# Subtracting from a negative fraction. (Signs in cross multiplication)

• Feb 23rd 2011, 01:27 AM
alyosha2
Subtracting from a negative fraction. (Signs in cross multiplication)
$\displaystyle -1 - \frac{1}{3}$

$\displaystyle - \frac{1 * 3}{1 * 3} - \frac{1}{3} = - \frac{3}{3} - \frac{1}{3} = - \frac{-3 - 1}{3} = - \frac{-4}{3} = ??$

does the negative sign on the faction mean that both the numerator and the denomanator shoud be treated as negative numbers when used in cross multiplication?
• Feb 23rd 2011, 01:47 AM
DrSteve
$\displaystyle -\frac{-4}{3}=(-1)\frac{-4}{3}=\frac{4}{3}$
• Feb 23rd 2011, 02:34 AM
Wilmer
-1-1/3 = -(1 + 1/3) = ?

Sometimes easier I find:
8 - 2/3 - 3/5 ; multiply by -1:
= -8 + 2/3 + 3/5
= -8 + 10/15 + 9/15
= -8 + 19/15
= -120/15 + 19/15 ; multiply by -1:
= 120/15 - 19/15
= 101/15
• Feb 23rd 2011, 02:36 AM
alyosha2
Quote:

Originally Posted by DrSteve
$\displaystyle -\frac{-4}{3}=(-1)\frac{-4}{3}=\frac{4}{3}$

so

$\displaystyle -1 - \frac{1}{3} = \frac{4}{3}$ !?!
• Feb 23rd 2011, 02:53 AM
alyosha2
$\displaystyle - 1 - \frac{1}{3} = - \frac{1}{1} - \frac{1}{3} = - ( \frac{-1 * 3 - 1 * - 1}{ -1 * 3}) = -(\frac{(-3) - (-1)}{-3}) = -(\frac{-2}{-3}) = -(\frac{2}{3})= - \frac{2}{3}$ ???
• Feb 23rd 2011, 02:56 AM
HallsofIvy
Quote:

Originally Posted by alyosha2
$\displaystyle -1 - \frac{1}{3}$

$\displaystyle - \frac{1 * 3}{1 * 3} - \frac{1}{3} = - \frac{3}{3} - \frac{1}{3} = - \frac{-3 - 1}{3} = - \frac{-4}{3} = ??$

does the negative sign on the faction mean that both the numerator and the denomanator shoud be treated as negative numbers when used in cross multiplication?

$\displaystyle -\frac{-4}{3}= \frac{4}{3}$

But "$\displaystyle - \frac{3}{3} - \frac{1}{3} = - \frac{-3 - 1}{3}$" is wrong.

You should have either $\displaystyle -\frac{3}{3}- \frac{1}{3}= \frac{-3- 1}{3}= \frac{-4}{3}$ or
$\displaystyle -\frac{3}{3}- \frac{1}{3}= -\frac{3+ 1}{3}= -\frac{4}{3}$
but not both!

And, of course, $\displaystyle -\frac{4}{3}= \frac{-4}{3}$.
• Feb 23rd 2011, 04:30 AM
DrSteve
Quote:

Originally Posted by alyosha2
so

$\displaystyle -1 - \frac{1}{3} = \frac{4}{3}$ !?!

No. I was just simplifying $\displaystyle -\frac{-4}{3}$ for you. There are errors in your computation as other posters have already explained.
• Feb 23rd 2011, 06:59 AM
alyosha2
Quote:

Originally Posted by HallsofIvy
$\displaystyle -\frac{-4}{3}= \frac{4}{3}$

But "$\displaystyle - \frac{3}{3} - \frac{1}{3} = - \frac{-3 - 1}{3}$" is wrong.

You should have either $\displaystyle -\frac{3}{3}- \frac{1}{3}= \frac{-3- 1}{3}= \frac{-4}{3}$ or
$\displaystyle -\frac{3}{3}- \frac{1}{3}= -\frac{3+ 1}{3}= -\frac{4}{3}$
but not both!

And, of course, $\displaystyle -\frac{4}{3}= \frac{-4}{3}$.

But I still don't understand how we are supposed to understand the negative sign in front of the fraction.

I understand what I'm supposed to have. I just don't understand why.
• Feb 23rd 2011, 07:04 AM
Plato
Quote:

Originally Posted by alyosha2
But I still don't understand how we are supposed to understand the negative sign in front of the fraction

$\displaystyle -\dfrac{a}{b}=\dfrac{-a}{b}=\dfrac{a}{-b}$

$\displaystyle -\dfrac{3}{4}=\dfrac{-3}{4}=\dfrac{3}{-4}$

Does that help?