TITLE: Prove that $\sqrt 2 +\sqrt 3$ is irrational.
QUESTION [22 upvotes]: Please prove that $\sqrt 2 + \sqrt 3$ is irrational.
One of the proofs I've seen goes:

If $\sqrt 2 +\sqrt 3$ is rational, then consider $(\sqrt 3 +\sqrt 2)(\sqrt 3 -\sqrt 2)=1$, which implies that $\sqrt 3 − \sqrt 2$ is rational. Hence, $\sqrt 3$ would be rational. It is impossible. So $\sqrt 2 +\sqrt 3$ is irrational.

Now how do we know that if $\sqrt 3 -\sqrt 2$ is rational, then $\sqrt 3$ should be rational?
Thank you.

REPLY [7 votes]: This is a solution in the style of "mathematics made difficult".
I'm assuming you know that $\sqrt{2}$ is irrational and of degree $2$ over $\mathbb{Q}$. It is easy to show that $\mathbb{Q}(\sqrt{2},\sqrt{3})= \mathbb{Q}(\sqrt{3}+\sqrt{2})$. If $\sqrt 3 + \sqrt 2$ is rational, then we would have $[\mathbb Q(\sqrt 3 + \sqrt 2): \mathbb Q]=1$, but $$1 = [\mathbb Q(\sqrt 3 + \sqrt 2): \mathbb Q]=[\mathbb Q (\sqrt 3, \sqrt 2):\mathbb Q(\sqrt 2)][\mathbb Q(\sqrt 2): \mathbb Q] \geq 2,$$
a contradiction.