$\begingroup$

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).

$\endgroup$ 6

2 Answers

$\begingroup$

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.

$\endgroup$ $\begingroup$

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.

$\endgroup$