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

 Messages: [ Previous | Next ]
 r3769@aol.com Posts: 347 Registered: 12/12/04
Re: HELP - SQRT(3)
Posted: Nov 21, 1997 1:54 AM

In article <3470035D.57530D9D@maths.ex.ac.uk>, Robin Chapman
<rjc@maths.ex.ac.uk> writes:

>
>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.
>
>

So assuming sqrt(r) is a real number the desired conclusion follows (rather
nicely). But isn't this a rather BIG assumption?

R.

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