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Math Help - Direct Proof

  1. #1
    Junior Member
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    Direct Proof

    Hi people,

    I think its true that for every positive natural number, lets call it n, can be put in form of 2^k * m (where m is an odd number).

    I think thats correct? Please confirm...

    How would you use direct proof to proof this?? I cant get grips of it. Its abit advanced for me


    Thanks.
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  2. #2
    MHF Contributor chisigma's Avatar
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    If we consider 1 an odd number your are right...

    Kind regards

    \chi \sigma
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  3. #3
    Junior Member
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    if n is odd choose k = 0 and m = n.
    if n is even then write n in its prime decomposition.

    n=2^rp_1p_2p_3\dots

    Choose k=r and choose m to be the product of primes, p_i. The product of primes is guaranteed to be odd because an odd times an odd is an odd and no primes other than 2 are even. If there are no p_is (ie. n is a power of 2), then take m=1.
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