.
as great as the
is at least
real numbers.
you that the integral part of the
be
it
Consider the real numbers 93 and
18.
It is
possible to find an
integer k such that
18k<93<
18(k+l).
given any real numbers a and
In fact,
b,
[Find
it.]
b > 0, there exists an
integer k such that
kb < a < (k + l)b
Prove
that such an integer exists.
[Hint.
example about
In the
3"*
93 and 18, did you see that the sought -for integer
2.
In
Exercise
1
is
(19
18
?]
you proved the following theorem:
Vx V
< x < (k + l)y
.
3, ky
0.
k '
y> n
Now prove
3.
(a)
Vx V
(b)
Vx V
Use
the cofinality principle on page 7-89 to prove:
y>0
3 k°^ X
^3
>
z
-
(x =
ky < V
ky +
and
z
Vx V
.
„ 3
n
y>
F.
1.
Graph both
*y =
2.
Graph y
-|[-x]|\
3.
Complete:
4
=
HxJT and
Vx -ff-xj =the
4.
*y =
tt
< z <
ny>x
y)
[Theorem
[Theorem
7
L
integer k such that
Find the formula for the postage
w
pounds.
121]J
-*]' on the same chart
x [Theorem 122]
The U.S. Post Office has a book rate of 9 cents for the
pound and 5 cents for each additional pound or fraction
which weighs
120]
c, in
first
of a
pound.
cents, for a parcel of books
[7-107]
[7.06]
G.
On a picture
number
of the
+
[5.36J
Since
,
for each x,
[xje
I
Line,
graph the solution set
[x]
=
'x -
[xj'
VV
'I
"|>
s
and
for each x, there is an integer k,
Instead of writing
V«*
it is
of
[5.36 + xj
^l"1N
[ x] <
namely
x < |[x]|+
ffx]|,
customary
1,
it
follows that,
< x
such that
-
k <
1.
to write '{[xj' [read as
'braces x' or as 'the fractional part of x'].
V x
The
=
[xj+
where
fxj,
set of ordered pairs,
and
is a function,
is
sometimes called
O^x
•Jt*
"T*
"l
x
J €
«
y =
{(x, y):
H.
[
I
and
< ffx} <
1
},
the fractional part function
<J^
*j x
1.
Graph
'y =
2.
State,
and prove, a theorem about the fractional part function
{fxj\
analogous to the theorem about the greatest integer function
which you proved
I.
J^.
Part C.
Solve.
1.
-sir
in
[5.36]) +
[x]
=
[5.36
+
xj
2.
[5.36J
+
[xj
< [5.36 +
Compute
75621'
10
1.
2.
10
75621
\L
10
75621
3.
10
io^
'
[
4.
I
7.
75621
10
Vk>
4
10
10
75621
5
5.
75621
10 -]]
5
10
Ox
10
6.
75621
10 6
10
xj
[7.06]
[7-108]
8.
Find the positive integer n which satisfies the following six sentences.
10
10
= 7
10
10
]
10
= 5
10
n
= 3
2
10
n
= 5
k>
10 5
10<
=
10
:
5
10
=
10 1
•j.
9.
Complete without doing any dividing or multiplying.
357829630572164
10 n
(a)
10.
10
1T68257934625TI
(b)
Repeat Exercises 1-6 with each
*10'
Recall the results
75621
=
of
l_
12.
Use
4
8'.
Then,
7.
Exercises 1-6 and that
+ 2« 10 + 6/ 10 2 +
Now, interpret the result
75621 =
jf
replaced by an
formulate a theorem like that in Exercise
11.
10
[I!
.
35
of
5_-
Exercise
10
10
3
+ 7- 10 4
.
by completing and checking
5 +
the facts that
15278 =
4+5-7
+ 3- 7 2 + 2
•
7
3
+ 6- 7 4
and that
15278
to
(a)
= 2 +
1
•
12 + 10
•
12 2 + 8- 12 3
complete the following:
15 27811
15278
-
(b)
12
(c)
15278
15278
(d)
12
nL
T*
Vl^
"J v
12
3
5
12
O-p
>
*'l
In doing the exercises in Part J you
discovered a technique for
'splitting" each positive integer into multiples of
powers
of a
given
[7-109]
[7.06]
integer greater than
When
1.
the given integer is 10, the splitting
As you
leads to the customary decimal notation for positive integers.
probably guessed, or perhaps know,
than 10 in a place value system of
the integer
whose name
in the
names
for integers.
decimal system
'223545' in the base-8 system.
integer.
possible to use bases other
it is
For example,
is '75621' is
Let's find the base-6
named
name
for this
According to the procedure illustrated in the exercises, the
units digit is obtained by simplifying:
I
ilnty
The sixes
2
3
'"
']
= 3
o"
digit is obtained by simplifying:
75621
62
Instead of computing 75621 v 36 directly by long division, we can
use
of the fact that
75621
4-
6 = 12603 +
75621
f
o
12603 +
2100 +
£
+ ^g-V
+
o"
6
3o"
-!
Now
there
is
a short cut here.
make
Instead
12603 +
of
=
computing
3
we could have computed
12603
[Check
this.
This suggests a very simple procedure for finding the digits.
]
3
[7.06]
[7-110]
6
75621
6
12603
2100
3
\
350
58
2
6
9
4
6
1
3
j
1
So, the base-6
name
75621
[Check
this.
for 75621 is
3+3-6
=
+
*
0-6 2
Remainders
13420 3 3' --that
+
2-6 3
is,
4-6 4 +3'65
+
+
1 •
6
6
]
The short
cut used here is a consequence of the
75621
P" l
1L6
theorem:
M
Similarly, the digits obtained in Exercises 1-6 of Part J by evaluating
'75621
10
10'
for the values
1,
2,
3,
.
.
.
,
6 of
4
can be obtained by evaluating, in-
p',
stead:
r~
75621
tT
p-1
]_10
J)
/no
10
Evaluate the two expressions for the value
The short
&
m
To understand
this
integer p. Since
(*)
depends on the theorem
cut, in general,
Vx V
c
—
3 of 'p'.
m
JIM.
m m
theorem, consider a real number
is
the
sum
of its integral
—
I**©
>>p
and
<
part and
c
and a positive
its
fractional part,
[7-111]
[7.06]
In
a
words: one can divide a real number
way
by a positive integer p in such
c
— L and
as to obtain an integral partial quotient ,
a nonnegative
remainder (f~T)p» less than p. [What has been said up to now holds
for any positive number p, integral or not.] The theorem says that the
,
integral part of the remainder on dividing a real
number by
a positive
integer is the remainder on dividing the integral part of the real
number
by the positive integer.
To see why
it
follows from
return to
this is so,
and note
that,
pel,
since
by Theorem 119, that
(*),
H-
(#*)
Now, since
(*)
< ((— ,p < p.
it
l
c
P
l<
<
< p, so that
<
follows that
i.
P
Hence, from
(**) ["dividing
<
by p"],
""
'
M
<
+
1
<
m
.
Conse-
P
gel
quently,
p
From
this
and
(**),
We combine
these two results into two theorems:
Theorem
124.
>-
Vx V
<
—
m
In a later unit
prove
that,
we
for each
i
(fi*i\
i l}m
=
E«l
_
X
mj m
shall use the second of these two
m
>
1,
theorems
to
each positive integer has a base-m place
value name.
Use the algorithm based on Theorem 123
for 75621.
to find the base-9
name
[7.06]
[7-112]
MISCELLANEOUS EXERCISES
A.
1.
2.
Complete and prove the generalizations about the positive integers
V
n(n
{b)
V
n(n * '><" + 2 >
+ (n + l)(n +
<c)
Vn n(n+l)(n+3)(nt4) + (ntl) nH)(nt3)
If,
l)
If,
1.
=
1)
1
2)
=
<
n+1 ><
for each n, g
= n(n + l)(2n + 1) then, for
)(
1
=
each
n,
g
then, for each p,
f
).
for each,.p,
f
=
K
)(
H
t
|
(
B.
+ (n +
t
(n+l)(
3.
t±±M
(a)
p(2p
-
l)(2p +
1)
).
The ancient Babylonians used the following incorrect formula
area-measure
for finding the
K
where K
ures
(a)
the
a quadrilateral:
(a + c)(b + d)
area-measure, and
a,
b,
and d are the meas-
c,
consecutive sides.
of
If
is
_
of
the formula is applied to a parallelogram or a trapezoid,
does
it
give an overestimate or an underestimate of the
area -measure?
(b)
2.
For which class
The number
this
means
lation,
3.
a certain city
is
48000.
If
is the
of
population of the city
the following
?
numerals stand
for the
same num-
Which one does not?
ber.
(A)
of television sets in
formula correct?
that there are 13.9 television sets per 1000 of popu-
what
All but one
of quadrilaterals is the
|
(B)
1.125
(C)
^
(D)
|i
(E)
||
[7-U3]
[7.06]
4.
The vertices
What
5.
6.
is the
of
a triangle are A(0,
area -measure
and C(5,
B(0, 4),
8),
1).
AABC?
of
Simplify.
(a)
(4&*b^(-3aV)
(b)
-3p2 q 7 *7p 4q 6
(c)
(20xy2 z 3 )(4x2 y 3 z)
(d)
-8r 2 s 5 t
If 1
child is born every x seconds,
•
-8s¥u
how many children are born
in a half -hour?
7.
By
selling an article at
a profit
%
of
8.
45 seconds
9.
Simplify.
is
30% above
the cost price, a dealer
of the selling price.
what per cent
hours?
of 2
2
(a)
0.6 *
|
makes
(b)
|
+
£
-
3L
(c)
i+ I
5
L^
+
(d)
^JL
8
10.
Solve this system of equations.
f
x
6.3*4*0
80 light bulbs cost k dollars,
11.
If
12.
A team
20%
it
13.
is
=
x + y + 48
1.474?9
=
y
+3
-
024
how many can you buy
for 75 cents?
scheduled to play 30 games during the season.
of the first 15
games, how many
of the
win to make the season win average
Mr. Sanchez plans
averages 40 miles per hour during the first
average during the remaining
3
hours?
loses
remaining games must
at least
to travel 350 miles in 8
If it
90%
hours
?
of driving.
5 hours,
If
he
what must he
[7-114]
[7.06]
4
for '*'.
14.
Solve the equation S =
15.
Suppose that one amoeba divides into two every 30 seconds.
these conditions,
16.
(a +
i)'
takes 24 hours to
it
with two amoebas,
j
how long would
^
, b)
TT
:T
ll-2x
fill
the
you start
same
test tube?
If
a+2
la+1
>-
* 2 -2
'
1
—
Solve the equation
2
*
<
''
4
18.
take to
If
Simplify.
