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Inequality Proof for Greatest Integer

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Date: 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

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/
```
Associated Topics:
High School Number Theory

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