Proof by ContradictionDate: 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/ |
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