(a)(r^r-^]x
x-4
x-3J
17,
it
a test tube.
fill
Under
—1
=
p
r
—
1
+
for
r'.
q
a car travels x miles in y hours, what is
its
rate in feet per
second?
19.
One method used by the Egyptians
circle
was
measure
to
area-measure
to find the
compute the area-measure
of a
8/9 of the diameter of the circle.
is
of a
square whose side-
Does
this
method
give an overestimate or an underestimate of the area-measure of
the circle?
'
20.
Solve the equation
21.
If
22.
23.
for
two numbers
what
is
their product?
of the
given numbers
<b)
V?
7f
10
is
4
b*.
and the sum
of their
reciprocals
3
is not =- ?
(c)
7f
(D)
i*?
Give a geometric argument to show that, for each a >
each b >
24.
d
of
Which
(A)
=
sum
the
is 5,
ab
—
0,
If
a first
is
125%
(E)
^
and for
Va* + b* < a + b.
number
is
of the first?
40%
of a
second, what per cent of the second
[7-115]
[7.06]
DIVISIBILITY
m
is
and only
if
You know
or:
m|n]
if
that
there
L
<=> 3
In
p
'
n=
And, since
and 6|l2, but 5^9-
I
m
[or:
divides n,
mp:
a positive integer p such that n =
is
Vm Vn [m
So, 2 |4
+
a factor of n with respect to
Theorem
V
n = mp]
rj
1
»n = n»
1,
it
follows that
125.
and n
[1 Jn
jn]
EXERCISES
A.
B^.
True -or -false?
1.
5 130
2.
l|78
6.
7|21
7.
Zl
11.
3|59
12.
15.
2
1
103
•
Prove each
168
j
3 (63 +
59
104
V
V
.
Solution
.
mq
80|82
8.
7
168
9.
8|80 + 888
m~
'•
p
Hint
101,
n.
Exercises
16.
6|94-95«96
17.
6
m
m
1
and
2
=>
1
103
•
104
•
say that
'
'
[Theorem 126a]
follows that there is a q such
It
•
m
I
'
is
104,
< q and, by
1
Hence, using Theorem 86e,
> 0.
1
<
mq
= n.
Vm V fm
L
n
In
So,
if
m
m jn then
< n].
'
m |p]
[Theorem
= n]
[Theorem 126c]
in the
[(mlnandmlp)
r
l
< n]
m=m
'
Use the theorem
m Vn Vp
V
3 |63
2|94«95
Consequently,
^
J
[(m |n and n |m)
.
10.
14.
Vm Vn V [(m|nandn|p)
L
,r
[
8| 17
7|14'8273591
By Theorem
follows that
it
V V
1
5.
13.
m Jn.
= n.
Theorem
3.
4.
[m |n=>
Suppose that
that
2.
82|80
of the following.
Sample
1.
3.
126b]
Sample.]
transitive and antisymmetric
m (n
+ p]
[Theorem 126d]
105
—
[7.06]
[7-116]
m In
4.
Vm Vn V [(m In and
5.
mp|np]
Vm Vn V [m|n=>
l
p
p
V
6.
[Theorem 126e]
[Theorem 126f]
'
2|n(n +
1)
[
Hint
(p + l)(p + 2) =
Vn
6|n(n + l)(n +
'8.
Vn
6 n(n
2
For part
.
p(p +
7.
THE
m |p]
+ p)
1)
(ii) of
+ 2(p +
an induction proof, note that
Now, use Theorem 126d, above.]
1).
2)
For each odd
*9.
+ 5)
DIVISIBILITY
The relation
much
has
j
in
common
with the relation
Vn In
Vn
Vn n In
Vn n
V™
m V«n ^p K m n and
n
l
'
Vm Vn
m
[(
l
I
n and n
m |p]
m) =i> m = n]
|
'
l
<
m n
Vm Vn
J
corresponds
with V
r
m Vn
4
(m
J
«:
is that
there
—> m
< n <
[m —
L
is
that
<.
3
<_
4, 4
' n + p]
+ p
rj
r —
mpr"_
*'
np]
r-j
< n +
J
which
1]'.
J
theorems about
|
do show
For example, those
<_
and
5,
5
<
6,
6
are
1,
2,
3,
4,
we may, knowing
5,
and
6.
<
is
that
Since
<
1
2,
reflexive and
transitive, indicate the < -order of these integers schematically by:
On
the other hand, the positive integers
2,
3,
and
6.
p]
rj
= n]
orders the positive integers which are less
which are less than or equal to
3,
m
no theorem about
than or equal to a given one in a very simple way.
<_
m
<,
orders the positive integers in an interesting way. To see
means, note
this
m
>
m)
< n or n —
<
"~
L
In spite of these differences, the first four
that, like <,
<:
column indicate two important differences between
Another difference
and <.
< n
~~
VV [m — n <=>
mnp
V V L_
Vmnp
[m < n <=>
i
in the left
:
n
<:
> mplnp]
r
rj
'
The two gaps
2
1)
l
'
Vm Vn Vp [mln
~<
1
<.
r=^
Vm Vn V [(m ~ n and n ~~ p)
r
p
< m) =>
V V f(m < n and n ~"
Ip)
V
what
-
RELATION
'
j
24|n(n 2
n,
'
Since
1
|2,
l|3,
2|6,
which are divisors
of 6
are
and 3|6, we may, knowing that
1,
j
is
[7-117]
[7.06]
reflexive and transitive, indicate the (-order of these integers by:
2
1
The diagram
transitive,
tells us that
we conclude
conclude that
1
1
1,
\Z,
that
1
1
J
6,
2|6, and 3 J6.
|3,
and, knowing that
Knowing
is
we
1
of
n [for example, for n =
same simple way
"most'* choices of n this
30,
|
is reflexive,
j
[linearly] in
8],
which
positive integers which are less than or equal to n.
whose factors are
that
2|2, 3J3 and 6 6
For certain choices
factors^of n in the
1
1,
2,
whose divisors are
is not the
3,
4,
2,
1,
6,
3,
8,
5,
case.
12,
2
orders the
<,
orders the
However, for
As examples, consider
24,
and 24:
10,
6,
j
15,
and 30
6.
1
^30
0-
15
and 60, whose factors are
1,
2,
3,
4,
5,
6,
10,
12,
15,
20,
30,
and 60
[7.06]
[7-118]
EXERCISES
A. Here are the divisibility-ordering diagrams for the divisors of 8
and for the divisors
of 243.
3
1
4
2
1
27
9
Give ten more examples
8
of positive
243
81
integers whose diagrams are
linear.
B.
The diagram for
Draw
and 24.
So are the diagrams for 12
two-dimensional.
6 is
the diagram for 48;
two-dimensional,
it is
Other two-dimensional diagrams are those for
54,
18,
also.
108,
36,
and 215.
C.
1.
Draw
the
diagram for 1125.
2.
Draw
the
diagram
3.
Give three more examples
[
Hint
1125 =
.
3
2
•
5
3
]
for 500.
of
positive integers
whose diagrams
have the same shape as the diagrams in Exercises
4.
Draw
5.
For each number
2
-7 3
(f)
2
(j)
(n)
(e)
5
(i)
13-17
«10 5
The diagrams
for 30
If
•
3
4
and
24
listed below, tell
2
2 -5 3
the positive integers
E_.
for 2 3
(b)
3
b
diagrams
-5 3
(a)
(m)4 3
D.
the
5
»
(g)
7
28
(k)
3
14 2 »7 3
(o)
l
-3 80
2.
.
3
2
and
4
how many factors
(c)
51
1
•
51
3
4
-11 80
17
4
-25 -3 2
and 60 are three-dimensional.
has.
it
4
(d)
2
(h)
23
(i)
43
(p)
2
4
-5 6
•
3
-3 5 -5 2
Characterize
whose diagrams are three-dimensional.
a positive integer has just two factors,
its
diagram looks
like
this:
1
[Such
numbers are,
of
D
course, prime numbers.]
If
a positive
[7-119]
[7.06]
integer has just four factors,
like:
p
1
p
2
p
its
diagram looks either
3
or like
^pq
^
1
:
q
[What are the smallest such positive integers?]
Give
all
possible diagrams for a positive integer which has just
1.
three factors
2.
five factors
3.
six factors
4.
eight factors
5.
twelve factors
6.
sixteen factors
a HIGHEST
It is
COMMON FACTOR
one
of the
important properties
integers, one is less than the other.
of
< that, given two positive
So, by transitivity, the positive
integers which are less than or equal to both of two positive integers
are just those which are less than or equal to the smaller of the two.
For example, since 12 < 30 [and since
V
[(p
<.
is transitive],
< 12 and p < 30)
The situation with regard
<=>
to the relation
Given two positive integers one
interesting.
j
p < 12].
is less
same argument used
above, just those which divide the smaller of the two.
the other.
To
it
is
not usually the case that one of
and 30 60,
in the
diagram on page 7-117 which shows how
1
60.
1
From
and the set
common
it
this
all the
them does divide
we see
j
orders the divisors
that the set of divisors of 12 is
of divisors of
30
is
divisors of 12 and 30
{1,
2,
is the
{1,
is just the set of
V
Since
divisors of 12 and all those of 30 are indicated
3,
5,
6,
10,
2,
divisors of 6.
15,
{1,
2,
30}.
3,
4,
3,
6}.
So,
[(p|l2 and p|30)
<=>
p|6].
of
6, 12}
The set
intersection of these two sets.
is
This
But, given two
see what happens, let's again consider 12 and 30.
12 60
more
which divides the other,
of
the positive integers which divide both are, by the
positive integers,
simple, and
of
So,
[7.06]
[7-120]
Just as 12 is the greatest integer less than or equal to both 12 and 30,
common
called the highest
GCD]
divisor,
Find the
33,
and
A
of 2
of 12
HCF
3
•
3
s
factor [HCF; or:
and 24,
of 15
and
2
7
•
3
Part B on page 7-115,
an
show
HCF
of
m
if
it is
HCF
of
m
cover an important property
and 4
of
use in searching for the
and
and n when
<=>
p|n)
p|d].
common
-part] that each
m
and
divisor of
m
and
Using Exercise 2
n.
easy to show that no two integers have
[Do so now, by showing that
and n then d 1
3
of 14
.
|d
and d 2
2
that each two integers do have an
Exercises
and 27,
of 21
50,
-part] that d divides both
more than one HCF.
is
common
4
Notice that this means [only
[if
and
of 25
positive integer d is the
n divides d and
the greatest
6 is
and 30.
V [(p|m and
of
divides both 12 and 30.
"most divisible" integer which
6 is the
highest
of
j
d.
x
.
if
each
of
d 1 and d 2
However, we have yet
]
HCF.
