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Prove log_2(m) is not an integer.

  1. Dec 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]m[/itex] be a positive integer that has an odd divisor greater than [itex]1[/itex]. Prove (by contradiction) that [itex]\log_2{m}[/itex] is not an integer.

    2. Relevant equations

    3. The attempt at a solution
    I'm not sure where to begin here. The book I'm working with provides a similar example, but it was more specific and straightforward than this one (prove [itex]\log_2{19}[/itex] isn't an integer).

    I think I'm having the most trouble with grasping the part saying "m has an odd divisor greater than 1." How do I interpret that?
     
  2. jcsd
  3. Dec 18, 2012 #2

    pasmith

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    m is divisible by an odd prime (since all odd numbers greater than 1 are either prime or products of odd primes).
     
  4. Dec 18, 2012 #3

    HallsofIvy

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    "m has an odd divisor greater than 1" simply means that m is NOT a power of 2.
    Given any number, we can factor out all factors of 2 to get the form [itex]2^n x[/itex] where x has no factors of 2 (and so is odd) and n is an integer. If x= 1, of course, that is a power of 2. If x> 1, it is not.

    If [itex]y= log_2(m)[/itex] then [itex]m= 2^y[/itex]. If y were an integer, then m would be a power of 2.
     
  5. Dec 18, 2012 #4
    Ah, I think I get it now. Here's the proof:

    Assume [itex]\log_2{m}[/itex] is an integer. Let [itex]y = \log_2{m}[/itex].
    Then, [itex]2^y = m[/itex].

    Since [itex]y[/itex] is assumed to be an integer, [itex]m[/itex] is a power of 2. This contradicts the fact that [itex]m[/itex] has an odd divisor greater than 1. Therefore, [itex]\log_2{m}[/itex] is not an integer.

    Thanks for the help!
     
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