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Math Help - Basic proof where x is a real number

  1. #1
    Member mybrohshi5's Avatar
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    Basic proof where x is a real number

    We just started proofs in my mathematical structures class so this is the first time i am dealing with proofs.

    The question i am having trouble proving is:

    Suppose x is a real number and x  \neq 0. Prove that if  \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x} then  x \neq 8.

    This is what i have started:

    Givens:
    x is a real number
    x  \neq 0
     \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x}

    Goals:
     x \neq 8.

    i feel like to complete this proof you would solve the equation  \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x} for x and then this would show that whatever this value of x is, it will not be equal to 8.

    The problem i am running into is that i cannot solve the equation  \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x} for x equaling a value, so i feel like i may be going about this wrong.

    Any help / tips for proofs would be very great

    Thank you
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by mybrohshi5 View Post
    Suppose x is a real number and x  \neq 0. Prove that if  \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x} then  x \neq 8.

    Prove:


     \dfrac{8^{1/3} + 5}{8^2+6} \neq \dfrac{1}{8}



    Fernando Revilla
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by mybrohshi5 View Post
    We just started proofs in my mathematical structures class so this is the first time i am dealing with proofs.

    The question i am having trouble proving is:

    Suppose x is a real number and x  \neq 0. Prove that if  \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x} then  x \neq 8.

    This is what i have started:

    Givens:
    x is a real number
    x  \neq 0
     \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x}

    Goals:
     x \neq 8.

    i feel like to complete this proof you would solve the equation  \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x} for x and then this would show that whatever this value of x is, it will not be equal to 8.

    The problem i am running into is that i cannot solve the equation  \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x} for x equaling a value, so i feel like i may be going about this wrong.

    Any help / tips for proofs would be very great

    Thank you
    Have you thought about the contra positive of the statement.
    Remember that If a, then b is logically equivilent to if not b, then not a

    If x=8, then \displaystyle  \frac{x^{1/3} + 5}{x^2+6} \ne \frac{1}{x}
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