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This question already has an answer here:

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?

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marked as duplicate by Watson, Chinnapparaj R, user10354138, Brahadeesh, ancientmathematician Nov 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.

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    $\begingroup$ The sum of two irrational numbers need not be irrational. $\sqrt{2} + (-\sqrt{2}) = 0$, for example. $\endgroup$ – Daniel Fischer Jul 25 '13 at 17:22
  • $\begingroup$ your first reasoning is incorrect. $\endgroup$ – DonAntonio Jul 25 '13 at 17:24
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    $\begingroup$ Very nice! Now what about the second part? $\endgroup$ – dotslash Jul 25 '13 at 17:25
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    $\begingroup$ If $a=2^{1/4}$ and $b=-2^{1/4}$, then $ab\not\in \mathbb Q$ but $a+b\in\mathbb Q$. $\endgroup$ – Etienne Jul 25 '13 at 18:03
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    $\begingroup$ See also math.stackexchange.com/questions/278935/… $\endgroup$ – user7530 Jul 31 '13 at 20:27
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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.

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

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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

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

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

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

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

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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.

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

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    $\begingroup$ I am not sure how this will help the OP. $\endgroup$ – Lord Soth Jul 25 '13 at 17:33
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    $\begingroup$ 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) $\endgroup$ – DonAntonio Jul 25 '13 at 17:35
  • $\begingroup$ @DonAntonio Yes, that was my point. $\endgroup$ – Lord Soth Jul 25 '13 at 17:37
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    $\begingroup$ For the record, I don't know Galois Theory. :) $\endgroup$ – dotslash Jul 25 '13 at 17:45
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    $\begingroup$ 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? $\endgroup$ – Pete L. Clark Jul 25 '13 at 18:44
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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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    $\begingroup$ $\sqrt{2},\,1-\sqrt{2}$ work too. I didn't downvote though. $\endgroup$ – user26486 Jul 5 '15 at 16:43
  • $\begingroup$ @user26486 $1-\sqrt{2}$ is not positive, but I get your point $\endgroup$ – Andrei Kh Nov 10 '15 at 20:16

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