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Are my proofs 'correct'?

  1. Aug 23, 2011 #1
    I can never seem to create proofs the way it is shown in every textbook I've seen. To be honest, I don't really know how to write the proofs correctly. I've seen sometimes my reasons are flawed and other times I go around aimlessly and get home after some unnecessary steps. So I would just like some feedback here:

    1. The problem statement, all variables and given/known data
    Prove that (log10 a)/(log10 b) is not rational if a and b are relatively prime and both are greater than 1.

    2. Relevant equations

    3. The attempt at a solution
    Suppose that (log10 a)/(log10 b) is rational, then they are expressed as fraction:
    (log10 a)/(log10 b) = p/q, where p and q are integers. q not 0. p not 0 because a is greater than 1.
    Let 10x = a
    and 10y = b
    x and y cannot be zero because a and b cannot be 1.
    Now if a and b are relatively prime, ie they have only 1 as their common factor then x and y too are relatively prime.

    Suppose that x and y share a common factor greater than 1, then:
    let x = kc
    and y = kz
    where k is the common factor, which is not negative or 0 because a and b are greater than 1 and c and z are also not 0 or negative.
    So then, 10kc = a
    and 10kz = b
    x = log10 a
    y = log10 b
    So, kc = log10 a
    and kz = log10 b
    But then, (10k)c = a
    and (10k)z = b
    And this contradicts the fact that a and b are relatively prime. So x and y are also relatively prime.

    So now we can write:
    x/y = p/q
    i.e, xq = py
    This cannot be true because x and y are relatively prime. This means x/y is not rational.
    And since x = log10 a
    and y = log10 b
    (log10 a)/(log10 b) is not rational.

    Last edited: Aug 23, 2011
  2. jcsd
  3. Aug 23, 2011 #2


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    Science Advisor
    Homework Helper

    Hi Thinker8921! :smile:

    hmm … this is really complicated :redface:

    you're introducing x and y for no particular reason. :confused:

    Just do …
    … then write that as q(log10 a) = p(log10 b), and carry on from there. :smile:

    (or better still use logb a, if you know how to do that)
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