On Wed, 14 Dec 2011 10:56:02 +0200, Jussi Piitulainen wrote: > Steven D'Aprano writes: >> On Mon, 12 Dec 2011 09:29:11 -0800, Eelco wrote: >> >> [quoting Jussi Piitulainen <jpiit...@ling.helsinki.fi>] >> >> They recognize modular arithmetic but for some reason insist that >> >> there is no such _binary operation_. But as I said, I don't >> >> understand their concern. (Except the related concern about some >> >> programming languages, not Python, where the remainder does not >> >> behave well with respect to division.) >> >> I've never come across this, and frankly I find it implausible that >> *actual* mathematicians would say that. Likely you are misunderstanding >> a technical argument about remainder being a relation rather than a >> bijunction. The argument would go something like this: > > (For 'bijunction', read 'function'.)
Oops, you're right of course. It's been about 20 years since I've needed to care about the precise difference between a bijection and a function, and I made a mistake. And then to add to my shame, I also misspelt bijection. > I'm not misunderstanding any argument. There was no argument. There was > a blanket pronouncement that _in mathematics_ mod is not a binary > operator. I should learn to challenge such pronouncements and ask what > the problem is. Maybe next time. So this was *one* person making that claim? I understand that, in general, mathematicians don't have much need for a remainder function in the same way programmers do -- modulo arithmetic is far more important. But there's a world of difference between saying "In mathematics, extracting the remainder is not important enough to be given a special symbol and treated as an operator" and saying "remainder is not a binary operator". The first is reasonable; the second is not. > But you are right that I don't know how actual mathematicians these > people are. I'm not a mathematician. I don't know where to draw the > line. > > A Finnish actual mathematician stated a similar prejudice towards mod as > a binary operator in a Finnish group. I asked him what is wrong with > Knuth's definition (remainder after flooring division), and I think he > conceded that it's not wrong. Number theorists just choose to work with > congruence relations. I have no problem with that. Agreed. [...] > (There is no way to make remainder a bijection. You mean it is not a > function if it is looked at in a particular way.) You're right, of course -- remainder cannot be 1:1. I don't know what I was thinking. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
