Proof by Contradiction

Date: 05/09/2003 at 07:36:23
From: Fred
Subject: Irrational

Show that the number

   log 7
      2
   -----
   log 6
      2

is irrational.

Date: 05/09/2003 at 10:31:27
From: Doctor Luis
Subject: Re: Irrational

Hi Fred,

In this problem, we'll assume log is always base 2.

In order to prove that log(7)/log(6) is irrational, we will be
assuming that it IS rational, then show this leads to an absurdity. 
This method is known "Proof by Contradiction" and is a pretty powerful 
way to prove a certain kind of theorem.

If the number log(7)/log(6) is rational, then we have to suppose that

 log(7)/log(6) = p/q

for some integers p,q such that p,q have no common factors.

Then, cross-multiplying

  q log(7) = p log(6)

  log(7^q) = log(6^p)

       7^q = 6^p

Here comes the trick. We know the number 7 divides 7^q (obviously). 
But according to the equation above, 7 would also have to divide 6^p, 
which means 7 would have to appear in the prime factorization of 6^p 
at least once. However,

  6^p = (2*3)^p = 2^p * 3^p

It follows that only the primes 2 and 3 appear in the prime 
factorization, so it is impossible for 7 to divide 6^p.

We have eliminated the choice of log(7)/log(6) being a rational 
number. Therefore it has to be irrational (the only possible second 
choice). End of proof.

I hope this helped. Let us know if you have any more questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/