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?
If $\sqrt{2} + \sqrt{3}$ is rational, then so is $(\sqrt{2} + \sqrt{3})^2 = 5 + 2 \sqrt{6}$. But this is absurd since $\sqrt{6}$ is irrational.
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.