On Wed, Jul 31, 2024 at 04:05:07PM +0200, Ralf Hemmecke wrote:
> Before I am going to fix this, I guess I need a discussion of what
> x/y or 1/x is actually supposed to means in a non-commutative ring.
>
> You probably realized that
>
> ((*$sTO)(rs1, s3) - rs2)$sTO
>
> gives the same "wrong" result.
It should be
((*$sTO)(s3, rs1) - rs2)$sTO
and this works without the patch.
> I have nothing against introducing right-/left- division, but / should be
> reserved for the case when x * y^(-1) = y^(-1) * x otherwise I would leave
> it undefined.
Well, division solves equation x = d*y, so we have d = x * y^(-1).
This definition is necessary to have sane interaction with multiplication.
To do it on the other side is a different operation. If you look
at definitions in the algebra, FreeDivisionAlgebra, Group and
XPolynomialRing are quite explicit and define x/y as x * y^(-1).
In most other cases order does not matter as corresponding
multiplication is commutative. Possibly the only unclear definition
is the one in StreamTaylorSeriesOperations.
> In other words we should have
>
> if A has CommutativeRing then
> "exquo" : (ST A,ST A) -> Union(ST A,"failed")
> "/" : (ST A,ST A) -> ST A
> recip : ST A -> UN
>
> Maybe we can a bit weaker for recip, but I would still require that
> x * recip(x) = recip(x) * x. Otherwise recip should fail.
We have:
recip : % -> Union(%,"failed")
++ recip(a) returns an element, which is both a left and a right
++ inverse of \spad{a},
++ or \spad{"failed"} if such an element doesn't exist or cannot
++ be determined (see unitsKnown).
So yes, x * recip(x) = recip(x) * x = e where e is the unit element
for multiplication.
And once you have recip "/" follows logically.
And we also have 'leftRecip' and 'rightRecip' for cases when inverse
works only on one side.
--
Waldek Hebisch
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