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$$\newcommand{\gcd}{\text{gcd}}$$

Prove: if $d=\gcd(m,n)$ so $\gcd\left(\frac{m}{d},\frac{n}{d}\right)=1$

Intuitively it is obvious, but I am having a hard time to formalize the proof, what I have came to this:

$d=\gcd(m,n)$ so $d|m$ and $d|n$ therefore $m=dx$ and $n=dy$ now if $\gcd\left(\frac{m}{d},\frac{n}{d}\right)\neq 1$ that mean that $m$ and $n$ have a common factor, after the division in $d$ which is the greatest common divisor is contradiction.

Martin Sleziak
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gbox
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3 Answers3

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If $d=\gcd(m,n)$, then $\exists \, x,y \in \mathbb{Z}$ such that $d=mx+ny$. From this we can also write $$1=\frac{m}{d}x+\frac{n}{d}y.$$ Thus $\gcd(m/d,n/d)=1$.

Anurag A
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Your proof seems fine although you could show the contradiction more evidently.

We want to prove that if $d = \gcd(m,n)$ then $\gcd(m/d, n/d) = 1$.

Assume for the sake of contradiction that this is not the case i.e. $d = \gcd(m,n)$ but $\gcd(m/d, n/d) = k$ where $k > 1$. Then we have:

$$\frac{m}{d} = kx$$

and

$$\frac{n}{d} = ky$$

for some $x$ and $y$, so $m = kdx$ and $n = kdy$ giving:

$$ \gcd(m, n) \geq kd > d $$

which is a contradiction since we assumed $\gcd(m,n) = d$.

gowrath
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  • I think that I had to show $p \rightarrow q$ and I proved $\neg q\rightarrow \neg p$ – gbox Dec 19 '16 at 19:47
  • @gbox Those are logically equivalent statements to prove :) It is the case that $p \rightarrow q \iff \lnot q \rightarrow \lnot p$. – gowrath Dec 19 '16 at 19:50
  • @gbox And in actuality you proved by contradiction which is more like saying if you want to prove $p \rightarrow q$, you assume $\lnot(p \rightarrow q)$ which is logically equivalent to $p \land \lnot q$ and derive a contradiction so you must have $\lnot\lnot(p \rightarrow q)$ which is $p \rightarrow q$. But now we're getting into the arcane world of logic – gowrath Dec 19 '16 at 19:53
  • So I showed $\neg q \rightarrow \neg p$, so that is ok? – gbox Dec 19 '16 at 19:57
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    @gbox Indeed (y) – gowrath Dec 19 '16 at 19:57
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Suppose that $gcd(\frac{m}{d},\frac{n}{d})=d'>1.$ Then $d' \mid m$ and $ d' \mid n$. It implies that $d'd \mid m$ and $ d'd \mid n$ and we find a common divisor $d d'$ which is greater then $d.$ Contradiction.

Leox
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