# Quick maths question

1. Nov 6, 2004

### roger

Hello,

please can someone help me with this :

Is 1 / (ab/d)^x/y the same as (d/ab) ^x/y ?

Does the order matter ?

Also, if I have to simplify root 10/root 160 and put it into surd form is 1/4
wrong ?

thanks

Roger

2. Nov 6, 2004

### da_willem

$$\frac{1}{(ab/d)^{x/y}}=(\frac{1}{(ab/d)})^{x/y}=(\frac{d}{ab})^{x/y}$$

So yes, it is the same

Also:

$$\frac{\sqrt{10}}{\sqrt{160}}=\sqrt{\frac{10}{160}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$$

So you are again correct

Last edited: Nov 6, 2004
3. Nov 6, 2004

### arildno

I assume you are talking about the expression:
$$\frac{1}{(\frac{ab}{d})^{\frac{x}{y}}}$$
By ordinary rules of arithmetic, we have:
$$\frac{1}{(\frac{ab}{d})^{\frac{x}{y}}}=\frac{(1)^{\frac{x}{y}}}{(\frac{ab}{d})^{\frac{x}{y}}}=$$
$$(\frac{1}{\frac{ab}{d}})^{\frac{x}{y}}=(\frac{d}{ab})^{\frac{x}{y}}$$

Secondly, we have for positive, real numbers a,b:
$$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$$
EDIT:
Hmm..dawillem beat me here..

4. Nov 6, 2004

### roger

Dear Arildno,

The bit that says by the ordinary rules of arithmetic we have....

Why did you apply the x/y to the top and bottom ?
I thought it only applies to whats inside the brackets at the bottom ?
Please can you explain this for me

Also for the last bit, on roots, if it was root minus x / root minus y what is the general answer for that ?

Thanks

Roger

5. Nov 6, 2004

### Zurtex

It just made it easier and 1a=1 for any value of a.

6. Nov 6, 2004

### CeeAnne

Response to your first question: My understanding is the expressions are equivalent, as would be
1/((ab^x/y)/d^x/y) and (d^x/y)/(ab^xy). You need to observe any hierarchy is all.

The second is numerically correct. Unless you need to express it as root 1/16 or 1/root 16 for some reason.