NetNews Usenet Archive 1992 #23/NN_1992_23.iso/spool/sci/math/12929

Text File | 1992-10-09 | 2.0 KB | 51 lines

Newsgroups: sci.math
Path: sparky!uunet!think.com!ames!sgi!wdl1!wdl39!mab
From: [e-mail] (Mark A Biggar)
Subject: Re: Question on real numbers
Message-ID: <[e-mail]>
Sender: [e-mail]
Organization: Loral Western Development Labs
References: <[e-mail]>
Date: Thu, 8 Oct 1992 23:32:40 GMT
Lines: 39

In article <[e-mail]> [e-mail] (Wolfgang von Hansen) writes:
>is it possible to express any real number x with the following term
>
>x = a + rb;  a, b \in Q; r \in R, r const.
>
>Some more words to explain what I mean:
>I was wondering if there is an analogon between the real numbers and
>the complex numbers.
>
>My idea is to write any real number as a pair of rational numbers
>one of them multiplied with a constant real number r (see above).
>a is not necessarily (spelling? ;-) different from null. A useful value
>for r may be \sqrt(n), n \in N, because r * r = n is easy to handle.
>This representation of the real numbers might improve the speed
>and/or accuracy of algorithms on computers since all calculations are
>done with rational numbers.
>
>There are some things left to do:
>1. Prove if it is (not) possible.

No you can't do this.  The set of all pairs of rationals (a,b) in only
countably infinite, therefore the set {a+rb} is countable, But the set
R in uncountably infinite, therefore R != {a+rb}.  As a counter example
to you suggestion that you use sqrt(n) lets choose sqrt(2), and lets try
to construct sqrt(3) using your system (I'm going to use s2 ans s3 for
sqrt(2) and sqrt(3) to save typing.

s3 = a + s2 * b                                  (a & b rational)
s3 = p/q + s2 * (m/n)                            (p,q,m,n integers)
3 = p^2/q^2 + 2*s2*(m/n) + 2(m^2)/(n^2)          (square both sides)
3 = (p^2/q^2 + 2(m^2)/n^2) + 2*s2*(m/n)

Thus 3 is the sum of a rational and a irrational, which is impossible.

--
Mark Biggar
[e-mail]