CHAPTER 0
Preliminaries
Well-Ordering Principle (WO)
Every non-empty subset of positive integers (natural numbers or N) contains a
smallest member.
()
If S ✓ N and S 6= ;,
9
m 2 S |{z}
3
m  k |{z}
8
k 2 S.
|{z}
there exists
for every
such that
Definition. Let t, s 2 Z (the integers), t 6= 0. Then t is a divisor of s if
9 u 2 Z 3 s = tu. We write t | s. If t is not a divisor of s, we write t - s.
A prime is a positive integer greater than 1 whose only positive divisors are 1
and itself.
s is a multiple of t 2 Z if 9 u 2 Z 3 s = tu.
Recall. 13 ÷ 5 has quotient 2 and remainder 3 since we can subtract 5
twice from 13 and have 3 left over, i.e.,
But also,
13 = 5 · 2 + 3 (checking).
13 = 5 · ( 3) + 28,
so why don’t we say 13 ÷ 5 has quotient 3 and remainder 28?
1
2
0. PRELIMINARIES
Theorem (0.1 — Division Algorithm). Let a, b 2 Z, b > 0.Then 9 unique
q, r 2 Z 3 a = bq + r with 0  r < b.
Proof.
(Existence) Consider
S = {a
bk | k 2 Z and a
bk
0}.
a
If 0 2 S, 9 k 2 S 3 a = bk, so b | a, q = k = , and r = 0.
b
Suppose 0 62 S. Then a 6= 0, since 0 b · 0 = 0 62 S. If a > 0, a b · 0 2 S,
and if a < 0, a b(2a) = a(1 2b) 2 S. Thus S 6= ;, so by WO the set S has
a least member r = a bq. Then a = bq + r and r 0.
[To show r < b] If r
b,
a
so a
b(q + 1) = a
bq
b=r
b(q + 1) 2 S. But, since b > 0,
r
b=a
b(q + 1) < a
b
0
bq = r,
the smallest member of S, a contradiction. Thus r < b.
(Uniqueness) Suppose also a = bq 0 + r0, 0  r0 < b. Without loss of generality
(WLOG), suppose r0 r. Then
bq + r = bq 0 + r0 and b(q
If r0
r > 0, q
q 0) = r0
r.
q 0 > 1 since b > 0. Then
0  r|0 {z r}  r0 |{z}
< b  b(q
a contradiction. Thus r0
r = 0 =) q 0
q 0) = r|0 {z r},
q = 0, so r0 = r and q 0 = q.
⇤
0. PRELIMINARIES
3
Example. For a Florida male with birth month m and birth day b, the
last three digits of the drivers license are 40(m 1) + b. For a female they are
40(m 1) + b + 500. The fourth and fifth digits from the end give the year of
birth. Determine the dates of birth and gender if the last five digits are 42218
and 53953. We first note that, using the largest possible numbers for m and b,
40(m
1) + b  40(11) + 31 = 471,
so the first is a male and the second a female.
218 = 40(5) + 18,
so the first person is a male with birth date June 18, 1942.
953
500 = 453 = 40(11) + 13,
so the second person is a female with birth date December 13, 1953.
Definition. The greatest common divisor (gcd) of two nonzero integers a
and b is the largest of all common divisors of a and b, and is denoted gcd(a, b).
If gcd(a, b) = 1, a and b are relatively prime.
4
0. PRELIMINARIES
Theorem (0.2 — gcd is a linear combination). For any nonzero integers a
and b, 9 s, t 2 Z 3 gcd(a, b) = as + bt. Moreover, gcd(a, b) is the smallest
positive integer of the form as + bt.
Proof. Consider
S = {am + bn | m, n 2 Z and am + bn > 0}.
S 6= ; since if am + bn < 0, a( m) + b( n) > 0. By WO, S has a smallest
element d = as + bt.
[To show d = gcd(a, b).]
Now, since d > 0, a = dq + r, 0  r < d. [What do we want r to be?]
If r > 0,
r=a
dq = a
(as + bt)q = a
asq
btq = a(1
sq) + b( tq) 2 S,
contradicting that d is the smallest element of S. Thus r = 0 an d | a.
