I hadn't heard of "complete residue systems" before. Perhaps another way of
saying the same thing is: f←{5J3|⍵} should be idempotent, so (f ⍵)≡f f ⍵
for any ⍵ ... ?Jay. On 19 June 2017 at 18:03, Frederick Pitts <fred.pit...@comcast.net> wrote: > Jürgen, > > With gnu apl (svn 961 on Fedora 25, Intel(R) Core(TM) i7-6700 > CPU), the residue function (∣) yields the following: > > 5J3 ∣ 14J5 > 1J4 > 5J3 | 1J4 > ¯4J1 > 5J3 | ¯4J1 > ¯4J1 > The above result means that two elements in the complete residue system > (CSR) for mod 5J3 are equal, i.e. 1J4 = ¯4J1 mod 5J3, which is not > allowed. None of the elements of a CSR can be equal modulo the CSR's > basis. > > Regards, > > Fred > >
