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 http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum