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What is the most easy way I can prove $\sqrt 8/2$ is equal to $\sqrt 2$

I have done $(\sqrt 8/2)^2$ but at the end it gives me $\pm \sqrt 2$ and not positive $\sqrt 2$ So how?

azimut
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user112724
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    Nobody ever asks for the most difficult way. – Will Jagy Dec 01 '13 at 21:45
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    Interesting! I also want most most of most difficult way – user112724 Dec 01 '13 at 21:46
  • @WillJagy Out of curiosity, what is the most difficult way? :) – Jeel Shah Dec 01 '13 at 21:46
  • The most difficult way is to find a polynomial that gives the same result for x=his two values. :) ok, just kidding, but 2sqrt2=sqrt8. – snowfall512 Dec 01 '13 at 21:50
  • I think there is no such thing as the most difficult way. You can probably always overcomplicate a solution. – N. S. Dec 01 '13 at 21:52
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    Quoted comment by me, in 2010: Latest paper, my co-author put in "but we will choose a more painful way, because there is nothing like pain for feeling alive" but the referee jumped on it. – Will Jagy Apr 23 '10...... http://mathoverflow.net/questions/22299/what-are-some-examples-of-colorful-language-in-serious-mathematics-papers – Will Jagy Dec 01 '13 at 21:53
  • And a "simple" complicated solution would probably be to find a complex integral where the residue theorem would Yields $\sqrt{2} \pi i$ while a parametrisation of the curve would Yield $\frac{\sqrt{8}\pi i}{2}$.... – N. S. Dec 01 '13 at 21:55
  • According to the last line of the question, all that you really needed was a proof that $\sqrt 8/2$ is positive. Well, $\sqrt8$ is positive, and half of a positive number is still positive. – Andreas Blass Dec 01 '13 at 23:03

4 Answers4

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$$\frac{\sqrt{8}}{2} = \frac{\sqrt{2\times4}}{2} = \frac{\sqrt{2}\times\sqrt{4}}{2} = \dots$$

Old John
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    In the last one you can cancel the 2's, for $$ \frac{\sqrt{ }\times\sqrt{4}}{ } $$ – Will Jagy Dec 01 '13 at 21:54
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    So we end up with $\sqrt{\sqrt{4}}$??? – Old John Dec 01 '13 at 21:57
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    @WillJagy Hence we have $\frac{\sqrt{}\times\sqrt{4}}{}=\sqrt{2}$? Ingeniuos! – Michael Hoppe Dec 01 '13 at 22:03
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    @MichaelHoppe, I actually did not think it through that far. I was just thinking of John's answer at http://math.stackexchange.com/questions/260656/cant-argue-with-success-looking-for-bad-math-that-gets-away-with-it/260662#260662 This one does seem to work out as well, which is pretty funny. – Will Jagy Dec 01 '13 at 22:11
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$$\sqrt 8 =\sqrt{4\cdot 2}=\sqrt4\cdot\sqrt2=2\sqrt2$$

Berci
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$\sqrt{8}=2\sqrt{2}\iff\sqrt{8}=\sqrt{2^2\cdot2}$

Michael Hoppe
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Multiply by the square root of 4 and divide by 2. It is the same thing as multiplying by 1.

LASV
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