Lecture 023

Number Theory Revisit

Division

a|b \land a | c \implies a| (bx+cy)

Euclidean Algorithm

Continuously divide remainder gives gcd(a,b) TODO: Proof

Bezout's Lemma

  1. gcd(a, b) can be written in gcd(a, b) = ax + by
  2. gcd(a,b) is the smallest positive integer written in multiples of a and b xa + yb TODO: Proof

Corollary to Bezout's Lemma

  1. t|a \land t|d \implies t|gcd(a,b) greatest divisor divides common divisor
  2. a number divisible by gcd(a,b) can be written as ax+by
  3. gcd(a, b)=1 \implies ax+by=1 if a,b coprime, then we can find some number of multiple that gets close but not 0. (special case for Bezout's Lemma)
  4. gcd(ma, mb) = m \times gcd(a,b)

Least Common Multiple

lcm[a,b]

Corollary to LCM

Number Theory on Friday

Euclid's Lemma (coprime)

gcd(a,b)=1 \land a|bc \implies a|c TODO: proof

When a is prime: p|bc \implies p|b \lor p|c

Fundamental Theorem of Arithmetic (FTA)

Every integer n > 1 can be written as a product of primes in a unique way.

The Infinitude of Primes

The next prime is a prime for P = (\prod_{i=1}^n p_i)+1

Divisors

ord_p(n) = max\{a\in \mathbb{N} | p^a|n\}

so n can be expressed as n=p_1^{ord_{p_1}(n)}p_2^{ord_{p_2}(n)}...

so a > b means that

  1. the power of b's factorization is smaller equal to the power of a's factorization. ((\forall p \in \mathbb{P})(ord_p(a) \leq ord_p(n)))
  2. we can calculate how many possible divisor of a and b

GCD-LCM Theorem

We can express a = \prod_{i=1}^n p_q^{a_i} b = \prod_{i=1}^n p_q^{b_i} Then gcd(a,b) = \prod_{i=1}^n p_i^{min\{a_i, b_i\}} lcm(a,b) = \prod_{i=1}^n p_i^{max\{a_i, b_i\}} TODO: proof

Corollary to GCD-LCM Theorem

ab = gcd(a,b) \times lcm[a,b]

Additional Things

gcd(a,b)|c iff a(kx)+b(ky)=c (ax+by=gcd(a,b)) d|a, d|b -> d|gcd(a,b)

Table of Content