In doing so,
common
we
to
shall dis-
factors.
Part B on page 7-115 suggest a procedure to
HCF
of
two positive integers.
For they
tell
us
that the divisors of
n + p and n
are just the divisors of
n and p.
So, the
HCF
of n + p
For example,
and n
HCF of n
HCF of 30
is the
let's find the
and p.
and
12.
HCF(30,
12)
=
HCF(12,
18).
HCF(18,
12)
=
HCF(12,
6).
Since 6J12,
HCF(12,
6)
=
6.
So,
HCF(30,
12)
=
6.
Since
18=12
+ 6,
In obtaining the first
12
from
two
of the
displayed equations, we subtracted
30 twice and found that
30
This told us that HCF(30,
12) =
-
12*2
6) = 6.
[Notice that 30-
So,
1
HCF(30,
+ 12
•
-2
=
=
HCF(12,
12-6*2
HCF(12,
Since 30 = 12 + 18,
=
6.
6).
0,
12) = 6.
HCF(30,
12).]
Since
[7-121]
[7.06]
Let's try this with
4147 X 2
=
1378,
HCF(9672, 4147)
=
HCF(4147, 1378).
=
13,
9672
Since
we see
that
-
4147
Since
it
some larger numbers, say 9672 and 4147.
1378 X
-
follows that HCF(4147,
1378
But,
So,
HCF(1378,
In finding
3
HCF(1378,
1378) =
13).
13 X 106 = 0.
-
Hence, HCF(9672, 4147)
13) = 13.
HCF(30,
12)
we discovered
= 13.
that there are integers
i
and
j
such that
HCF(30,
[These integers are
is
and -2, respectively.]
1
an integral linear combination
of 30
and
1
words, HCF(30,
In other
are there integers
is,
and
i
j
such that 13
=
of
9672 and 4147?
Our
9672i + 4147j?
method
investigation of this question will suggest a
12)
2
HCF(9672, 4147) an integral linear combination
Is
That
12) = 30i + 12j.
of
proving that each
two positive integers do have an HCF.
We
begin by noting that both 9672 and 4147 are integral linear
binations
We
of
com-
9672 and 4147.
9672
=
9672 X
4147
=
9672 X
1
+ 4147
X
+ 4147
X
1
also note that 1378 is the remainder upon dividing 9672 by 4147, and
that 13 is the
remainder upon dividing 4147 by 1378.
So,
that 13 is an integral linear combination of 9672 and 4147
if
we can show
remainder upon dividing one integral linear combination
that the
two numbers by a second such linear combination
linear combination of the two
Let's show this.
integers
i,
j,
k,
and
l
m
and
'
*C
IL
traction,
€
U1
it
I
and since
n.
such that p
q £E] = (mi + »j>
-
these
again, an integral
is,
numbers [Explain.]
remainder upon dividing p by q
p
of
Suppose that p and q are integral linear combina-
some two integers
tions of
Since
we can show
I
is
-
That
is,
=
mi
+
nl)|[£]
+ nj
suppose that there are
and q
=
mk
+ ni
.
The
is
(mk
.
m(i
-
k[| ]) +
»(j
-
||[£ }).
closed with respect to multiplication and sub-
follows that the remainder is an integral linear combination
[7-122]
of
m
[7.06]
and
n.
remainder upon dividing 9672 by 4147, both
So, since 1378 is the
which are integral linear combinations
And, since 13
of
9672 and 4147, so
have proved that there are integers
and
i
Our examples suggest
common
two integers.
HCF,
let's
to us that
factor then
To show
is
it
As we have seen,
m
m
and
m
and n do have an
integers which are integral
of all positive
n.
3^-
n
and
e
such that
an integral linear combination of the
S = {p:
e S,
is 13.
two positive integers have a
that two positive integers
consider the set S
linear combinations of
if
j
which
of
[Find two such integers.]
13 = 9672i + 4147j.
highest
is 1378.
remainder upon dividing 4147 by 1378, both
is the
are integral linear combinations
We
9672 and 4147, so
of
of
S,
mi
p =
if
+ nj}
p and q belong to S then
if
the
remainder upon dividing p by q is positive, it also belongs to S.
Since S is a nonempty set of positive integers, it has a least member
d.
Suppose that p
any
is
member
of S.
Since d
This remainder
or belongs to S.
dividing p by d is either
remainder upon
the
e S,
is
p
-
d
II
~
II.
Since
0<p-d[|]<d,
it
follows that the remainder is less than d.
of S, the
remainder
So, d|p.
We have shown
d |m and d
On
That
is 0.
member
member
of S.
In particular,
|n.
e S,
d =
common
divisor of
m
mi
there are integers
In fact,
HCF(m,
n)
i
and
and n divides
m Vn
m
i
j
HCF(m,
common
and n do have a highest
is the least
3. 3.
such that
d.
member
d of the set S
positive integers which are integral linear combinations of
V
j
+ nj.
Hence, each two positive integers
factor.
is the least
is,
that d divides each
the other hand, since d
So, each
Since d
n) =
mi
+ njj
m
of
and
n.
[Theorem
\.
those
127]j
[7-123]
[7.06]
The algorithm for computing the highest common factor
m
positive integers
and
n:
m
-
nq 2
n
-
ri q 2 = r 2
r
is
nothing
members
more
of
of this set.
i "
than a convenient
=
^3=
way
r
r
1
3
smaller and smaller
of getting
S until you finally find the one which divides each
This
Example
is the last
1
Solution
nonzero remainder.
Find HCF(3705, 2618).
.
.
1_
2618) 3705
2618
3705
-
2618*1
=
1087
2618
-
1087-2
=
444
1087
-
444
1087
2
1087)2618
2174
444
2
444) 1087
888
2 = 199
•
199
2
199)144
444-199-2
=
46
398
46
46) 199
184
199
-
46-4
= 15
15
_3
15
)
46
45
1
two
of
46
-
15
•
3 =
1
-<
HCF
member
3
1
.
[7.06]
[7-124]
Example
Example
In
2.
we found
1
that
factor of 3705 and 2618.
1
is the
highest
we know
So,
that
common
is
1
integral linear combination of 3705 and 2618.
integers
and
i
j
Find
such that
3705i + 2618j
= 1.
The required integers can be found by using the
equations obtained in the Solution for Example 1
Solution.
One way
an
from the
substitute
is to
second ['3705
-
2618*
V
first equation into the
for '1087'], then
second equation into the third,
=
3705
2618
=
3705
+ 2618- -1
2618-
1
=
1087
=
3705
1
2618
-
1087-
2 =
444
=
3705
-2
1087
-
444-2
=
199 = 3705
5 +
444
.
199-2
=
46
199
-
46-4
=
15 = 3705
46
-
15-3
=
1
=
3705
following:
+ 2618-
-
3705
is the
2618-0
+
1
3705
=
the first and
etc.
Another way, perhaps easier,
3705
from
+ 2618-
2618- -7
-12
+
2618-17
53 + 2618- -75
-171 + 2618-242
t
t
^EXERCISES
A. Use the algorithm to find the highest
of positive
integers.
B.
7469, 2387
3.
8749,
1.
2387
7459
common
11143
Reduce each fraction
2.
factor for each pair
[This algorithm is sometimes called Euclid's
algorithm for finding greatest
1.
common
to lowest
4389
5320
divisors.]
2.
5320, 4389
4.
4147, 10672
terms.
3.
8749
11143
4.
2574
5049
[7-125]
[7.06]
C.
For each
satisfy
following equations, find a pair of integers which
of the
it.
2.
38i + 43j =
1
270i + 170j = 10
4.
38i
43j =
1
=11
6.
38i + 43j = 2
1.
27i + 17j =
3.
5.
270i + 170j
1
-
^EXPLORATION EXERCISES
A, For each
of
the following linear equations, tell whether
contains a point with integral coordinates.
dinates of one such point;
B.
does, give the coor-
1.
13x+10y=l
2.
42x + 60y
= 6
3.
13x + lOy =
2
4.
42x + 60y
= 5
5.
525x + 231y
=
6.
7469x + 2387y
Each
of the
42
following linear equations is satisfied by
1.
6x + lOy =
3.
13x+17y
5.
17x+13y
1.
(a)
Is
it
(b)
It is
3x + 5y =
=
4.
13x
=
6.
1322x
-
4
6x + lOy
=
(0,
=
0).
For
which satisfy
it.
17y =
-
1263y
the case that, for each integer k, (10k,
=
-6k) satisfies
0'?
the case that, for each ordered pair of integers which
satisfies
4
6x + lOy =
ordered pair
(c)
of integers
2.
the equation
graph
doesn't, say so.
if it
each equation, find several other pairs
C^.
If it
its
Repeat
(a)
0*,
is (10k,
and
(b) for
there
is
an integer k such that the
-6k)?
the equation *3x + 5y = 0' and *(5k, -3k)*..
•^
*"i^
In
of all
answering Exercise
1(a) of
ordered pairs (10k, -6k)
Part C, you probably saw that the set
is a
subset of the set of ordered pairs of
.
[7-126]
[7.06]
integers which satisfy *6x + lOy = 0'.
3k
{(x, y):
But,
when you
2.
3.
and y = -6k)}
C
you probably saw that
is,
{(x, y) € I
X
I:
6x + lOy
= 0}.
did Exercise 1(b), you should have discovered that
3k
{(x, y):
For example,
(x = 10k
That
(x = 10k
(5, -3)
and y
= -6k)}
/*
{(x, y) e
I
XI: 6x + lOy
= 0}.
belongs to the second-named set but not to the first.
True-or -false?
(a)
{([x,y):
3 k (x= 12k and y
(b)
{(>,
3 k (x= 12k and y= -15k)}
(c)
{([x, y):
3
<d)
{(Ix, y):
3 k (x=-4k and y
<e)
{([x,y):
3 k (x = 4k and y
(£)
{(;x,y):
3 k (x
y):
Change the
k
(x =
=
= -15k)}
4k and y= -5k)} =
C
= {(x, y) €
{(x, y) €
left side of
I
= 5k)} = {(x,y) 6 I
= 5k)} = {(x, y) €
584k and y= -626k}
IXI: 15x+12y = 0}
{(x,y)e
I
I
X
X
I:
I:
15x+12y = 0}
15x + 12y = 0}
X I: 15x + 12y = 0}
XI: 15x- 12y
= {(x, y) e
the sentence in Exercise
I
=
0}
XI: 626x + 584y=0}
2(f)
to
make
a
true sentence.
4.
Complete the following statement:
{(x, y):
[
_D.
1.
Hint
.
and y =
3 k (x=
HCF(1162, 938)
How many ways
)}
= {(x, y) 6
I
X
I:
1
162x + 938y = 0}
= 14]
can you make change for 75£ using nothing but
nickels and dimes?
2.
Repeat Exercise
1
for 73£.
3.
Repeat Exercise
1
for $3.75, but this time use nothing but quarters
and halves
[7-127]
[7.06]
^DIOPHANTINE PROBLEMS
In
order to solve the exercises
you had to find how
many ordered
of
Part
D
of the
Exploration Exercises
pairs (x, y) of nonnegative integers
satisfy:
and
(1)
5x + lOy
= 75
(2)
5x + lOy
=
73
(3)
25x + 50y
=
375
:
to finding the integral solutions of equations are
Problems which amount
Greek mathematician, Diophantus
called 'Diophantine problems' after the
[
circa 250 A.D.].
Equation
(1) is
equivalent to 'x
= 15
-
2y'
and
it is
obvious, by sub-
stitution, that the nonnegative integral solutions are (15, 0),
(11, 2), (9,
3),
(7,
4),
(5,
5),
(3,
and
6),
(1,
7).
(13,
1),
So, there are eight
ways to make change for 75 £ using nothing but nickels and dimes.
In
do
Exercise
much
2 of
thinking, that you can't
nickels and dimes.
pairs
Part D, you probably answered, without having to
(x, y) of
Looking
(3) is
Since
for 73£ using nothing but
you see that, for
5/[73,
all
ordered
there are no integers
= 73.
equivalent to equation (1)- -making change for 75£
using nickels and dimes
is
for $3.75 using quarters
Here
at equation (2)
integers, 5J5x + lOy.
x and y such that 5x + lOy
Equation
make change
essentially the
same problem
and halves.
is
a fourth exercise like those of Part D:
A
bit is
a coin worth 12y£.
make change
as making change
In
how many ways can you
for 75£ using only nickels and bits?
This problem amounts to finding the number of nonnegative integral
solutions of:
5x +
This equation
(4).
=
75
is equivalent to:
2x + 5y = 30
(4)
As we
12~y
shall see,
we can
find all the integral solutions of equation
Then, we can solve the nickel-and-bit problem by counting the
.
[7.06]
[7-128]
number
of
that (3,
-1) is a solution of the equation:
2x + 5y
So, (3
•
The
nonnegative integral solutions.
30,
follows that
-1
if
•
=
1
Since 2
30) is a solution of (4).
2x + 5y
= 30
first step is to notice
•
90 + 5
•
-30
= 30,
it
then
2x + 5y
=
2*90 + 5* -30.
So,
2(x
(5)
Now,
in
if
-
90) + 5(y + 30) = 0.
x and y are integers, then so are x
Part C
-
Your work
90 and y + 30.
Exploration Exercises suggests that, since HCF(2,
of the
5) = 1,
the integral solutions of (5) are just the solutions of:
3 k (x
In
order that x >
-
90 = 5k and y + 30 = -2k)
we must have 90
and y >
must be an integer such
--that is, k
and -30
+ 5k >
-
2k >
that
-18 < k < -15.
So, there are four
ways
change 75£ using nothing but nickels and
to
bits
'• x
The key points
finding
some
*"|
H
*v
in finding the integral solution of (4)
were,
first,
solution of:
2i + 5j =
1
and, second, finding all solutions of:
2i + 5j =
We have
already proved that, because H(2,
does have solutions.
finding one.
[It is
fall
We now need
equation
is
it is
the first equation
way
of
guess solutions for
to use the algorithm --but, one
back on.
to
1,
the Euclidean algorithm gives us a
often, as in this case, easier to
such equations than
algorithm to
And
5) =
always has the
]
check
that,
again because H(2,
equivalent to the sentence:
3,
(i
=
5k and
j
=
-2k)
5) =
1,
the second
[7-129]
[7.06]
This
is
a consequence of
Theorem
129.
V
L
m Vn rHCF(m,
[mi +
V. V.
In
n) =
1
order to prove Theorem 129,
Theorem
m
integers are
that
if
nk and
j
=
and n have no
commonly
prove, first:
1
m Ink) =>
and
m|k]
m
and n that H(m,
common
n) =
amounts
it
to
Two such
prime. Theorem 128 says
factor other than
said to be relatively
1
1.
a positive integer divides a product and is relatively
one factor then
-mk)]]
'
two positive integers
saying that
is helpful to
it
n) =
L
of
=
128,
Vm Vn Vk [(HCF(m,
To say
<=> 3 k (i
nj =
prime
to
divides the other factor.
Notice that, in view
129,
we have extended
is to
make sense.
application we wish to
of the
meaning
the
we
In fact,
of
'
make
of
so that, for example,
'
|
shall define
4
|'
so that
it is
Theorem
4
3j— 6*
true that
3|-6:
V
m V.
[m
=> 3
k
1
,J
j
L
j
Now,
i
and
=
mk]
prove Theorem 128.
let's
Suppose that HCF(m,
integers
J
j
n) =
1
m |nk.
and that
It
follows that there are
such that
mi
(*)
+ nj =
1
and an integer k Q such that
nk
(**)
From
(*) it
From
this
follows that mki + nkj
and
(**),
mki
+
mk Q
k
Hence [since
I
is
j
=
=
mk
.
= k.
= k.
So,
m(ki + k
j).
closed with respect to multiplication and addition],
m
k.
[7.06]
[7-130]
We can now prove Theorem
Vm Vn [HCF(m.
129:
1=>
n) =
<-
[mi + nj
V. V.
To begin
we notice
with,
<=> 3
k
=
that,
any integers
3,
and
i
n) =
mnk
= 0.
j
Now, suppose that HCF(m,
mi
it
n
=
m Jmi,
-j and, since
•
Theorem
follows, by
integer k such that -j
m
Hence, since
^
0,
mi
Combining
it
mk.
n) =
-
1
^ mi
1
=
mk)]]
or not,
m jn
Since
-j.
°
nj = 0.
HCF(m,
and mi +
nj = 0,
mi
-
=
n) = 1,
mnk
Consequently [assuming that HCF(m,
(i
Then,
So, by definition, there is an
-mk
=
j
+ nj = 0.
Suppose that mi +
.
mj-j.
^=> 3,
+ nj =
1
follows that
Since
= nk.
i
-mk)
=
128, that
=
mnk
=
j
j,
nk and
=
(i
nk and
=
whether HCF(m,
m(nk) + n(-mk)
So, for
(i
nk and
n) = 1],
-mk).
=
j
= 0.
and generalizing, we
the two results displayed above,
have:
V. V.
1
[mi +
<=> 3
^
nj =
J
So, recalling the
assumption
that
(i =
HCF(m,
nk and
n) =
1
j
,
=
-mk)]
we see
that
we have
proved Theorem 129.
J,
T»
We have
+.K
.A.
.y.
..,>
seen how the Euclidean algorithm and Theorem 129 can
be used to find all integral solutions
(x, y) of
linear equations of the
form:
mx
+ ny = p
Such an equation has an infinite number
of integral solutions if
HCF(m,
if
and no integral solutions
n) |p,
Example
.
Solution
Solve:
.
.
]
35i
-
42j = 14
The given equation
(1)
5i +
is
6--j
/
n)/j p.
[Finding
sometimes called solving linear
integral solutions of such equations is
Diophantine equations
HCF(m,
equivalent to:
= 2,
[7-131]
[7.06]
and HCF(5,
6) =
1
we
First,
.
6--j
5i +
(2)
find a solution of:
=
1
Using this solution, we can immediately find a particular
solution of (1).
Then, Theorem 129 gives us
6- -j =
5i +
(3)
all solutions of:
By combining these with
the particular solution of
we
(1),
get
all solutions of (1).
By
inspection [or the Euclidean algorithm],
(2')
5- -1 + 6-
(1')
5- -2 +
1
= 1.
6-2
= 2.
So,
Thus, (-2, -2)
a particular solution of
is
By Theorem
129, (3) is equivalent to:
3k
(3')
(1).
(i
=
6k and
Consequently, the solutions
(-2 + 6k, -2
-
j
-5k)
=
of (1)
5k),
are
for k
€ I.
^EXERCISES
A. Find
1.
B_.
65i + 77j = 200
each
of the
2.
50i
For each
of
are pairs
of positive integers.
1.
C^.
all solutions of
78x
-
following equations.
-
63j = 75
how many
the following equations tell
117y = 97
3.
2.
33i +
1
9j =
250
of its solutions
51x + 85y = 1037
Find the least integer p for which *5x + 7y
= p*
has just nine
nonnegative integral solutions.
JD.
Solve the following problems.
1.
Mr. Schwartz hands a storekeeper a dollar
when paying
for a 49-cent article.
bill
and 4 pennies
The storekeeper has
7 nickels
[7.06]
[7-132]
and
2.
6
dimes.
In
how many ways can
that he
He purchased an
bank.
A
article for $3.50 and then noticed
had twice as much money
For what amount had
left
as the check had called for.
the check been written?
stick is divided into sections, all of the
36 red lines across
lines.
it.
It
is
same
comes
A
length, by drawing
also divided uniformly by 28 green
Counting from a given end
the stick, which red line
of
Which red
the least distance after one of the green lines?
4.
arranged?
When Mr. Perez cashed a check at the bank, the teller mistook
the number of dollars for the number of cents, and vice versa.
Without noticing the error, Mr. Perez pocketed the money and
left the
3.
the change be
comes
line
the least distance before one of the green lines?
manufacturer uses three kinds
of
components --googels, goggels,
and gaggels--in assembling a biggel.
the unit cost is a whole
number
goggels, together, cost 25£
of
For each kind
component,
Four googels and
cents.
more than three gaggels. Two
and three gaggels cost 38£.
four goggels,
of
five
googels,
Each biggel contains
What
seven googels, eleven goggels, and six gaggels.
is the total
cost of the components of a biggel?
5.
Three robbers
stole a
to divide their loot,
sack
of
spade guineas.
l/2 to the first man,
As they were about
l/3 to the second, and
1/6 to the third, they were surprised by the King's foresters.
robbers were able to escape with
a portion of
it.
over a part
of
Meeting again
all the
The
money, each man carrying
in a safer place,
what he had carried away; the
each robber turned
first
man
kept half
of
what he had, the second, two-thirds, and the third, five-sixths.
This
out,
money was then shared equally among all three. As it turned
each robber got, in all, the same amount as he would have, if
the foresters
had not interfered.
What
is
the smallest
coins which the sack could have contained?
number
of
[7-133]
REVIEW EXERCISES
1.
