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I'm trying to show an alternative proof to Bezout's Lemma (let $a, b \in \mathbb{Z}$, then there exists $x,y \in \mathbb{Z}$ such that $gcd(a, b) = ax + by$). Heres one of the steps in proving it:

a) Show the case where $a=b=0$. Then assume $a$ and $b$ are not both zero. Consider the set $S = $ {$s \in \mathbb{N} | \exists m,n \in \mathbb{Z}, s = ma+ nb $}. Prove that $s \not= \emptyset $ by showing that at least one of $|a|$ or $|b|$ is an element of $S$.

My work:

I'm assuming for $a=b=0$ we can just say that since $gcd(0,0)$ = $0$ and since $0x + 0y = 0$ then $gcd(0,0) = ax + by$ for $a=b=0$. And for the second part can you simply say if $a \not= 0$ and $b \not= 0$ and since $s = ma + nb$ then $|a|$ or $|b|$ must be in $S$?.. Thanks for your help!

Alex Chavez
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  • $|a|=(\pm1)\cdot a+0\cdot b$, where the sign is chosen depending on $a\geq0$ or $a<0$. – OR. Mar 24 '14 at 01:49
  • would it be the same for |b| ? @ABC – Alex Chavez Mar 24 '14 at 01:53
  • More generally if nonempty $,S\subset \Bbb Z,$ is closed under subtraction then it has a nonnegative element. Being nonempty there exists $,a\in S,$ so either $,a\ge 0,$ or $, (a-a)-a, =, -a \ge 0,,$ i.e. closed under subtraction $\Rightarrow$ closed under negation. – Bill Dubuque Mar 24 '14 at 02:13
  • so what does that mean @BillDubuque ? Why does being closed under subtraction mean anything? – Alex Chavez Mar 24 '14 at 03:12
  • @Alex Your set $,S,$ is closed under subtraction, and this is the essence of the matter in Bezout's Lemma (as one learns in abstract algebra when one studies ideals). See this answer. – Bill Dubuque Mar 24 '14 at 03:23
  • Thanks @BillDubuque , so does that mean what you wrote earlier was the proof for this, cause as you can see I was no where close to that – Alex Chavez Mar 24 '14 at 03:24
  • @Alex That's one way. Another is to use $,(a,b)= (a,b-a) = ax+(b-a)y = a(x-y)+by\ $ as the induction step. – Bill Dubuque Mar 24 '14 at 03:28
  • @BillDubuque couldn't we just pick m = 2 and m = 3, and this would show that it is non-empty, wouldn't finding a counter example suffice? – user3182418 Mar 24 '14 at 08:58
  • @user3182418 S is nonempty if there exists integers $,m,n,$ such that $,ma+nb \in\Bbb N.$ – Bill Dubuque Mar 24 '14 at 12:44

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