Math 214 Foundations of Higher Mathematics
C.K. Li
Chapter 11 Number Theory
We practice proof techniques using number theory problems.
A goal: Prove the Fundamental Theorem of Arithmetic.
11.1 Divisibility
Definition A positive integer p ≥ 2 is a prime if 1 and p are the only positiver integer factor (divisor) of p.
A positive integer n ≥ 2 is a composite number if it is not a prime.
Lemma A positive integer n ≥ 2 is composite if and only if n = ab with a, b ∈ N such that 1 < a < n and
1 < b < n.
Theorem Let a, b, c be integers such that a 6= 0.
(1) If a|b, then a|bc.
(2) If a|b and b|c, then a|c.
(3) If a|b and a|c, then a|(b + c).
Suppose b is also nonzero.
(a) If a|b and b|a, then a = b or a = −b.
(b) If a|b, then |a| ≤ |b|.
Proof. (1) Assume b = am with m ∈ Z. Then bc = a(mc).
(2) Assume b = am and c = bn with m, n ∈ Z. Then c = bn = a(mn).
(3) Assume b = am and c = an for some m, n ∈ Z, then b + c = a(m + n).
(a) Suppose b = ma and a = nb with m, n ∈ Z. Then a = a(mn) so that mn = 1.
Hence, (m, n) = (1, 1) or (m, n) = (−1, −1). So, a = b or a = −b.
(b) Suppose b = ma with m ∈ Z. Then |b| = |m||a| ≥ |a|.
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11.2 Division Algorithm
Theorem Suppose a, b ∈ N. Then there are unique integers a and r such that b = aq + r with 0 ≤ r < a.
Proof. Consider S = {b − ax : x ∈ Z, b − ax ≥ 0}. Then S has a unique smallest number r by the well
ordered property of N. So, r = b − aq ≥ 0 for some q ∈ Z. If r ≥ a, then
0 ≤ r − a = b − aq − a = b − a(q + 1) ∈ S,
which is a contradiction. So, 0 ≤ r ≤ a, and b = aq + r, where q = (b − r)/a is also uniquely determined. 2
Corollary (General form of the Division Algorithm) Suppose a, b ∈ Z and a 6= 0. Then there exist
unique integers q and r such that b = aq + r with 0 ≤ r < |a|.
Partition of integers in remainder classes
Definition/notation Let n ≥ 2 be a positive integers.
Zn = {[0], [1], . . . , [n − 1]}
with [k] = {nx + k : x ∈ Z} is a partition of Z. We say that
a ≡ b (mod n)
if
a − b is divisible by n.
11.3 Greatest common divisors
Definition Suppose a, b ∈ Z are not both zero. Then their greatest common divisor d is the largest common
divisor of a and b. That is: (1) d is a divisor of a as well as b, (2) if d˜ is a common divisor of a and b, then
d˜ ≤ d.
Theorem Let a, b ∈ Z are not both zero. Then the following condition are equivalent.
(1) d is the greatest common divisor of a and b.
(2) d is the smallest element in the set
S = {ax + by : x, y ∈ Z, ax + by ∈ N}.
(3) d is a common divisor of a and b, and c|d for any common divisor c of a and b.
Proof. (2) ⇐⇒ (3). Suppose d = ax0 + by0 is the smallest element in S. We show that d =gcd(a, b).
If d does not divide a, then a = dq + r with 0 < r < d. Then
r = dq − a = (ax0 + by0 )q − a = a(x0 q − 1) + by0 q
is an element in S smaller than d, which is a contradiction. Similarly, we can prove that d divides b.
Now, if c is a common divisor of a and b, then a = cm and b = cn for some m, n ∈ Z. Thus, d =
ax0 + by0 = cmx0 + cny0 = c(mx0 + ny0 ), i.e., c divides d.
(1) ⇐⇒ (2). The above proof shows that the minimum element d of S is a common divisor of a and b,
and is larger than any other common divisor of a and b.
11.4 Euclidean Algorithm
Lemma Let a, b ∈ N. If b = aq + r with 0 ≤ r < a, then gcd(a, b) = gcd(r, a).
Consequently, there are positive integers r1 > r2 > r3 · · · > rn−1 such that
gcd(a, b) = gcd(r1 , a) = gcd(r1 , r2 ) = · · · = gcd(0, rn−1 ) = rn−1 .
Example gcd(374, 946) = gcd(374, 198) = gcd(176, 198) = gcd(176, 22) = gcd(0, 22). One can write
946 − 374(2) = 198,
374 − 198 = 176,
198 − 176 = 22.
Thus,
22 = 198 − 176 = 198 − (374 − 198) = (2)198 − 374 = (2)(946 − (2)374) − 374 = (2)946 + (−5)374.
11.5 Relatively Prime Integers
Definition Let a, b ∈ Z, not both zero. Then they are relatively prime if gcd(a, b) = 1.
