Irrational NumbersDate: 01/13/98 at 05:53:41 From: Chew zonghui Subject: Irrational numbers How do you prove that (sqrt)2 is irrational? Given two irrational numbers, a and b, is it possible to get an a^b that is rational? Date: 01/16/98 at 13:48:26 From: Doctor Joe Subject: Re: Irrational numbers Dear Zonghui, The first question is easy. I shall show you the first few steps: Let r be the sqrt(2). Suppose that r = a/b reduced to its lowest terms (i.e. we are supposing that r is rational). Now, squaring both sides, we have: r^2 = a^2/b^2 => 2*b^2 = a^2 (remember that r^2 = 2) Since 2 is a prime number, 2 must divide a^2 (denoted by 2|a^2, which means that a^2 is a multiple of 2. It follows that a is even and thus may be written as a = 2a'. Substituting into the previous equation, 2*b^2 = (2a')^2 => 2*b^2 = 4(a')^2 => b^2 = 2a' Using the same argument earlier, try to show that a and b thus have a common factor other than 1 (technically called not coprime) (i.e. a/b is not reduced to the lowest terms as we assumed). Hence, we arrive at a contradiction, meaning that the original supposition, that r is rational, is wrong. Hence, r is irrational. (QED) Regarding the second question, let a = e (Euler's natural base = 2.71828...) let b = ln 2 (the natural logarithm of 1, based e) Both a and b are irrational, but a^b = 2 is clearly rational! Cheers. -Doctor Joe, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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