Inequality Proof for Greatest IntegerDate: 10/27/2001 at 23:43:59 From: Lisa G Subject: Proof for greatest integer [x] in n<=x<n+1 The question I've been struggling with for the past week is: If x is an arbitrary real number, prove that there is exactly one integer n that satisfies the inequalities n<=x<n+1. I understand the concept of the inequalities, and the greatest integer function; but it seems so obvious to me that I don't know how to go about proving it. Thank you for any and all help you can give me. Lisa Date: 10/29/2001 at 04:48:52 From: Doctor Floor Subject: Re: Proof for greatest integer [x] in n<=x<n+1 Hi, Lisa, Thanks for writing. Suppose there are two integers n and m with n <= x < n+1, m <= x < m+1, or equivalently n <= x < n+1, m+1 > x >= m. We will show that this means n = m. By subtraction we get n-m-1 < 0 < n-m+1 (*). Since adding and subtracting integers gives integers again, we see that n-m-1 and n-m+1 are integers. Also the successor of n-m-1 is n-m, and the successor of n-m is n-m+1. That means that there is only one integer x satisfying n-m-1 < x < n-m+1 and that is x = n-m. Together with (*) this shows that n = m. If you have more questions, just write back. - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/