Prove that $\sqrt{2}+\sqrt{3}$ is irrational. Problem: Prove that $\sqrt{2}+\sqrt{3}$ is irrational. 
The book where I encountered this problem had the following hint:

We make a polynomial with integer coefficients called $f(x)$ that $f(\sqrt{2}+\sqrt{3})=0$. (Why?)

Accepting this I solved the problem like this:
If $x=\sqrt{2}+\sqrt{3}$ then $x^2=5+2\sqrt{6}$ and so $(x^2-5)^2=24$ thus:
$$x^4-10x^2-1=0$$
But I want to know the reason that we should do this.

More generally,
suppose
$r
=\sqrt{a}+\sqrt{b}
$
is rational,
where
$a$ and $b$
are positive integers.
Then
$r(\sqrt{a}-\sqrt{b})
=a-b
$
so
$\sqrt{a}-\sqrt{b}
=\dfrac{a-b}{r}
$
is also rational.
Adding and subtracting these,
$\sqrt{a}$ and $\sqrt{b}$
are rational.
Therefore,
if either or both
of
$\sqrt{a}$ and $\sqrt{b}$
are irrational,
then
$\sqrt{a}+\sqrt{b}$
is irrational
(and similarly for
$\sqrt{a}-\sqrt{b}$).
This might form a basis
for a proof that
$\sum_{k=1}^n \sqrt{a_k}
$
is irrational
under suitable assumptions on the
$a_k$.
