[ Funny how you have to see something in print before you recognize that it's wrong... :-/ ]
On 6/10/2014 1:42 PM I said: > IANAM, so this is more of a "inclination by a mathematically > challenged member" ... Apparently true. > The 'remainder' for two numbers 'a' & 'b' is 'x=a%b' with the sign of > 'a' preserved in 'x'. The 'modulus' OTOH, preserves the sign of 'b'. Patently false. That should have been 'x=a-((a%b)*b)' I'm afraid. I'm still not sure if there is useful purpose (besides orthogonality) for a "remainder of complex numbers" operator. -Chip- ------------------------------------------------------------------------------ HPCC Systems Open Source Big Data Platform from LexisNexis Risk Solutions Find What Matters Most in Your Big Data with HPCC Systems Open Source. Fast. Scalable. Simple. Ideal for Dirty Data. Leverages Graph Analysis for Fast Processing & Easy Data Exploration http://p.sf.net/sfu/hpccsystems _______________________________________________ Oorexx-devel mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/oorexx-devel