By a symmetric argument, d | b, so d is a common divisor of a and b.
Now suppose d0 is another common divisor of a and b. Then a = d0h and
b = d0k for some h, k 2 Z. Then
d = as + bt = (d0h)s + (d0k)t = d0(hs + kt),
⇤
so d0 | d. Thus d = gcd(a, b).
Note. Although this is a good theoretical result, it is not too helpful for
computation. We refer to an argument of this type as non-constructive.
Corollary. If a and b are relatively prime, then 9 s, t 2 Z 3
as + bt = 1.
0. PRELIMINARIES
5
Euclidean Algorithm
To find gcd(a, b) and gcd(3256, 256):
a = bq1 + r1
b = r1q2 + r2
r1 = r2q3 + r3
...
rk 3 = rk 2qk 1 + rk
rk 2 = rk 1qk + rk
rk 1 = rk qk+1 + 0
1
0 < r1 < b
0 < r2 < r1
0 < r3 < r2
...
0 < rk 1 < rk
0 < rk < rk 1
3256 = 256 · 12 + 184
256 = 184 · 1 + 72
184 = 72 · 2 + 40
2
72 = 40 · 1 + 32
40 = 32 · 1 + 8
32 = 8 · 4 + 0
Then gcd(a, b) = rk and gcd(3256, 256) = 8, in each case the last nonzero
remainder. From WO, the algorithm must terminate.
From above, we also have:
184 = 3256 + 256( 12)
72 = 256 + 184( 1)
40 = 184 + 72( 2)
32 = 72 + 40( 1)
8 = 40 + 32( 1) = 40 + [72 + 40( 1)]( 1) = 40 · 2 + 72( 1)
= [184 + 72( 2)]2 + 72( 1) = 184 · 2 + 72( 5)
= 184 · 2 + [256 + 184( 1)]( 5) = 184(7) + 256( 5)
= [3256 + 256( 12)] · 7 + 256( 5)
= 3256 · 7 + 256( 89).
Therefore, s = 7 and t =
89.
6
0. PRELIMINARIES
To show rk = gcd(a, b): rk | rk
rk | r1 =) rk | b =) rk | a,
1
=) rk | rk
2
=) rk | rk
3
=) · · · =)
so rk is a common divisor of a and b.
If r is any divisor of a and b, then
r | r1 by the first equation=)
r | r2 by the second equation=) · · · =)
r | rk by the k’th equation=)
Thus, rk = gcd(a, b).
Theorem (Euclid’s Lemma). If p is a prime that divides ab, then p | a
or p | b.
Proof. Suppose p is prime, p | ab, but p - a. Then gcd(a, p) = 1 =)
9 s, t 2 Z 3 1 = as + pt =) b = abs + ptb.
Since p | (abs + ptb), p | b.
⇤
Note. The requirement that p is prime is necessary, since 6 | (4 · 3), but
6 - 4 an 6 - 3.
Theorem (Fundamental Theorem of Arithmetic). Every integer greater
than 1 is prime or a product of primes. This product is unique, except for
the order in which the factors appear. Thus, if
n = p1p2 · · · pr and n = q1q2 · · · qs,
where the p’s and q’s are primes, then r = s, and, after renumbering the
q’s, pi = qi for all i.
Definition (Least Common Multiple). The least common multiple of two
nonzero integers a and b is the smallest positive integer that is a multiple of
both a and b, and is denoted lcm(a, b).
0. PRELIMINARIES
7
Modular Arithmetic — an application of the division algorithm.
If
a = qn + r,
we say
a mod n = r or a = r mod n.
That is, r is the remainder when dividing a by n.
A Bit of Logic
The implication or conditional:
(P =) Q)
(if P , then Q)
(P implies Q)
is logically equivalent to its contrapositive:
(not Q) =) (not P ).
The converse of P =) Q is Q =) P .
The double implication or biconditional is
(P () Q)
(P |{z}
if and|only
{z if} Q)
(=
(P i↵ Q)
=)
(P () Q) is true precisely when P and Q are either both true or both false.
