On Sun, 20 Jul 2008 19:41:54 +0100, Jack <jj@[EMAIL PROTECTED]
>
wrote in <news:QFLgk.14194$9S6.11255@[EMAIL PROTECTED]
> in
alt.algebra.help:
> "Brian M. Scott" <b.scott@[EMAIL PROTECTED]
> wrote in message
> news:hmapmaf61g2e$.1dk3xpc4s4kgu.dlg@[EMAIL PROTECTED]
>> On Sat, 19 Jul 2008 19:58:13 +0100, Jack <jj@[EMAIL PROTECTED]
>
>> wrote in <news:8Pqgk.9112$X72.6423@[EMAIL PROTECTED]
> in
>> alt.algebra.help:
[...]
>>> The integer b is a member of B if and only if any p or q
>>> divide n in [x,y] such that n=(y+1)/2-b, or n=
>>> (y+1)/2+b.
>> This is incomprehensible.
> Why?
Because it makes no sense as English, let alone as
mathematics.
> It's exactly as you had it: B(x, p, q) = {b \in |N :
> p | x + b or q | x - b}.
No, it isn't: what I wrote is meaningful.
> If I might do my best to give a tantalising insight into
> my gobbledegook world, this is the way I formulated -
> for what it's worth - what I wrote:
> We take only ('such that') those 'n' for which ('n=...') if you subtract
b
> from (y+1)/2 ('....=(y+1)/2-b') and get p,q|n ('if any p or q divide
n')
> then you get a member of B ('integer b is a member of B if and only
if...').
Let's look at that without the parenthetical comments and
incorrect quotation marks[*]:
We take only those n for which if you subtract
b from (y + 1)/2 and get p,q | n then you get a
member of B.
This also isn't English.
[*] When you put quotation marks around the letter, you're
referring to the symbol itself, the lower-case en, not to
its value.
> My only issues are how I convey to the reader that the
> members of B(x, p, q) are always integers n in [1,y];
> and that x is always (1+y)/2; and that I want to be able
> to refer to a set in which every possible p and q, i.e.
> every member of J, is considered in such a fa****on.
The first two are trivial and have been addressed elsewhere.
The last remains incomprehensible.
[...]
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