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Math Help - Two Enumeration Questions

  1. #16
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    are we finished with f(a).b and when a is odd f(a)=-(a+1)/2 and when a is even f(a)=a/2 ?
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  2. #17
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    Quote Originally Posted by DrSteve View Post
    \mathbb{N}\rightarrow \mathbb{N}\times \mathbb{N}\rightarrow \mathbb{Z}\times \mathbb{N}\rightarrow \mathbb{Q}.

    The first is the inverse of the pairing function. The second is the map (a,b)\rightarrow (f(a),b) where f is the function we've previously discussed, and the third is the map (a,b)\rightarrow a.b.
    For the function suggested by Plato, one can take p_1(n)=\max\{k\in\mathbb{N}\mid\text{$2^k$ divides $n$}\}+1 and p_2(n)=(n/p_1(n)+1)/2. (One must check whether p_1(n) and p_2(n) need to assume 0 and whether they actually do.) Then n\mapsto(p_1(n),p_2(n)) is a bijection \mathbb{Z}^+\to\mathbb{Z}^+\times\mathbb{Z}^+, which is the first step in the chain at the top of this post.
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  3. #18
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    Thanks emakarov - I didn't remember which direction was technically called the pairing function.

    gutnedawg, you essentially got my idea right. The bijection F:\mathbb{N}\rightarrow A is given by F(n)=f(a).b where f is the function you described and \pi(a,b)=n, where \pi is the pairing function.

    (Here A is the set of rational numbers with terminating decimal expansions.)
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