Theorem Let a, b ∈ Z, not both zero. Then gcd(a, b) = 1 if and only if there are x, y ∈ Z such that
ax + by = 1.
Example Note that gcd(33, 64) = 1 and 1 = (16)64 − (31)33.
Question How to find x and y? Use Euclidean algorithm.
Theorem (Euclid’s Lemma) If a|(bc) and gcd(a, b) = 1, then a|c.
Proof. Let bc = am with m ∈ N, and ar + bs = 1. Then
c = c(ar + bs) = car + cbs = car + ams = a(cr + ms).
So, a|c.
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Corollary Let p is a prime. If p|bc, then p|b or p|c. More generally, for any integers a1 , . . . , an with n ≥ 2,
if p|a1 · · · an then p|ai for some i ∈ {1, . . . , n}.
Theorem Let a, b, c ∈ Z such that ab 6= 0 and gcd(a, b) = 1. If a|c and b|c, then (ab)|c.
Proof. Let c = am and c = bn. Then a|bn so that a|n. Let n = ak. Then c = bn = bak. So, ab|c.
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11.6 The Fundamental Theorem of Arithmetic
Theorem Every integer n ≥ 2 is a product of primes p1 · · · pm . The factorization is uniques up to a
rearrangement of p1 , . . . , pm . That is: if n is a product of primes q1 · · · qr , then r = m, and q1 , . . . , qm is a
rearrangement of p1 , . . . , pm .
Proof. Let P (m): If a positive integer n = p1 · · · pm , where p1 , . . . , pm are primes, and if n = q1 · · · qr
such that q1 , . . . , qr are primes, then r = m and q1 , . . . , qm are rearrangement of p1 , . . . , pm .
Proof by induction on m with m ∈ N. If m = 1, then the p1 = q1 · · · qr implies that the prime numbers
qi can only be p1 , and there can only be one of them. So, r = 1 and p1 = q1 .
Assume the result holds for integers that are product of m − 1 primes with m − 1 ≥ 1. Suppose
n = p1 · · · pm = q1 · · · qr . Then p1 |qi for some i. So, q1 = qi by Euclid’s Lemma. Relabeling q1 , . . . , qr ,
we may assume that qi = q1 . Then p2 · · · pm = q2 · · · qr . Then by induction assumption, q2 , . . . , qr are
rearrangement of p2 · · · pm .
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By the Principle of M.I., the conclusion holds.
Theorem Every integer n ≥ 2 has a prime factor. If n is not a prime, then there is a prime factor of size
√
not larger than n.
Proof. Suppose n = p1 · · · pm are the prime factors such that p1 ≤ · · · ≤ pm and m ≥ 2. Then
√
√
p1 p2 ≤ n.
2
p1 ≤
Theorem If n ∈ N such that n 6= k 2 for some k ∈ N, then
r1s1
√
n is irrational.
such that r1 , · · · , rm are primes and s1 , . . . , sm ∈ N. Since n 6= k 2 . At
√
s /2
s /2
least one of si ∈ N is odd. Else, k = r11 · · · rmm ∈ N satisfies k 2 = n. If n = p/q, then q 2 n = p2 .
Proof. Suppose n =
sm
· · · rm
Note that all the prime factors of p2 and q 2 appear in even number of times. Thus, the prime factors in q 2 n
cannot be the same as the prime factors in p2 , a contradiction.
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Theorem There are infinitely many prime numbers.
Proof. If p1 , . . . , pm are all the primes, then q = p1 · · · pm +1 is a product of primes including one different
from p1 , . . . , pm .
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Remark We only need to show that P (n): every n ∈ N larger than 1 is a product of primes, which can be
proved by induction. [Uniqueness is not needed.]
Proof. P (2) holds. Assume P (m) holds for m ≤ n − 1 with n − 1 ≥ 2. If n is a prime, done. If not,
n = ab with 1 < a, b < n. By induction assumption, a and b are product of primes, and so is n.
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Some open problems in number theory
Conjecture There are infinitely pair of twin primes, i.e., (p, p + 2) pairs such that both p and p + 2 are
primes.
Example (3, 5), (5, 7), (11, 13).
Goldbach conjecture
Every even number n > 4 is the sum of two prime numbers.
Example 4 = 2 + 2, 6 = 3 + 3, 8 = 3 + 5, 10 = 3 + 7 = 5 + 5, . . . .
11.7 Sum of divisor
Definition An integer n ≥ 2 is a perfect number if n is the sum of its proper factors, i.e., factors strictly
smaller than n.
Example 6, 28, 496.
Questions Is there an odd perfect number?
Are there infinitely many even perfect numbers?
Definition A pair of integers n, n + 1 are Ruth-Aaron pair if the sums of their prime divisors are equal.
Theorem (Erdös) There are infinitely many pairs of Ruth-Aaron pairs.