Sign problem on complex fractions

**Alienis Back** · May 2nd 2009, 05:37 AM

Ok. Here is my problem.

$\frac{\frac{1}{a-1}+1}{\frac{1}{a+1}-1}$

And my aproach

$=\frac{\frac{1}{a-1}+\frac{a-1}{a-1}}{\frac{1}{a+1}-\frac{a+1}{a+1}}$

$=\frac{\frac{1+a-1}{a-1}}{\frac{1-a+1}{a+1}}$

$=\frac{\frac{a}{a-1}}{\frac{-a}{a+1}}$

$=\frac{a}{a-1}*\frac{a+1}{-a}$

$=\frac{(a)(a+1)}{(-a)(a-1)}$

I'm having some trouble to solve for these signs...mmm...The book I'm using has the answers to odd problems only. This is an even one. But anyway, I want to have it right. The istruction says "Simplify as much as possible" And I've got the feeling this can be worked further. Any idea?

PS: I thought about multiplicating by -1, but then It would be the same as the numerator'd turn into negatives.

**running-gag** · May 2nd 2009, 05:49 AM

Hi

$\frac{a}{a} = 1$ therefore you get $-\frac{a+1}{a-1}$

**Alienis Back** · May 2nd 2009, 05:51 AM

Mmm...Sorry, but I can't see the point...

**running-gag** · May 2nd 2009, 06:02 AM

From this

$\frac{(a)(a+1)}{(-a)(a-1)}$

you can simplify by "a", since "a" is a factor of both numerator and denominator

**Alienis Back** · May 2nd 2009, 06:18 AM

How would you do that?

**VonNemo19** · May 2nd 2009, 02:37 PM

How many times will 1 go into -1 (-1 times right?)

How many times will 5 go into -5 (-1 times right?)

how many times will 10,000 go into -10000 (-1 times right?)

So, how many times will a go into -a

You guessed it -1 TIMES!

So the two a's cacel to become -1! Got It?

**Alienis Back** · May 7th 2009, 07:42 PM

according to what running-gag said and following the approach I started using, the development should follow as:

$=\frac{(a)(a+1)}{(-a)(a-1)}$

$=\frac{(a)(a+1)}{(-1)(a)(a-1)}$

$=\frac{a+1}{-1(a-1)}$

$=\frac{a+1}{-a+1}$

$=\frac{a+1}{1-a}$

And there it was...!!! Ops...wait...wait...wait...What if this $=\frac{\frac{1+a-1}{a-1}}{\frac{1-a+1}{a+1}}$ was this $=\frac{\frac{1+a-1}{a-1}}{\frac{2-a}{a+1}}$ , which to me seem more correct. Then:

$=\frac{\frac{1+a-1}{a-1}}{\frac{2-a}{a+1}}$

$=\frac{a}{a-1} * \frac{a+1}{2-a}$

$=\frac{(a)(a+1)}{-a^2+3a-2}$

$=\frac{(a)(a+1)}{(a-1)(2-a)}$

$=\frac{(a)(a+1)}{(-1)(a-2)(a-1)}$

Now my question is which one is the right one:

a) $=\frac{a+1}{1-a}$ or

b) $=\frac{(a)(a+1)}{(-1)(a-2)(a-1)}$

c) None. Then which______________________________?

**pickslides** · May 7th 2009, 08:24 PM

$<br /> \frac{\frac{1}{a-1}+\frac{a-1}{a-1}}{\frac{1}{a+1}-\frac{a+1}{a+1}}<br />$

$=$

$\frac{\frac{1+a-1}{a-1}}{\frac{1-a-1}{a+1}}$

$<br /> \frac{\frac{1}{a-1}+\frac{a-1}{a-1}}{\frac{1}{a+1}-\frac{a+1}{a+1}}$

$\neq$

$<br /> \frac{\frac{1+a-1}{a-1}}{\frac{1-a+1}{a+1}}$

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Sign problem on complex fractions

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