Solutions to Homework 2
Mathematics 503
Foundations of Mathematics
Spring 2014
§1.2: 1. (a) Here is a short table.
Step
Know
Reason
P
m is an even integer
Hypothesis
P1
There is an integer q such that m = 2q
Definition of even integer
P2
m + 1 = 2q + 1
Algebra
Q1 There is an integer k such that m + 1 = 2k + 1
Let q = k
Q
m + 1 is an odd integer
Definition of odd integer
Proof. If m is an even integer, then there is an integer q such that
m = 2q.
If we add 1 to both sides of this equation, we obtain
m + 1 = 2q + 1
This equation implies that m + 1 is an odd integer.
(b) Here is a short table.
Step
Know
Reason
P
m is an odd integer
Hypothesis
P 1 There is an integer q such that m = 2q + 1
Definition of odd integer
P2
m + 1 = 2q + 2
Algebra
P3
m = 2(q + 1)
Algebra
P4
q + 1 is an integer
Closure property of the integers
Q1 There is an integer k such that m + 1 = 2k
Let k = q + 1
Q
m + 1 is an odd integer
Definition of even integer
Proof. If m is an odd integer, then there is an integer q such that
m = 2q + 1.
When we add the number 1 to both sides of this equation and do some “algebra”
, we obtain
m + 1 = 2q + 2 = 2(q + 1).
The preceding equation implies that m is even, for q + 1 is an integer.
2. (a) Here is a short table.
Step
Know
Reason
P
x and y are even integers
Hypothesis
P 1 There are integers p and q such that x = 2p and y = 2q
Definition of even integer
P2
x + y = 2p + 2q
Algebra
P3
x + y = 2(p + q)
Algebra
P4
p + q is an integer
Closure property of the integers
Q1
There is an integer k such that x + y = 2k
Let k = p + q
Q
x + y is an even integer
Definition of even integer
Proof. Suppose that x and y are even integers; then, there exist integers p and q
such that
x = 2p and y = 2q.
When we add x and y, we obtain
x + y = 2p + 2q = 2(p + q).
This equation implies that x + y is an even integer, for p + q is an integer.
(b) Here is a short table.
Step
Know
Reason
P
x is an even integer and y is an odd integer
Hypothesis
P 1 There are integers p and q such that x = 2p and y = 2q + 1 Definition of even, odd integers
P2
x + y = 2p + 2q + 1
Algebra
P3
x + y = 2(p + q) + 1
Algebra
P4
p + q is an integer
Closure property of the integers
Q1
There is an integer k such that x + y = 2k + 1
Let k = p + q
Q
x + y is an odd integer
Definition of odd integer
Proof. If x is an even integer and y is an odd integer, then there exist integers p
and q such that
x = 2p and y = 2q + 1.
When we add x and y and do some algebra, we obtain
x + y = 2p + 2q + 1 = 2(p + q) + 1.
Now, p + q is an integer. It follows from the definition of odd integer that x + y
is an odd integer.
(c) Here is a short table.
Step
Know
Reason
P
x and y are even integers
Hypothesis
P 1 There are integers p and q such that x = 2p + 1 and y = 2q + 1
Definition of odd integer
P2
x + y = 2p + 2q + 2
Algebra
P3
x + y = 2(p + q + 1)
Algebra
P4
p + q + 1 is an integer
Closure property of the integers
Q1
There is an integer k such that x + y = 2k
Let k = p + q + 1
Q
x + y is an even integer
Definition of even integer
Proof. If x and y are both odd integers, then there exist integers p and q such
that
x = 2p + 1 and y = 2q + 1.
When we add x with y and do some algebra, we obtain the following equation:
x + y = 2p + 2q + 2 = 2(p + q + 1).
Since (p + q + 1) is an integer, it follows that x + y is an even integer.
3. (b) Here is a short table.
Step
Know
Reason
P
n is an even integer
Hypothesis
P 1 There is an integer k such that n = 2k
Definition of even integer
P2
n2 = (2k)2
Substitution
2
2
P3
n = 4k
Algebra
2
2
P4
n = 2(2k )
Algebra
P5
2k 2 is an integer
Closure Property of the Integers
Q1 There is an integer q such that n2 = 2q
Let q = 2k 2
Q
n2 is an even integer
Definition of even integer
Proof. Suppose that n is an even integer; then, there is an integer k such that
n = 2k.
We square n, and note that
n2 = (2k)2 = 4k 2 .
The second equation follows from some properties of the integers. Finally, we
note that
4k 2 = 2(2k 2 ).
This implies that
n2 = 2(2k 2 ).
This last equation implies that n2 is even.
