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I am having trouble rationalizing $\frac{2}{\sqrt{2-\sqrt{2}}}$

I have tried multiplying the fraction by $\sqrt{2 + \sqrt{2}}$ and got $\frac{2\sqrt{2+\sqrt{2}}}{\sqrt{2}}$ I am not sure if that is correct or not but I then multiplied by $\sqrt{2}$ but got stuck...

Assuming that I did the problem correctly so far, how would I multiply $2\sqrt{2+\sqrt{2}}$ with $\sqrt{2}$ ?

  • 0
    But $\sqrt{2 - \sqrt 2}\sqrt{2 + \sqrt 2} = \sqrt{4 - 2} = \sqrt 2$. How did you get $4- \sqrt 2$? And $2 \sqrt{2 + \sqrt 2}\sqrt 2 = 2 \sqrt{4 + 2\sqrt 2}$.2012-11-05
  • 0
    Could you explain how you multiplied $\sqrt{2}$ with $2\sqrt{2+\sqrt{2}}$?2012-11-05

1 Answers 1

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Instead of multiplying top and bottom by $\sqrt{2}$, note that $2=\left(\sqrt{2}\right)^2$, so that $$\frac{2\sqrt{2+\sqrt{2}}}{\sqrt{2}}=\sqrt{2}\sqrt{2+\sqrt{2}}=\sqrt{4+2\sqrt{2}}.$$

  • 0
    Does this mean that 'rationalizing' means getting rid of radicals only in the denominator, but not in the whole expression?2012-11-06
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    Yes. We talk about "rationalizing the denominator" of a fraction with radicals in the denominator. Having a rational (integer, really) denominator makes arithmetic easier. We can't necessarily get rid of *all* the radicals, though, as we saw in this case. *If we're lucky*, we can get rid of all of the radicals--for an easy example, consider $\frac{\sqrt{8}}{\sqrt{2}}$--but most of the time we'll still have some hanging around.2012-11-06
  • 0
    Ok that's cool.2012-11-06