Re: Rational and irrational numbers

deepkdeb@xxxxxxxxx writes in article <1125252423.614978.59470@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> dated 28 Aug 2005 11:07:03 -0700:
>Correction and modification:
>
>Given situations:
>
>sqrt(a) + sqrt(b) = u^k (1); sqrt(a) - sqrt(b) = v^k (2)
>uv is an integer, odd k > 5, both a and b are nonsquare integers > 0.
>
>Assertion: u and v must be of the form (3) and (4) where
>u = A*sqrt(g) + B*sqrt(h) (3); v = A*sqrt(g) - B*sqrt(h) (4)
>g, h are nonsquare integers and A, B are integers > 0.

Still incorrect. I am not going to provide a specific counterexample
(although I believe (pi,1/pi) suggested by another poster still stands),
just an argument based on cardinality.

The set of ordered pairs (x,y) s.t. xy=z has the cardinality of R. Remove
the rational points and it's still |R|.

Your expression of x and y as functions of a finite number of (now 4)
integers has the cardinality of N^4 which is equal to that of N. There must
be some solutions missing.

--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
.