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Rational/Irrational Numbers

Date: 03/12/99 at 14:29:06
From: Jim Melbourne
Subject: Intermediate Nature of rational/irrational Numbers 

I have been asked to prove that between any two numbers, there is a 
rational number and an irrational number and, therefore, infinite 
numbers of each. Can you help me? 

Date: 03/15/99 at 13:02:07
From: Doctor Nick
Subject: Re: Intermediate Nature of rational/irrational Numbers 

There are quite a few ways to prove this; here is one.

Suppose your two numbers are x and y, with x < y. If y - x > 1, then 
there is an integer r between x and y, and integers are rational, and 
we are done. If y - x is not greater than 1, then consider the 
   10x to 10y
   100x to 100y
   1000x to 1000y

and so forth. Eventually, one of these intervals will be longer than 1, 
since each one is ten times longer than the previous one. In other 
words, for some n, we will have 

   (10^n)y - (10^n)x > 1

This means there must be an integer in the interval: there will be an 
integer r so that

   (10^n)x < r < (10^n)y

Now, divide through by (10^n) and we get

   x < r/(10^n) < y

Since r is an integer, r/(10^n) is rational.

To find an irrational number, you can do a similar thing. When 
(10^n)y - (10^n)x > the square root of 2 = sqrt(2), the interval 
contains an integer multiple of sqrt(2). That is, for some integer n, 
and some integer t, we have

   (10^n)x < t*sqrt(2) < (10^n)y

and hence

   x < (t/(10^n))*sqrt(2) < y

Since sqrt(2) is irrational, every non-zero rational multiple of 
sqrt(2) is irrational; hence, (t/(10^n))*sqrt(2) is irrational.

Make sure that every step makes complete sense to you. Write back if 
you do not understand any of it.

- Doctor Nick, The Math Forum   
Associated Topics:
College Analysis
High School Analysis

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