Consider the number system consisting
0,
1,
and
2,
and the two operations
following tables
1
2
2
and
ft
<I>
S of three
numbers
which are defined in the
:
a
l
of the set
1
2
1
2
4>
1
2
1
1
2
2
2
1
2
1
1
In the following questions,
domain
(a)
'y',
and
'z'
are variables whose
is S.
True-or-false?
(1
(3
(4
(5
Vx V x &
y
y7 = y
7
VVV
x
x y z
VVV
x
x y z
VVV (x
(6
VX V VZ x
(7
V V V
Z
X
(x
3 x V v7
y
ft
(8
ft
z) = (x ft v)
ft z
7
«$•
(y
7
<t>
z)' =
ft
<J>
y
<fc
x
(14
L
z = (x
<fr
z) ft
(y
<d>
z)
<t>
y)
ft
(x
<J>
z)
ft z)
<t>
(y
ft
z)
(x
z = (x
'
7
4>
(11)
3
x
V
y
y7
4>
VV
3 x
x y z
x
4>
=
y7
z = y
7
z = y
7
=
x
z
y =
z]
=
x<tz
y =
z]
ft
Notice that
VX VV
x
ft
y = the
Complete:
r
Vx V x
<$>
v7 -
y
z
(9)
z = v
7
VVV
y
x y z
VVV
[x4>y
x y
z
y)
3 '
<!>
7
7
4>
V7
=
VV
x<ty=y4>x
x y
(x
y
z) =
ft
y)
V/
V3 x
x/ n y z
ft
&
(y
(12
ft
4>
y)
V V 3 x
x y z
[x
(2)
(y
7
(10
(13
x
ft
ft
y
(b)
'x\
remainder on dividing x + y by
3.
[7-134]
2.
Suppose that S
each ye
3,
2,
1,
4,
ft
y = the remainder on dividing x + y by
x & y
and
xe
S and
remainder on dividing xy by
= the
6,
6.
Complete the following tables
ft
2
1
4
3
*
5
1
1
2
2
3
3
4
4
5
1
2
4
3
5
5
.
(b)
for each
5} and that,
S,
x
(a)
= {0,
,
.
.
(1)
Is
ft
commutative?
(2)
Is
<i>
commutative?
(3)
Is the
.
,
same number
in S listed twice in
any row
of the table
for ft?
(4)
Your answer
for part (3) tells you of a principle for
ft.
What
principle?
each number in S listed in each row
(5)
Is
(6)
Your answer for part
(5) tells
you
of
of the table for ft?
a theorem about
ft,
State
the theorem:
Vx V 3 z
y
(7)
Find
(i)
(8)
Is
5
all solutions of
<t>
x
= 3
each
(ii)
of the
2
* x
following equations.
=
there a cancellation principle for
(iii)
4> ?
2
4>
x
= 3
[7-135]
3.
4.
Simplify.
3x2 + 2x
(a)
(7x3
(b)
2
(8x 4 + 2x
(c)
(8y
(d)
(9a 3
(e)
(£)
-
-
-
x +
+ (2x3 + 7X2
5)
3)
+ (9x 4 + 3x 3
3
4
2y + 4y
-
1)
+ (7
-
3a2 + 2a
-
5)
-
(5x 4
-
3x3
(3x2
-
2x + l)(5x2 + 3x
-
-
2)
+ (7x 3
2x +
-
2y* + Qy*
-
(7a 3 + 5a2
-
3x 2 +
-
5x +
-
2)
(8X2
-
(3x
-
7)
7y
6a +
-
1)
-
1)
-
4
)
7)
2x3 +
-
7)
2
2 + 5x2 )(3x
-
2x +
1)
Expand and simplify.
(a)
(7x2
(c)
(2y
(e)
(5x3 + 3x +
6
-
2x + l)(5x
-
3)
+ 7y 3 + 3y)(4y 3 +
2)(
2^
+
7)
10x 2 +4)
x2 + 3)(2x 3
(b)
(8x 4
(d)
(5a 4 + 3a + 4)(7a 3 + 2a 2 +
(f)
(4x 4 + 3X2 + x
-
-
-
3x +
1)
2)
4)(5x3 + 8x +
2)
Simplify by using the division -with -remainder algorithm.
lOx5
-
(g)
15x 4 + 4x3 - 12x2 + 23x- 21
2x - 3
5.
If
2x
6.
Show
that
4
x
2' is
-
.
is
3
the
...
(h)
4x 4 + 2X3 -8X2 + 2
2TT1
volume-measure
(16x 3
-
28x2
of the
base
of the
height?
a factor of *3x 3
-
is
24x +
-
2x
-
3,
2X2 + 7x
-
of a circular
45)tt
what
cone
and the radius
is
the
measure
30' without using the
division-with-remainder algorithm.
6x 4 + 15x 3
7.
One factor
8.
The equation 6x 4
9.
l
of
4
and -4.
-
x3
-
-
8x2
-
Use
definitions such as '4 = 3 +
(a)
+ 5'.
1'
and algebra theorems
following.
3 + 2 = 5
What
is
Two
another?
of
them
What are the other two?
3
of the
is *2x
82x2 + 81x + 36' has four roots.
are
each
6x + 35'
(b)
5
-
2 €
P
to
prove
[7-136]
10.
Prove these generalizations.
(a)
V
q(q + 4)(q +
(b)
v
(q
v
q(q
,
(d)
ya)
(e)
11.
+
-
l)(2q +
q
q(q +
q 2(2q +
l)
Vq
1)
5)
+ {q +
1}
2 .
x
. (q + l)(q + 5)(q + 6)
(
2q + l)(2q +
of the listed sets,
4)
2)
= (q +
=
q(q+lHq+2)(q+3)
DqUq
+ 7)
o
_ (q + l)(q + 2)
2(2q + 3)
(q + IV
+
'
5)
2
q(6q +15q+ll) +
(3q +
For each
+
+ 2)
l>q(q + l)(q + 2)
+ q(q + 1)(q +
-
6
V
5)
3)
+ D[6(q -H) 2 +
= (q
2
whether or not
tell
it is
15(q+l) + 11]
closed with
respect to each of the given operations.
+
12.
(a)
the set of positive integers
(b)
the set of negative integers
(c)
the set of nonzero rational
(d)
the set of nonnegative
(e)
the set of irrational
(f)
the set of nonzero integers
(g)
the set of real
(a)
Suppose the domain
means
(b)
the
thing as
Theorems
(2)
Does Theorem 87
Suppose the domain
same
and
'is
86a,
still
thing as
Theorems
Which
(2)
Does Theorem 87
4
y' is
the set of all people and '>'
86b,
and and 86c
still
hold?
hold?
4
y' is
is the
86a,
still
4
older than'.
and
of 'x'
(1)
of
numbers
numbers
of 'x'
Which
the
-r
numbers
(1)
means
-
numbers
same
of
X
the set of all people and *>'
father of.
86b, and 86c still hold?
hold?
[7-137]
13.
Prove each
(a)
<
14.
b>
the following theorems.
of
Vx V Vu Vv
1
Vx>0 Vy>0
<=>
<
- v)(x - y)
[(u
L
'
y
-1
=>
(v
-
u)(x
-
y)
> 0]
> X
y
True-or-false?
(a)
V V
not both u > v and v > u
(b)
Vx V
[x - 3 >
u
v
y=>
x
-
y > 0]
y
(c)
Vx V Vz
(d)
> x
Vx 2x ~~
(e)
yx^O
(f)
V < -2, V .>
x<
y
(g)
V
(h)
x
VU V
'
(i)
0)
15.
Jl < I
x2 - x
(u + 2)
2
(c)
> y]
4-x2
x
1
>
4-x2
xy
2
> uv
u 2 + v 2 ""
V
u>0 Vv>0
=>
<
V
u
"
v
>
u
"
v
l
VU J
VV
Vu VvCu2> V2
V
n
yn
1
-
n
n2 +
I
n
U > V > -u
<*-
+
*
.
(n +
\)
c
<
1
-
1
BDEF
is
ABCD
a square.
the rectangle is less than the
that
n+
(b)'
Vn ~
n
(d)
yn n+
< -4t
-r
n*
n+ 1
-
1
Suppose that quadrilateral
Prove
1
1
quadrilateral
17.
z
"]
.
> u
=>
z)
Prove:
(a)
16.
u
y7 and x <
>
[(x
L
y
if a,
ad + be < ab.
b,
c,
is a
1
(n+
1
l)
2
<
1
n+2
nonsquare rectangle, and that
Show
that the
area-measure
area-measure
of the
and d are nonnegative and
of
square.
c + d <
a < b then
[7-138]
18.
2
that the equation 'x +
Prove
= 0'
1
not satisfied by any real
is
number.
19.
Prove:
Vx V x 4
20.
Arrange
in order
1
l
22.
Solve each
2 > 9
3x
(c)
x 2 + 7x > 44
-
In a basket of
2
2xV
>
~~
'
"6
n
29'
'
9
7
14
25*
30'
50'
5x
-
2
9
1
-7'
17'
(b)
5(x
(d)
^j
-
2)
2
< 3(4
were doubled, there would
but
if
oranges.
of
of grapefruit
How many
would be more than a dozen.
the
x)
of grapefruit is
number
of
oranges
be less than a dozen pieces of fruit
still
number
the
If
-
< 2
oranges and grapefruit, the number
greater than the number
in the basket,
to smallest.
following inequations.
of the
(a)
4
from largest
-5'
5'
21.
+ y
3
y
were doubled, there
oranges and grapefruit are
there in the basket?
23.
In
each
ping
f.
of the
In
following exercises you are given a set S and a
each case consider the generalization:
Ve [eeS=^
and answer
€ S],
'true' or 'false'.
(a)
S is the set of even numbers;
(b)
S = {x:
(c)
S = {m: 3
(d)
S = {m:
(e)
S =
{i:
3
(f)
S =
{i:
V if
(g)
S
= {x:
f (e)
3
x
n
Vn
i
2n
=
-
1};
= 3n};
*
f
:
m
£ 3n};
f
:
= 3j};
x > 5};
3j};
f
:
n
—
n + 20
n
-*
n + 20
—
i
+ 6
f
-*
i
-
x
i
—
n -* n + 12
—
:
£:i
:
:
n
f
m
f
x +
1
6
n + 12
map-
[7-139]
(h)
S = {x: x is a prime number};
(i)
S =
(j)
S=
(k)
S = {x: x < 0};
(i)
S = {x: 3
f
x
:
x+
-*-
2
+ x
:x- i~-^
1
1,
3;
(n)
f:x-I±*.
{1};
(m) S = {x:
f
3
p
V
/
f
q^Q
:
qx
qx 4
Q
—
x
-i-j-*-
= p};
I};
f
:
f
-
x
:
—
x
i-±-*-
*
2
and y > 0}; f (x, y) - (x, y + 2)
[Now, repeat part (n) for Quadrant II, Quadrant III, and
S = {(x, y)
:
x >
:
Quadrant IV.]
