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Prove that if $d$ is a common divisor of $a$ & $b$, then $d=\gcd(a,b)$ if and only if $\gcd(\frac{a}{d},\frac{b}{d})=1$

I know I already posted this question, but I want to know if my proof is valid: So for my preliminary definition work I have: $\frac{a}{d}=k, a=\frac{dk b}{d}=l,b=ld $

so then I wrote a linear combination of the $\gcd(a,b)$, $$ax+by=d$$ and substituted:

$$dk(x)+dl(y)=d d(kx+ly)=d kx+ly=1 a/d(x)+b/d(y)=1$$

Is this proof correct? If not, where did I go wrong? Thanks!

Mr Mathster
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Lil
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  • @NasuSama I said in my question it was a duplicate. I just want to know if my solution is correct. I'm not interested in alternative solutions at the moment – Lil Mar 02 '14 at 20:38
  • @Lil The recommended way to do that on MSE is to post an answer to your prior question and ask if it is correct, or edit the proof into your prior question. Please don't post duplicates - it complicates site organization, among other detriments. – Bill Dubuque Mar 02 '14 at 21:13
  • ok no problem; sorry I'm new to the site so I wasn't aware. – Lil Mar 02 '14 at 21:14

2 Answers2

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It seems fine however this proof can be reduced massively. To prove '$\Leftarrow$',we know $gcd(\frac{a}{d},\frac{b}{d})=1$ therefore we can write this as a linear combination. $$\frac{a}{d}x+\frac{b}{d}y=1$$ Now multiply through by $d$: $$ax+by=d$$.Therefore $gcd(a,b)=d$ To prove'$\Rightarrow$' is mostly the same logic, try reducing yours.

George1811
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  • wait for if and only if statements you have to prove both sides? – Lil Mar 02 '14 at 21:04
  • and for the forward implication I can write ax+by=d and then divide every term by d and have a/d(x)+b/d(y)=1 as a proof correct? – Lil Mar 02 '14 at 21:06
  • Yes for if and only if statement $\Rightarrow$ and $\Leftarrow$ must be shown. However the forward implication only says $d$ is a common divisor, not the greatest common divisor. So some more thought is needed. – George1811 Mar 02 '14 at 22:46
  • so if you prove d to be the greatest common divisor it is not implied that d is a common divisor? – Lil Mar 02 '14 at 23:43
  • No sorry my bad, I read the question wrong the first time round. Yeah you can just reverse the proof for the other implication. And no if d is the gcd then d is definitely a common divisor. – George1811 Mar 02 '14 at 23:55
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Let us look at the logic of your proof.

It uses the fact that if $d$ is the greatest common divisor of $a$ and $b$, then there exist integers $x$ and $y$ such that $ax+by=d$. Then you conclude correctly that $\frac{a}{d}x+\frac{b}{d}y=1$, and therefore $\frac{a}{d}$ and $\frac{b}{d}$ are relatively prime.

But you are asked to prove an "if and only if" statement, and you have not shown that if $d$ is a common divisor of $a$ and $b$, and $\frac{a}{d}$ and $\frac{b}{d}$ are relatively prime, then $d$ is the gcd of $a$ and $b$. You can prove this result in various ways. One way close to how you handled the first part is to use the fact that there exist integers $s$ and $t$ such that $\frac{a}{d}\cdot s+\frac{b][d}\cdot t=1$.

André Nicolas
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