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  $\sqrt{t} \in \mathbb{N}$ 
 A: There is a classic proof on the irrationality of $\sqrt{2}$.  You should be able to modify it to achieve this result.  Give it a shot and maybe you can answer your own question.
A: 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.
A: Imagine we have $\frac{m^2}{n^2}=x,\quad x,m,n\in\mathbb{N},n\neq 1, GCD(m,n)=1$.
Now we know there is a prime $p$ so that it appears in the factorization of $x$ with odd exponent $a$ - else $x$ would be a perfect square. Let us say $x=p^\alpha r$.
Now we know that $m^2=xn=p^{a}rn^2$.
From this we get $p|m$ (why?).
Since $GCD(m,n)=1$, we also know that $p$ is not a prime factor of $n$.
Now take a look at the prime factorization of the left and the right side.
What is the power of $p$ in the left side? Is it even? What about the right side?
A: Since only things related to multiplication and not addition are involved in this problem, looking at the prime factorization of things should prove useful.
(we can extend prime factorization to rational numbers: e.g. $4/6$ has prime factorization $2 \cdot 3^{-1}$)
