Download JANUARY `10 TBL MATH 28A — LECTURE 2 — OUTLINE NOTES

JANUARY ’10
TBL
MATH 28A — LECTURE 2 — OUTLINE NOTES
§1. Integers and integer factorization—
Our main goal in this section is to show that every integer can be factorized as a product of primes in
just one way. The fact that every number is a product of primes is easy.
Proposition 1. Every integer n ≥ 2 is a product of prime numbers.
Proof. If n is prime, we are done. Otherwise, we can write n = ab where 1 < a, b < n. By strong induction,
we may assume that a = p1 . . . pr and b = q1 . . . qs . Then ab = p1 . . . pr q1 . . . qs , and we are done.
We often actually consider 1 to be an ‘empty’ product of primes, and if we do, we can replace n ≥ 2
with n ≥ 1.
This is a cool application of Prop 1:
Proposition 2. There are infinitely many prime numbers.
Proof. Suppose not. Then there are finitely many prime numbers, and we can list them; p1 , . . . , pr , say.
Then let N = p1 . . . pr + 1. Then no pi divides N , so N has no prime factors, contradicting prop 1.
Definition 3. Let a, b be integers. We say that a natural number c is the highest common factor of a and
b iff:
• c|a and c|b
• If d is a natural number such that d|a and d|b then d|c.
Note that it is then clear that c is at least as big as any common factor of a and b. Note also that it
is not obvious that there is an hcf!
Proposition 4 (Division with remainder). Suppose a and b are natural numbers. Then we can write
a = bq + r where q, r ∈ Z≥0 and r < b.
Proof. We work by induction on a. a = 1 is OK. If a > 1, we write a − 1 = bq + r (0 ≤ r ≤ b − 1). If
r = b − 1 then a = b(q + 1) + 0. Otherwise a = bq + (r + 1) works.
Using this fact, we can compute the hcf of two numbers using something called the Euclidean algorithm.
Here is an example on the right, and the general case on the left (in the example a = 372, b = 162).
Write a = q1 b + r1 . (0 ≤ r1 < b)
372 = 2 × 162 + 48
Write b = q3 r1 + r2 . (0 ≤ r2 < r1 )
162 = 3 × 48 + 18
Write r1 = q3 r2 + r3 . (0 ≤ r3 < r2 )
48 = 2 × 18 + 12
...
18 = 1 × 12 + 6
Keep going until rn−1 = qn+1 rn + rn+1 with rn+1 = 0. 12 = 2 × 6 + 0.
Output rn .
Answer is 6.
Proposition 5. Euclid’s algorithm will always eventually stop. When it stops, the output is the hcf of a
and b.
Proof. Since b > r1 > r2 > · · · > rj , it can last at most b steps, so it will indeed stop.
We have rn |rn−1 (last line). So rn |rn−2 (second to last line); so rn |ri for all i (technically, by induction).
The first two rows then say rn |b and rn |a.
On the other hand, if d|a and d|b then d|r1 (by the first line). Then d|r2 (second line) and (again,
technically by induction) d|ri for all i. In particular, d|rn .