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?
$\endgroup$ 179 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ $\begingroup$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
$$\sqrt6=\frac{2p'}q\implies 2\mid q\;\;\;\;\text{also , and this is a contradiction}$$
$\endgroup$ 5 $\begingroup$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.
$\endgroup$ $\begingroup$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: $$x^4-10x^2+1=0.$$
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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 $\sqrt 2+\sqrt 3\notin\mathbb Q$.
$\endgroup$ 5 $\begingroup$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.
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