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From: ams@hou4b.UUCP
Newsgroups: net.math
Subject: Trivial Proof Needed
Message-ID: <1124@hou4b.UUCP>
Date: Thu, 30-Aug-84 19:42:32 EDT
Article-I.D.: hou4b.1124
Posted: Thu Aug 30 19:42:32 1984
Date-Received: Fri, 31-Aug-84 03:17:46 EDT
Organization: AT&T Information Systems Laboratories, Holmdel, NJ
Lines: 33

I need a proof of the following trivial assertion:

Let (- denote "is a member of"
Let |R denote the set of real numbers
Let |N denote the set of natural numbers

Let there be two numbers n and x, n (- |N and x (- |R.
Then let n be the greatest number such that n <= x.
We can denote functions that give the largest integer n <= x:
	n = int (x) = |_ x _| <= x

Now let R (x) denote the rounding function:
	n = R (x) = |_ x + 1/2 _|, where n (- |N and x (- |R.

Assertion:
	I maintain that:
			 
	            |  a + |_ b / 2 _| |
	R (a / b) = |  --------------- |; a,b (- |N, b != 0.
                    |___     b      ___|

For the case where b is even this is trivial, since int (b/2) = b/2
and int ((a + b/2)/b) = (a/b + 1/2) = R (a/b).

However, I don't seem to have a satisfactory proof for the case
where b is odd.

Any leads would be appreciated: I am sure the problem is really
trivial and I am not looking at it in the right way.

		Andrew Shaw AT&TISL
		834-4085 HO 1C-412A
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