A Simple Truth Table:
P
T
T
F
F
Q
T
F
T
F
P =) Q
T
F
T
T
Proof by Contradiction:
If C is a contradiction (a statement that is always false, e.g., 1 = 0), then
P =) Q and [P and (not Q)] =) C
are logically equivalent.
8
0. PRELIMINARIES
Problem (Page 24 #7). If a, b 2 Z and n 2 Z+, then
a mod n = b mod n () n | (a
b).
Proof. By the division algorithm 9 unique qa, ra, qb, rb 2 Z 3
a = nqa + ra and b = nqb + rb with 0  ra, rb < n.
(=)) a mod n = b mod n =) ra = rb. Then
a
b = n(qa
((=) We have a
qb) + (ra
b = n(qa
9 unique q and r = 0 3 a
0 = ra
rb) = n(qa
qb) + (ra
qb) =) n | (a
rb). Since n | (a
b).
b),
b = nq + 0. Thus
rb =) ra = rb =) a mod n = b mod n.
⇤
p
Complex Numbers — These are numbers of the form a+b
1 where a, b 2 R.
p
p
1 is defined
the property that (
1)2 = 1. We customarily use i
p to have
to denote
1, so i2 = 1. The standard form of a complex number is a + bi.
A complex number a + bi can be represented in the plane by the point (a, b).
We refer to the horizontal axis as the real axis and the vertical axis as the
imaginary axis.
p
The distance of a + bi from the origin is r = a2 + b2 and is denoted |a + bi|.
0. PRELIMINARIES
9
If ✓ is the angle formed by the line segment joining the origin to a + bi and the
positive real axis, then a + bi = r(cos ✓ + i sin ✓), the polar form of the complex
number.
Theorem (0.4 — Properties of Complex Numbers).
1. Closure under addition:
(a + bi) + (c + di) = (a + c) + (b + d)i.
2. Closure under multiplication:
(a + bi)(c + di) = (ac) + (ad)i + (bc)i + (bd)i2 = (ac
bd) + (ad + bc)i.
3. Closure under division( c + di 6= 0):
a + bi (a + bi)(c
=
c + di (c + di)(c
di)
=
di)
(ac + bd) + (bc
c2 + d2
4. Complex conjugation:
(a + bi)(a
ad)i
=
(ac + bd) (bc ad)
+ 2
i.
c2 + d2
c + d2
bi) = a2 + b2.
5. Inverse: For every complex number a + bi there is a complex number
c + di such that (a + bi)(c + di) = 1. (That is, (a + bi)
1
exists in C).
10
0. PRELIMINARIES
6. Powers: For every complex number a + bi = r(cos ✓ + i sin ✓) and every
positive integer n, we have
(a + bi)n = [r(cos ✓ + i sin ✓)]n = rn(cos n✓ + i sin n✓).
7. Radicals: For every complex number a + bi = r(cos ✓ + i sin ✓) and every
positive integer n, we have
1
1
1
(a + bi) n = [r(cos ✓ + i sin ✓)] n = r n (cos
✓
✓
+ i sin ).
n
n
Example.
(4 + 7i) + ( 8 + 2i) =
4 + 9i.
(4 3i)(6 + 5i) = 39 + 2i.
5 2i 5 2i 3 6i 3 36i
3
36
1
4
=
·
=
=
i=
i.
3 + 6i 3 + 6i 3 6i
45
45 45
15 5
(8 3i)(8 + 3i) = 64 + 9 = 73.
1
1
4 + 2i 4 + 2i 1
1
(4 2i) 1 =
=
·
=
= + i.
4 2i 4 2i 4 + 2i
20
5 10
3
4
1
To find (4 + 3i)5 and (4 + 3i) 5 , we first note that if ✓ = arctan , then cos ✓ =
4
5
3
and sin ✓ = . Then
5
1
1
✓
✓
(4 + 3i)5 = 55(cos 5✓ + i sin 5✓) and (4 + 3i) 5 = 5 5 (cos + i sin ).
5
5
0. PRELIMINARIES
11
Theorem (0.5 — First Principle of Mathematical Indunction (MI)).
