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

(N/A) In this method,we assume that the given statement is false. That is,we assume that $\sqrt{7}$ is rational.
This means that there exist positive integers $a$ and $b$ such that $\sqrt{7} = \frac{a}{b}$,where $a$ and $b$ have no common factors.
Squaring the equation,we get $7 = \frac{a^2}{b^2} \Rightarrow a^2 = 7b^2$.
This implies that $7$ divides $a^2$,and since $7$ is prime,$7$ must divide $a$.
Therefore,there exists an integer $c$ such that $a = 7c$.
Substituting this into $a^2 = 7b^2$,we get $(7c)^2 = 7b^2$ $\Rightarrow 49c^2 = 7b^2$ $\Rightarrow b^2 = 7c^2$.
This implies that $7$ divides $b^2$,and consequently,$7$ must divide $b$.
Thus,$7$ is a common factor of both $a$ and $b$,which contradicts our earlier assumption that $a$ and $b$ have no common factors.
This shows that the assumption that $\sqrt{7}$ is rational is false.
Hence,the statement $p: \sqrt{7}$ is irrational is true.