On Sun, Apr 1, 2012 at 10:26 PM, Aaron Meurer <asmeu...@gmail.com> wrote: > On Sun, Apr 1, 2012 at 12:08 PM, Ronan Lamy <ronan.l...@gmail.com> wrote: > >> Besides, how can a and b know which they belong to? This would require >> having several different versions of e.g. sqrt(6): one in QQ[sqrt(6)], >> one in QQ[sqrt(2), sqrt(3)], one in RR, one in CC, ... IIUC, Sage does >> exactly that, but it feels profoundly wrong. > > I agree with this. Doing it this way feels like just too much > unnecessary work (for the user). I think it makes sense to just > define sqrt(6) in one way, namely, as 6**(1/2), and let each ring or > field figure out how it makes sense in itself (or doesn't). This > feels natural because sqrt(6) is actually implemented as Pow(6, > Rational(1, 2)). Maybe if you are treating QQ<sqrt(6)> as a vector > space, you can internally convert sqrt(6) to something like [0, 1]. > But the coercion rules should make it so that you never need to even > care about that, unless you are specifically working with code that > requires this implementation, or if you are in the coercion code.
I think we absolutely must distinguish between concrete rings of numbers and abstract rings. I absolutely agree that having the user explicitly specify which sqrt() to apply to a given number goes against usual practise. However, just as Tom says, I don't see absolutely any problem with the fact that an *abstract* ring will have to explicitly define sqrt(), should this definition make any sense. >> It should clearly be the other way round: R.mul(a, b) should be defined as >> a*b. >> And since we'd like ring operations to be symbolic objects, we should >> even have R.mul = Lambda((x, y), x*y). > > So supposing that we had QQ<sqrt(6)>, and we wanted to define a*b as > [a1*b1 + 6*a2*b2, a1*b2 + a2*b1], where a = [a1, a2], b = [b1, b2], > and a = [a1, a2] means that a = a1 + a2*sqrt(6). Would it make sense > to make R.mul equal to Lambda((x, y), x.as_vector()*y.as_vector())? This would make sense *iff* R would have a large number of useful properties (for example, if it were a vector space). Attempting to transform the arguments of R.mul to vectors imposes a lot of unnecessary restrictions on R (like that it should be isomorphic to a vector space). I hope I understood your statement correctly. Sergiu -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to sympy@googlegroups.com. To unsubscribe from this group, send email to sympy+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
