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From: Elementary Number Theory: Primes, Congruences, and Secrets by William Stein

${Lemma}$ $1.1.9.$ For any integers a and b, we have gcd(a, b) = gcd(b, a) = gcd(±a, ±b) = gcd(a, b − a) = gcd(a, b + a).

$Proof.$ We only prove that $gcd(a, b) = gcd(a, b − a)$, since the other cases are proved in a similar way. Suppose $d | a$ and $d | b$, so there exist integers $c_1$ and $c_2$ such that $dc_1 = a$ and $dc_2 = b$. Then $b−a = d(c_2−c_1)$, so $d | b − a$. Thus $gcd(a, b) ≤ gcd(a, b − a)$, since the set over which we are taking the max for $gcd(a, b)$ is a subset of the set for $gcd(a, b − a)$. The same argument with a replaced by $−a$ and $b$ replaced by $b − a$, shows that $gcd(a, b − a) = gcd(−a, b − a) ≤ gcd(−a, b) = gcd(a, b),$ which proves that $gcd(a, b) = gcd(a, b − a)$.

I don't understand how they came to this conclusion:

Thus $gcd(a, b) ≤ gcd(a, b − a)$, since the set over which we are taking the max for $gcd(a, b)$ is a subset of the set for $gcd(a,b-a)$.

Vidyanshu Mishra
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horizon
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  • Since $\gcd(a,b)$ divides both $a$ and $b-a$ it is clear that $\gcd(a,b)≤\gcd(a, b-a)$ (any common divisor is $≤$ the gcd) – lulu Sep 30 '19 at 18:29

1 Answers1

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It says that if any number $d$ divides $a,b$ (and hence $(a,b)$ then it also divides $(a,b-a)$. So all the divisors of $(a,b)$ ultimately lie in the set of divisors of $(a,b-a)$ and hence (set of all divisors of $(a,b)$ is a subset of $(a,b-a)$)

Now notice that $(a,b)$ and $(a,b-a)$ are highest numbers in the respective sets. Thus $(a,b)\le (a,b-a) $

Here $(a,b)=GCD(a,b)$

Vidyanshu Mishra
  • 10,143