<s>[INST] 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.
 [/INST] Directly:
If
$r = \sqrt{2}+\sqrt{3}$
is rational,
$\begin{array}\\
\dfrac1{r}
&=\dfrac1{\sqrt{2}+\sqrt{3}}\\
&=\dfrac1{\sqrt{2}+\sqrt{3}}\dfrac{-\sqrt{2}+\sqrt{3}}{-\sqrt{2}+\sqrt{3}}\\
&=-\sqrt{2}+\sqrt{3}\\
\end{array}
$
is also rational.
Adding and subtracting these,
$\sqrt{2}$ and $\sqrt{3}$
are rational.
</s>