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Thread: 3 of 3 prove the following contradiction.

  1. #1
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    3 of 3 prove the following contradiction.

    There does not exist a largest negative rational number.

    Assume, that there is a largest negative rational number; call it k.

    Then 2k, which is the product of two rational numbers, is a negative rational number.

    Since 2k is farther from 0 than k is, we have that 2k < k, which implies that k < k/2.

    Now, k/2 = 1/ 2 k is the product of two rational numbers.

    so k/2 is a negative rational number that islarger than k,

    resulting in a contradiction if we sub a value in for K .

    Is that ok ?

    I'm off to bed so forgive if I don't reply straight away but will when I wake up tomorrow cheers,
    Last edited by bee77; Aug 23rd 2017 at 06:43 AM.
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  2. #2
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    Re: 3 of 3 prove the following contradiction.

    Quote Originally Posted by bee77 View Post
    There does not exist a largest negative rational number.
    Assume, that there is a largest negative rational number; call it k.
    Then 2k, which is the product of two rational numbers, is a negative rational number.
    Since 2k is farther from 0 than k is, we have that 2k < k, which implies that k < k/2.
    Now, k/2 = 1/ 2 k is the product of two rational numbers.
    so k/2 is a negative rational number that islarger than k,
    resulting in a contradiction if we sub a value in for K .
    Is that ok ?
    To echo Prof Ivey, it is very rough correct.

    You should try to prove that "between any two numbers there is a rational number."
    That theorem gives this "There is no largest negative number, and no smallest positive number".
    Thanks from bee77
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  3. #3
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    Re: 3 of 3 prove the following contradiction.

    Thanks Plato I'll work on it ..
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  4. #4
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    Re: 3 of 3 prove the following contradiction.

    Why not just multiply by 1/2 from the start?
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