From: Bill Dubuque <wgd@nestle.ai.mit.edu>
Subject: Re: CubeRoot(2)
Date: 08 Mar 2000 15:53:53 -0500
Newsgroups: sci.math
Summary: [missing]

The Ridge <pmytnn@nottingham.ac.uk> wrote:
: Is there a proof for proving that CubeRoot(2) is irrational?  Would it
: go along the same lines as trying to prove that Sqrt(2) is irrational?

Rob Johnson <robjohn9@idt.net> wrote:
| A more general result that answers all of these sorts of questions 
| is that all algebraic integers are either integers or irrational. 
| See  http://idt.net/~robjohn9/math/ratalint.html  for a proof of this.

Les Wright <leslie_wright@yahoo.com> wrote:
> Run, don't walk, to Rob's link. The proof is brief, elegant, and 
> requires only very basic number theory as a prerequisite [...]

The proof is actually much simpler and known since high-school, namely

THEOREM  If a reduced rational a/b is root of a polynomial P(x) with
integer coeffs,  P(X) =  d X^n + ... + c,  then  a|c  and  b|d.

Proof:  0 = b^n P(a/b) = d a^n + ab(...) + c b^n  with (...) an integer

                       = d a^n + b[a(...)+ c b^(n-1)]

thus  b|(d a^n)  =>  b|d  since  gcd(a,b)=1 (because  a/b is reduced).

That  a|c follows symmetrically.  QED   Note: a|c  means  a divides c


Specializing to the case where the leading coeff  d = +-1  yields

COROLLARY.  A rational root of a monic integral poly is an integer.


This specialized to  P(X) = X^n - m  yields a rationality test

COROLLARY. The n'th root of an integer if not integral is irrational.


Take close note how the above proof depends crucially on divisibility
properties of the integers (in particular, employing gcd's).  For
generalizations to other domains see some of my prior sci.math posts.

-Bill Dubuque