Verify by the method of contradiction. $p:\sqrt{7}$is irrational.

Solution: Let us assume $p:\sqrt{7}$ is a rational.
$\Rightarrow \sqrt{7}=\frac{a}{b}$
It means, ,$a$ and $b$ have no common factor.
Also, $\frac{a^2}{b^2}=7\Rightarrow a^2=7b^2\to \left(1\right)$
Above equation means, $a^2$ has a factor of $7$.
So, $a$ will also have a factor of $7$.We can also write it as,
$a=7c\to \left(2\right)$ where $c$ is another real number.
From (1) and (2),
$49c^2=7b^2\Rightarrow b^2=7c^2$
Above means, that $b$ is also having a factor of $7$.
But, earlier we proved $a$ and $b$ have no common factor.
So, our assumption is incorrect that $\sqrt{7}$ is a rational number.
It means, $\sqrt{7}$ is an irrational number.