If $2$ numbers are co-prime, does it imply that their difference is also prime to those numbers?

Let $q$ and $p$ be coprime. And without loss of generality, as $p$ and $q$ are interchangeable, let $p>q$, $p=q+d$.

If $p$ and $q$ are coprime, the fraction cannot be simplified. Therefore, we can rewrite $p/q$ as $(q+d)/q$, and we obtain $1+d/q$. As the fraction for $p/q$ cannot be simplified, $d/q$ can also not be simplified, therefore $d$ is also prime to $q$. [We can do same arguement for $p$ too]. Is this correct?

• Yes but perhaps you should explain why: If $d/q$ $could$ be simplified to $d'/q'$ with $|q'|<|q|$, then $p/q=(q'+d')/q'$ is a simplification of $p/q$.... In general if $p,q$ are co-prime then $p, \pm (pn+q)$ are co-prime for any $n\in \Bbb Z.$ – DanielWainfleet Aug 6 '18 at 20:27

Yes, if $d$ and $q$ had a common factor, it would also be a factor of $d+q=p$. This is the heart of the Euclidean algorithm for greatest common divisor.

• Alright, thanks :) – Dean Yang Aug 6 '18 at 18:20

Of course it is. If $q$ and $d$ have a common factor $f$:

$q = Qf$
$d = Df$

then

$p = q + d = (Q +D)f$

and more generally

$kq\pm md = (kQ \pm mD)f$

will have that factor, too.