# $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational

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.

• It looks good to me. By the way , shouldn't the question be to prove $\sqrt6 - \sqrt2 -\sqrt3$ is irrational – The Demonix _ Hermit Oct 8 '19 at 11:02
• I think you need to prove $x \neq 1$ – David Oct 8 '19 at 11:04
• $\sqrt{6}-\sqrt{2}-\sqrt{3}=(\sqrt{3}-1)(\sqrt{2}-1)-1$ which is clearly less than zero. – user655800 Oct 8 '19 at 11:09
• @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 – David Oct 8 '19 at 11:34
• For a somewhat stronger result, see this. – Robert Israel Oct 8 '19 at 12:45

$$\sqrt{6}-\sqrt{2}-\sqrt{3}$$ 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.

• Very nice . Thank you. – user655800 Oct 8 '19 at 11:16
• See also the last part of math.stackexchange.com/a/1203120/589 for a slightly different take. – lhf Oct 8 '19 at 11:31

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.

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