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Density Property of Rational Numbers

Date: 09/21/2001 at 11:12:36
From: Benjamin Sutton
Subject: Density Property of Rational Numbers  

How was the density property of rational numbers proven?

Date: 09/21/2001 at 12:37:09
From: Doctor Paul
Subject: Re: Density Property of Rational Numbers  

If by the density property of rational number you mean the statement:

"between any two real numbers a and b there exists a rational number r 
such that a < r < b"

then the proof is as follows:

We first need the Archimedean Property:

if x > 0 and y > 0 are real numbers, then there exists a positive 
integer n such that n*x > y.  This tells us that even if a is quite 
small and b is quite large, some integer multiple of a will exceed b.  
Another way to think of this is, "given enough time, one can empty a 
large bathtub with a small spoon."

The Archimedean property has a proof, but I won't state it here 
because it requires a bit more math than you've probably had. If 
you're interested, write back and I'll provide it. Otherwise we'll 
just claim that it is inherently obvious.

Now back to the denseness of the rational numbers.

We need to show that given real number a and b we can construct 
a < m/n < b for some integers m and n.

We can assume that n > 0 without loss of generality. If our fraction 
m/n is to be negative, just let m be the negative number.

Since n > 0 we can multiply everything by n and we don't have to worry 
about flipping the inequality. Thus we need an < m < bn.

Since b > a we have b-a > 0 and so, by the Archimedean property, there 
exists a postive integer n such that n(b-a) > 1 (pick x = b-a and 
y = 1 in the statement of the Archimedean property above; we can do 
this since b-a and 1 are both greater than zero).

Since bn-an > 1 it is fairly evident that there is an integer m 
between an and bn. Does this make sense? If the difference between two 
numbers is greater than one unit (i.e., if two numbers are more than 
one unit apart), there must be an integer between them, since the 
integers are spaced *exactly* one unit apart. If this isn't clear, try 
to pick two numbers that are more than one unit apart such that there 
isn't an integer between them. I think you'll find yourself unable to 
do so.

Thus there exist integers m and n such that an < m < bn, which implies 

   a < m/n < b

- which was to be shown. Notice that this proof doesn't tell us how to 
find m/n. All it does is show that m/n exists. Such proofs are very 
common in higher mathematics.

- Doctor Paul, The Math Forum   
Associated Topics:
High School Number Theory

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