Re: square root denominators

From: Bill Dubuque (wgd_at_nestle.csail.mit.edu)
Date: 02/24/05 

Date: 24 Feb 2005 18:06:03 -0500

aegis <aegis@mad.scientist.com> wrote:
>
> If you consult a typical math book you will undoubtedly find
> the steps needed to rationalize a denominator; however, you
> will most likely not find a rationale as to why you need
> to rationalize the denominator. So I pose to you readers
> of sci.math, what is the reasoning behind doing so?

It often leads to a "simplification", since it replaces
an irrational denominator by a "simpler" rational one.

As a simple example, consider the number theory of the
Gaussian integers := {m + n i : m,n integers}. Given
a Gaussian rational (m + n i)/(j + k i) it may not be
a priori clear whether or not it's a Gaussian integer.
However, multiplying the denominator by its conjugate
rationalizes it, resulting in a number having the form
(b + c i)/d. In this form the Gaussian integer test is
much easier, one need only test that d divides b and c.

In fact this idea of reducing divisibility of algebraic
integers to that of rational integers lies at the heart
of Kronecker's divisor theory - a powerful generalization
of number theory from rational to algebraic numbers, e.g.
see the introduction in Harold Edwards book: Divisor Theory.

For higher degree algebraic extensions one rationalizes
by multiplying by all other conjugates, i.e. by taking
the Norm. The Norm often yields a simpler image of a
problem which, moreover, preserves the multiplicative
structure since N(ab) = (Na)(Nb). This in turn is a
special case of an algebraic method of problem solving,
i.e. studying a complex structure in terms of its simple
images under structure-preserving maps (homomorphisms),
a kind of "divide and conquer" used in mathematics.

--Bill Dubuque

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