Prove that $\sqrt{3}+\sqrt{5}$ is irrational.

(N/A) Let us assume that $\sqrt{3}+\sqrt{5}$ is a rational number.
Let $\sqrt{3}+\sqrt{5} = a$,where $a$ is a rational number.
Then,$\sqrt{3} = a - \sqrt{5}$.
Squaring both sides,we get:
$(\sqrt{3})^2 = (a - \sqrt{5})^2$
$3 = a^2 + 5 - 2a\sqrt{5}$ (using the identity $(x - y)^2 = x^2 + y^2 - 2xy$).
Rearranging the terms to isolate $\sqrt{5}$:
$2a\sqrt{5} = a^2 + 5 - 3$
$2a\sqrt{5} = a^2 + 2$
$\sqrt{5} = \frac{a^2 + 2}{2a}$.
Since $a$ is a rational number,$\frac{a^2 + 2}{2a}$ must also be a rational number.
This implies that $\sqrt{5}$ is a rational number,which contradicts the fact that $\sqrt{5}$ is an irrational number.
Therefore,our initial assumption is incorrect,and $\sqrt{3}+\sqrt{5}$ is an irrational number.