Question

Prove that for any prime positive integer p, $\sqrt{p}$ is an irrational number.

Answer 1

Let $\sqrt{p}$ be a rational number.
Also a and b are rational.
then,$\sqrt{p}=\frac{a}{b}$
on squaring both sides,we get,
$(\sqrt{p}{)}^2=\frac{a^2}{b^2}$
→p = $\frac{a^2}{b^2}$
→b² = $\frac{a^2}{p}$ [p divides a² so,p divides a]
Let a= pr for some integer r
→b² = $\frac{\left(pr\right)²}{p}$
→b² = $\frac{p^2{r}^2}{p}$
→b² = pr²
→r² = $\frac{b²}{p}$ [p divides b² so, p divides b]
Thus p is a common factor of a and b.
But this is a contradiction, since a and b have no common factor. ( any rational number can be written in the form of $\frac{a}{b}$ where a and b are co prime ie they dont have any common factor other than 1.)
This contradiction arises by assuming √p a rational number.
Hence,√p is irrational number