(o)
S = {(x,
y)
:
(p)
S = {(x,
y)
:
(q)
S
= {(x, y):
(r)
S
=
x
= 7}; f
:
x < y and 3. y
2
y > x };
Quadrant
[Repeat part
f
I;
(r)
:
f
-
(x, y)
= 2i};
-
(x, y)
:
—
(x, y)
(x,
(x
- 2,
(t)
S = {(x, y):
2y-3x=
(u)
S = {(x,
y)
:
4x
-
5y
= 7
(v)
S = {(x,
y)
:
4x
-
5y
= 7} r\ {<x, y)
(w)
S = {(x,
y)
:
2y-3x
4x
p
= 7};
-
7}
S = {(x, y): y > |x|};
(y)
S
(z)
S = {(x, y): xy =
x 2 + y2
f
:
y
w
{(x, y)
or 4x
f
:
-
(x, y)
= l}; f
:
-
(x, y)
5y = 7p};
(x)
= {(x, y):
(x, y)
:
-
y +
(x,
2)
- 3)
for each of the other three quadrants.]
S=
3
f
2)
(2x, 4y)
(s)
{(x, y):
y +
-
f
-
:
4x
-
(x, y)
:
(x, y)
or y2 = x2 };
3x
- 2,
:
y
- 3)
2y = 9};
f
(x, y)
:
—
(x
5y = 14};
f
(x, y)
-
y
= 14}; f
5y
f
:
(x
:
(x, y)
(x
- 2,
:
-
- 2,
y
-
- 3)
(-x, 2y)
-
*jJL\
(^L+JL,
(x, y)
-
4=I
££
(
>
^j^)
(x
- 2,
y
- 3)
- 3)
(x - 2, y
- 3)
[7-140]
24.
Here are the
numbers
first four
•
in the sequence C,
» •
13
(a)
Study the formation pattern and try to describe
it
in a recursive
definition of C.
(b)
Guess an explicit definition
induction
25.
Here are the
from
and derive
of C,
it
by mathematical
the recursive definition.
first four
numbers
in the
sequence D.
• • •
••
•
• •
•
•
•
••
••••••••
••••••••••
••••••
• • •
(a)
%
'
21
8
i
Dl
=
1
= S
Dn+in+i
and, for each n,
,
the sequences of square
spectively.
40
+
4T
n
,
where S and T are
numbers and triangular numbers, re-
Use the recursive
definitions of S and
recursive definition of D:
Di
Vn D
(b)
26.
Let
n+
1
=
Dn
of
D.
i
Derive an explicit definition
f
=
+
be a function which is defined recursively by:
V
n
f
=
.
n+
l
f
n
+ 3
T
to find the
[7-141]
and
(a)
Compute
(b)
Discover an explicit definition
f
f
3,
6,
f
- f
12
10
.
of
f
and derive
from the recur
it
sive definition.
27.
Given n points on a circle, how many chords have these points
(a)
as end points ?
Given
(b)
such that no three
7 points on a circle,
chords which
of the
have these as end points are concurrent inside the circle, how
many
28.
intersections do the chords have inside the circle?
(c)
Repeat part
(a)
In
with
'n'
in place of
how many ways can you
pockets
(b)
(b)
4
7'.
put seven silver dollars in three
?
Solve part
under the additional condition that no pocket be
(a)
empty.
29.
Show
30.
Give an example
31.
of
n,
135
n > -«-
a set of real
if
and only
a lower bound but no upper bound;
(b)
a lower bound but no least
Does
—n
1
3
Suppose that
f
f
-
member? A
x
= 4
of the
Theorem
(a)
Vn°n—
g >
/
Vn
1^
g
fe
n +L
=
l
3g
&n
following theorems by mathematical induction
114].
2
member?
Bl = 2
f
and:
L
[or
greatest
&
and g are defined by:
Vf
nn+i =2f+3
n
Prove each
n > 5
member.
j—} have a least
x =
n
n* '
least upper bound?
{x:
1
if
numbers which has
(a)
A
32.
each
that, for
(b)
*
'
V
w
&n
n> 4 g
- f
n
>
1
[7-142]
33.
Compute.
(a)
13.11
(c)
[9. 28]] + 49.28}}
+
12.9]
34.
What ordered pairs belong
35.
Solve:
[4.8]
(a)
36.
Vx [xl
x+
X+
<
37.
Solve: n|n 2 +
38.
For each number
(a)
2
-
J
(d)
H»]H
+
€1*11
4y
-
3x
0} r\ {(x,
=
(b)
J2.891
(b)
V
+
Ixl
y)
y
:
[xj}?
=
< [2.89 +
xl
ail
40.
What
is the
41.
A
how many
listed, tell
highest
common
heads and legs
and more than
15
1
wax
it
has.
-
3
2
•
5
2
•
72
= 6'.
25y
factor of 298 and 746?
of just
699.
cows, how many
a 10-ounce can of car
$1.43, how
is
factors
(c)
the integer solutions of *31x
of
< |[xl +
2
1000
farmer's livestock consists
number
X+
1
(b)
Find
If
4-
2 3 '3 4
39.
43.
7.6
+ 11*1 <
:
J -3. 5 J
Prove:
(a)
42.
to {(x, y)
2*1
(b)
If
of
cows and chickens.
he has
more
The
total
than 200 chickens
each animal does he have?
costs 60 cents and a 26 -ounce can costs
much do you save per ounce by buying
the larger size?
Simplify.
(b)
i-x
2y+
1
3
[7-143]
44.
Solve the equation:
45.
If
46.
Suppose that
r
and
7
2
>
2x
. \)Z
"
(2x
are positive integers, which
s
+ ^ = °
- 1)
AABC is right-angled at C.
from A and B are 13 and 2V 34
medians
If
is the
measure
of
s
1
-
s
or
measures
the
r
r t
?
,
i
of the
respectively, what is the
,
perimeter and what
—+—
larger,
is
the third median?
<->
47.
yS^"
A
Given
"^Xt*
:
A
handtruck can hold either 10 boxes
If
a stock clerk wants to load
ries, what is the largest
49.
24=0.25%(x)
PT
= 6
or 15 boxes of cherries
boxes
of
of
plums as cher-
each kind he can load?
~%(l600)=x
(b)
o
Simplify,
(a)
—
—
/Tl +
£*Ll
V 48
/8
vT2
y
51.
of
= 12,
many boxes
with as
number
plums
PA
a tangent,
Solve these equations:
(a)
50.
it
of
is
PB
Find:
48.
PT
+
The width
of
a rectangle is
/u\"V/l
(b)
V25
Lb
75%
1
"
,
(c)
TTq
lb9
of the length.
increased in such a way that the new rectangle
\
If
is
one, but its perimeter is increased by 21 and its
i-i.i5
9
the dimensions are
similar to the old
area-measure
increased by 63, find the dimensions of the original rectangle.
52.
Solve these equations:
5
1+418
is
[7-144]
53.
The hypotenuse
of a right triangle is
What
legs and its perimeter is 6.
54.
is its
area?
Simplify.
,
.
(a)
55.
foot longer than one of the
1
3
5
II
"
4
+
7
{b)
20
There are n players
How many
T
+
—
3m
1.2 X 0.
(c)
To"
in an elimination -type
6
checkers tournament.
must be played
individual matches
1
2
to
determine the
winner?
56.
r-
The
^-srT-^
latitude of Chicago is approxi-
mately 42°
,
and the radius
of the
earth is about 3960 miles.
is
57.
Pete walks from
from B
to
A
A
to
is the
far
Chicago from the earth's axis?
at the rate of 4
at the rate of 7
took Sj hours, what
58.
B
How
miles per hour and rides
miles per hour.
distance between
If
A
the round trip
and B?
Find four consecutive integers such that the product
smallest differs from the product
of the
of the
two largest by the
two
sum
of
the four.
59.
Solve the system
60.
What
is the
area
of
an isosceles triangle whose base
is 10
inches
long and whose congruent sides are each 13 inches long?
61.
A man
inherits
l/5 of
it.
How many
1/3 of his father's estate, and his sister inherits
The balance
of
dollars did the
$14000 was
man
left to
inherit?
a charitable institution,
[7-145]
BASIC PRINCIPLES AND THEOREMS
Commutative principles
+ y
VVx
x y
and multiplication
VVxy=yx
x y
y7 + x
=
7
for addition
7
7
Associative principles for addition and multiplication
VVVx+y+z
xyz
7
VVV
xyz xyz
x+(y+z)
'
=
7
= x(yz)
7
'
Distributive principle [for multiplication over addition]
VVV
xyz
xz + yz
7
=
(x + y)z
7 '
and
Principles for
V x
x
+
VX
X
=
Xl
=
=
+
-
x
1/0
X
Principle of Opposites
VX x
1
Principle for Subtraction
Vx V x
=
y
-
y = x + -y
P: rinciple of Quotients
Vx V /o X y
y
7
=
X
*
1-
Vx V V x(y
z
+
y
z) =
2.
Vx
3-
vxv Y.a V.d V.c ax
4'
Vx Vy Va Vb ax >< b y>
5-
Vx Vy Va Vb
6'
Vx V Vz f x
xy + xz
lx = x
[2-61]
+ bx + ex = (a + b + c)x
<
<
=
a + x> +
=
<
(t>
y==> x +
y
7'
Vx V V2 t x+
z = y +
Vx V Vz [ x
y
y
8*
[page 2-60]
y
=
ab )< xY)
[2-61]
+ y) = (a + b) + (x + y)
[2-61]
z = y + zl
[2-64]
z=>
=:> z +
[2-61]
*=
x
= y]
[2-65]
z + y]
[2-66]
yy
[7-146]
9.
10.
11.
12.
VVV
x y
z
Vx V
y
=5>
= y
[x
L
7
VVV
x y z
VVV
x y
z
y^^
x = z +
fz
+
L
-x
7
x
[page 2-66]
= yl
' J
= -y]
/J
[2-66]
=^>
xz
= yzl
7
J
[2-66]
=£»
fx = y
7
zx
= zyl
7J
[2-66]
fx = y
7
L
L
=>
13.
Vu Vv Vx V
=
[(u
L
v and x
14.
Vu Vv Vx V
L
f(u =
v and u + x = v +
15.
Vx
16.
V V
x
L
17.
VX
X
18.
Vx V -(x
=
+ y)
7
+
-y
[2-69]
19.
Vx V -(x
= y
+ -y)
+
7
7
-x
[2-69]
20.
V V
-(xy) = x
'
21.
