Prove that $\sqrt{2}+\sqrt{3}$ is irrational.

Let us assume that $\sqrt{2}+\sqrt{3}$ is rational. Let $\sqrt{2}+\sqrt{3} = a$,where $a$ is a rational number.
Then,$\sqrt{2} = a - \sqrt{3}$.
Squaring both sides,we get:
$(\sqrt{2})^2 = (a - \sqrt{3})^2$
$2 = a^2 + 3 - 2a\sqrt{3}$
Rearranging the terms to isolate $\sqrt{3}$:
$2a\sqrt{3} = a^2 + 3 - 2$
$2a\sqrt{3} = a^2 + 1$
$\sqrt{3} = \frac{a^2 + 1}{2a}$
Since $a$ is a rational number,$\frac{a^2 + 1}{2a}$ must also be a rational number. This implies that $\sqrt{3}$ is rational.
However,this contradicts the known fact that $\sqrt{3}$ is an irrational number. This contradiction has arisen because of our initial assumption that $\sqrt{2}+\sqrt{3}$ is rational.
Therefore,we conclude that $\sqrt{2}+\sqrt{3}$ is irrational.