MOD function with negative numbers

[Joe Boyle]

I always assumed that it meant 'return the remainder of integer i, after
multiples of j are removed', and given that i = (n * j + k) * -1 when i is
negative, that mod(i,j) would return k * -1, which is what you are getting
from PH; given that -1 = ( 0 * j + k ) * -1, for k = 1.

Perhaps, if you specify 7 as -7 in your expression (I feel the presence of
the 'E universal constant' error approaching), you might get what you
require ?

Regards, Joe.



-----Original Message-----
From: powerh-l-bounces+atla38=dsl.pipex.com at lists.sowder.com
[mailto:powerh-l-bounces+atla38=dsl.pipex.com at lists.sowder.com] On Behalf Of
roger32909 at bellsouth.net
Sent: 01 March 2006 21:36
To: powerh-l at lists.sowder.com
Subject: MOD function with negative numbers

If I use the mod function in Excel, say =mod(-1,7), the result is 6.

If I google "-1 modulo 7", the result is 6.

But if I use the PH mod function for MOD(-1,7) the result is -1

$ QUIZ

Q U I Z   (7.10.G1)
Copyright 1998 COGNOS INC. (ALPHA)
Licensed PH-AXP-RUNTIME 

> DEFINE D-1 = MOD(1,7)
> DEFINE D-2 = MOD(2,7)
> DEFINE D-3 = MOD(3,7)
> DEFINE D-4 = MOD(4,7)
> DEFINE D-5 = MOD(5,7)
> DEFINE D-6 = MOD(6,7)
> DEFINE D-7 = MOD(7,7)
> DEFINE D-1N = MOD(-1,7)
>  DEFINE D-2N = MOD(-2,7)
>  DEFINE D-3N = MOD(-3,7)
>  DEFINE D-4N = MOD(-4,7)
>  DEFINE D-5N = MOD(-5,7)
>  DEFINE D-6N = MOD(-6,7)
>  DEFINE D-7N = MOD(-7,7)
> SAVE TEST-MOD
> REPO ALL
> GO
03/01/06                                       System            PAGE   1

    D-1      D-2      D-3      D-4      D-5      D-6      D-7      D-1N

    D-2N     D-3N     D-4N     D-5N     D-6N     D-7N

        1        2        3        4        5        6        0       -1
       -2       -3       -4       -5       -6        0

Are there two different modular arithmetic systems for negative numbers?  Or
am I doing something wrong?

-- 
= = = = = = = = = = = = = = = = = = = = = = = = = = = =
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