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 intervals 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 http://mathforum.org/dr.math/
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