Insert an irrational number between $\sqrt{5}$ and $\sqrt{7}$.

consider the squares of $\sqrt{5}$ and $\sqrt{7}$

${(\sqrt5)^2}$ = 5 and ${(\sqrt7)^2}$ = 7

As, $5 \lt 6\lt 7$ , it follows that

$\sqrt{5} \lt \sqrt{6} \lt \sqrt{7}$ therefore , $\sqrt{6}$ lie between $\sqrt{5}$ and $\sqrt{7}$

Hence, irrational number between $\sqrt{5}$ and $\sqrt{7}$ is $\bold{\sqrt{6}}$ .