For which natural numbers $n$ is $\sqrt n$ irrational? How would you prove your answer? Can someone help me out here? I'm trying to prove this the same way as whether if $\sqrt 2$ is irrational, but i'm not sure what am i doing. 
Can someone show me how to prove this?
 A: In fact, every integer which is not a square has an irrational square root. Can you see how to generalize the result for $\sqrt{2}$ to prove this? 
A: Once again,
here is a proof
that if $n$ is a positive integer that is
not a square of an integer,
then $\sqrt{n}$ is irrational.
Let $k$ be such that
$k^2 < n < (k+1)^2$.
Suppose $\sqrt{n}$ is rational.
Then there is a smallest positive integer $q$ such that
$\sqrt{n} = p/q$.
Then $\sqrt{n} = \sqrt{n}\frac{\sqrt{n}-k}{\sqrt{n}-k}
= \frac{n-k\sqrt{n}}{\sqrt{n}-k}
= \frac{n-kp/q}{p/q-k}
= \frac{nq-kp}{p-kq}
$.
Since $k < \sqrt{n} < k+1$,
$k < p/q < k+1$,
or $kq < p < (k+1)q$,
so $0 < p-kq < q$.
We have thus found a representation of
$\sqrt{n}$ with a smaller denominator,
which contradicts the specification of $q$.
Note: This is certainly not original - 
but I had fun working it out
based on the proof I know
that $\sqrt{2}$ is irrational.
Note 2: It is interesting that this does not use
any divisibility properties.
A: 
For $\forall n \in \mathbb{N}$, either $ \sqrt{n}  \in \mathbb{N}$ or
  $\sqrt{n} \in \mathbb{R} \setminus \mathbb{Q}$.

There are two possible cases:


*

*If $n = p^2$, $p \in \mathbb{N} \Rightarrow \sqrt{n}  = p ∈ N$

*If $n \ne p^2$ or there is no such $p \in \mathbb{N}$ that would satisfy $n = p^2$, then $\sqrt{n}$  is irrational. Let's suppose the contrary, i.e. there $\exists p,q \in \mathbb{N}$ and $\gcd(p,q) = 1$ such that $$\sqrt{n}=\frac{p}{q} \Leftrightarrow n\cdot q^2 = p^2$$
From $\gcd(p,q) = 1 \Rightarrow \gcd(p^2,q^2) = 1$. Using Bézout’s theorem $\exists z,t \in \mathbb{Z}$ such that 
$$z·p^2 + t·q^2 = 1 \Leftrightarrow z·n·q^2 + t·q^2 = 1 \Leftrightarrow
q^2·(z·n + t) = 1$$
which means $q^2$ divides $1$, but this is possible only if $q = 1$. As a result $n = p^2$ - contradiction with the initial assumption.
This proves the statement (also available here).
A: Take $\sqrt a=\frac{p}{q}$ with gcd(p,q)=1 and a is prime square both sides...
$$
a=\frac{p^2}{q^2}
$$
then
$$
2q^2=p^2
$$
then 
$$
a|p^2
$$
a contradiction
then think about what would happen if a was not prime
