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Prove that the cuberoot of 2 is irrational

  1. Jan 27, 2008 #1
    [SOLVED] Prove that the cuberoot of 2 is irrational

    1. The problem statement, all variables and given/known data
    Prove that the cuberoot of 2 is irrational

    3. The attempt at a solution
    Assume it is rational, and find a contradiction.
    2^(1/3) = a/b, where a, b are integers, where a/b is in lowest terms, and where b != 0.
    2 = a^3 / b^3
    2b^3 = a^3

    Aaaand stuck.
    Can someone give me a little push?
  2. jcsd
  3. Jan 27, 2008 #2


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    It works just like the square root proof. What happens if a is not divisible by 2?
  4. Jan 27, 2008 #3
  5. Jan 27, 2008 #4
    To be completely honest, neither of your answers really make sense to me :(
    I mean, I'm sure they're completely valid and correct answers, but I'm having trouble grasping what you're getting at.

    Can someone elaborate a little? I realize it has something to do with the definitions of a and b (Being integers, lowest terms, and b != 0), but I don't see what.

    And NateTG, I have the squareroot proof in my book, in front of me... but it uses the fact that a^2 and b^2 = even. I can't use that fact here, as a^3 will be of the same polarity as a, and b^3 the same polarity as b. I can't rely on it being even, nor odd.
  6. Jan 27, 2008 #5
    since 2b^3 = a^3 don't you know that a^3 is even? therefore making a even?

    then a = 2c for some integer c, and subbing back in you get 2b^3=(2c)^3=8c^3?

    does this help?
  7. Jan 27, 2008 #6
    If a^3 = 6, then a = cuberoot(6), which in theory would be irrational (However proving this would just be another huge circle).

    If a^3 is even, why is a even?
    Ooohh... 'cause a has to be an integer...

    Yeah, I think I see now.
  8. Jan 27, 2008 #7
    And then because a and b are both even, they're both divisible by 2, which breaks the hypothesis of "lowest common terms".
  9. Jan 27, 2008 #8
    If a is odd what's a^3, odd or even?
  10. Jan 27, 2008 #9


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    Yeah, which is the same proof as for square root 2.
  11. Jan 27, 2008 #10
    Yep, solved the problem now. Wanna know what was confusing me? For some reason, I thought that a^2 = Even, regardless of value of a. But then I sat there and thought... 3^2=9...wtf?

    Yeah... sorry for the stupid questions, and thanks for your help everyone :)
    With your help, I've not only completed this question, but blasted straight through the next one ^^;
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