# Rationalize the denominator

• Jan 11th 2010, 10:32 AM
Mukilab
Rationalize the denominator

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

Give answer in the form $\displaystyle \frac{\sqrt{2}}{p}$ Where p is a positive interger
• Jan 11th 2010, 10:36 AM
earboth
Quote:

Originally Posted by Mukilab

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

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

$\displaystyle \frac{1}{8\sqrt{8}} = \frac{1}{8\cdot 2 \sqrt{2}} = \frac{1}{8\cdot 2 \sqrt{2}}\cdot \ \color{red}\frac{\sqrt{2}}{\sqrt{2}}$

• Jan 11th 2010, 10:43 AM
Mukilab
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 AM
e^(i*pi)
Quote:

Originally Posted by Mukilab
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

For $\displaystyle \sqrt{8} = 2\sqrt{2}$ you need to know that $\displaystyle \sqrt{a^2} = a$ and $\displaystyle \sqrt{ab} = \sqrt{a}\sqrt{b}$

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

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

---------------------------------

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 AM
Mukilab
so $\displaystyle \frac{1}{8\cdot 2\sqrt{2}}=\frac{1}{2^{4}\sqrt{2}}???$

:o
• Jan 11th 2010, 11:17 AM
Yes, but that is not the form requested, Mukilab.
Look at Earboth's solution.
• Jan 11th 2010, 11:23 AM
Mukilab
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 AM
e^(i*pi)
Quote:

Originally Posted by Mukilab
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.

Earboth has outlined the method - your task is to multiply $\displaystyle \frac{1}{8 \cdot 2\sqrt{2}}$ by $\displaystyle \frac{\sqrt2}{\sqrt2}$
• Jan 11th 2010, 11:35 AM
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 AM
Mukilab
Edit: I saw Mr Meade's post. Thank you very much
• Jan 11th 2010, 11:49 AM
bigwave
Quote:

Originally Posted by Mukilab

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

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

here is another way to look at it

$\displaystyle \frac{1}{8\sqrt{8}} \times \frac{\sqrt{8}}{\sqrt{8}} \Rightarrow \frac{\sqrt{8}}{8*8} \Rightarrow \frac{\sqrt{2}\sqrt{4}}{64} \Rightarrow \frac{2\sqrt{2}}{64} \Rightarrow \frac{\sqrt{2}}{32}$
• Jan 11th 2010, 12:58 PM
Mukilab
Quote:

Originally Posted by bigwave
here is another way to look at it

$\displaystyle \frac{1}{8\sqrt{8}} \times \frac{\sqrt{8}}{\sqrt{8}} \Rightarrow \frac{\sqrt{8}}{8*8} \Rightarrow \frac{\sqrt{2}\sqrt{4}}{64} \Rightarrow \frac{2\sqrt{2}}{64} \Rightarrow \frac{\sqrt{2}}{32}$

Even better, thanks :o
• Jan 11th 2010, 01:23 PM
Or...

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

$\displaystyle p(1)=8\sqrt{2}\sqrt{8}$
• Jan 11th 2010, 01:29 PM
Oloria
Quote:

Originally Posted by e^(i*pi)
you need to know that $\displaystyle \sqrt{a^2} = a$

Precisely $\displaystyle \sqrt{a^2} = |a|$
• Jan 11th 2010, 01:43 PM