Prove that the following number is irrational: $\sqrt{5}-\sqrt{3}$.

(N/A) Let us assume,to the contrary,that $\sqrt{5}-\sqrt{3}$ is a rational number.
Let $\sqrt{5}-\sqrt{3} = r$,where $r$ is a rational number.
Then,$\sqrt{5} = r + \sqrt{3}$.
Squaring both sides,we get: $(\sqrt{5})^2 = (r + \sqrt{3})^2$.
$5 = r^2 + 3 + 2r\sqrt{3}$.
$5 - 3 - r^2 = 2r\sqrt{3}$.
$2 - r^2 = 2r\sqrt{3}$.
$\sqrt{3} = \frac{2 - r^2}{2r}$.
Since $r$ is a rational number,$\frac{2 - r^2}{2r}$ must also be a rational number.
This implies that $\sqrt{3}$ is a rational number.
However,this contradicts the fact that $\sqrt{3}$ is an irrational number.
Therefore,our assumption that $\sqrt{5}-\sqrt{3}$ is rational is incorrect.
Hence,$\sqrt{5}-\sqrt{3}$ is an irrational number.