Let S ✓ Z 3
(1) a 2 S;
(2) if a  n 2 S, then n + 1 2 S.
Then {n 2 Z|n
a} ✓ S.
Theorem (Corollary — Applied Version of MI).
Let a 2 Z and P (n) be a staement for all n
a, n 2 Z. Suppose
(1) P (a) is true.
(2) 8n
a, “P (n) true” (the induction hypothesis) =) “P (n+1) true.”
Then P (n) is true for all n
a.
Example. Prove that 5|n
4n
P (n)
Proof. Let S ✓ Z+ 3 5n
1 2 S since 5
4
1 is{zdivisible by 16} 8n 2 N.
4n
1 is divisible by 16.
1 = 0 is divisible by 16.
Suppose n 2 S, i.e. 5n
4n
1 is divisible by 16 (induction Hypothesis).
Then
5n+1
5n+1
4(n + 1)
4n
1=
5=
(5n+1 20n 5) + 16n =
5 (5n 4n 1) +16n
|
{z
}
divisible by 16
is divisible by 16, so n + 1 2 S.
Thus, by the first principle of math induction, S = Z+.
⇤
12
0. PRELIMINARIES
4, n 2 N.
n
Example. Prove 2 < n! 8n
Proof. Let P (n) be (2n < n!).
P (4) is true since 24 = 16 < 24 = 4!.
Suppose P (n) is true, i.e., 2n < n!, n
4.
Then
2n+1
= 2n · 2
< n! · 2
< n!(n + 1)
= (n + 1)!,
so P (n + 1) is true.
Thus P (n) is true 8n
4, n 2 N.
⇤
Theorem (0.6 — Second Principle of Mathematical Induction).
Let S ✓ Z 3
(1) a 2 S;
(2) if {a, a + 1, , . . . , n} ✓ S, then n + 1 2 S.
Then {n 2 Z|n
a} ✓ S.
Notation. A\B = {x 2 A|x 2
/ B}. This is also sometimes written as
A B.
0. PRELIMINARIES
13
Theorem (1). W O () M I () SI.
Proof.
(W O =) M I).
Suppose S ✓ Z 3 a 2 S and (n 2 S =) n + 1 2 S) [so we need to prove
{n 2 Z|n a} ✓ S], but {n 2 Z|n a} 6✓ S.
Then {n 2 Z|n
a}\S 6= ;,
so by W O {n 2 Z|n
a}\S contains a least element, say m.
Since a 2 S, m 6= a, so m > a and m
⇥
⇤
Where is m 1?
Since m
m
1 < m and m is the least element of {n 2 Z|n
1 2 S.
Then (m
1 2 {n 2 Z|n
a}.
a}\S,
1) + 1 = m 2 S, contradicting that m 2 {n 2 Z|n
Since our assumption that {n 2 Z|n
{n 2 Z|n a} ✓ S, proving M I.
a}\S.
a} 6✓ S led to this contradiction,
14
0. PRELIMINARIES
(M I =) W O)
Suppose ; 6= S ✓ N
[we need to prove S has a least element] and S does not have a least element.
Let T be the set of natural numbers that every element of S is greater than or
equal to. [We use induction to show T = N]
1 2 T , since if x 2 S, x 2 N, and so x
1.
Suppose n 2 T (induction hypothesis).
Then 8x 2 S, x
n [so where is n + 1?],
so n 62 S (since it would be a least element if it were in S).
Thus, 8x 2 S, x > n =)
x
n + 1 =) n + 1 2 T.
Then, by M I, N ✓ T =) T = N
Since S 6= ;, 9 n 2 S =) n 2 N =) n + 1 2 N = T .
But then, since n 2 S and n + 1 2 T , n > n + 1, a contradiction.
Since assuming S does not have a least element led to this contradiction, S
does have a least element and W O is proved.
0. PRELIMINARIES
15
(M I =) SI)
We are given that M I is true.
Let S ✓ Z 3 a 2 S and ({a, a + 1, a + 2, . . . , n} ✓ S =) n + 1 2 S).
