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Math Help - Condition for equality to hold

  1. #1
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    Condition for equality to hold

    Question is

    Are this equal? 1/(a+b) =1/a + 1/b


    Does this have a condition for equality?
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  2. #2
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    Re: Condition for equality to hold

    You can go ahead and add the right side: \frac{1}{a}+ \frac{1}{b}= \frac{a+ b}{ab}.

    So you are asking for a and b that satisfy \frac{1}{a+ b}= \frac{a+ b}{ab}
    That is the same as (a+ b)^2= a^2+ 2ab+ b^2= ab.

    Solve the quadratic equation a^2+ ab+ b^2= 0.

    By the quadratic formula a= \frac{-b\pm\sqrt{b^2- 4b^2}}{2}= \frac{-1\pm\sqrt{-3}}{2}b
    Since that is not a real number, no, there are no real numbers a and b such that \frac{1}{a}+ \frac{1}{b}= \frac{1}{a+ b}.
    Last edited by HallsofIvy; August 8th 2014 at 06:26 AM.
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  3. #3
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    Re: Condition for equality to hold

    $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{a + b} \implies \dfrac{a + b}{ab} = \dfrac{1}{a + b} \implies a^2 + ab + b^2 = 0 \implies$

    $a= \dfrac{- b \pm b\sqrt{-3}}{2} \implies a,\ b \not \in \mathbb R.$

    In short, the equality does not apply to any pair of real numbers.
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  4. #4
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    Re: Condition for equality to hold

    Thank you!
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