Do Rational and Irrational Numbers Alternate?

Date: 10/13/2000 at 10:43:05
From: Joe Cooper
Subject: Do rational and irrational numbers alternate?

It is known that:

   1. Any two non-equal real numbers "contain" an irrational.
   2. Any two non-equal real numbers "contain" a rational.

These real numbers can of course be rational or irrational. Does this 
mean rational and irrational numbers alternate?

Date: 10/13/2000 at 11:44:07
From: Doctor Rob
Subject: Re: Do rational and irrational numbers alternate?

Thanks for writing to Ask Dr. Math, Joe.

Yes, both statement 1 and statement 2 are true. The answer to your 
question, however, is "No." If that were so, the two sets would have 
the same size, and they don't (the irrationals are uncountable, the 
rationals are countable).

The proofs of these depend on the Archimedean Ordering Principle: if 
r > 0 and s > 0 are real numbers, then no matter how small r is and 
how large s is, there is a positive integer n such that n*r > s.

To prove the statement 2, for example, let the two unequal real 
numbers be x < y. Then let r = y - x > 0, and s = 1 + r. Then by the 
A.O.P., there is a positive n such that

     n*r > 1 + r,
     n*y - n*x > 1 + r,
     n*x < n*y - r - 1.

Now consider integer m = [n*y-r]. (Here [x] means the greatest integer 
less than or equal to x, that is, the integer you get by rounding down 
from x. It satisfies x-1 < [x] <= x.) Then we can see that

     n*x < n*y - r - 1 < m <= n*y - r < n*y,
     x < m/n < y.

Thus m/n is a rational number in the interval (x,y).

To get an irrational number in the interval (x,y), let z be any 
positive irrational number. Now use the above argument to find a 
rational number m/n in the interval (x/z,y/z). Then z*m/n will be in 
(x,y) and irrational.

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