Rationalize the denominator

**Mukilab** · January 11th 2010, 10:32 AM

Method and WIP please ^^

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

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

**earboth** · January 11th 2010, 10:36 AM

Originally Posted by Mukilab

Method and WIP please ^^

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

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

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

And now it's your turn.

**Mukilab** · January 11th 2010, 10:43 AM

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

**e^(i*pi)** · January 11th 2010, 10:52 AM

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

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

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

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

The $\frac{\sqrt{2}}{\sqrt{2}}$ comes about because we can multiply a fraction by 1 and leave it unchanged.

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

**Mukilab** · January 11th 2010, 11:05 AM

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

**Archie Meade** · January 11th 2010, 11:17 AM

Yes, but that is not the form requested, Mukilab.
Look at Earboth's solution.

**Mukilab** · January 11th 2010, 11:23 AM

But I don't get how this

Goes to

Wait. Is that even the answer? I'm confused.

**e^(i*pi)** · January 11th 2010, 11:28 AM

Originally Posted by Mukilab

But I don't get how this

Goes to

Wait. Is that even the answer? I'm confused.

Earboth has outlined the method - your task is to multiply $\frac{1}{8 \cdot 2\sqrt{2}}$ by $\frac{\sqrt2}{\sqrt2} $

**Archie Meade** · January 11th 2010, 11:35 AM

Here's a more obvious way...

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

$=\frac{\sqrt{2}}{8\sqrt{2(8)}}=\frac{\sqrt{2}}{8\s qrt{16}}$

$=\frac{\sqrt{2}}{8(4)}=\frac{\sqrt{2}}{32}$

**Mukilab** · January 11th 2010, 11:36 AM

Edit: I saw Mr Meade's post. Thank you very much

**bigwave** · January 11th 2010, 11:49 AM

Originally Posted by Mukilab

Method and WIP please ^^

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

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

here is another way to look at it

$\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} $

**Mukilab** · January 11th 2010, 12:58 PM

Originally Posted by bigwave

here is another way to look at it

$\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

**Archie Meade** · January 11th 2010, 01:23 PM

Or...

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

$p(1)=8\sqrt{2}\sqrt{8}$

**Oloria** · January 11th 2010, 01:29 PM

Originally Posted by e^(i*pi)

you need to know that $\sqrt{a^2} = a$

Precisely $\sqrt{a^2} = |a|$

**Archie Meade** · January 11th 2010, 01:43 PM

No, that's only one of them.

Thread: Rationalize the denominator

LinkBack

Thread Tools

Display