Irrationality of Expressions

Date: 09/01/2003 at 00:30:56
From: Shan
Subject: rational or irrational?

How do you know whether (3 sqrt(2) - 1) is rational or irrational?
The square root makes it difficult to determine.

Date: 09/04/2003 at 15:30:27
From: Doctor Marshall
Subject: Re: rational or irrational numbers

Dear Shan,

Statements about the rationality of complicated expressions are best
dealt with in their most general form.

We can start with a classic proof that sqrt(2) is irrational.

Proof.  Let sqrt(2) be expressed as a ratio between two integers m,n
and choose m,n such that at most one is even.  (Such a choice is
always possible, since any ratio of even numbers can be reduced.)

                   m/n = sqrt(2)

Then,  
                   m^2
                   --- = 2
                   n^2

So
                   m^2 = 2 * n^2

Hence m^2 is even, hence m is even, hence m^2 is divisible by 4.

Now let m/2 = k,

so        m^2
          --- = 2*k^2 = n^2 (from above).
           2

This shows that n is even which contradicts our original assumption
(that at most one of the numbers we chose could be even).  The
conclusion is that sqrt(2) CANNOT be expressed as the ratio of two
integers.

Next we can investigate the general expression nx where n is rational 
and x is irrational.

Suppose nx is rational and let n=a/b and nx=p/q.

             nx     p/q     pb
Then    x = ---- = ----- = ----.  
             n      a/b     qa

Hence x is rational, which contradicts our choice of x, therefore nx 
IS NOT rational.

We can also investigate whether x + n is rational.  Again, assume it 
is.  Let n = a/b and x+n = p/q. 
                                   pb     aq     pb-ab
Then  x = x + n - n = p/q - a/b = ---- - ---- = -------
                                   qb     bq       bq

Since both numerator and denominator are integers, we see that x is 
rational which again contradicts our choice of x.  Hence, x+n is 
IRRATIONAL.

I hope you find this answers your question, please feel free to write 
back with more questions if you have any.

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