In article <KeUYj.44844$iD4.8530@[EMAIL PROTECTED]
>, Jack <jj@[EMAIL PROTECTED]
>
wrote:
> 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.
No, just the opposite. It is the factors of n which are _not_ factors
of n'.
> 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?
see part 2
> >> 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?
Sure it is done all the time.
> >> 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.
Let's kill off this thread.
--
Paul Sperry
Columbia, SC (USA)
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