Lecture 3

GCD

Example: GCD(15, 20) = ?

Positive divisors of 15: $15, 1, 5, 3$
Positive divisors of 20: $20, 1, 10, 2, 5, 4$
Common divisors: $1, 5$
GCD(15,20): $1 \times 5 = 5$

Manually writing out the divisors and finding the common divisors is quite slow.

Euclid’s Algorithm

Put $a_{0} = a \land b_{0} = b$
$b_{0} = q_{0} a_{0} + r_{0}$ where $0 \leq r_{0} < a_{0}$
Check $r_{0}$ value
- If $r_{0} = 0$ then $GC D (a, b) = a_{0}$
- Otherwise $0 < r_{0} < a_{0}$ ; put $a_{1} = r_{0}$ and $b_{1} = a_{0}$ and loop

To me: Remember to fix this algorithm description to be generic to i

Example: GCD(2024, 70) = ?

$2024 = 28 \times 70 + 64$ ( $b_{0} = 2024, a_{0} = 70$ )
$70 = 1 \times 64 + 6$ ( $b_{1} = 70, a_{1} = 64$ )
$64 = 10 \times 6 + 4$ ( $b_{2} = 64$ , $a_{2} = 6$ )
$6 = 1 \times 4 + 2$ ( $b_{3} = 6$ , $a_{3} = 4$ )
$4 = 2 \times 2 + 0$ ( $b_{4} = 4$ , $a_{4} = 2$ )

Algorithm terminates since $r_{4} = 0$ .

Proof of Correctness

Firstly, $a_{i} \in Z_{\geq 0}$ is strictly decreasing so the process eventually terminates.

Extended Euclidean Algorithm

Run Euclid’s Algorithm backwards, and we get $GC D (a, b)$ as a linear combination of $a, b$ . This is called the Extended Euclidean Algorithm.

2024 = 28x70 + 64
70 = 64 + 6
64 = 10x6 + 4
6 = 1x4 + 2
4 = 2x2 + 0

2 = 6 - 4
2 = 6 - (64 - 10x6) = 11x6 - 64
2 = 11(70-64) - 64 = 11x70 - 12x64
2 = 11x70 - 12(2024 - 28x70)
2 = 347x70 - 12x2024

Bezout’s Lemma

(Useful for solving diophantine eqs)

$f$

Proof is by induction

Corollary

If $d ∣ a$ and $d ∣ b$ then $d ∣ (a, b)$ . This is because we can write $(a, b)$ as $x a + y b$ , and $d ∣ (x a + y b)$ .

Euclid’s Lemma

Let $m, n \in Z$ , and let $p$ be a prime divisor of $mn$ . Then $p ∣ m$ or $p ∣ n$ .

Ed's Notes

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