# surds fraction question

• Oct 30th 2009, 05:53 AM
renlok
surds fraction question
ok bit of a silly question but i really dont understand how to rearange

$\displaystyle \frac{\sqrt{2}}{2\sqrt{y}}$
to get
$\displaystyle \frac{1}{\sqrt{2y}}$
• Oct 30th 2009, 06:09 AM
e^(i*pi)
Quote:

Originally Posted by renlok
ok bit of a silly question but i really dont understand how to rearange

$\displaystyle \frac{\sqrt{2}}{2\sqrt{y}}$
to get
$\displaystyle \frac{1}{\sqrt{2y}}$

$\displaystyle \frac{\sqrt{2}}{(\sqrt{2})^2}$. It's of the form $\displaystyle \frac{a}{a^2} = \frac{1}{a}$
• Oct 30th 2009, 06:37 AM
renlok
but would you not have to square the $\displaystyle \sqrt{y}$ aswell?
• Oct 30th 2009, 06:43 AM
e^(i*pi)
Quote:

Originally Posted by renlok
but would you not have to square the $\displaystyle \sqrt{y}$ aswell?

No, you can treat this sum as $\displaystyle \frac{\sqrt2}{2} \times \frac{1}{\sqrt{y}}$

As $\displaystyle \frac{\sqrt2}{2} = \frac{\sqrt{2}}{\sqrt{2} \sqrt{2}} = \frac{1}{\sqrt2}$ then it becomes $\displaystyle \frac{1}{\sqrt2 \sqrt{y}}$

As $\displaystyle \sqrt{a} \sqrt{b} = \sqrt{ab} \: , \: a,b \geq 0$ we can combine the square roots in the denominator:

$\displaystyle \frac{1}{\sqrt{2y}}$
• Oct 30th 2009, 07:14 AM
rowe
Nice and quick, but little motivation:

$\displaystyle \frac{\sqrt{2}}{2\sqrt{y}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

$\displaystyle =\frac{2}{2\sqrt{y}\sqrt{2}}$

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

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