Prove that $\sqrt 2 + \sqrt 3$ is irrational I have proved in earlier exercises of this book that $\sqrt 2$ and $\sqrt 3$ are irrational. Then, the sum of two irrational numbers is an irrational number. Thus, $\sqrt 2 + \sqrt 3$ is irrational. My first question is, is this reasoning correct?
Secondly, the book wants me to use the fact that if $n$ is an integer that is not a perfect square, then $\sqrt n$ is irrational. This means that $\sqrt 6$ is irrational. How are we to use this fact? Can we reason as follows:
$\sqrt 6$ is irrational
$\Rightarrow \sqrt{2 \cdot 3}$ is irrational.
$\Rightarrow \sqrt 2 \cdot \sqrt 3$ is irrational
$\Rightarrow \sqrt 2$ or $\sqrt 3$ or both are irrational.
$\Rightarrow \sqrt 2 + \sqrt 3$ is irrational.
Is this way of reasoning correct?

As already pointed out, the sum of
two irrational numbers can be rational,
so your proof is invalid.
This is even true if both numbers
are positive, as the following shows:

Let $a = 0.12112111211112...$
  and form $b$ by changing every $1$ in $a$
  to a $2$ and every $2$ to $1$.
So $b = 0.21221222122221...$

Clearly $a$ and $b$ are irrational,
but $a+b = 0.33333... = \frac 13$,
which is a rational number.
