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!

If $\dfrac p q =\sqrt{47}$ then $p^2=q^2\cdot 47$.  That means the number of $47$s in the prime factorization of $p^2$ is one more than the number of $47$s in the prime factorization of $q^2$.  But that cannot happen because $p^2$ and $q^2$, both being squares, must both have an even number of $47$s in their factorizations.
(This relies on the theorem that tells us that an integer cannot have more than one prime factorization.)
