# Thread: Sign problem on complex fractions

1. ## Sign problem on complex fractions

Ok. Here is my problem.

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

And my aproach

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

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

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

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

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

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.

2. Hi

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

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

4. From this

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

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

5. How would you do that?

6. ## A little Clarification

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?

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

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

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

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

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

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

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

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

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

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

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

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

Now my question is which one is the right one:

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

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

c) None. Then which______________________________?

8. $\displaystyle \frac{\frac{1}{a-1}+\frac{a-1}{a-1}}{\frac{1}{a+1}-\frac{a+1}{a+1}}$

$\displaystyle =$

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

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

$\displaystyle \neq$

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