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Prove that $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational

My attempt:- Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational, then for some $x\in\mathbb{Q}$ we have $$\sqrt{6}-\sqrt{2}-\sqrt{3}=x$$ Rewriting this equation as $$\sqrt{6}-x=\sqrt{2}+\sqrt{3}$$ and now squaring this we get $$ 6-2x\sqrt{6}+x^2=5+2\sqrt{6}$$. This implies that $$\sqrt{6}=\frac{x^2-1}{2+2x}$$ but this is absurd as RHS of the above equation is rational but we know that $\sqrt6$ is irrational. Therefore , $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational. Does this look good? Have I written it properly? Is there any other proof besides this..like one using geometry? Thank you.

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    $\begingroup$ It looks good to me. By the way , shouldn't the question be to prove $\sqrt6 - \sqrt2 -\sqrt3 $ is irrational $\endgroup$ – The Demonix _ Hermit Oct 8 '19 at 11:02
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    $\begingroup$ I think you need to prove $x \neq 1$ $\endgroup$ – David Oct 8 '19 at 11:04
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    $\begingroup$ $\sqrt{6}-\sqrt{2}-\sqrt{3}=(\sqrt{3}-1)(\sqrt{2}-1)-1$ which is clearly less than zero. $\endgroup$ – user655800 Oct 8 '19 at 11:09
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    $\begingroup$ @HVxvejjw Good point! I think we can also go to the second-to-last expression and easily check there that indeed $x \neq 1$, then go on with the next step $\endgroup$ – David Oct 8 '19 at 11:34
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    $\begingroup$ For a somewhat stronger result, see this. $\endgroup$ – Robert Israel Oct 8 '19 at 12:45
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$\sqrt{6}-\sqrt{2}-\sqrt{3}$ is a root of $x^4 - 22 x^2 - 48 x - 23$.

By the rational root theorem, $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is either irrational or an integer.

But $$ 1.4 < \sqrt 2 < 1.5 \\ 1.7 < \sqrt 3 < 1.8 \\ 2.4 < \sqrt 6 < 2.5 \\ $$ imply $$ -0.9 < \sqrt{6}-\sqrt{2}-\sqrt{3} <-0.6 $$ and so $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is not an integer. Therefore, it is irrational.

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  • $\begingroup$ Very nice . Thank you. $\endgroup$ – user655800 Oct 8 '19 at 11:16
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    $\begingroup$ See also the last part of math.stackexchange.com/a/1203120/589 for a slightly different take. $\endgroup$ – lhf Oct 8 '19 at 11:31
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In your proof, after $6-2x\sqrt{6}+x^2=5+2\sqrt{6}$ we have that $$x^2+1=2(x+1)\sqrt{6}$$ If $x=-1$ then, from the above equation, it follows that $2=0$. Therefore $x$ is a rational number different from $-1$. After dividing by $2(x+1)\not=0$ we get $$\sqrt{6}=\frac{x^2+1}{2(x+1)}\in \mathbb{Q}.$$ Contradiction! Hence $x=\sqrt{6}-\sqrt{2}-\sqrt{3}$ is not a rational number.

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    $\begingroup$ @HVxvejjw In your proof it should be $\sqrt{6}=\frac{x^2+1}{2(x+1)}$ $\endgroup$ – Robert Z Oct 8 '19 at 11:28
  • $\begingroup$ ,oh yes. Thanks for pointing out. Thanks a lot. $\endgroup$ – user655800 Oct 8 '19 at 11:44

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