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Thread: Can't figure out this identity problem

  1. #1
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    Can't figure out this identity problem

    Hello, I'm studying for the GRE and i'm trying to figure out how (1/b-1/a)=(a-b/ab). If anyone can demonstrate the steps of how these two expressions are identical I would greatly appreciate it. Thanks.
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    Re: Can't figure out this identity problem

    common denominator is $ab$ ...

    $\dfrac{1}{b} - \dfrac{1}{a} = \dfrac{a}{ab} - \dfrac{b}{ab} = \dfrac{a-b}{ab}$
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    Re: Can't figure out this identity problem

    No, no one can explain "how these two expressions are identical" because they are not! What is true is that 1/b- 1/a= (a- b)/ab. To do that, get "common denominators" in the two fractions: (a/a)(1/b)- (b/b)(1/a)= a/ab- b/ab= (a- b)/ab.
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    Re: Can't figure out this identity problem

    Quote Originally Posted by rgragsda View Post
    Hello, I'm studying for the GRE and i'm trying to figure out how (1/b-1/a)=(a-b/ab).
    Quote Originally Posted by HallsofIvy View Post
    No, no one can explain "how these two expressions are identical" because they are not! What is true is that 1/b- 1/a= (a- b)/ab. To do that, get "common denominators" in the two fractions: (a/a)(1/b)- (b/b)(1/a)= a/ab- b/ab= (a- b)/ab.
    Both of you need grouping symbols around your product denominators.

    1/b - 1/a = (a - b)/(ab)

    (a/a)(1/b) - (b/b)(1/a)= a/(ab) - b/(ab) = (a - b)/(ab)
    Last edited by greg1313; Dec 30th 2016 at 02:17 PM.
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    Re: Can't figure out this identity problem

    Quote Originally Posted by greg1313 View Post
    Both of you need grouping symbols around your product denominators.

    1/b - 1/a = (a - b)/(ab)

    (a/a)(1/b) - (b/b)(1/a)= a/(ab) - b/(ab) = (a - b)/(ab)
    ... a plug for Latex?
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