Prove that if $gcd(a, b) = d$, then $gcd(\frac{a}{d}, \frac{b}{d}) = 1$.

Corollary: If $gcd(a, b) = d$, then $gcd(\frac{a}{d}, \frac{b}{d}) = 1$.

Proof:  Since $d = gcd(a,b)$, we have  $d ~|~ a$ and $d ~|~ b$. Hence $\frac{a}{d},~ \frac{b}{d}$ are integers. Since $d = gcd(a,b)$, Theorem, there exist integers $x$ and $y$ such that $d = gcd(a, b) = a \cdot x+ b \cdot y$. By dividing both side of this equation by $d$, we get $1 = \frac{a}{d} \cdot x + \frac{b}{d} \cdot y$. By Theorem, $gcd(\frac{a}{d}, \frac{b}{d}) = 1$.

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