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

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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?

## marked as duplicate by Watson, Chinnapparaj R, user10354138, Brahadeesh, ancientmathematicianNov 26 '18 at 10:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• The sum of two irrational numbers need not be irrational. $\sqrt{2} + (-\sqrt{2}) = 0$, for example. – Daniel Fischer Jul 25 '13 at 17:22
• your first reasoning is incorrect. – DonAntonio Jul 25 '13 at 17:24
• Very nice! Now what about the second part? – dotslash Jul 25 '13 at 17:25
• If $a=2^{1/4}$ and $b=-2^{1/4}$, then $ab\not\in \mathbb Q$ but $a+b\in\mathbb Q$. – Etienne Jul 25 '13 at 18:03
• – user7530 Jul 31 '13 at 20:27

## 9 Answers

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.

• Love the usage of the word absurd! – zerosofthezeta Aug 10 '13 at 17:25
• May be a proof by contradiction... – Sufyan Naeem Apr 11 '15 at 11:11

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.

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

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

• The accepted answer gives the OP a fish. Your answer teach him to fish. – Isaac Jul 26 '13 at 8:17
• one small nit detail - we need to say p,q are co prime as well [i.e. p/q is in simplest form]. – Fakrudeen Jul 26 '13 at 10:13
• I don't understand, should it not be $2\sqrt6$ instead of just $\sqrt6$? I am referring to when you squared both sides of $\sqrt2 + \sqrt3 = \frac{a}{b}$ – ashimashi Jan 21 '15 at 2:53
• @ashimashi I most second this stance. This answer seems to involve incorrect arithmetic! – Brevan Ellefsen Feb 24 '17 at 21:55
• @ashimashi Yes, it seems to be lacking a rational factor of two there, but (1) it doesn't affect the outcome, and (2) this was so much time ago...I shall edit it, however. THanks – DonAntonio Feb 25 '17 at 0:17

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.

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

• Or use $(x^2-5)^2=24$ – Mark Bennet Jul 25 '13 at 17:45

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.

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.

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):

$\sqrt 2+\sqrt 3$ is a primitive element of the Galois field of the polynomial $(x^2-2)(x^2-3)$, which has degree $4$ over $\mathbb{Q}$. It follows that $\sqrt 2+\sqrt 3\notin\mathbb Q$.

• I am not sure how this will help the OP. – Lord Soth Jul 25 '13 at 17:33
• Well @LordSoth, if the OP knows Galois Theory then the above answers his question...of course, if he knows Galois theory then asking this question is completely anachronical (mathematicalwise) – DonAntonio Jul 25 '13 at 17:35
• @DonAntonio Yes, that was my point. – Lord Soth Jul 25 '13 at 17:37
• For the record, I don't know Galois Theory. :) – dotslash Jul 25 '13 at 17:45
• It seems to me that this answer just asserts without proof a stronger statement than what the OP asked, namely that the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$ has degree $4$ (the OP asked for a proof that it does not have degree $1$). In order for this to be a reasonable answer, shouldn't you say something about how to prove this stronger statement? – Pete L. Clark Jul 25 '13 at 18:44

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.

• $\sqrt{2},\,1-\sqrt{2}$ work too. I didn't downvote though. – user26486 Jul 5 '15 at 16:43
• @user26486 $1-\sqrt{2}$ is not positive, but I get your point – Andrei Kh Nov 10 '15 at 20:16