[To show {n 2 Z|n
a} ✓ S by using M I.]
Suppose n 2 S. Now a 2 S =) {a} ✓ S =) a + 1 2 S =)
{a, a + 1} ✓ S =) a + 2 2 S =) {a, a + 1, a + 2} ✓ S. Continuing, within a
finite number of steps, we get {a, a + 1, . . . , n} ✓ S =) n 2 S =) n + 1 2 S
(by M I).
Thus, by M I, {n 2 Z|n
a} ✓ S and SI is proved.
(SI =) M I)
We are given that SI is true.
Let S ✓ Z 3 a 2 S and (n 2 S =) n + 1 2 S).
[To show {n 2 Z|n
a} ✓ S by using SI.]
Suppose {1, 2, . . . , n} 2 S. Then n 2 S =) n + 1 2 S.
Thus {n 2 Z|n
a} ✓ S by SI, so M I is proved.
⇤
As a result of this equivalence, once one of the three is taken as an axiom, the
two others follow as Corollaries.
16
0. PRELIMINARIES
Example.
Every integer greater than 27 can be written as a · 5 + b · 8 for a, b
Proof.
Let Z0 = {n 2 Z|n
0.
28} and S = {n 2 Z0|n = a · 5 + b · 8 for a, b
0}.
Clearly, 28 2 S since 28 = 4 · 5 + 1 · 8.
[Two proofs.]
[Using MI] Suppose n 2 S, n = a · 5 + b · 8 with a, b
Then a
If a
If b
3 or b
0.
3.
3,
n + 1 = (a · 5 + b · 8) + ( 3 · 5 + 2 · 8) = (a
3,
n + 1 = (a · 5 + b · 8) + (5 · 5
3) · 5 + (b + 2) · 8.
3 · 8) = (a + 5) · 5 + (b
In either case, n + 1 2 S, so Z0 ✓ S by MI.
3) · 8.
[Using SI] Note 28 = 4·5+1·8, 29 = 1·5+3·8, 30 = 6·5+0·8, 31 = 3·5+2·8,
and 32 = 0 · 5 + 4 · 8 are in S.
For n > 32, assume {k 2 Z0|28  k < n} ✓ S.
[Show n 2 S.] Since n
5 2 S, 9 a, b
03 n
n = (a + 1) · 5 = b · 8 =) n 2 S, so Z0 ✓ S by SI.
5 = a · 5 = b · 8 =)
⇤
0. PRELIMINARIES
17
Equivalence Relations
Example.
Although they are all distinct, we consider the class of all vectors in the plane
having the same same magnitude and same direction (a plane vector and all of
its parallel translates) as being the same “free” or “displacement” vector. Any
of the distinct vectors in the class may be used to “name” the class. This is an
example of an equivalence relation.
Definition. An equivalence relation on a set S is a set R of ordered pairs
of elements of S 3
1. (a, a) 2 R for all a 2 S (reflexive property).
2. (a, b) 2 R =) (b, a) 2 R (symmetric property).
3. (a, b) 2 R and (b, c) 2 R =) (a, c) 2 R (transitive property)
Note.
(1) We often write aRb for (a, b) 2 R.
(2) Since equivalence is a generalization of equality, we also often use a ⇡ b,
a ⌘ b, or a ⇠ b for (a, b) 2 R. Then the three properties of an equivalence
relation become, for instance, a ⇠ a, a ⇠ b =) b ⇠ a and a ⇠ b and b ⇠
c =) a ⇠ c.
(3) If ⇠ is an equivalence relation on S, and a 2 S, [a] = {x 2 S|x ⇠ a} is
called the equivalence class of S containing a.
Example.
Let S = {fR|f is di↵erentiable on [a, b]}. Define f ⇠ g if f 0 = g 0 on [a, b].
Then [f ] = f 0(x) dx = {f + c|c 2 R} is the antiderivative of f 0.
Example.
na
o
a
c
a
c
Let S =
a, b 2 Z, b 6= 0 . Define ⇠ if ad = bc, i.e., ⇠ if they are
b
b
d
b
d
the same in lowest terms.
