1. ## 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

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

3. 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

4. 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.

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

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

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

7. But I don't get how this

Goes to

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

8. 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 $\displaystyle \frac{1}{8 \cdot 2\sqrt{2}}$ by $\displaystyle \frac{\sqrt2}{\sqrt2}$

9. Here's a more obvious way...

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

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

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

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

11. 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

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

12. 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

13. Or...

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

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

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

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

15. No, that's only one of them.