• January 5th 2008, 01:26 AM
Peacewanters007
1/underroot 8

plz explain the way u adopt as I know nothing about it
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• January 5th 2008, 02:55 AM
earboth
Quote:

Originally Posted by Peacewanters007
I assume that you know: $\sqrt{a} \cdot \sqrt{a} = a$ and
$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$ . Then:
$\frac1{\sqrt{8}}=\frac1{\sqrt{8}} \cdot \frac{\sqrt{8}}{\sqrt{8}}= \frac{\sqrt{8}}{8} = \frac{\sqrt{4} \cdot \sqrt{2}}{8}=\frac{\not 2^1 \cdot \sqrt{2}}{\not 8_4}=\frac{\sqrt{2}}{4}$