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Let $a, b \in \mathbb{Z} \setminus \{0\}$ and $d = \gcd(a, b)$. Show that $gcd(\frac{a}{d}, \frac{b}{d}) = 1$.

I tried proving this by contradiction and showing that otherwise $d$ isn't the gcd of $a$ and $b$, but it didn't work. Could someone please give me a hint on what the proof should look like?

Martin Sleziak
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Angie
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2 Answers2

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If $d'>1$ divides both $\frac{a}{d}$ and $\frac{b}{d}$ then $dd'> d$ divides $a$ and $b$, contradicting the fact that $d$ is the greatest common divisor of $a$ and $b$.

fretty
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Recall Bezout's identity and the definition of greatest common divisior that comes along with it.

If $(a,b)=d$ then there are integers $u$ and $v$ such that $$ au + bv = d$$

What happens when you divide $a$ and $b$ by $d$?

Alternatively, if you prefer, just use the fundamental theorem of arithmetic and see with your own eyes they have no common divisor.

proofromthebook
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