|
|
Re: Open Call for Conjectures
Posted:
Aug 11, 2001 7:41 AM
|
|
Copyright 1998 James Wanless
"Randy Poe" <rpoe@nospamatl.lmco.com> wrote in message
news://3B742E65.275DF96C@nospamatl.lmco.com...
> James Wanless wrote:
> >
> > So I guess you haven't seen it then... :-)
>
> Curious about this exchange, I looked at the link in question:
>
> > BERTRAND'S THEOREM
> >
> > For all n, there exists a prime p, s.t. n<=p<2n
> >
> > Proof:
> > Either: For all n, there exists m s.t. prime, p = n+m, with: 0<=m<n
> > Or: For all m, there exists n s.t. prime, p = m+n, with: 0<=n<m
> >
> > Therefore
> > Either: n<=p<2n
> > Or: m<=p<2m
> > From which result follows
>
> The result doesn't follow of course.
>
> Consider a numerical example, say n = 1000. The conjecture
> is that there is a prime between 1000 and 2000.
>
> You have said: For this n, there is some prime p=n+m. Either
> m < n, i.e. p < 2000, or m > n, i.e., p > 2000 and lies
> between m and 2m.
>
> But that certainly doesn't imply we can find a prime in
> [1000, 2000].
>
> - Randy
|
|
