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From this, I know that $d=sa+tb$ for $s,t\in \mathbb{Z}$. Also, that $d|a$ and $d|b$. I don't know where to go from there.

I understand that if $a/d$ and $b/d$ were not relatively prime, then $d$ would not be the GCD.

Xam
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Mario
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  • Your statement that if a/d and b/d were not relatively prime, then d would not by the GCD is a proof by contradiction, which is a perfectly valid way of proof. – D.R. Sep 14 '17 at 01:55
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    "I understand that if $a/d$ and $b/d$ were not relatively prime, then $d$ would not be the GCD." Formalize that with a proof and you'll have proven your original statement by contrapositive. – JMoravitz Sep 14 '17 at 01:55
  • @JMoravitz My problem is that I don't know how to formalize that. – Mario Sep 14 '17 at 02:00
  • Suppose that $gcd(a/d, b/d)=x>1$. Then $dx$ divides both of $a$ and $b$ and $dx>d$. Does this cause a problem? (remember the importance of the word "Greatest" in "Greatest common divisor") – JMoravitz Sep 14 '17 at 02:03
  • I'm pretty sure this has been asked several times on this site. You should make a better searching. – Xam Sep 14 '17 at 02:03

1 Answers1

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If $gcd(a,b)=d$ then we know that $a=sd$ and $b=td$, for $s,t\in \mathbb{Z}$.

Since $d$ contains all the common factors of $a$ and $b$, $s$ and $t$ must be relatively prime.

Therefore:

$gcd\left(\frac{a}{d},\frac{b}{d}\right)=gcd\left(\frac{sd}{d},\frac{td}{d}\right)gcd=\left(s,t\right)=1$. Thus, $\frac{a}{d}$ and $\frac{b}{d}$ are coprime.