How to check this number $\sqrt{47}$ is irrational Prove that $\sqrt{47}$ is irrational number. 
I know that a rational number is written as $\frac{p}{q}$ where $p$ & $q$ are co-prime numbers. But I do not have any idea to prove it irrational number. 
Thank you very much!

$47$ is a prime, hence you may just mimic the proof of the irrationality of $\sqrt{2}$. Let: 
$$\nu_{47}(m) = \max\{n\in\mathbb{N}: 47^n\mid m\}.$$
Assuming $\sqrt{47}=\frac{p}{q}$ with $\gcd(p,q)=1$, it follows that:
$$ p^2 = 47\, q^2 $$
but $\nu_{47}$ of the LHS is even while $\nu_{47}$ of the RHS is odd, contradiction.
