Prove that $\sqrt 2 + \sqrt 3$ is irrational [duplicate]

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?

Solution 1:

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.

Solution 2:

If $\sqrt 3 +\sqrt 2$ is rational/irrational, then so is $\sqrt 3 -\sqrt 2$ because $\sqrt 3 +\sqrt 2=\large \frac {1}{\sqrt 3- \sqrt 2}$ . Now assume $\sqrt 3 +\sqrt 2$ is rational. If we add $(\sqrt 3 +\sqrt 2)+(\sqrt 3 -\sqrt 2)$ we get $2\sqrt 3$ which is irrational. But the sum of two rationals can never be irrational, because for integers $a, b, c, d$ $\large \frac ab+\frac cd=\frac {ad+bc}{bd}$ which is rational. Therefore, our assumption that $\sqrt 3 +\sqrt 2$ is rational is incorrect, so $\sqrt 3 +\sqrt 2$ is irrational.

Solution 3:

Hints:

Suppose there exist coprime $\,a,b\in\Bbb Z\,$ s.t.

$$\sqrt2+\sqrt3=\frac ab\implies \sqrt6=\frac{a^2}{2b^2}-\frac52=\frac{a^2-5b^2}{2b^2}$$

If you already know $\,\sqrt6\,$ is irrational then you're already done, otherwise prove it as with $\,\sqrt2\,$ , say:

$$\sqrt6=\frac pq\;,\;\;(p,q)=1\implies 6q^2=p^2\implies 2\mid p$$

and thus we can write

$$\sqrt6=\frac{2p'}q\implies 2\mid q\;\;\;\;\text{also , and this is a contradiction}$$

Solution 4:

If $\sqrt2+\sqrt3 =r \in \mathbb{Q}$, then $\frac{r^2-5}{2}=\sqrt6 \in \mathbb{Q}$. Contradiction! This could be a way of your proof.

Solution 5:

Note that $\sqrt{2}+\sqrt{3}$ is a solution to the equation: $$x^4-10x^2+1=0$$Does this polynomial have any rational roots?

Edit: To find this polynomial, note that if $x=\sqrt{2}+\sqrt{3}$, then: $$x^2=5+2\sqrt{6}$$and: $$x^4=49+20\sqrt{6}.$$You need $-10x^2$'s to get rid of the $20\sqrt{6}$ in $x^4$, and $x^4-10x^2=-1$, so you get: $$x^4-10x^2+1=0.$$