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Math Help - Proof by Contradiction

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    Question Proof by Contradiction

    Struggling with this problem:
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by somestudent2 View Post
    Struggling with this problem:
    a) 0 cannot be part of S because if it were then all three of r \in S, -r \in S, and r = 0 are true. But only one of them is allowed to be true.

    b) I will simply note that the positive integers are closed under addition and multiplication. You can supply the rest of the details.

    c) I would like to say the same thing about c) as I did about b): the positive rationals are closed under addition and multiplication. However, this set excludes the negative rationals, and I'm not sure exactly how to go about doing that. The exclusion of 0 from the set is going to be important, I can tell you that much.

    -Dan
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  3. #3
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    Notice first that if a∈S then a+a=2a∈S, and by induction na∈S for every positive integer n.

    If –1∈S then (–1)(–1)=1∈S, and this contradicts the defining property of S. So –1∉S and therefore 1∈S. By the previous paragraph, n∈S for every positive integer n. Therefore n∉S for every negative integer n.

    If a/b is a negative rational number, where a is a negative integer and b is a positive integer then, by the first paragraph, b(a/b)∈S ... .
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  4. #4
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    Thank You very much guys, this really helps
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