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Math Help - Contradiction and rational numbers

  1. #1
    Senior Member I-Think's Avatar
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    Contradiction and rational numbers

    I have a set S\subset{\mathbb{Q}} that is closed under addition and multiplication. Additionally, for any r\in{\mathbb{Q}}, one of the three is true
    r\in{S}, -r\in{S}, r=0

    I have to prove that all positive integers are in S.
    So I assume there exists a positive integer z\not\in{S}, hence -z\in{S}.
    Not sure which direction to go in to get the contradiction.
    Help please?
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  2. #2
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    Re: Contradiction and rational numbers

    Quote Originally Posted by I-Think View Post
    I have a set S\subset{\mathbb{Q}} that is closed under addition and multiplication. Additionally, for any r\in{\mathbb{Q}}, one of the three is true r\in{S}, -r\in{S}, r=0
    If you can show that 1\in S then you are done because S is closed by addition.

    You know that 1\ne 0. So if 1\notin  S then -1\in S.

    What is (-1)(-1)~?. What is the contradiction there?
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  3. #3
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    Re: Contradiction and rational numbers

    Quote Originally Posted by I-Think View Post
    I have a set S\subset{\mathbb{Q}} that is closed under addition and multiplication. Additionally, for any r\in{\mathbb{Q}}, one of the three is true
    r\in{S}, -r\in{S}, r=0

    I have to prove that all positive integers are in S.
    So I assume there exists a positive integer z\not\in{S}, hence -z\in{S}.
    Not sure which direction to go in to get the contradiction.
    Help please?
    You can't prove it. It is not true.
    For example S= {0} satisfies all of the conditions but does not include all of the integers.

    You need some additional condition, perhaps the "1 in S" that Plato suggested.
    Thanks from johng
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    Re: Contradiction and rational numbers

    Quote Originally Posted by I-Think View Post
    Additionally, for any r\in{\mathbb{Q}}, one of the three is true
    r\in{S}, -r\in{S}, r=0
    Quote Originally Posted by HallsofIvy View Post
    For example S= {0} satisfies all of the conditions but does not include all of the integers.
    I don't think S = {0} satisfies \forall r\in\mathbb{Q}.\,r\ne0\to r\in S\lor -r\in S.
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    Re: Contradiction and rational numbers

    Quote Originally Posted by emakarov View Post
    I don't think S = {0} satisfies \forall r\in\mathbb{Q}.\,r\ne0\to r\in S\lor -r\in S.
    You are right. I was thinking the condition was " r\in S".
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