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Re: Independent sqaure roots???
Posted:
Apr 30, 2003 8:50 AM
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On Wed, 30 Apr 2003 06:51:52 -0500, David C. Ullrich
<ullrich@math.okstate.edu> wrote:
>Are the square roots of the positive square-free integers
>(including 1) linearly independent over Q?
>
>Hmm, in case this looks like homework I should say how
>far I got: Say we have a dependence among sqrt(n_0),
>... sqrt(n_N) (say n_0 = 1 to make the notation nice
>below). To make things look nice, we can assume by
>adding more n_j that in fact n_0, ... n_N is a list of
>all the square-free products of a certain finite set of
>primes. Now we have
>
>(*) a_0 sqrt(n_0) + ... + a_N sqrt(n_N) = 0
>
>for integers a_0, ... a_N, which we can take to have
>no common factor.
Oh: Say n_j "appears" here if a_j <> 0. There must
be a prime p such that p is a factor of at least one
of the n_j which appears but not all of them
(if not then we could divide the whole thing by
sqrt(p) and get a counterexample with fewer primes).
Put all the terms where p appears on the left
side and the other terms on the right side. Square
both sides. After squaring both sides, p will not
be a factor of any of the n_j which appear on
either side. So after squaring, by induction,
the coefficients on both sides must be the same.
This is impossible because ___(???)
(After squaring, all the coefficients on the left
side are divisible by p. It's impossible for all the
coefficients on the right to be divisible by p
because ___???)
>By our hypothesis on the n_i's,
>if we repeatedly square (*) we get other relations
>that look the same (it's not clear to me whether or
>not the coefficients in the new relations can have
>common factors.) So we have Av = 0, where
>v is the (column) vector consisting of the sqrt(n_i)
>and A has as many rows as we want. _If_ we
>could show that A eventually has rank N+1 then
>we could solve for the sqrt's, showing that they're
>rational. Does the rank of A have to increase
>when we add more rows by repeated squaring
>of (*)? (Or taking other powers of (*), come
>to think of it...)
>
>******************
>
>David C. Ullrich
******************
David C. Ullrich
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