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