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. firstname.lastname@example.org 2 dificult 2 work out." http://www.maths.ex.ac.uk/~rjc/rjc.html Iain M. Banks - Feersum Endjinn