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Required To Prove: Let $a$ and $b$ be integers. Let $S$ = {$ma + nb$ : $m,n \in \mathbb{Z}$ and $ma + nb > 0$}, and $d$ be the least element. Show that $d$|$a$.

The following is how I attempted my proof:

Consider $d$ is not a divisor of $a$, i.e. by the Division Algorithm, $a = qd + r$ for $r \in \mathbb{Z}, 0 < r < d$. (Note that $0 < r$ since we are considering that $d$ is not a divisor, i.e. there must be a remainder of sorts).

Since $d \in S$, $d >= 1$.

It follows that $0 < r < 1$, since $d >= 1$.

However, $r \in \mathbb{Z}$, and since $0 < r < 1$, we have a contradiction. By definition of the remainder then, $0 <= r < 1$ i.e. $r = 0$.

Therefore, $d$ must be a divisor of $a$.

I am not entirely sure about the proof, however I have some doubts about the question as well. Consider the case where $a = 5$ and $b = 3$. In this case. Then the least element is 2 ($m = 1$, $n = -1$). In this case, 2|5 cannot be the case. I am not sure if the question is asking for some specific values of $a$ and $b$ in the set $\mathbb{Z}$, or if for all values of $a$ and $b$.

Regards, Xandru

Xandru Mifsud
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  • In your last paragraph, you say that $2$ is the minimal element, however note that $5 - 2 \cdot 2 = 1 < 2$ – Zestylemonzi Dec 11 '16 at 17:10
  • @Zestylemonzi oh good catch! Thanks for pointing that out. In that case, consider for example a = 11 and b = 7...the least element would be 4 in this case. – Xandru Mifsud Dec 11 '16 at 17:26
  • Again, $2 \cdot 11 - 3 \cdot 7 = 1 < 4$. In fact, the extended Euclidean algorithm shows that given $a,b \in \mathbb{N}$ then there exists $n,m \in \mathbb{Z}$ such that $an + bm = \text{gcd}(a,b)$. Hope this helps. – Zestylemonzi Dec 11 '16 at 17:29
  • See this answer for a conceptual viewpoint on such proofs. – Bill Dubuque Dec 11 '16 at 20:38

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You're not using that $d\in S$, so your proof is surely flawed. In particular the fact that $d\ge1$ does not allow you to argue that $0<r<1$; for instance, if $a=6$ and $b=9$, we have $d=3$, so you can only say $0<r<3$.

Since $d\in S$, you can write $d=ma+nb$ for some $m,n\in\mathbb{Z}$. Therefore $$ a=dq+r=mqa+nqb+r $$ and $$ r=(1-mq)+(-nq)b $$ Also $0\le r<d$. If $r>0$ we'd have…

Note that for $a=5$ and $b=2$, we have $d=1$, because $1=1\cdot 5+(-2)\cdot 2$. Indeed, $d$ is the greatest common divisor of $a$ and $b$, in general.

egreg
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  • However I did use that $d \in S$...in fact I wrote $d >= 1$, since if we let $d = ma + nb$ (as you have said), then $d > 0$, since by the definition of the set $S$, $ma + nb > 0$. – Xandru Mifsud Dec 11 '16 at 17:22
  • @XandruMifsud The fact that $d\ge1$ does not allow you to conclude that $0<r<1$. – egreg Dec 11 '16 at 17:25
  • How come? If $r < d$ and $d >= 1$, then $r < 1$, no? – Xandru Mifsud Dec 11 '16 at 17:29
  • Yes I was aware of that particular case, and many others like it. However, if the definition of the set says that the smallest possible divisor is 1, then for what reason is my reasoning flawed? Why is saying $r < 1$ wrong is $r < d$? – Xandru Mifsud Dec 11 '16 at 17:36
  • @XandruMifsud Suppose $d=42$; then $0<23<42$. You could conclude if you had $d\le1$, but unfortunately it is not the case. – egreg Dec 11 '16 at 17:37