5. (a) Here is a short table:
Step
Know
Reason
P
m is an even integer
Hypothesis
P1
There is an integer k such that m = 2k
Definition of even integer
2
2
P2
3m + 2m + 3 = 3(2k) + 2(2k) + 3
Substitution
P3
3(2k)2 + 2(2k) + 3 = 2(6k 2 + 2k + 1) + 1
Algebra
2
P4
6k + 2k + 1 is an integer
Closure Properties of the Integers
2
Q1 There is an integer q such that 3m + 2m + 3 = 2q + 1
Let q = (6k 2 + 2k + 1)
2
Q
3m + 2m + 3 is an odd integer
Definition of odd integer
Proof. Suppose that m is an even integer; then, there is a k ∈ Z such that
m = 2k.
This means that
3m2 + 2m + 3 = 3(2k)2 + 2(2k) + 3.
It follows that
3m2 + 2m + 3 = 2(6k 2 + 2k + 1) + 1.
Since 6k 2 + 2k + 1 is an integer, it follows that 3m2 + 2m + 3 is odd.
6. (a) There are many possible answers. For instance, let x and y be the real numbers
under consideration. If, for each real number > 0, we find that
− < x − y < ,
we can conclude that x − y = 0.
(b) There are many possible answers. For instance: let x be the real number under
consideration. If we can show that, for every real > 0, that
− < x < ,
then we can conclude that x = 0.
(c) There are many possible answers. One might come from coordinate geometry. In
this situation, the two lines are defined by two linear equations: ax + by = c and
dx + ey = f . The two lines would be parallel if b = e = 0, or if − ab = − de . One
could also think of constructing vectors u and v parallel to the given lines, and
examing the dot product of the two vectors as a means of determining the cosine
of the angle between them. If one thinks of classical geometry, two lines l1 and l2
are parallel provided that when the two lines are intersected with a transversal,
the corresponding angles at the intersections are equal.
(d) There are many possible answers. One could show that the given triangle possessed at least two sides of equal length.
7.
Proposition. If m is a real number and m, m + 1, and m + 2 are the lengths of the
three sides of a right triangle, then m = 3.
Proof. Let m be a real number. Suppose that m, m + 1, and m + 2 are the lengths of
the three sides of a right triangle. Since m + 2 > m + 1 > m, the side of length m + 2
must be the hypotenuse of the triangle. Thus, by the Pythagorean Theorem,
m2 + (m + 1)2 = (m + 2)2 .
Expanding the powers of the binomials, we obtain
m2 + m2 + 2m + 1 = m2 + 4m + 4.
Collecting all of the terms, we obtain
m2 − 2m − 3 = 0.
Factoring this quadratic expression gives
(m − 3)(m + 1) = 0.
It follows that either m − 3 = 0 or m + 1 = 0, i.e., m = 3 or m = −1. Since m is the
length of one of the sides of the triangle, m > 0. It follows that m 6= −1, so m = 3.
9. (a) 1 = 3 · 0 + 1, 4 = 3 · 1 + 1, −2 = 3 · (−1) + 1, and −5 = 3 · (−2) + 1 are all type 1
integers.
(b) 2 = 3 · 0 + 2, 5 = 3 · 1 + 2, −1 = 3 · (−1) + 2, and −4 = 3 · (−2) + 2 are all type 2
integers.
(c) Yes, it does appear that the product of two type 1 integers is of type 1. In fact,
if a = 3n + 1 and b = 3m + 1 for some integers m and n, then
ab = (3n + 1)(3m + 1) = 9mn + 3n + 3m + 1 = 3(3mn + n + m) + 1,
so ab must be of type 1 if a and b are.
10. (a) The proposition in question is:
Proposition. If a and b are type 1 integers, then a + b is a type 2 integer.
Proof. If a and b are type 1 integers, then there exist integers m and n such that
a = 3m + 1 and b = 3n + 1.
If we add a and b, we obtain
a + b = (3m + 1) + (3n + 1).
After some algebra is performed, we get
a + b = 3(m + n) + 2.
§2.1: 1. Daisy’s statement is true. The hypothesis in this case is false, which means that the
conditional statement she made is true.
3. Note that the conditional P → Q is false when P is true and Q is false.
(a) True
(b) True
(c) True
4. If Q is false and ¬P → Q is true, then it must be that ¬P is false. This means that
P is true.
(a) True
(b) True
(c) Since P is true, P ∧ R will be true when R is true, and false when R is false.
(d) The conditional R → ¬P will be true when R is false, and false when R is true,
since ¬P is false.
5. (a)
P Q P →Q
T T
T
T F
F
F T
T
F F
T
(b)
P Q Q→P
T T
T
T F
T
F T
F
F F
T
(c)
P Q ¬P ¬Q ¬P → ¬Q
T T F
F
T
T F F
T
T
F T T
F
F
F F T
T
T
(d)
P Q ¬P ¬Q ¬Q → ¬P
T T F
F
T
T F F
T
F
F T T
F
T
F F T
T
T
Note that the truth tables for (a) and (d) are the same. The truth tables for (b)
and (c) are the same.