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This question already has an answer here:

I have proven that the sum of any two irrational numbers is not always irrational and that a rational plus an irrational is irrational, however I am not sure how to prove this specific case.

I was thinking along the lines of: $\sqrt{2}+\sqrt{3}=n$ where $n$ is an element of rationals. By squaring both sides

$$(\sqrt{2}+\sqrt{3})^2 = n^2$$ $$7 + 2\sqrt{2}\sqrt{3} = n^2$$ I have proven the sum of a rational and irrational is irrational yet it is too long and complicated to prove that root two and root three make root six. Moreover, it will be an additionally difficult to prove that two times root six is irrational so I was wondering if there was a quicker less complex way to finish this proof?

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marked as duplicate by user236182, Parcly Taxel, Lord Shark the Unknown, Guy Fsone, Dietrich Burde Oct 16 '17 at 19:13

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    $\begingroup$ Also see math.stackexchange.com/questions/1516482/… $\endgroup$ – user236182 Oct 16 '17 at 18:56
  • $\begingroup$ You should have in your tool box the proof that $\sqrt 2$ is irrational. And it is pretty easy to generalize that proof to show that the square root of a non-perfect square is irrational. So proving the irrationality of $\sqrt 6$ should not be your problem. $\endgroup$ – Doug M Oct 16 '17 at 18:57
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The novelty in this question is the request for a proof that doesn't involve knowing (or proving) the irrationality of $\sqrt6$. So here's a proof that only assumes the irrationality of $\sqrt2$ and/or $\sqrt3$.

If $\sqrt2+\sqrt3$ is rational, then so is $\sqrt3-\sqrt2$, since $(\sqrt2+\sqrt3)(\sqrt3-\sqrt2)=3-2=1$. But then so is

$$\sqrt2={(\sqrt2+\sqrt3)-(\sqrt3-\sqrt2)\over2}$$

(and also $\sqrt3=((\sqrt2+\sqrt3)+(\sqrt3-\sqrt2))/2$).

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Assume that, $$(\sqrt{2}+\sqrt{3})=\frac{p}{q}$$ Then we have $$(\sqrt{2}+\sqrt{3})=\frac{p}{q}\in\Bbb Q \implies 5+2\sqrt{6} =(\sqrt{2}+\sqrt{3})^2 =\frac{p^2}{q^2}\in\Bbb Q\\\implies \sqrt{6} =\frac 52+\frac{p^2}{2q^2}\in\Bbb Q ~~(Impossible )$$

since $\sqrt 6\not \in\Bbb Q$

Hence $$\sqrt{2}+\sqrt{3}\not \in\Bbb Q$$

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You can show that $\sqrt{2}+\sqrt{3}$ is a zero of the polynomial $x^4-10x^2+1$. By the Rational Root Theorem, the only possible rational roots of this polynomial are $1$ and $-1$.

So $\sqrt{2}+\sqrt{3}$ is irrational.

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It is enough to generalize a little one proof of $\sqrt{2}\not\in\mathbb{Q}$, namely the following one.

Given a prime $p$ and a positive integer $n$, we may denote as $$\nu_p(n) = \max\{m\in\mathbb{N}: p^m|n\}.$$

Claim: if $\nu_p(n)$ is odd, then $\sqrt{n}\not\in\mathbb{Q}$.

Proof: The assumption $\sqrt{n}=\frac{a}{b}$ with $a,b\in\mathbb{N}^+$ leads to $$ a^2 = n b^2 $$ which is impossible, since $\nu_p(a^2)$ is even while $\nu_p(nb^2)=\nu_p(n)+\nu_p(b^2)$ is odd.

Corollary: if $n$ is a positive integer, $\sqrt{n}\in\mathbb{Q}$ only if $n$ is an integer square.

Corollary: $\sqrt{6}\not\in\mathbb{Q}$.

Corollary: $\sqrt{2}+\sqrt{3}\not\in\mathbb{Q}$, since the assumption $\sqrt{2}+\sqrt{3}\in\mathbb{Q}$ leads to $\sqrt{6}\in\mathbb{Q}$, contradicting the previous point.

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  • $\begingroup$ This proof is nice and deserved only to expert:)+1 Very nice $\endgroup$ – Guy Fsone Oct 17 '17 at 12:32
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after squaring we get $$2\sqrt{6}=\frac{m^2}{n^2}-5$$ the right-hand side is rational, and the left-hand side irrational.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – kjetil b halvorsen Oct 16 '17 at 19:27
  • $\begingroup$ it is the same like above, how ever, i know your phrases $\endgroup$ – Dr. Sonnhard Graubner Oct 16 '17 at 19:31
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Suppose $\sqrt{2}+\sqrt{3} = q$ for some $q\in \mathbb{Q}$. Then $$2+2\sqrt{6}+3 =q^2$$

and so $$\sqrt{6} = {q^2-5\over 2}$$

Since all elementary operation $(x,-,\cdot, :)$ are closed in $\mathbb{Q}$, the right side is rationa and the left is not. A contradiction.

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