On Wed, Oct 24, 2018 at 4:08 PM Daniel Krenn <kr...@aon.at> wrote: > we have > > sage: C.<z> = CyclotomicField(60) > sage: abs(z) > 1.00000000000000 > sage: abs(z).parent() > Real Field with 53 bits of precision > > What is the reason, why this returns an inexact result and not something > in an exact ring like QQbar? >
certainly for any cyclotomic number a+ib one can compute its complex conjugate a-ib exactly, and thus the square a^2+b^2 of its abs() exactly. IMHO this won't work for a general algebraic number, as you'd need to know its embeddings into complex numbers, something that's not readily available. Disclaimer: I am not a computational number theorist, and am puzzled by all this. What's the computational complexity status of this question, computing the square of abs() of a algebraic number? Dima > Best wishes > > Daniel > > -- > You received this message because you are subscribed to the Google Groups > "sage-devel" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to sage-devel+unsubscr...@googlegroups.com. > To post to this group, send email to sage-devel@googlegroups.com. > Visit this group at https://groups.google.com/group/sage-devel. > For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "sage-devel" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-devel+unsubscr...@googlegroups.com. To post to this group, send email to sage-devel@googlegroups.com. Visit this group at https://groups.google.com/group/sage-devel. For more options, visit https://groups.google.com/d/optout.
