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Math Help - Counting numbers/set proof

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    Counting numbers/set proof

    Prove by contradiction that there is no set B of counting numbers such that if n is an element of B, then (n-1) is an element of B.
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  2. #2
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    Proof by induction and contradiction

    Hello noles2188
    Quote Originally Posted by noles2188 View Post
    Prove by contradiction that there is no set B of counting numbers such that if n is an element of B, then (n-1) is an element of B.
    In fact, B = \oslash is such a set (*see the Note below), so we must assume that B \ne \oslash.

    So, let us assume that \exists B \subseteq \mathbb{N}, (B \ne \oslash), and (n \in B) \Rightarrow (n-1) \in B. (In other words, let's assume the opposite of what we want to prove.)

    Then let P(n) be the propositional function: n \in B

    Then by hypothesis P(n) \Rightarrow P(n-1)

    Now B \ne \oslash \Rightarrow P(k) is true for some k \in B

    So, by induction P(n) is true \forall n \le k, n \in \mathbb{Z}

    \Rightarrow P(0) is true \Rightarrow 0 \in B. But 0 \notin \mathbb{N}. Contradiction. Therefore the original assumption is false.

    Grandad

    *Note: the proposition (p \Rightarrow q) is true whenever p is false. So if B = \oslash, the proposition (n \in B \Rightarrow (n-1) \in B) is true, because n \in B is always false. It seems a bit crazy at first sight, but it is OK!
    Last edited by Grandad; March 9th 2009 at 06:05 AM. Reason: Improved solution
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