Re: Newbie question(s)...

From: Jonathan Baker (jonathanrbaker_at_yahoo.com)
Date: 09/17/03 

Date: 17 Sep 2003 11:52:52 -0700

"Ernst Lippe" <ernstl-at-planet-dot-nl@ignore.this> wrote in message news:<3f685380$0$491$48b97d01@reader22.wxs.nl>...

> It is a simple proof: when sqrt(p) is a positive rational number it
> can be expressed as the ratio n/m, where n and m are both natural
> numbers such that the greatest common divisor of n an m is 1.
>
> We then have p = n^2/m^2 and consequently p * m^2 = n^2. It is
> obvious that p divides n^2, but because p is a prime number it must
> also be the case that p divides n (because the exponent for each prime
> in the factorization of n^2 must be even). Now we also have that p^2
> divides n^2, i.e. n^2 = p^2 * k, for some integer k. From this it
> follows that p * m^2 = n^2 = p^2 * k and consequently m^2 = p *
> k. Thus p is a divisor of m^2 and using the same reasoning as above
> that also means that p is also a divisor of m.
>
> Now we have reached the contradiction because we assumed that n and m
> had no common factor.
Ernst! Thank You!
I knew somebody could come up with an elegant proof! Kudos!

Jon

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