question about the proof about the square root of natural numbers Could someone please help me to prove that for $t \in \mathbb{N}$ , $\sqrt{t} \in \mathbb{Q} $ if only if  $\sqrt{t} \in \mathbb{N}$ 

The easiest way to do this is to take a rational, written in lowest terms, and square it. In particular,
$$\left(\frac{p}{q}\right)^2=\frac{p^2}{q^2}$$
where the fraction on the right is also in lowest terms, since $\gcd(p^2,q^2)=\gcd(p,q)^2=1$. However, any integer must have a denominator of $1$, thus its square root must have denominator $q$ satisfying $q^2=1$ and thus itself be an integer, implying that non-integer rationals square to non-integer rationals, so the only integers with rational square roots have integer square roots.