Vx V
-(xy) =
22.
Vx V
=
[x
L
23.
Vx V —x — y
24.
Vx V — xy
25.
Vx V Vz
-x(y +
VVV
x y
-x(-y7 + -z)
y
y
= y)
7
u + x
xO =
y
=> -x
=
[x + y
}
= y]
J J
X
=
y
—x
•
—
7
*
'
7
y
z
-xy
=
x
=
•
[2-70]
—'
[2-70]
z) =
-(xy) + -(xz)
[2-70]
xy7 + xz
[2-70]
28.
VX -x
29.
Vx V
30.
VV
(x + y)-y = x
x y
-X
=
'
=
[2-70]
-lx
(x + y)
+
7
[2-69]
xyJ
Vx x
y
= yl
7J
'
1
[2-68]
[2-69]
27.
=
[2-66]
= y]
[2-69]
]
— yJ => -x
y
y
26.
x
[2-69]
y
y
—
[2-66]
[2-66]
y
x y
y)
J
v + y]
[2-70]
-y7
7
=
x
[2-71]
[2-71]
[7-147]
31.
VVV
x
x y z
32.
Vx V x
33.
Vx V -(x
-
y
-
= x + -yz
yz
'
]
yJ + yJ
x
=
y) =
-
y7
"
y
[page 2-71]
[2-72]
x
-
[2-72]
VVV
x y
VVV
x y
VVV
x y
y-z
[2-73]
x-(y+z)=x-v-z
[2-73]
x-(y-z)=x-y+z
w
[2-73]
37.
Vx V Vz
x+(y-z)=x-z
[2-73]
38.
VV
x(y
39.
Vx Vy Vz
40.
Vx V V x
34.
35.
36.
z
z
z
y
Vz
x+(y-z)=x
w
y
'
7
7
/
7
(x
-
-
xz
[2-74]
y)z = xz
-
yz
[2-74]
z) =
(-y
-
+ y
'
'
xy
-
z
y
+
#
=
- z)
x + y + z
[2-74]
41.
Vx V V V X
z u
42.
Vx
-
43.
VX x
-
44.
Vx V V X+
z
"
^+
z)
=
x
"
y
[2-75]
45.
Vx V Vz X
Z
"
(y
z) =
x
"
V
[2-75]
46.
Va V V V
b c d
47.
Vx V Vz' Z+y
x
Z
"
= x
" U)
=
x
[2-75]
"
'
(a " b) + (c " d) = (a + c
>
=
X=^
z =
x
-
48.
vx v v
y z/o
[xz = y z=>
49.
Vx V , V
y/0 z
X
= x=*> z =
[zy
L
y
rt
51.
Vo
52.
=
VX
X
-
1
x
=
y]
y
x
l
[2-74]
[2-75]
y
x
w —
V
x 1
V + z + u
-x
y
50.
"
=
z
y
(y
"
y
1
J
y]
"
<
b + d)
[2-75]
[2-89]
[2-90]
[2-91]
[2-91]
=
1
[2-91]
[2-91]
'
[7-148]
-
53.
V
54.
Vx V n [y/0 L y
55.
Vx V
[(x
56.
Vx V
[xy =
,
n
[page 2-91]
=
/
y
y
=
0=> x=
/0andy/0)^>xy/0]
=>
(x =
ww
57.
ww
58.
ww
x y/=0
u
x
y
vfO
ww
x y/*0
vp
u
Vx Vy^O
,„ V V
u v^O
60.
Vz/cOyz
V
Vxy/0
62.
63.
64.
/
y^O
x
x
w
x y/*0 y
[2-94]
U-95]
[2-96]
[2-97]
[2-97]
y
=
y+z
^
£
vxyz/Oz
v v n £
68.
Vx V /
y/0
69.
Vx V Vz/tOn i
z
-
70
V
V V
,
V
+
V VV
z^O u vpQ yz
,
/
xVo V<)
Vx V Vz^O
,_ x+
y
X=
* * 2? u
(y
-f
[2-98]
z)v
[2-99]
+
u
xv + uy
vz
yvz
^
W0
z
[2-99]
[2-100]
z
i= *£±X
z
(
iUlX
z
U
=
^±x
z
=
z
/
[2-97]
'
V V j V ,
u v^O z/*0 yv
67.
71.
—
y
Vx V
-
[2-93]
yv
v
1
66.
y
[2-92]
ly
,
'
uy
y
Vx/ n V Vz xX±2£E
x
y
y/*0
-
yv
v
=
65.
,
xv
z
*
x y/*0
u
[2-92]
yT"z
=
,
y
xv + uy
yv
:
z^O y
Vx V VZ/eOn *£
z
[2-91]
v
u
i
y
—
,
[2-91]
or y = 0)]
x
y
59.
61.
[2-91]
0]J
u
vz
xv - uv
yvz
[2-100]
[2-100]
[7-149]
V
72.
Vx V
73
fJ<
Vx Vy^O Vu^O Vv^O —
y
74.
V,--!-=2-x
V,
x ^ y/=Ox/y
/n
/n x-r
=
*-
z^
y^
[page 2-101]
XV
yu
u
/
/
/
xv
[2-101]
[2-101]
;=
V
x
yz
75.
V V
76.
Vx Vy/O
/
n
x
-X
y
y
V V
X
X
y
^y
[2-103]
X
y
[2-103]
77.
78.
,
y/= n
x
,.
*-
y
Z/=
y
x y/*0n
X
Vx Vy/O
,
n
•y
z
-r
[2-101]
[2-103]
79.
Vx [x^O=> -x/*0]
[7-18]
80.
-0=0
[7-18]
OxN
Ox
Ox
"j"
"i"
'i
=>
(Px
)
V^ [x ^
(P2
)
V^ not both x
(P^ )
s'
V^
V„ [(x
x-yi
(P4
Vx V
)
either x
P and -x
[(xe
P
P)
=>
x
P andyeP)
>
e
Of/P
82.
1
Ox
"4 X
Ox
V V
x y
[y
LJ >
x
y
Jx
',*
83.
V
84.
V V
[x >
[y
+ y e P]
€
P]
==»
> x
x €P]
<=>
y
-
Ox
*.,«.
-
x
Ox
-v>
[7-23]
[7-24]
Ox
VxeP x^O]
[i
*-,x
[7-22]
[7-22]
xy
f
(G)
P]
'4 V
P
e
€
y
Ox
T»
81.
-x
or
e
c
P and-yy
€
P
€
> o]
[7-23]
[7-23]
Ox
„,*
e
P]
[7-30]
Ox
„,.»
[7-30]
x >
0]
[7-31]
[7-150]
<=> -x
[x <
85.
V
86
a.
c
y
V V not both x > y and y > x
b.
—
x y
Vx V V z f(x > y and y > z)=>
c.
—
e.
—
VVV
x y
VVV
x y
L
'
>
[(z
z
x/x
88.
Vx V
89.
[x +
VVV
x y z
90.
V x
91.
Vx V Vu Vv
92.
VVV
x y
93.
V V
94.
> -y <=> y >
Vx V [-x
L
95.
a.
<=> x^
—x
z > y + z
L
(x =
y=> x^
[7-33]
y)]
[7-33]
y]
<=> x
[7-33]
> y]
[7-35]
> x
1
[(x
y
z
x y
V
x y
[y
L/ >.
+
xz > yz]
[V
Vx
> y and
u>v)=>x + u>y
=>
f(x
"
>
> y and y """
z)
>
—
> x)
y' and y —
=>
[(x
x
x >
[7-35]
+ v]
[7-35]
z]
[7-35]
= y]
[7-35]
x]
y
^'
96
y) =*>
and x >
87.
x
[7-32]
z]
[x>y=>x+z>y+z]
z
y
x >
L
y
d.
—
>
[x/y^>(x>yory>x)]
Vx V
—
[page 7-32]
> 0]
Vx V Vz>
y
Vx V Vz <
Q
y
a.
Vx V
—
y
Vx Vy
b.
—
[xz > yz
> y]
[7-36]
XZ * yZ <=^> x > y l
f
>
[xy
L
7
<
[xy
L
<=> x
<=>
(
[x >
and y >
0] or [x
<
and y <
0]
)
]
(
[x >
and y <
0] or [x <
and y >
0]
)
]
'
[7-36]
97
a.
±'
b.
Vvx/0 x 2 >
Vx V [x/*y=S> x2
y
c,
—
V^_x+i>2
x>
x —
[7-38]
+ y 2 > 2xy]
[7-39]
[7-40]
[7-151]
b
—
-*
_
Vx>0 Vy>ot X,2
98 *•
2
Vx V >
y
Vx>^
c
—
-
V
n
y
x=>
>
[y
7
x
=>
> x2
n [y
.
-
,,2
y > x > -y]
y7
> x2
2
^
> £
Vx V Vz/£ n [L
z
z
y
a.
99 —
*\
= y]
,
\
]
J
[page 7-38]
->
xz > yz]
[7-41]
Vx
£"
([i >
<=> x
^
V
Vx>0 V
y/0 z>
100.
[
y>
Z
^
j'I
[domain
4
of 'm',
n',
N
(V)
Vn n
<V>
V
g
4
[7-41]
?J
o„x
'l
q' is
+
I
]
+
+
[7-49]
e I
1
€
[( 1
S and
Vr
[n € S
=>
n +
„^
^,X
C P
I
102.
V
103.
Vm Vn
104.
>
Vn n —
105.
V
m Vn
L
106.
Vm Vn
l
m Vn m
Vn n
1
e
S]
)
=> V n
n
+ n
xU
P]
[7-49]
[7-56]
[7-84]
1
[n >
m
=>
[n >
m
+
1
m
+
—
Vm Vn
€
[7-56]
€ I*
mneT
f-
€ S]
*•,>.
[V n
n
101.
b.
—
vu
«V
and
*p*,
Oy
*-j>.
107 —
a.
I>
€l +
1
+
<=> x <0])
> 0] and [~ <
n
-
mc
+
I
]
J
n > m]
[7-84]
[7-86]
1
<
[n
L
[7-86]
1
<=>
n < m]J
—
[7-152]
108.
Each nonempty
[V
set of positive integers has a least
C I*=> 3 m
[0 / S
V
n
s
O,x
0'
109.
Vx 3 n n
(I)
Vx [xel
V [xe
x
[domain
[page 7-88]
J
"i
x
[7-89]
> x
+
(x
<=> 3_ 3 x
m n
I
< n]
-J,
^r
"i
(C)
m
s
e
member.
=
V,
of
e I
s.1,
s,«,
si,
•V
'J
'l
m
*j\
or -x
or x =
-
%
n
and
€ I
+
[7-94]
) ]
x
[7-94]
]
'k' is I]
*\
110 a.
