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Math Help - Opposite number proof

  1. #1
    Junior Member DIOGYK's Avatar
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    Opposite number proof

    Hello, I am stuck with this problem from Gelfand's Algebra book:
    Problem 126. Prove that if \frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c  }, then there are two opposite numbers among a, b, c (i.e: a=-b, a=-c, b=-c)
    To be honest, I don't understand the question.
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  2. #2
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    Re: Opposite number proof

    Hello, DIOGYK!

    I think I got it . . .


    \text{Prove that if }\,\frac{1}{a+b+c}\:=\:\frac{1}{a}+\frac{1}{b}+ \frac{1}{c}

    \text{then there are two opposite numbers among }a, b, c.

    \text{That is: }\,a=\text{-}b\,\text{ or }\,b=\text{-}c\,\text{ or }\,a=\text{-}c.

    We have: . \frac{1}{a+b+c} \;=\;\frac{1}{a} + \frac{1}{b} + \frac{1}{c}

    . . . . . . . . \frac{1}{a+b+c} \;=\;\frac{bc + ac + ab}{abc}

    . . . . . . . . . . . . abc \;=\;(a+b+c)(bc+ac+ab)

    . . . . . . . . . . . . abc \;=\;abc + a^2c + a^2b + b^2c + abc + bc^2 + ac^2 + abc

    . . a^2b + a^2c + ab^2 + 2abc + ac^2 + b^2c + bc^2 \;=\;0

    n . a^2(b + c) + a(b^2 + 2bc + c^2) + bc(b+c) \;=\;0

    . . . . . . . a^2(b+c) + a(b+c)^2 + bc(b+c) \;=\;0

    Factor: . . . . . (b+c)\big[a^2 + a(b+c) + bc\big] \;=\;0

    . . . . . . . . . . . (b+c)\big[a^2 + ab + ac + bc\big] \;=\;0

    . . . . . . . . . . (b+c)\big[a(a+b) + c(a+b)\big] \;=\;0

    Factor: . . . . . . . . . (a+b)(b+c)(a+c) \;=\;0


    We see that the product of three factors equals zero.
    Hence, at least one of the factors must be zero.

    Therefore: . \begin{Bmatrix}a+b \:=\:0 & \Rightarrow & a \:=\:\text{-}b \\ & \text{or} \\  b+c \:=\:0 & \Rightarrow & b \:=\:\text{-}c \\ & \text{or} \\ a+c\:=\:0 & \Rightarrow & a \:=\:\text{-}c \end{Bmatrix}
    Thanks from DIOGYK
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  3. #3
    Junior Member DIOGYK's Avatar
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    Re: Opposite number proof

    Quote Originally Posted by Soroban View Post
    Hello, DIOGYK!

    I think I got it . . .



    We have: . \frac{1}{a+b+c} \;=\;\frac{1}{a} + \frac{1}{b} + \frac{1}{c}

    . . . . . . . . \frac{1}{a+b+c} \;=\;\frac{bc + ac + ab}{abc}

    . . . . . . . . . . . . abc \;=\;(a+b+c)(bc+ac+ab)

    . . . . . . . . . . . . abc \;=\;abc + a^2c + a^2b + b^2c + abc + bc^2 + ac^2 + abc

    . . a^2b + a^2c + ab^2 + 2abc + ac^2 + b^2c + bc^2 \;=\;0

    n . a^2(b + c) + a(b^2 + 2bc + c^2) + bc(b+c) \;=\;0

    . . . . . . . a^2(b+c) + a(b+c)^2 + bc(b+c) \;=\;0

    Factor: . . . . . (b+c)\big[a^2 + a(b+c) + bc\big] \;=\;0

    . . . . . . . . . . . (b+c)\big[a^2 + ab + ac + bc\big] \;=\;0

    . . . . . . . . . . (b+c)\big[a(a+b) + c(a+b)\big] \;=\;0

    Factor: . . . . . . . . . (a+b)(b+c)(a+c) \;=\;0


    We see that the product of three factors equals zero.
    Hence, at least one of the factors must be zero.

    Therefore: . \begin{Bmatrix}a+b \:=\:0 & \Rightarrow & a \:=\:\text{-}b \\ & \text{or} \\  b+c \:=\:0 & \Rightarrow & b \:=\:\text{-}c \\ & \text{or} \\ a+c\:=\:0 & \Rightarrow & a \:=\:\text{-}c \end{Bmatrix}
    Sorry for the late reply and thank you for your help, I appreciate it. Your explanation was very clear. : )
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