In article <DLyYj.36248$1q4.21586@[EMAIL PROTECTED]
>, Jack <jj@[EMAIL PROTECTED]
>
wrote:
> Paul,
>
> <<What you need is something like
> sum(|R(n,n')|; n in [x,y]) i.e. tell us _which |R(n,n')|'s are being
> added up. >>
>
> The semi-colon is not a typo? Have only ever encountered colons in
formal
> expressions.
Actually, a comma is used in Maple and Mathematica but use whatever you
want.
> <<What would make sense is if your t(n,m) was the _set_ of primes less
> than or equal to m which are factors of n. You could then reference
> |t(n,m)| if you wanted the count of those primes i.e. get your original
> t(n,m).>>
>
> Please remember that my *first* objective is to prove propositions that
have
> nothing to do with factorisation.
Well, OK but |t(n,m)| (with my t(m,n) ) is what _you_ defined.
> <<With the new t(n,m) you could let R(n,n') = t(n,m) - t(n',m) = the set
> of primes in P(m) that are factors of n but not of n' or
> R(n,n') = |t(n,m) - t(n',m)| = how _many_ such primes there are or
> R(n.n') = |t(n,m)| - |t(n',m)| = how many more (or less) such primes
> divide n that divide n'. Which, if any, of those you want I don't
know.>>
>
> 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).
> <<We shall be constructing a binary r by s matrix M such that
> sum(M(i,n); i = 1 .. r) = max(i : M(i,n) =/= 0). [Here, =/= means "not
> equal (remember M is binary), M(i,j) is the entry in the i-th row and
> j-th column.]>>
>
> 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.
> << > Let P(m) be the first m primes.
> >
> > Let J be a subset of P(m).
> >
> > Given an interval [x,y] of integers, define N(n) to be the set of
positive
> > entries in a column indexed by an integer n in [x,y], N(x,y) to be the
set
> > of positive entries in [x,y], and N(x,y,|J|) to be the set of positive
> > entries in [x,y] such that the greatest value of |N(n)| in [x,y] is
|J|.
>
> Is the matrix still M?>>
>
> Yes.
>
> << You've got problems : N(n) = {1} or N(n) = {}.>>
>
> The *members* of N(n) are either 1 or 0. My ultimate interest is the
> cardinality of N(n).
That isn't what you said - "positive" generally means greater than 0.
> <<Moreover what if M only has 10 columns and [x,y] = [100,110]?>>
>
> I don't know; how do I get round this? I had learnt previously that one
can
> define a matrix's extents and speak quite safely of columns outside of
it,
> as though it will be protracted ad infinitum.
>
> <<I was guessing you meant x >= 0.>>
>
> Yes.
>
> <<No? If so N(x,y) = y - x + 1 or (in
> case x = 0) y-x.>>
>
> 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.
> <<|N(n)| is either 0 or 1.>>
>
> But I thought I had defined it as the column sum.
No, you didn't.
> <<Suppose M turned out to be
>
> 1 0 1 1
> 0 1 1 0
> 0 0 1 1
> 1 1 1 0
>
> What would you want N(1), N(2), N(3) and N(4) to be?>>
>
> {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.
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.
--
Paul Sperry
Columbia, SC (USA)
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