Proof: log(x) is irrational

  • Thread starter mattmns
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  • #1
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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?
 
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  • #2
Hurkyl
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mattmns said:
Do I now try to prove that [tex]\sqrt[n]{2^m}[/tex] is irrational
You just need to prove it's not 7.
 
  • #3
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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]
 
  • #4
Curious3141
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mattmns said:
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]
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.
 
  • #5
Tide
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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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