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.
 A: An easy way to see this is to notice that
 $$(\sqrt 2 + \sqrt 3)^2 = 5+2\sqrt 6.$$
Now, if $\sqrt 2 + \sqrt 3$ is rational, then so is its square. This implies that $(\sqrt 2 + \sqrt 3)^2 - 5 = 2\sqrt6$ is rational - a contradiction. Hope this helps.
A: The rational root theorem, states that if a rational number solves a given polynomial equation with integer coefficients, then it must be possible to write that number as $\frac pq$ where $p$ is an integer that divides the constant term, and $q$ is an integer that divides the highest-degree coefficient.
In this case we only need to check $\pm 1$, which clearly does not solve the equation. Thus there are no rational numbers that solves $x^4 - 10x^2 - 1 = 0$.
A: 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.
A: 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
$a_k$.
