2015/041) Show that if square root of an integer is not integer then it is irrational

Let the number be r and square root be rational and not ain integer let it be $\dfrac{p}{q}$ where $gcd(p,q) = 1$ that is it is in lowest form
$\sqrt{r}= \dfrac{p}{q}$
so
$p = q\sqrt{r}\cdot(1)$
and
$p\sqrt{r}= qr \cdot(2)$
as $gcd(p,q) =1$ there exists m and n as per Bezout Identity such that
$mp + nq = 1$
so $\sqrt{r} = \sqrt{r}.1= \sqrt{r}(mp+nq) =mp\sqrt{r} + nq\sqrt{r} = mqr + np$
which is integer.
this leads to contradiction and hence it is irrational.