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Re: Integer pairs in sum of reciprocals
Posted:
Jan 2, 2001 1:08 PM
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On Tue, 2 Jan 2001, Ross A. Finlayson wrote:
> Jan Kristian Haugland wrote:
>
> > "Ross A. Finlayson" wrote:
> >
> > > saxon970@yahoo.com wrote:
> > >
> > > > Hello. I have some questions about how to approach the problem below:
> > > >
> > > > How many pairs of positive integers a and b are there such that a < b
> > > > and
> > > > 1/a + 1/b = 1/2001 ?
> > > >
> > > > End of problem.
> > > >
> > > > Is trial and error and making a manual list the way to go?
> > > >
> > > > Thanks.
> > > >
> > > > Sent via Deja.com
> > > > http://www.deja.com/
> > >
> > > Set a to 2001 or greater and b to a large value. The result is very close
> > > to 1/2001, where 2001 is called c,
> > >
> > > 1/a + 1/b = 1/c, or
> > >
> > > a ^-1 + b ^-1 = c^-1
> > >
> > > There are probably infinite solutions.
> >
> > What does this mean? (1) There are infinitely
> > many triplets (a, b, c) with 1/a + 1/b = 1/c,
> > or (2) there are infinitely many pairs (a, b)
> > with 1/a + 1/b = 1/2001? If a, b and c are
> > restricted to the positive integers then (1)
> > is trivially true and (2) is trivially false.
> >
> > > Ross
> > >
> > > --
> > > Ross Andrew Finlayson
> > > Finlayson Consulting
> > > Ross at Tiki-Lounge: http://www.tiki-lounge.com/~raf/
> > > "The best mathematician in the world is Maplev in Ontario." - Pertti L.
> >
> > --
> >
> > Jan Kristian Haugland
> > http://home.hia.no/~jkhaug00
>
> If you assign c a sufficiently large value, then for some value a slightly
> greater than c and b greater than a then there is a solution.
Do you have a proof for this statement?
> If there is one there is probably infinite.
There is one positive integer less than 2. Does that mean there are an
infinite number of them?
Steve L
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