pouzdrotf

2022-07-06

Prove that $\sqrt{47}$ is irrational number.

lofoptiformfp

Expert

If $\frac{p}{q}=\sqrt{47}$ 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.