How can I prove that $\sqrt{2} + \sqrt{3}$ is irrational using the fact that $\sqrt{2}$ is irrational?

Question

If I start with $\sqrt{2}+\sqrt{3}=\frac{a}{b}$, I can solve for $\sqrt{2}$ to get: $$\sqrt{2} = \frac{a-\sqrt{3}b}{b}$$

However, this doesn't feel like its getting to any contradictions.

Ideas?

Answer 1

Let $\sqrt3=p-\sqrt2$ where $p$ is rational

Squaring both sides and on rearrangement we get, $$\sqrt2=\dfrac{p^2+2-3}{2p}$$ which is rational unlike $\sqrt2$

This is called Proof by contradiction

Answer 2

As $\sqrt{2}^2-\sqrt{3}^2=-1=(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})$

if $\sqrt{2}-\sqrt{3} \in\mathbb{Q}$ then $(\sqrt{2}+\sqrt{3})\in\mathbb{Q}$

$\dfrac{\sqrt{2}-\sqrt{3}}{2}+\dfrac{\sqrt{2}+\sqrt{3}}{2}=\sqrt{2}\in\mathbb{Q}\quad $ contradiction