V. -j e
I
V.
Vk k
+
c.
y.
vk k
- j € i
d.
V.Vk
b.
jel
>
kj C
Vk [k>
j
[k+
1
I
J
k-j€l*]
111.
V.
112.
V. V.
113.
Each nonempty
has a least
V
>
<=> k
j
eS and Vfc >
115.
Each nonempty set
has a greatest
V.
V
g
[(j €
[7-96]
j]
[7-98]
member.
V.
116.
>
[7-96]
set of integers which has a lower bound
114.
[(j
[7-95]
.
of
[k e S
=> k
+
1 e
S]
)
=> V
k>
.
k
e S]
integers which has an upper bound
[7-99]
[7-100]
member.
S and
Vk<
.
[k €
S=> k-
1
e
S})=> V
R<
.
k
e S]
[7-100]
[7-153]
117.
Vg[(0
S and
e
Vk [k
S^=> (k+
€
1
€
S and k
Vk kes]^
- 1 e S)])
[page 7-100]
Ox
Ox
*,H.
*,%
V
|[x J = the
Vx «xj
=
greatest integer k such that
VX V.K
[k
Vx Vk [k> IxJ <=^ k>
c.
Vx Vk
[k >
d.
—
VX V,K
[k
e.
VX V.K
—
[k =
> x]
<=> k
+
1
< x]
ffxfl
[xj <=> k
VV,
3,3
xy>0kz
121.
Vx Vy>.
122.
V
123.
Vx Vm
124.
V *
x m
J]]
=
ft
'
>
=
[x
L
[7-105]
ky7 + z and
<
—
z < y]
Ji
[7-106]
3 ny > x
On
A
[7-106]
7
-xj
= the least integer
Mm
x
m
o<S¥} m
V
m V.
j
[m
l
I j
tJ
=
<=>
[xj
3,
k
j
J
V"
and n
In)
k such that k > x
m m
and
*'x
(1 In
[7-105]
J
1]
[xj+j
x
Ox
*V
=
-m
m m
-
"i
V
"N
< x < k +
Oxx
125
[7-104]
x]
1
120.
j[
[7-103]
+
Vx V.([x
-
Ox
X-,%
[xj <=> k
119.
+
Ox
#-p
< x]
b.
<
[7-102]
[7-107]
[xj <=> k
<
k< x
x-[xj
Ox
<-,N
a.
118 •—
O.
«y.
Ox
m
m
[7-111]
[7-111]
Ox
T»
mk]J
"<"*
^
[7-106]
[7-115 and 7-129]
Ox
*V
[7-115]
[7-154]
Vm V [mln^^
126 a.
n
L
'
Vm Vn V
b.
[
l
p
"\
m
< n]J
~~
(m n and n
I
p) ==>
,r
I
m irj
p]
I
'
>
Vm Vn
£.
[ (
lx
m
I
[
p
(m n and
l
m Vn Vp ^ m
V
e.
'
'
'
Vm Vn V
d.
I
n and n m) =>
l
n and
m lp)
m
l
m
==c>
n +
P)
[page 7-115]
= n]J
m
l
n +
^m
y
Pi
lpl
[7-116]
Vm Vn Vp [m|n=> mp|np]
£.
127.
128.
129.
V
m Vn
3.
3.
l
j
HCF(m,
Vm Vn V.k [(HCF(m,
L
Vm Vn [HCF(m,
L
V. V.
1
J
n) =
'
1
mi
[7-122]
+ nj
m Ik]
and mink)
'
n) =
[mi +
n) =
[7-129]
'
1
nj =
<=> 3
K
(i
=
nk and
j
=
-mk)]
]
[7-129]
[7-155]
TABLE OF TRIGONOMETRIC RATIOS
sin
cos
tan
Angle
sin
cos
.0175
.0349
.0175
.0349
3°
.05 23
4°
5°
.0698
.0872
.9998
.9994
.9986
.9976
.9962
.0699
.0875
46°
47°
48°
49°
50°
.7193
.7314
.7431
.7547
.7660
.6947
.6820
.6691
.6561
.6428
1.
2°
1045
1219
1392
15 64
1736
.9945
.9925
.9903
.9877
.9848
105
51°
.7771
.7880
.7986
.8090
8192
.6293
.6157
.6018
.5878
.5736
1.2349
1.2799
1. 3270
1.3764
1.4281
.8290
.8387
8480
.8572
.8660
.5592
.5446
1.4826
1.5399
1. 6003
i. 6643
1. 7321
.4848
.4695
.4540
.4384
.4226
1.8040
1. 8807
1.9626
2. 0503
Angle
1°
6°
7°
.
.05 24
.
1
5 2°
tan
0355
1.0724
1. 1106
1. 1504
1. 1918
.
1228
1405
15 84
1763
.
1944
5 6°
.2126
.2309
57°
2493
.2679
59°
.2867
.305 7
.3249
.3443
.3640
61°
62°
63°
64°
65°
.8746
.8829
.8910
.8988
.9063
66°
67°
68°
69°
70°
.9135
.9205
.9272
.9336
.9397
.4067
.3907
.3746
.3584
.3420
2460
2.3559
2.4751
.90 63
.3839
.4040
.4245
.445 2
.4663
.4384
.4540
.4695
.4848
.5000
.8988
.8910
.8829
.8746
.8660
.4877
.5095
.5317
.5543
.5774
71°
72°
73°
74°
75°
.9455
.9511
.9563
.9613
.9659
.3256
.3090
.2924
.2588
2.9042
3.0777
3.2709
3.4874
3.7321
3J e
32°
33°
34°
35°
.5150
.5299
.5446
.5592
.5736
.85 72
.
6009
6249
6494
6745
76°
77°
78°
79°
80°
.9703
.9744
.9781
.9816
.9848
.2419
.2250
.2079
1908
.1736
4. 0108
4.3315
4. 7046
5. 1446
5.6713
36°
37°
38°
39°
40°
.5878
. 6018
.6157
.6293
.6428
.8090
.7986
.7880
7771
.7660
7265
.7536
.7813
81°
82°
83°
84°
85'
.9877
.9903
.9925
.9945
.9962
1564
.1392
1219
.1045
.0872
3138
1154
8. 1443
9.5144
11.4301
41°
42°
43°
44°
45°
.65 61
.7547
.7431
.7314
.7193
.7071
.8693
.9004
86°
87°
88°
89°
.9976
.9986
.9994
.9998
.0698
.0523
.0349
.0175
14.3007
19.0811
28. 6363
57. 2900
8°
.
.
9°
.
10°
.
11°
12°
13°
14°
15°
1908
.2079
.2250
.2419
.25 88
.9816
.9781
.9744
.9703
.9659
16°
17°
18°
19°
20°
2756
.2924
.3090
.3256
.3420
.9613
21°
22°
23°
24°
25°
.3584
.3746
.3907
.4067
.4226
.9336
.9272
.9205
.9135
26°
27°
28°
29°
30°
.
.
.6691
6820
.6947
.7071
.
.95 63
.9511
.9455
.9397
8480
.8387
.8290
.8192
.
.
.
.
.
.
.
.
.
.7002
.
.8098
.8391
.93 25
.9657
1.0000
53°
54°
55°
5 8°
60°
.
.
.5 299
.5150
.5000
.275 6
.
.
.
2.
1445
2.
2.
605
2.
7475
1
6.
7.
[7-156]
TABLE OF SQUARES AND SQUARE ROOTS
n
yflOn
1
1
2
4
3
9
4
16
5
6
25
7
8
9
36
49
64
81
n
1.000
1.414
1.732
2.000
162
51
4.472
5.477
6.325
52
53
54
236
2.449
2.646
2.828
3.000
7.071
7.746
8.367
8.944
9.487
55
162
10.000
10.488
10.954
11.402
11.832
60
61
62
2.
3.
56
57
58
59
VTOn
n
2601
2704
2809
2916
7.141
7.211
7. 280
7.348
22.583
22.804
23.022
23.238
3025
3136
3249
3364
3481
7.416
7.483
7.550
7.616
7.681
23.452
23.664
23.875
24.083
24.290
3600
3721
3844
3969
4096
7.746
7.810
7.874
7.937
8.000
24.495
24.698
24.900
8.062
8.246
8.307
25.495
25.690
25.884
26.077
26.268
3.
12
13
14
100
121
144
169
196
15
225
12.247
65
16
17
18
19
256
289
324
361
3.873
4.000
12. 649
123
13.038
13.416
13.784
66
67
68
69
4225
435 6
4489
4624
4761
20
21
22
23
4.472
4.583
4.690
4.796
4.899
14.
24
400
441
484
529
576
142
14.491
14.832
15. 166
15.492
70
71
72
73
74
4900
5041
5184
5329
5476
8.367
8.426
8.485
8.544
8. 602
26.458
26. 646
26.833
27.019
27.203
25
26
27
28
29
625
5.000
5.099
15.811
75
5625
676
729
784
841
16. 125
196
16.432
16.733
17.029
76
77
78
79
5776
5929
6084
8.660
8.718
8.775
8.832
8.888
27.386
27.568
27.749
27.928
17.321
17.607
17.889
18. 166
18.439
80
6400
6561
6724
6889
8.944
9.000
9.055
9.110
9.165
28.284
28.460
28.636
28.810
28.983
18. 708
85
29. 155
18.974
19.235
19.494
19.748
86
87
88
89
7744
7921
9.220
9.274
9.327
9.381
9.434
20.000
90
91
92
93
94
8100
8281
8464
8649
8836
9.487
9.539
9.592
9.644
9.695
30.000
95
9025
9216
9409
9604
9801
9.747
9.798
9.849
9.899
9.950
30.822
30.984
31. 145
10000
10.000
31.623
10
11
3.317
3.464
3.606
3.742
4.
4.243
4.359
5.
5.292
5.385
63
64
900
961
1024
1089
1156
5.477
36
37
38
39
1225
1296
1369
1444
1521
5.916
6.000
40
41
42
43
44
1600
1681
1764
1849
1936
6.325
6.403
6.481
6.557
45
46
47
48
49
2025
2116
2209
2304
2401
6.708
6.782
6.856
6.928
7.000
21.213
21.448
21.679
21.909
22. 136
96
97
98
99
50
2500
7.071
22.361
100
30
31
32
33
34
35
5.5 68
5. 657
5.745
5.831
6.083
6.
6.
6.
164
245
633
20. 248
20.494
20.736
20.976
81
82
83
84
6241
705 6
7225
7396
75 69
8.
124
8. 185
25. 100
25.298
28. 107
29.326
29.496
29.665
29.833
30. 166
30.332
30.496
30.659
31.305
31.464