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Math Help - Proving that the cube root of 5 is irrational

  1. #1
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    Unhappy Proving that the cube root of 5 is irrational

    I am trying to prove or disprove that the cube root of 5 is an irrational number.

    I have tried using a proof by contradiction, although am not convinced that I have a good answer.

    Can anyone help please?
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  2. #2
    Super Member
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    Try using the Rational Root Theorem :-)

    x = \sqrt[3]{5}

    x^3 = 5

    x^3 - 5 = 0
    Cube root of 5 is a root. Find all the possible rational roots. If they don't satisfy the equation, then there are no rational roots => cube root of 5 is irrational.
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  3. #3
    Junior Member
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    you do the proof via contradiction. assume that there exists a/b=sqrt5 where a and b are integers and are at their lowest common form (eg not 4/2 or 12/3). then using use that equation, prove that both a and b are even, thus they are divisible.
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  4. #4
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    The question says prove that the cube root of 5 is irrational

    Thank you for your post, but the question is prove that the cube root is irrational not the square root.
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  5. #5
    Senior Member JaneBennet's Avatar
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    If you are not familiar with the rational-root theorem, you can still tackle the problem directly.

    Assume that \sqrt[3]{5} is rational, so \sqrt[3]{5}=\frac{a}{b} where a,b\in\mathbb{Z}^+ and \gcd(a,b)=1.

    Thus \frac{a^3}{b^3}=5\ \Rightarrow\ a^3=5b^3

    \color{white}.\hspace{20mm}. \Rightarrow\ 5\mid a^3

    \color{white}.\hspace{20mm}. \Rightarrow\ 5\mid a

    \color{white}.\hspace{20mm}. \Rightarrow\ a=5c for some a,b\in\mathbb{Z}

    If you then plug this back into the equation, you would find that 5 divides b as well, contradicting the assumption that a and b are coprime.
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