# Thread: Minus on denominator again

1. ## Minus on denominator again

Every time I think I understand how to deal with a minus in a question, something different crops up. I thought I could do this but I need a little help on the last stage.

Rewrite the expression with a rational denominator. $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+4}$

$\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}-4)}{(\sqrt{3}+4)(\sqrt{3}-4)}$

= $\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}-4)}{-13}$

Now what do I do to get rid of the minus on the denominator?

I know the answer is $-\frac{(\sqrt{3}+2)(\sqrt{3}-4)}{13}$ but cannot fathom how they got there.

If I multiply top and bottom by -13, don't I end up with $-13(\sqrt{3}+2)(\sqrt{3}-4)$

2. Any fraction of the form $\frac{a}{-b}$ or $\frac{-a}{b}$ is equivalent to $-\frac{a}{b}$. I'm not going to bother with a general proof, but in your case:

$\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}-4)}{-13}$

$=\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}-4)}{-1 \cdot 13}$

$=\frac{1}{-1}\cdot\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}-4)}{13}$

$=-1\cdot\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}-4)}{13}$

$=-\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}-4)}{13}$

3. $\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}-4)}{-13} = \frac{-(\sqrt{3}+\sqrt{2})(\sqrt{3}-4)}{13} = \frac{-\left (3-4\sqrt{3}+\sqrt{6}-4\sqrt{2} \right )}{13}$

4. Thanks both

Are you saying that I should multiply top and bottom by -1?

Marilyn

5. All we're saying is that $\frac{-a}{b} = \frac{a}{-b} = -\frac{a}{b}$. So, you can "move" the negative factor "around" the fraction itself.

In the case of your problem, you've already got the answer, it's just the answer you've been given has been written in an different (but equivalent) way.

6. Originally Posted by rowe
All we're saying is that $\frac{-a}{b} = \frac{a}{-b} = -\frac{a}{b}$. So, you can "move" the negative factor "around" the fraction itself.

In the case of your problem, you've already got the answer, it's just the answer you've been given has been written in an different (but equivalent) way.

Ohhhh!!! Got it, thank you. This is another of those 'it just is - so learn it for now' things. Maybe I will learn enough to understand the explanation one day - then again, maybe not.

7. Originally Posted by Meggomumsie
Ohhhh!!! Got it, thank you. This is another of those 'it just is - so learn it for now' things. Maybe I will learn enough to understand the explanation one day - then again, maybe not.
Well, it's fairly simple, I detailed it above, but here's a more general proof...

$\frac{-a}{b}$

$=\frac{-1 \cdot a}{b}$

$=\frac{-1}{1}\cdot\frac{a}{b}$

$=-1\cdot\frac{a}{b}$

$=-\frac{a}{b}$

Does this make sense? The same rules apply for a negative factor in the denominator, except we "extract" the reciprocal of -1, $\frac{1}{-1}$. Of course, this is still equal to -1.

8. Originally Posted by rowe
Well, it's fairly simple, I detailed it above, but here's a more general proof...

$\frac{-a}{b}$

$=\frac{-1 \cdot a}{b}$

$=\frac{-1}{1}\cdot\frac{a}{b}$

$=-1\cdot\frac{a}{b}$

$=-\frac{a}{b}$

Does this make sense? The same rules apply for a negative factor in the denominator, except we "extract" the reciprocal of -1, $\frac{1}{-1}$. Of course, this is still equal to -1.
Yes it makes complete sense and I can now 'see' it. I don't seem to be able to work these things out for myself yet. Thanks ~ Marilyn