Proof by counterexample 
If $n$ is prime, then $\sqrt{n}$ is irrational. Prove this statement.

If I were to prove this using proof by contradiction, I would do:
Suppose $n$ is prime and $\sqrt{n}$ is rational. Let $\sqrt{n}=\frac{a}{b}$ where $a,b\in \mathbb{R}$ and have no common factor other than 1.
And then I would go on and get a contradiction like $a$ is even and $b$ is even.
However, I was wondering if I could also use proof by counterexample:
($n$ is prime) $\Rightarrow$ ($\sqrt{n}$ is irrational)
The negation of this statement is:
($n$ is prime) $\land$ ($\sqrt{n}$ is rational)
A counter example to this is $n=2$.
Therefore, this statement is disproved and hence, its negation is proven to be true.
Is this also a valid proof?

The claim is in plain text
"For every prime $n$, $\sqrt{n}$ is irrational."
The negation is in plain text
"There is some prime $n$, such that $\sqrt{n}$ is rational."
This is not a statement about all primes, but only about some prime. Therefore we cannot disprove it by a single counterexample.
To see the flaw : Assume $\sqrt{5}$ would be rational. Then the statement would be false since $5$ is prime, but $\sqrt{5}$ would not be irrational. You see, that the case $n=2$ is not enough.
