Re: Question on sums of roots of integers
- From: "Dave L. Renfro" <renfr1dl@xxxxxxxxx>
- Date: Mon, 2 Mar 2009 06:18:40 -0800 (PST)
Gauss Euler wrote:
suppose a_1, a_2, a_3, ..., a_n are positive integers, and
k_1, k_2, k_3, ..., k_n are also positive integers. Suppose
also that x_i = (a_i)^(1/k_i). Assume the a_i are not perfect
k_i powers. How does one show the sum x_1 + x_2 + x_3 + ... + x_n
is not rational? I know that if it was rational, it would have
to be an integer, but does that help at all? I would really
love an algebraic proof: using polynomials, getting a
contradiction by looking at roots/coefficients. I know
that is very vague, but I figured this might be a well-known
enough problem that there exists an elegant/pretty solution.
A much stronger result was proved by Besicovitch (the fractal
theory guy) in a way that I think fits what you're looking for.
I happen to have a .pdf file of Besicovitch's paper (one of the
many papers I obtained, and which anyone else here could have
obtained, back when I posted the following on 27 December 2006:
<http://groups.google.com/group/sci.math/msg/1c8a64a16c6be33d>),
if you're interested.
What remains is from a post I made in another discussion group
(31 Oct. 2007, <http://mathforum.org/kb/message.jspa?
messageID=5972698>).
**************************************************************
Here's what Besicovitch proved in the paper
I previously cited [1]:
THEOREM: Let k and n be positive integers.
Let p_1, p_2, ..., p_k be distinct prime
numbers. Then the following collection of
n^k many positive real numbers is linearly
independent over the rational numbers:
[ (p_1)^(m_1) * (p_2)^(m_2) * ... * (p_k)^(m_k) ] ^ (1/n),
where m_1, m_2, ..., m_k each varies over
the n many integers 0, 1, 2, ..., n-1.
If you play around with this some to see just
how inclusive it is, I think you'll find it
says that pretty much anything not "obviously
dependent" (such as sqrt(12) and sqrt(3))
is taken care of.
Note that being linearly independent implies more
than just that none of these n^k many numbers is a
linear combination of the other numbers. It also
implies that any nonzero linear combination of
any two of these numbers is not a linear combination
of the remaining numbers, any nonzero linear
combination of any three of these numbers is
not a linear combination of the remaining numbers,
and so on.
Proofs of this result can also be found in Gaal [2]
(pp. 234-237) and Richards [4]. An elementary
discussion of about one page in length can be
found in Richards [5] (Example 6, pp. 260-261).
Simpler versions in which only square roots are
involved are proved in Roth [6] and (I believe)
in Flanders [3].
[1] Abram S. Besicovitch, "On the linear independence
of fractional powers of integers", Journal of the
London Mathematical Society 15 (1940), 3-6.
[2] Lisl Gaal, "Classical Galois Theory With Examples",
Markham Publishing Company, 1971, x + 250 pages.
[3] Harley Flanders, "Solution to Problem #4797",
American Mathematical Monthly 67 #2 (February 1960),
188-189.
[4] Ian J. Richards, "An application of Galois theory
to elementary arithmetic", Advances in Mathematics
13 (1974), 268-273.
[5] Ian J. Richards, "Impossibility", Mathematics
Magazine 48 #5 (Nov./Dec. 1975), 249-262.
[6] Richard L. Roth, "On extensions of Q by square
roots", American Mathematical Monthly 78 #4
(April 1971), 392-393. [Reprinted on pp. 211-212
of "The Raymond W. Brink Selected Papers on Precalculus",
Mathematical Association of America, 1977.]
**************************************************************
Dave L. Renfro
.
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