# 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?

• The sum of two irrational numbers need not be irrational. $\sqrt{2} + (-\sqrt{2}) = 0$, for example. Jul 25, 2013 at 17:22
• your first reasoning is incorrect. Jul 25, 2013 at 17:24
• Very nice! Now what about the second part? Jul 25, 2013 at 17:25
• If $a=2^{1/4}$ and $b=-2^{1/4}$, then $ab\not\in \mathbb Q$ but $a+b\in\mathbb Q$. Jul 25, 2013 at 18:03
• Jul 31, 2013 at 20:27

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! Aug 10, 2013 at 17:25
• May be a proof by contradiction... Apr 11, 2015 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. Jul 26, 2013 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]. Jul 26, 2013 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 $\sqrt2 + \sqrt3 = \frac{a}{b}$ Jan 21, 2015 at 2:53
• @ashimashi I most second this stance. This answer seems to involve incorrect arithmetic! Feb 24, 2017 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 Feb 25, 2017 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 $(x^2-5)^2=24$ Jul 25, 2013 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.

$$\alpha=\sqrt 2+\sqrt 3$$ 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 $$\sqrt 2+\sqrt 3\notin\mathbb Q$$.

• I am not sure how this will help the OP. Jul 25, 2013 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) Jul 25, 2013 at 17:35
• @DonAntonio Yes, that was my point. Jul 25, 2013 at 17:37
• For the record, I don't know Galois Theory. :) Jul 25, 2013 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? Jul 25, 2013 at 18:44

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.

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. Jul 5, 2015 at 16:43
• @user26486 $1-\sqrt{2}$ is not positive, but I get your point Nov 10, 2015 at 20:16