On Jun 16, 7:22 am, Mel <chemmyg...@gmail.com> wrote: > I do want to convert to an integer in this particular case...but > integers are rationals, so I don't understand why I don't just get the > integer 344 when I try the following code:
If you are sure that the element can be represented as an integer, then just convert to an integer. That will always work: > sage: K=Qp(7,5) > sage: Q=QQ > sage: a=344 > sage: b=K(a) > sage: a=Q(b) #your error sage: ZZ(a) 344 sage: ZZ(K(1/5)) 6723 sage: c = K(1/21) sage: c sage: ZZ(c) ValueError: negative valuation If it's not, you can make sure you can: sage: ZZ(K.prime()^(-c.valuation())*c)*K.prime()^(c.valuation()) 11205/7 sage: Q(c) 1/21 sage: K(1/21)-K(11205/7) O(7^4) The default choice of conversion from K to Q is apparently to call "rational reconstruction" (see c.rational_reconstruction) which, as you observed, tries to find a rational number N/D close to your element with |N|, |D| bounded by sqrt(p^precision), i.e., it's trying to use the same number of bits, but equally distributed over the numerator and denominator. For an arbitrary p-adic that happens to be meaning to represent a rational number, this is almost always what you want. If you know you mean an integer, you're wasting bits on a denominator you want to take 1 anyway, so it's not what you want. I'm a little on the fence whether rational reconstruction is the best default conversion for p-adics to Q, since your example shows sometimes people just want some rational number that lies close to their p-adic number. If you want to have your cake and eat it too (i.e., nice reps if they exist and some rep otherwise), just double the apparent precision: sage: QQ(Qp(7,2*K.precision_cap()) (b).lift_to_precision(2*b.precision_relative())) 344 -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
