Paul,
>> R(n.n') = |t(n,m)| - |t(n',m)|, as long as it implies that there are
>> individual members, r, of R(n,n').
>
> It won't do that but R(n,n') = t(n,m) - t(n',m) will with the "new"
> definition of t(n,m).
Surely t(n,m) - t(n',m), under the new definition, will require that
t(n,m)
contains all (or some of?) the same prime factors as t(n',m)? I don't want
there to be any such stipulation.
How would I ensure that there are there are individual members, r, of
R(n,n'), for the case where I am not using prime factors to determine the
distribution of positive entries?
>> Do the r and s here refer to the members r of R(n,n') and s of S(n,n')?
>> If
>> not, as I strongly suspect, then surely I've got trouble?
>
> No, they just refer respectively to the number of rows and columns of M.
But am I OK using r and s for both those purposes in a single paper?
>> But if x=0 and y =10, there are surely 11 columns in the matrix.? I
want
>> zero to be an index value.
>
> But for a _matrix_ it is not.
Some how I've got to get zero incor****ated in such a way....
>> {1,1}, {1,1},{1,1,1,1},{1,1} respectively. Cardinalities 2, 2, 4, 2
>> respectively.
>
> Oh dear, that's what I was afraid of. The sets {1,1} and {1,1,1,1} are
> all the same and are all equal to {1}. Just because you mention "1"
> twice doesn't mean {1,1} has two elements.
This throws me considerably. Oh, heck.
> I've started a new thread "Algebra Questions (part 2)" in which I
> propose a new starting point for you which I hope will help with your
> notational difficulties. At least it introduces some terminology which
> may be useful.
Many thanks - it looks like a very useful post and I shall go away and
study
it.
Cheers.
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