Prove square root of 16 is rational I am student and i understand that the square root of any perfect square is a rational number but i'am trying to prove it (e.g for 16). 
 A: As the other commentors and answerer have noted, this is essentially by definition, but if you'd like something that looks perhaps more "mathematical":  suppose that $m = n^2$ where $n\in {\mathbb Z}$ is a perfect square and suppose that its square root is not rational.  Then there must exist $p \in {\mathbb Z}$ and $q\in {\mathbb N}$ such that 
$$ \left| \sqrt{m} - \frac{p}{q} \right| > 0$$
If that ever equalled zero then we'd have found a rational number that was the square root of $m$, and we're claiming we can't do that.
Since $\sqrt{m} = n$ because $m$ is a perfect square, we have
$$ \left| n - \frac{p}{q} \right| > 0$$
for all $p,q$.  In particular, this must hold when $q=1$, i.e.
$$ \left| n - \frac{p}{1} \right| > 0$$
But $n$ and $p$ both lie in $\mathbb Z$ so we can take $p=n$, which is a contradiction.  So our hypothesis that the square root of $m$ was not rational must have been false.
A: How can it not be?
By definition a perfect square is a square of an integer.
And by definition the square root of a number is the non-negative number which the number of is a square of.
And since a perfect square is the square of an integer, the number it is  square of is an integer.
And all integers are rational.
