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

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Robin Chapman

Posts: 695
Registered: 12/6/04
Re: HELP - SQRT(3)
Posted: Nov 17, 1997 3:42 AM
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Emanuel Binnun wrote:
>
> Please help me to proove that Square root of 3 is an irrational number
>


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).

We derive a contradiction.

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







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