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Topic: HELP - SQRT(3)
Replies: 16   Last Post: Dec 1, 1997 7:24 PM

 Messages: [ Previous | Next ]
 Robin Chapman Posts: 695 Registered: 12/6/04
Re: HELP - SQRT(3)
Posted: Nov 17, 1997 3:42 AM

Emanuel Binnun wrote:
>
>

Let r be a positive integer, and suppose
(i) sqrt(r) is not an integer
(ii) sqrt(r) is rational with denominator D (necessarliy D > 1).

From (i) we can choose an integer a with a < sqrt(r) < a + 1.
Thus 0 < sqrt(r) - a < 1. By induction, for each positive integer n,
(sqrt(r) - a)^n has the form B + C sqrt(r), for some integers
B and C (dependent on n). Thus for each N we can express (sqrt(r) - a)^n
as a positive rational with denominator D. Thus (sqrt(r) - a)^n >= 1/D
for all n. But this is impossible; for large enough n, as
|sqrt(r) - a| < 1 then (sqrt(r) - a)^n < 1/D. Contradiction.

--
Robin Chapman "256 256 256.
Department of Mathematics O hel, ol rite; 256; whot's
University of Exeter, EX4 4QE, UK 12 tyms 256? Bugird if I no.
rjc@maths.exeter.ac.uk 2 dificult 2 work out."
http://www.maths.ex.ac.uk/~rjc/rjc.html Iain M. Banks - Feersum Endjinn

Date Subject Author
11/14/97 EmanuelBinnun
11/14/97 Norman Walker
11/17/97 Robin Chapman
11/21/97 r3769@aol.com
11/21/97 Robin Chapman
12/1/97 Mike McCarty
11/15/97 Jean-Pierre MERX
11/16/97 William L. Bahn
11/17/97 Sam @ The NIMP Team
11/17/97 Wilbert Dijkhof
11/17/97 Sam @ The NIMP Team
11/17/97 William L. Bahn
11/18/97 Sam @ The NIMP Team
11/17/97 Wilbert Dijkhof
11/17/97 Pete Vermeire
11/18/97 Brian Hutchings
11/21/97 Zdislav V. Kovarik