Prove that $\sqrt 2 + \sqrt 3$ is irrational

Question

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?

Answer 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.

Answer 2

You'd have to prove that the sum of two irrational numbers yields an irrational number first. Note that its not true though. So to your first question, your reasoning is incorrect.

Answer 3

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.

Answer 4

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

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

Answer 5

Your reasoning is not correct when you go from $\sqrt 2 $ or $ \sqrt 3$ or both are irrational to $\sqrt 2 + \sqrt 3$ is irrational.

I would say: assume $\sqrt 2 + \sqrt 3$ is rational. Then its square is rational, because multiplying rationals gives a rational. But $(\sqrt 2 + \sqrt 3)^2=2+2\sqrt 6 +3$ is irrational because the sum of an irrational and a rational ($5$) is irrational, so we have a contradiction.

Answer 6

If you know anything about Galois theory, here is a very roundabout way of proving this (in other words, the other answers are better ways to think about this problem):

$\alpha=\sqrt 2+\sqrt 3$ is a primitive element of the Galois extension $[\Bbb Q(\alpha):\Bbb Q]$, with minimal poylnomial $(x^2-2)(x^2-3)$, which has degree $4$ over $\mathbb{Q}$. It follows that $\sqrt 2+\sqrt 3\notin\mathbb Q$.

Answer 7

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.

Answer 8

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: $$x^4-10x^2+1=0.$$

Answer 9

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.