surds fraction question

**renlok** · October 30th 2009, 05:53 AM

ok bit of a silly question but i really dont understand how to rearange

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

**e^(i*pi)** · October 30th 2009, 06:09 AM

Originally Posted by renlok

ok bit of a silly question but i really dont understand how to rearange

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

$\frac{\sqrt{2}}{(\sqrt{2})^2}$ . It's of the form $\frac{a}{a^2} = \frac{1}{a}$

**renlok** · October 30th 2009, 06:37 AM

but would you not have to square the $\sqrt{y}$ aswell?

**e^(i*pi)** · October 30th 2009, 06:43 AM

Originally Posted by renlok

but would you not have to square the $\sqrt{y}$ aswell?

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

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

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

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

**rowe** · October 30th 2009, 07:14 AM

Nice and quick, but little motivation:

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

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

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

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

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