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Math Help - Sign problem on complex fractions

  1. #1
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    Sign problem on complex fractions

    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.
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  2. #2
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    Hi

    \frac{a}{a} = 1 therefore you get -\frac{a+1}{a-1}
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  3. #3
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    Mmm...Sorry, but I can't see the point...
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  4. #4
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    From this

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

    you can simplify by "a", since "a" is a factor of both numerator and denominator
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  5. #5
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    How would you do that?
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  6. #6
    No one in Particular VonNemo19's Avatar
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    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?
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  7. #7
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    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______________________________?
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  8. #8
    Master Of Puppets
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    <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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