Re: Rational and irrational numbers

On 29 Aug 2005 15:06:16 -0700, deepkdeb@xxxxxxxxx wrote:

>Attention: quasi
>
>If you have time and interest kindly consider the following situation:
>
>u = Asqrt(g) + Bsqrt(h) (1) and then u^7 = Csqrt(g) + Dsqrt(h)
>(2)
>
>then [Csqrt(g) + Dsqrt(h)]^1/7 = Asqrt(g) + Bsqrt(h)
>
>Therefore, one can obtain (2) from (1) and (1) from (2).
>There is one additional element v which is a conjugate of u.
>
>After reviewinf my proof I believe the proof of the assertion is simple
>and self evident.
>But I could be wrong.

Yes, (1) => (2), but the reverse direction is not automatic.

What you are saying is that for fixed g,h, then each pair (A,B)
induces a pair (C,D), in other words (1) => (2). Fine, I accept that.
And of course you can go back if you take a pair (C,D) that came from
a pair (A,B). But how do you know that from an arbitrary pair (C,D)
you can go back. Not everything of that form is a 7th power of the
same form, unless it is, in which case, all you are saying is that if
(C,D) corresponds to a 7th power of some (A,B), then the (real) 7th
root has the same form, but of course it does since you created the
pair (C,D) from a pair (A,B).

Bottom line -- I think you should settle for (1) => (2). The fact that
the mixed radicals drop out is interesting, and can be explained by
applying the binomial theorem -- I assume that's what you did, right?

But a point should be raised here. Why did you specify odd k, k>5? You
do need k to be odd so as to avoid mixed radicals, but any odd
positive integer is just fine for k, even the trivial case, k=1.

quasi
.

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