How to show if $\sqrt{n} $ is rational number then $n$ is a perfect square? How to show if $\sqrt{n} $ is rational number then $n$ is a perfect square?
I got this far let $\sqrt{n}=\frac{b}{a}$
then $\sqrt{n}\sqrt{n}=\frac{b}{a}\sqrt{n}$
then
$n=\frac{b}{a}\sqrt{n}$
then 
$$n(a)=b\sqrt{n}$$
then we can say
$\sqrt{n}=\frac{n(a)}{b}$
But now I am a bit stuck.
 A: A perfect square is usually used to indicate squares of natural numbers; for example $2^2 = 4$ and so on.
With this definition the assertion is false, as $n = \frac{4}{9}$ results in $\sqrt n = \frac 23 \in \mathbb Q$
Unless you didn't mean $n \in \mathbb N$ in the first place. Then it is true, as the other answers show
A: An elementary proof (generalisation of a proof for $\sqrt 3$):
Suppose $x=\sqrt n$ is rational. Let $q$ be the smallest positive integer such that $qx$ is an integer.
; set $q'= q(x-m)$. Note it is a natural number since $qx$ is; furthermore
$$q'x =qx^2-mqx=qn-mqx$$
is  a natural number. 
However, since $m\le x< m+1$, we know $0\le x-m <1 $, so that
$$0<q'<q$$
Since $q$ is minimal among the positive integers such that $qx$ is an integer, this implies $q'=0$, in other words, $x=m$.
A: Assume that $n\in\mathbb{N}^+$ is not a square and $\sqrt{n}=\frac{a}{b}$ with $\gcd(a,b)=1$. There is some prime $p$ that divides $n$ with an odd multiplicity, i.e. $p^{2k+1}\| n$, but then
$$ a^2 = b^2 n $$
is impossible, since $p$ divides the LHS with an even multiplicity while $p$ divides the RHS with an odd multiplicity, leading to $a^2\neq b^2 n$. So the only chance for $\sqrt{n}$ to be an element of $\mathbb{Q}$ is that $n$ is a square, and that clearly happens is $n$ is a square.
A: Assuming $a,b$ are coprime and $n$ is a positive integer: since $n=a^2/b^2$, we have $b^2|a^2$.  But $a^2,b^2$ coprime implies $b=1$.
