Irrational Numbers

Date: 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/