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Proof: log(x) is irrational

  1. Aug 28, 2005 #1
    I am to prove that [tex]\log_{2} 7[/tex] is irrational. So I started by saying that what if [tex]\log_{2} 7[/tex] is rational. Then it must be in the form of [tex]\frac{m}{n}[/tex] where m and n are integers. So now [tex]\log_2 7 = \frac{m}{n}[/tex] So I took the 2^ up of each and now [tex]7 = 2^{\frac{m}{n}}[/tex] Then [tex]7 = \sqrt[n]{2^m}[/tex] But now I seem to be lost. Do I now try to prove that [tex]\sqrt[n]{2^m}[/tex] is irrational, or what do I need to do. Any ideas? Maybe proove that [tex]2^{\frac{m}{n}} \neq 7[/tex] by 2^{anything rational} must be something?
    Last edited: Aug 28, 2005
  2. jcsd
  3. Aug 28, 2005 #2


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    You just need to prove it's not 7.
  4. Aug 28, 2005 #3
    How about:

    [tex]7 = 2^{\frac{m}{n}}[/tex]

    So, [tex]7^n = 2^m[/tex]

    which is a contradiction because, [tex]7^n[/tex] is always odd while [tex]2^m[/tex] is always even, for n and m as integers and [tex]n \neq 0[/tex]
  5. Aug 28, 2005 #4


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    In this case, that approach is fine. But it wouldn't work, for example if the question asks you about base 3 logs. In that instance, use the uniqueness of prime factorisation.
  6. Aug 28, 2005 #5


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    More generally, if you wanted to prove it for a different base (e.g. 3) you could simply invoke the prime factorization theorem at that last step.
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