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Abstract Algebra Proof (factorization?)

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    If n is a nonzero integer, prove that n cam be written uniquely in the form n=(2^k)m, where k is greater than or equal to zero, and m is odd

    2. Relevant equations

    It is in the primes and unique factorization chapter so maybe that every integer n (except 0 and 1) can be written as a product of primes

    3. The attempt at a solution

    I am not sure, but I tried to do a proof by induction on n

    Let n=1
    so, 1=(2^k)(m)
    1=(1)(1)=1 where k=0 and m=1
    so the statement holds for n=1

    Now assume the statement holds for n=(r-1) for any integer (except 0 and 1)
    (r-1)=(2^k)(m) where k is greater than or equal to zero and m is odd

    Now let n=r
    so r=(2^k)(m)

    This is where I don't know what to do.... that is why i'm thinking there is an easier way to do this besides induction. Any help would be appreciated =)
  2. jcsd
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