Prove that the following number is irrational: $\sqrt{7}-\sqrt{2}$.

(N/A) Let us assume,to the contrary,that $\sqrt{7}-\sqrt{2}$ is a rational number.
Let $\sqrt{7}-\sqrt{2} = r$,where $r$ is a non-zero rational number.
Then,$\sqrt{7} = r + \sqrt{2}$.
Squaring both sides,we get: $(\sqrt{7})^2 = (r + \sqrt{2})^2$.
$7 = r^2 + 2 + 2r\sqrt{2}$.
$7 - r^2 - 2 = 2r\sqrt{2}$.
$5 - r^2 = 2r\sqrt{2}$.
$\sqrt{2} = \frac{5 - r^2}{2r}$.
Since $r$ is a rational number,$\frac{5 - r^2}{2r}$ must also be a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the fact that $\sqrt{2}$ is an irrational number.
Therefore,our assumption is incorrect,and $\sqrt{7}-\sqrt{2}$ is an irrational number.