Paul,
> You _still_ haven't learned the difference between t, the function, and
> t(n), a value of the function.
>
>> and not the number of primes that divide n. That way
>> I can easily assume (stipulate?) in my Proposition 0 that sum(t(n)) for
n
>> in
>> [x,y] is equal to sum(t(n)) for n in [x', y']. It just bothers me that
a)
>> I
>> don't know if the stipulation specifically has to be built into the
terms
>> of
>> the definition of the bijection
>
> Yes, it must.
But if my function takes the form f : G -->G' (as it does), how can one
sneak into that expression the principle that for [x,y] and [x',y'],
sum((g,n) in G) is equal to sum (h(g,n) in G'))?
>
>> and b) that when I am using my function that
>> counts that number of primes that divide n, I refer to 'by the proof of
>> Proposition 0'.
>
> Won't work. There is a BIG difference between a function, like t(n) = 5
> for all n, assumed to give equal sums, and your t_J which will only
> give equal sums (as far as I know) for special types of interval pairs.
But it is evident from the outset that what is valid for t : N---> N is
valid for t_J N -->N. Otherwise I wouldn't have bothered with it.
With thanks.
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