5

$ \sqrt{8+2\sqrt{10+2\sqrt{5}}} - \sqrt{8-2\sqrt{10+2\sqrt{5}}} $

I have tried to raising it to the square, but I can't obtain the result.

$ \sqrt{8+2\sqrt{10+2\sqrt{5}}} - \sqrt{8-2\sqrt{10+2\sqrt{5}}}= k $

$ 2\sqrt{10+2\sqrt{5}} -2\sqrt{10+2\sqrt{5}} = k^2 $

$2(\sqrt{(10+2\sqrt{5})(10-2\sqrt{5}})=2\sqrt{80}=8\sqrt{5}=k^2$

Is this a good lead?

@EDIT one more thing, how to show that $ \sqrt{8+2\sqrt{10+2\sqrt{5}}} + \sqrt{8-2\sqrt{10+2\sqrt{5}}} $ equals to $\sqrt{10}+\sqrt{2}$

Martin Sleziak
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Mark
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  • I'm not sure I follow all the steps, but the conclusion appears to be correct : If this number is rational, then you can conclude that $\sqrt{5}$ is rational, which it is not. – Prahlad Vaidyanathan Nov 25 '13 at 17:13
  • Using this (http://math.stackexchange.com/questions/438362/evaluate-cos-18-circ-without-using-the-calculator),

    $$8+2\sqrt{10+2\sqrt5}=8(1+\cos18^\circ)=16\cos^29^\circ$$

     – lab bhattacharjee Nov 25 '13 at 17:17
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    Two out of your three steps are wrong. – Kaster Nov 25 '13 at 17:27

1 Answers1

5

So, your steps aren't quite valid; if you have a statement of the form $$ \sqrt{a} - \sqrt{b} = c $$

Then, while it's true that $$ \left(\sqrt a - \sqrt b\right)^2 = c^2 $$

this unfortunately doesn't simplify to $$ a - b = c^2 $$

rather, you get $$ a - 2\sqrt a\sqrt b + b = c^2 $$

(check this by "foil"-ing).

What is true however, is that

$$ (\sqrt a - \sqrt b)(\sqrt a + \sqrt b) = a - b $$ (again, check by "foil"-ing), and that $\sqrt a - \sqrt b$ is rational if and only if $\sqrt a + \sqrt b$ is (and so, as a consequence of this assumption, $a - b$ would be rational as well).

So

$$\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}} - \sqrt{8-2\sqrt{10+2\sqrt{5}}}\right)\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}} + \sqrt{8-2\sqrt{10+2\sqrt{5}}}\right)\\ = 8+2\sqrt{10+2\sqrt{5}} - (8-2\sqrt{10+2\sqrt{5}})\\ = 4\sqrt{10+2\sqrt{5}}$$

Now, you might try squaring things.

BaronVT
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  • I understand your method but I don't know what to do next. I know that $a-b$ isn't rational but I'm looking for $\sqrt{a}+\sqrt{b}$ – Mark Nov 25 '13 at 17:48
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    Continuing with Baron's solution

    $$4\sqrt{10+2\sqrt5}\in\Bbb Q\implies \sqrt{10+2\sqrt5}=\frac pq\in\Bbb Q\implies$$

    $$10+2\sqrt5=\frac{p^2}{q^2}\in\Bbb Q\implies\sqrt5\in\Bbb Q$$

     – DonAntonio Nov 25 '13 at 18:11
  • ok, now I see, thank you – Mark Nov 25 '13 at 18:14