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Math Help - Proofs about irrational numbers.

  1. #1
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    Proofs about irrational numbers.

    Hi,

    Prove or disprove that

    a. sum of two distinct irrational numbers is irrational.
    b. Product of two distinct irrational numbers is irrational
    Last edited by mr fantastic; August 28th 2011 at 05:41 AM. Reason: Re-titled.
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  2. #2
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    Re: I need help for these questions

    Quote Originally Posted by MathsNewbie0811 View Post
    Hi,

    Prove or disprove that

    a. sum of two distinct irrational numbers is irrational.
    b. Product of two distinct irrational numbers is irrational
    a) Use a simple counterexample, like \displaystyle \sqrt{2} + \left(-\sqrt{2}\right)

    b) Use a simple counterexample, like \displaystyle \sqrt{2} \times \sqrt{2}...
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  3. #3
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    Re: I need help for these questions

    Quote Originally Posted by Prove It View Post
    a) Use a simple counterexample, like \displaystyle \sqrt{2} + \left(-\sqrt{2}\right)

    b) Use a simple counterexample, like \displaystyle \sqrt{2} \times \sqrt{2}...
    I suspect the OP's use of the word 'distinct' implies 'different'.

    @OP: 1 + \sqrt{2} and 1 - \sqrt{2} provide the necessary counter example for disproving the statements in both (a) and (b).
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  4. #4
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    Re: I need help for these questions

    Or simply \sqrt{2}*\frac{1}{\sqrt{2}}= \sqrt{2}\frac{\sqrt{2}}{2} which was probably what Prove It intended.
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  5. #5
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    Re: I need help for these questions

    I've always liked this proof that an irrational to an irrational power need not be irrational: consider \sqrt2^{\sqrt2}. If that's rational, then we have our counterexample; otherwise, a counterexample is \left(\sqrt2^{\sqrt2}\right)^{\sqrt2}={\sqrt{}2}^{  \sqrt{2}\sqrt{2}}=\sqrt2^2=2.
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