
Re: Integer pairs in sum of reciprocals
Posted:
Jan 2, 2001 11:59 AM


Jan Kristian Haugland wrote:
> "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 > > > > > > > > > > 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. If there is one > > there is probably infinite. > > Huh? If c is a fixed positive integer, and a and b > are positive integers with 1/a + 1/b = 1/c, then > clearly b <= c^2 + c (corresponding to a = c + 1). > Are you saying that b might be even bigger, or that > c is not fixed, or that not all of a, b, c need to > be integers? If c is not fixed, are you arguing for > the trivial (1) above? >
I don't understand your objection.
I think you are saying that it is possible to derive an ideal boundary where solutions would exist. You use the term c squared plus c, that can be derived various ways.
Here, the point is that if there are solutions for any numbers, then there are probably solutions for many values of c, that being an integer value of c greater than 1.
I am saying that c is not fixed.
Now, I feel more confident that I understand.
> > > Why I think this is so is that as a and b are each greater in this way than c, > > then > > > > 1/a + 1/b < 1/c > > > > because as soon the left side is equal then the search for a solution can stop. > > > > So, as b approaches a from being much larger than a, then > > > > lim 1/a + 1/(b>a) = 2/a > 1/c > > > > The relationship does show that the value of the left side can go form greater > > to less than 1/c, the right side. If so, there are more solutions for high > > values of c. > > > > Ross > > > >  > > Ross Andrew Finlayson > > Finlayson Consulting > > Ross at TikiLounge: http://www.tikilounge.com/~raf/ > > "The best mathematician in the world is Maplev in Ontario."  Pertti L. > >  > > Jan Kristian Haugland > http://home.hia.no/~jkhaug00
Ross
 Ross Andrew Finlayson Finlayson Consulting Ross at TikiLounge: http://www.tikilounge.com/~raf/ "The best mathematician in the world is Maplev in Ontario."  Pertti L.

