[SOLVED] Fractions with surds in the denominator

• Mar 16th 2009, 08:06 PM
waven
[SOLVED] Fractions with surds in the denominator
Simply by rationalising the denominator

$\displaystyle \frac {\sqrt 3} {\sqrt 6}$

thanks!
• Mar 16th 2009, 08:26 PM
Quote:

Originally Posted by waven
Simply by rationalising the denominator

$\displaystyle \frac {\sqrt 3} {\sqrt 6}$

thanks!

multiply it by $\displaystyle \frac{\sqrt{6}}{\sqrt{6}}$
• Mar 16th 2009, 08:35 PM
waven
Quote:

multiply it by $\displaystyle \frac{\sqrt{6}}{\sqrt{6}}$

$\displaystyle \frac {\sqrt 18} 6$ what do i do next?

btw the answer is $\displaystyle \frac {\sqrt 2} {2}$
• Mar 16th 2009, 10:11 PM
Prove It
Quote:

Originally Posted by waven
$\displaystyle \frac {\sqrt 18} 6$ what do i do next?

btw the answer is $\displaystyle \frac {\sqrt 2} {2}$

Notice that $\displaystyle \sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$.

So $\displaystyle \frac{\sqrt{18}}{6} = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2}$.