18
0. PRELIMINARIES
Example.
Let S = Z and n 2 N. Define
a ⌘ b if a mod n = b mod n (i.e., n|a
(1) a ⌘ a since n|a
(2) a ⌘ b =) n|a
b).
a.
b =) n|b
a =) b ⌘ a.
(3) a ⌘ b and b ⌘ c =) n|a b and n|b c =) a b = rn and b c = sn =)
a c = rn + sn = (r + s)n =) n|a c =) a ⌘ c.
So this is indeed an equivalence relation.
Note.
For n = 7,
[a] = {a + kn|k 2 Z}.
[3] = {. . . , 11, 4, 3, 10, 17, . . . }.
Definition. A partition of a set S is a collection of nonempty, disjoint
subsets of S whose union is S.
0. PRELIMINARIES
19
Theorem (0.7 — Equivalence Classes Partition).
The equivalence classes of an equivalence relation on a set S constitute a
partition of S. Conversely, for any partition P of S, there is an equivalence
relation on S whose equivalence classes are the elements of P .
Proof. Let ⇠ be an equivalence relation on S. For any a 2 S, a ⇠ a, so
a 2 [a] and S = [ [a]. Now suppose [a] 6= [b]. [To show [a] \ [b] = ;.]
a2S
By way of contradiction, suppose c 2 [a] \ [b]. [We show [a] = [b]. Recall
A ✓ B and B ✓ A =) A = B.]
Let x 2 [a]. Then c ⇠ a, c ⇠ b, and x ⇠ a, so a ⇠ c and thus x ⇠ c and
x ⇠ b. We now have x 2 [b] =) [a] ✓ [b].
Analagously, [b] ✓ [a] =) [a] = [b], a contradiction. Thus [a] \ [b] = ;, giving
us a partition of S.
For the second part, let P be a partition of S. Define a ⇠ b if a and b belong to
the same subset in the partition. To finish the proof, show ⇠ is an equivalence
relation.
⇤
Functions (Mappings)
Definition (Function (Mapping)).
A function (or mapping) from a set A (the domain of ) to a set B (the
range of ) is a rule that assigns to each element a of A exactly one element b
of B. : A ! B means is a mapping from A to B.
If assigns b 2 B to a 2 A, we write (a) = b or
image of A under .
: a ! b, and say b is the
(A) = {b 2 B| (a) = b for some a 2 A} is the image of A under .
(A) ✓ B.
20
0. PRELIMINARIES
Definition (Composition of Functions).
Let : A ! B and : B ! C. The composition
to C defined by ( )(a) = ( (a)) for all a 2 A.
is the mapping from A
Note.
In calculus we used
instead of
with (
)(a) = ( (a)).
Definition (One-to-One Function).
A function
: A ! B is one-to-one (1-1) if for every a1, a2 2 A,
(a1) = (a2) =) a1 = a2.
Note.
(1) Every b 2 (A) has a unique preimage in A.
(2) Such functions are also called injective.
0. PRELIMINARIES
21
Definition (Function from A onto B). A function : A ! B is said to
be onto B if each element of B is the image of at least one element of A, i.e.
(A) = B.
is onto if 8b 2 B, 9 a 2 A 3
Note.
(a) = b.
(1) Such functions are also called surjective.
(2) Every function
: A ! B is onto (A).
Theorem (0.8 — Properties of Functions).
Given functions ↵ : A ! B,
(1) ( ↵) = (
: B ! C, and
)↵ (Associativity)
(2) If ↵ and
are 1-1, so is ↵.
(3) If ↵ and
are onto , so is ↵.
: C ! D. Then
(4) If ↵ is 1-1 and onto, then there exists a function ↵
also 1-1 and onto such that
Proof.
1
: B ! A that is
(↵ 1↵)(a) = a 8a 2 A and (↵↵ 1(b) = b 8b 2 B.
(1) Let a 2 A. Then
( ( ↵))(a) = (( ↵)(a)) = ( (↵(a))) = (
)(↵(a)) = ((
)↵)(a).
⇤
Maple. See prelim.mw or prelim.pdf.