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Show that if $d$ is the smallest element in the set $S = \{s \in \mathbb{N} | \exists m,n \in \mathbb{Z}, s = ma+ nb \}$ such that $d = ax + by$ then $\gcd(a,b) |d $ and hence $\gcd(a, b) \leq d$

Martin Sleziak
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user3182418
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    Should this also be labeled "homework"? – marty cohen Mar 24 '14 at 04:17
  • yeah sure, it is homework, I am not gonna lie, but they're just practice questions @martycohen – user3182418 Mar 24 '14 at 04:22
  • Think about how you write $a$ and $b$ in terms of $\gcd(a,b)$. The rest should follow from there. – ruler501 Mar 24 '14 at 04:33
  • well $a = q_{1}b + r_{1}$ and $b = q_{2}r_{1} + r_{2}$, how does that show that gcd(a,b)|d ? @ruler501 – user3182418 Mar 24 '14 at 04:37
  • $gcd(a,b)|a \implies a=q_1gcd(a,b)$ similarly for $b$. How can you rewrite the equation $ma+nb$ with that information? – ruler501 Mar 24 '14 at 04:40
  • well sure, could it be $ma + nb$ = $mgcd(a,b) + ngcd(a,b)$ but how does that mean that $gcd(a,b)|d$ ? is beacuse $d$ is of that form and therefore, they must divide eachother? @ruler501 – user3182418 Mar 24 '14 at 04:47
  • I just wrote it up as an answer since it makes more since that way now. – ruler501 Mar 24 '14 at 05:12
  • Maybe some answer to this question might help you: http://math.stackexchange.com/questions/321061/proving-that-gcda-b-as-bt-i-e-gcd-is-a-linear-combination (And it might also be useful to have a look at links which you can find there.) – Martin Sleziak Mar 24 '14 at 13:55

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You can write $a$ as $q_1\gcd(a,b)$ and $b$ as $q_2\gcd(a,b)$ therefore $ma+mb=(mq_1+nq_2)\gcd(a,b)$ and it is obvious that for every $m,n$ it will be divisible by $\gcd(a,b)$. I'll leave it to you to prove that $\gcd(a,b)$ is the lowest value greater than zero it can take.

ruler501
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