• then $\gcd (a,b)=\gcd (b,r)$

• $d|a\wedge d|b⟹d|r$
  • as $r=a-q\cdot b$
• $d|b\wedge d|r⟹d|a$
  • as $a=q\cdot b+r$

Pseudocode gcd(a,b):
    if b|a:
        return b
    let q,r be s.t. a = q\cdot b + r
    return gcd(b,r)

int gcd(int a, int b) {
    if (a % b == 0)
        return b
    return gcd(b, b % a)
}

• 👨‍🏫 quick!

• $1800\%366=336$
• $366\%336=30$
• $336\%30=6$
• $30\%6=0$
• or, the other way:
  • $336=1800-4\cdot 366$
  • $30=366-336=5\cdot 366-1800$ $$\begin{array}{rl}6=336-11\cdot 30& =1800-4\cdot 366-11\cdot (5\cdot 366-1800)\\ & =1800-4\cdot 366-55\cdot 366+11\cdot 1800\\ & =12\cdot 1800-59\cdot 366\end{array}$$

• $a|l\wedge b|l$
• $\forall l^{\prime },a|l^{\prime }\wedge b|l^{\prime }⟹l|l^{\prime }$
  • or $l\le l^{\prime }$

• $d:=\gcd (a,b)$
  • $d|a,d|b,d=ax+by$
  • $l:=\frac{a\cdot b}{d}$
  • show: $l$ satisfies properties of $\text{lcm}$
• using $\gcd (a/d,b/d)=1$
  • $\exists x,y,s.t.(a/d)x+(b/d)y=1$
• $l=a\cdot (b/d)=b\cdot (a/d)$ where $b/d,a/d\in \mathbb{N}$
  • thus $l$: common multiple of $a,b$
• suppose: $l^{\prime }$ a common multiple of $a,b$
  • given $a|l^{\prime },b|l^{\prime }$, then $l|l^{\prime }$ $$\begin{array}{rl}\frac{l^{\prime }}{l}& =\frac{l^{\prime }d}{ab}\\ & =\frac{l^{\prime }(ax+by)}{ab}\\ & =\frac{l^{\prime }ax}{ab}+\frac{l^{\prime }by}{ab}\\ & =\frac{l^{\prime }x}{b}+\frac{l^{\prime }y}{a}\\ & =\frac{l^{\prime }}{b}\cdot x+\frac{l^{\prime }}{a}\cdot y\end{array}$$
  • and $l^{\prime }/b,l^{\prime }/a$: integers
    • thus $l|l^{\prime }$
    • and $\gcd (a,b)\cdot \text{lcm}(a,b)=a\cdot b$

• trivial: as $a\perp b⟺\gcd (a,b)=1$

• notice that: only positive integer excluding $1$ is eligible for prime

• $p|a\vee p|b$

• $\gcd (p,a)\ne p$, as it would mean $p|a$
• thus, $\gcd (p,a)$ must be 1, as $\gcd >0$ and $\gcd \ne p$
• by Euclid's lemma: $p|b$

• then $\exists i,p=q_i$

• $n=p_1{p}_2\dots p_k$ where every $p_i$: a prime, and $p_1\le p_2\le \cdots \le p_k$
  • not necessarily distinct (e.g. $4=2\cdot 2$)
• we call this: prime factorization of $n$
  • and this is unique for each number

• base case: $2=2$
• consider: $\forall m\in [2,n),\exists$ prime factorization
• case 1: if $n$ prime: $n=n$
• case 2: if $n$ composite: $\exists 1<a<n,a|n$
  • let $n=a\cdot b$
  • $a<n\wedge a<n$, thus prime factorization for $a,b$ exists
    • let $a=p_1{p}_2\dots p_k$
    • and $b=q_1{q}_2\dots q_t$
    • then $c=p_1{p}_2\dots p_k{q}_1{q}_2\dots q_t$ (sorted)

• suppose $n=p_1\cdot p_2\cdot p_k=q_1\cdot q_2\cdot q_s$
• then $\exists i,q_i=p_1$
  • e.g. $p_1$: can divide all sides
• from corollary, $\exists i,p_1=q_i$
  • eliminate $p_1,q_i$ from both sids
• that way, you can continue to eliminate all $k$ prime factors.
  • 👨‍🎓 in the end: they must be equal, and you realize that both sides were same from the first place

• e.g. $\gcd (a,b)\cdot \text{lcm}(a,b)={\prod }_{i=1}^k{p}_i^{\min ({\alpha }_i,{\beta }_i)+\max ({\alpha }_i,{\beta }_i)}={\prod }_{i=1}^k{p}_i^{{\alpha }_i+{\beta }_i}=a\cdot b$

• let $\sqrt{2}=\frac{a}{b}$ and $a\perp b$, then $2=\frac{a^2}{b^2}$
• $2b^2=a^2$
  • as $a,b\in \mathbb{Z}$, $2|a\wedge 2|b$
  • thus $\gcd (a,b)\ge 2>1$: thus contradiction

• suppose: $p_1,p_2,\dots ,p_n$: all possible primes
• then $q=p_1\cdot p_2\dots p_n+1$
  • $p_i$
• f

• $p_n\le 2^{2^{n-1}}$
• 👨‍🏫 prove it using induction, and show it holds for all previous primes
• as we know upper-bound of next prime