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Math Help - Prove (2^(1/2)) + (2^(1/3)) is irrational

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    Prove (2^(1/2)) + (2^(1/3)) is irrational

    This is what I have

    2^(1/2) + 2^(1/3) = r with r being a rational number
    2^(1/2) = r - 2^(1/3) then square both sides
    2 = r^(2) - 2r(2)^(1/2) + 4^(1/3)
    2r(2)^(1/2) = r^(2) + 4^(1/3) -2
    2^(1/2) = [r^2 + 4^1/3 -2 ]/2r. Here is my problem, how can I make the RHS rational?
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  2. #2
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    Re: Prove (2^(1/2)) + (2^(1/3)) is irrational

    Hey dave52.

    Square both sides and show that 4^(1/3) is irrational and since r^2 - 2 is also rational and irrational + rational = rational, then you are done.
    Thanks from topsquark
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  3. #3
    Junior Member Nehushtan's Avatar
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    Re: Prove (2^(1/2)) + (2^(1/3)) is irrational

    Quote Originally Posted by dave52 View Post
    This is what I have

    2^(1/2) + 2^(1/3) = r with r being a rational number
    2^(1/2) = r - 2^(1/3) then square both sides
    2 = r^(2) - 2r(2)^(1/2) + 4^(1/3)
    2r(2)^(1/2) = r^(2) + 4^(1/3) -2
    2^(1/2) = [r^2 + 4^1/3 -2 ]/2r. Here is my problem, how can I make the RHS rational?
    You have a typo in the third line. You want

    2 = r^(2) - 2r(2)^(1/3) + 4^(1/3)

    I would suggest however that you put the 2^{\frac13} on the left-hand side and cube it:

    2^{\frac13}=r-2^{\frac12}
    2=r^3+6r-(3r^2+2)\sqrt2

    Now itís easy to see the required contradiction that \sqrt2 is rational.
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