Re: Rational and irrational numbers

On Mon, 29 Aug 2005 17:20:10 +0000 (UTC), klewis@xxxxxxxxxxxxxxxx
(Keith A. Lewis) wrote:

>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.
>

No -- cardinality arguments won't work. All variables in the problem
are either integers or algebraic numbers, so the relevant sets are all
countable.

To clarify, let me try to restate the corrected version of the
problem:


-------------------------------------------------------------------------------------------
Let a,b be relatively prime, non-square positive integers, and let k
be an odd integer, k>5.

If u,v are real and such that:

sqrt(a)+sqrt(b)=u^k

sqrt(a)-sqrt(b)=v^k

u*v=1

Then u,v can be expressed in the form:

u = A*sqrt(g) + B*sqrt(h)

v = A*sqrt(g) - B*sqrt(h)

for some positive integers A,B,g,h and where g,h are non-square.

---------------------------------------------------------------------------------------------

The above is how I read the intent of the corrected problem.

Note that all the variables are positive integers except for u,v
which, though irrational, are still algebraic, hence there is no
cardinality mismatch.

However the counterexample I provided does show that the conclusion is
too strong, and I agree -- it's way too strong. At heart, the
conclusion implies that there are elements of arbitrarily high degree
over Q but which are also contained in a field of degree 4 over Q,
which is, of course, impossible.

So there will be lots of counterexamples, but still only countably
many.

quasi
.

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