good chemistry is complicated, and a little bit messy -LW |
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PerlMonks |
Re: 0 illegal modulus?by Zaxo (Archbishop) |
on Jun 16, 2001 at 12:31 UTC ( [id://89021]=note: print w/replies, xml ) | Need Help?? |
Everybody, please pay attention to jepri. He's got it right. The definition of Abel he mentions is that (perlishly): my $num = $y*int($n) + $modulus;This relation is uniquely satisfied by !$y && $modulus == $num It does not involve division, allowing it to apply to algebrae lacking a multiplicative inverse (those are called modules). It applies to floats too. Think $y = 2*$pi. Modulus is phase in that case. Perl's % operator is really just remainder on integers, a bogosity which does not generalize well. Don't fall into the trap of thinking your intuition or experience is a valid basis for modifying mathematical definitions. You will come to grief when a "proven correct" program crashes. You may blame math, but you will be wrong. After Compline, Update: Rereading this, I realize that it sounds directed at nella. Not the case: this is a broadcast rant. nella started a fine thread, and seems to be on the side of the angels.
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