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Thread: Denumerable Set

  1. #1
    Sep 2009

    Denumerable Set

    My book shows that the set of integers, \mathbb{Z}, is denumerable by expressing the elements of the set as infinite sequence

    \mathbb{Z}=\{0, 1,-1,2,-2,...\}, and then shows that the function

    f: \mathbb{N}\rightarrow \mathbb{Z} is bijective, where

    f is defined as


    The picture shows that there is one-one correspondence, such as this:

    1 \rightarrow 0

    1 \rightarrow 1

    2 \rightarrow -1

    3 \rightarrow 2

    4 \rightarrow -2


    Now I want to prove that f(n)=\frac{1+(-1)^n(2n-1)}{4} bijective, but (-1)^n in the expression is making it very clumsy.

    To prove the one-to-one isn't too bad, since f(a_1)=f(a_2) seems to look clean, but now when I get to the part where I need to prove it an onto function, I got a huge mess, so I am wondering whether it be alright to split the proof in two parts such that

    f(n) = g(s) \cup h(t) and define g(s) \subset \mathbb{Z}^- and h(t) \subset \mathbb{Z}^+ and by so doing, I plan to prove that

    there exists an odd positive integer p\in \mathbb{N} such that g(s)=\frac{1+(-1)^s(2s-1)}{4} where s=2n-1, n \in \mathbb{N}

    and there also exists an even positive integer q\in \mathbb{N} such that h(t)=\frac{1+(-1)^t(2t-1)}{4} where t=2n, n \in \mathbb{N}


    1. Is there a simpler way than to prove the onto-function in two parts?

    2. Is there a better way with handling (-1)^n expression ?
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  2. #2
    MHF Contributor

    Aug 2006
    If x=0 then f(1)=0.
    If x>0 then f(2x)=x.
    If x<0 then f(2|x|+1)=x.
    So it is onto.
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