Method and WIP please ^^

$\displaystyle \frac{1}{8\sqrt{8}}$

Give answer in the form $\displaystyle \frac{\sqrt{2}}{p}$ Where p is a positive interger

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- Jan 11th 2010, 10:32 AMMukilabRationalize the denominator
Method and WIP please ^^

$\displaystyle \frac{1}{8\sqrt{8}}$

Give answer in the form $\displaystyle \frac{\sqrt{2}}{p}$ Where p is a positive interger - Jan 11th 2010, 10:36 AMearboth
- Jan 11th 2010, 10:43 AMMukilab
Yes it is. How did you get 8*2\sqrt2?

Method for that

and from that step how did you get to the two square routes of 2? Method for that too please - Jan 11th 2010, 10:52 AMe^(i*pi)
For $\displaystyle \sqrt{8} = 2\sqrt{2}$ you need to know that $\displaystyle \sqrt{a^2} = a$ and $\displaystyle \sqrt{ab} = \sqrt{a}\sqrt{b}$

$\displaystyle 8 = 2^3 = 2^2 \cdot 2$

$\displaystyle \sqrt{8} = \sqrt{2^2 \cdot 2} = \sqrt{2^2} \cdot \sqrt{2} = 2\sqrt{2}$

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The $\displaystyle \frac{\sqrt{2}}{\sqrt{2}}$ comes about because we can multiply a fraction by 1 and leave it unchanged.

$\displaystyle \frac{\sqrt{2}}{\sqrt{2}} = 1$ - Jan 11th 2010, 11:05 AMMukilab
so $\displaystyle \frac{1}{8\cdot 2\sqrt{2}}=\frac{1}{2^{4}\sqrt{2}}???$

:o - Jan 11th 2010, 11:17 AMArchie Meade
Yes, but that is not the form requested, Mukilab.

Look at Earboth's solution. - Jan 11th 2010, 11:23 AMMukilab
But I don't get how this http://www.mathhelpforum.com/math-he...73e88dbd-1.gif

Goes to http://www.mathhelpforum.com/math-he...2d0eb15a-1.gif

Wait. Is that even the answer? I'm confused. - Jan 11th 2010, 11:28 AMe^(i*pi)
- Jan 11th 2010, 11:35 AMArchie Meade
Here's a more obvious way...

$\displaystyle \frac{1}{8\sqrt{8}}=\frac{\sqrt{2}}{\sqrt{2}}\ \frac{1}{8\sqrt{8}}=\frac{\sqrt{2}}{8\sqrt{2}\sqrt {8}}$

$\displaystyle =\frac{\sqrt{2}}{8\sqrt{2(8)}}=\frac{\sqrt{2}}{8\s qrt{16}}$

$\displaystyle =\frac{\sqrt{2}}{8(4)}=\frac{\sqrt{2}}{32}$ - Jan 11th 2010, 11:36 AMMukilab
Edit: I saw Mr Meade's post. Thank you very much

- Jan 11th 2010, 11:49 AMbigwave
- Jan 11th 2010, 12:58 PMMukilab
- Jan 11th 2010, 01:23 PMArchie Meade
Or...

$\displaystyle \frac{1}{8\sqrt{8}}=\frac{\sqrt{2}}{p}$

$\displaystyle p(1)=8\sqrt{2}\sqrt{8}$ - Jan 11th 2010, 01:29 PMOloria
- Jan 11th 2010, 01:43 PMArchie Meade
No, that's only one of them.