On 2016-01-20 11:17, Volker Braun wrote:
On Wednesday, January 20, 2016 at 8:28:23 AM UTC, Jeroen Demeyer wrote:
Well, you cannot have a fully consistent floor division in any case:
Either you make floor division on QQ consistent with ZZ or with QQ[x]
but you cannot have both. Personally, I would prefer making it
consistent with ZZ.
As I said in the previous email (the piece that you graciously left
out), this isn't really consistent with the division operation on ZZ.
In QQ, I would define a//b = floor(a/b).
Why do you consider this not "really consistent with the division
operation on ZZ."? Neither your previous post nor this one explains that.
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