Re: Rational and irrational numbers

On Mon, 29 Aug 2005 15:15:35 -0700, quasi <quasi@xxxxxxxx> wrote:

>On Mon, 29 Aug 2005 15:04:01 -0400, A N Niel
><anniel@xxxxxxxxxxxxxxxxxxxxx> wrote:
>
>>Counterexample: a=8,b=7,k=7
>>
>>
>>> 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.
>>>
>>>
>>
>>u*v=1 implies (u^k)*(v^k)=1, so we have
>>(sqrt(a)+sqrt(b))*(sqrt(a)-sqrt(b))=1, that is a-b=1.
>>
>>So, can we say that u = (sqrt(b+1)+sqrt(b))^(1/k)
>>must be of the form A*sqrt(g) + B*sqrt(h) ?
>>
>>No. (sqrt(8)+sqrt(7))^(1/7) is not of that form.
>>It has degree at least 28 over the rationals, since it is a root of the
>>degree 28 irreducible polynomial x^(28) - 30*x^(14) + 1 .
>>But A*sqrt(g) + B*sqrt(h) has degree at most 4.
>
>Fine, but your counterexample is analagous to the counterexample I
>provided yesterday:
>
> a=3, b=2, k=7
> u=(sqrt(3)+sqrt(2))^(1/7)
> v=(sqrt(3)-sqrt(3))^(1/7)
>
>quasi

whoops, typo ... should have been:

a=3, b=2, k=7
u=(sqrt(3)+sqrt(2))^(1/7)
v=(sqrt(3)-sqrt(2))^(1/7)

My keyboard must be defective, it happens too often (can't be me).

quasi
.

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