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Thread: One more proof

  1. #1
    Senior Member
    Jan 2009

    One more proof

    Prove that for all natural numbers k,m, Nk X Nm is finite.

    I know the lemma "for every k belong to N, every subset of Nk is finite", could we just say that Nk X Nm is a subset of Nk, so it is finite....?

    The hint is "consider the function f: NkXNm--->Nkm given by f(a,b)=(a-1)m+b. Use the Division Algorithm to show f is a one-to-one correspondence. This is a formal version of the product rule that says the number of elements in NkXNm is km

    This makes it more complicated than I thought.
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  2. #2
    MHF Contributor

    Aug 2006
    I suggest this mapping: $\displaystyle \Phi :N_k \times N_m \mapsto N_{\left( {2^k } \right)\left( {3^m } \right)} \,,\,\Phi (g,h) = \left( {2^g } \right)\left( {3^h } \right)$.
    Of course $\displaystyle N_{\left( {2^k } \right)\left( {3^m } \right)} \,$ is finite.
    It is easy to show that $\displaystyle \Phi$ is one to one.

    Here are theorems that you need to do these sorts of cardinality problems.
    T1 There is an injection from A to B if and only if there is a surjection from B to A.

    T2 If there is an injection from A to B then the cardinality of A at most the cardinality of B.
    i.e. $\displaystyle |A|\le |B|$.

    T3 If there is an surjection from B to A then the cardinality of B at least the cardinality of A.
    i.e. $\displaystyle |B|\ge |A|$.
    Last edited by Plato; Apr 19th 2009 at 03:28 PM.
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