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From: steve@jplgodo.UUCP (Steve Schlaifer x3171 156/224)
Newsgroups: net.lang.c,net.arch
Subject: Re: Integer division
Message-ID: <561@jplgodo.UUCP>
Date: Fri, 31-Jan-86 15:25:31 EST
Article-I.D.: jplgodo.561
Posted: Fri Jan 31 15:25:31 1986
Date-Received: Sun, 2-Feb-86 05:42:24 EST
References: <332@ism780c.UUCP> <11603@ucbvax.BERKELEY.EDU> <11610@ucbvax.BERKELEY.EDU>
Distribution: net
Organization: Jet Propulsion Labs, Pasadena, CA
Lines: 55
Xref: watmath net.lang.c:7725 net.arch:2453
Summary: where does it say 0<=a%b<b in number theory?

In article <11610@ucbvax.BERKELEY.EDU>, gsmith@brahms.BERKELEY.EDU (Gene Ward Smith) writes:
> In article <11603@ucbvax.BERKELEY.EDU> weemba@brahms.UUCP (Matthew P. Wiener) writes:
> >In article <332@ism780c.UUCP> tim@ism780c.UUCP (Tim Smith) writes:
> >>[Which is preferred: (-a)%b == - (a%b) or (-a)%b >= 0 always?]
> >>Are there any good mathematical grounds for choosing one alternative over
> >>the other here?  Note that I am not asking from a hardware point of view
> >>which is better.  I want to know mathematically if one is better than
> >>the other.
> >
> >I have NEVER seen an instance where the first one is preferable.  Not
> >only is it not preferable, it is just incorrect.  Why such a routine
> >has been allowed to be 50% inaccurate in every existing language all
> >these years is beyond me.
> >
> >[Whether CS people should even be *allowed* to make such mathematical
> >decisions is another question.  In C on UNIX, for example, one has
> 
> >ucbvax!brahms!weemba	Matthew P Wiener/UCB Math Dept/Berkeley CA 94720
> 
>   Matthew has just about said it all, but since this has been my absolute
> *pet* gripe for some time now, I can't resist adding another $0.02 to the
> bill. When mathematicians define functions in a certain way, it is almost
> always for good reasons. I can think of only a few cases where doing 
> .......
> Then why mess around with quotient and remainder
> functions when you don't have a clue on God's green Earth what you are
> doing?
> 
> 
>      Signed
> 
>      An Angry Number Theorist

If you think of % as returning the *mathematical* remainder of a/b then
it should return a value >=0.  On the other hand, to be consistent with this
view, the quotient operator (/) will also have to be modified to preserve
the formulae

	b=qa+r (0<=r<a)
	q=b/a

i.e. (-3)/2 must be -2 if (-3)%2 is 1. But this then means that (|a|)/b is not
the same as |a/b| for a<0.  Maybe *An Angry Number Theorist* wants this, but it
seems to me to be a trap just waiting for the unwary to fall into.

As for why the restriction of 0<=r<a was decided on, my only guess is that
it then always produces a unique (q,r) for any given (a,b); this is a useful
property when you are proving theorems or doing theoretical investigations.
-- 

...smeagol\			Steve Schlaifer
......wlbr->!jplgodo!steve	Advance Projects Group, Jet Propulsion Labs
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