From: maejohns@bronze.ucs.indiana.edu (Mark E. Johnson)
Subject: Re: Help me deal w/ infinity
Organization: Indiana University
Date: Tue, 22 Dec 1992 20:02:13 GMT

In <BzL73K.9xr@usenet.ucs.indiana.edu> mkohlhaa@silver.ucs.indiana.edu (mike) writes:

>Some friends and I have recently been having some discussions/arguments 
>about infinity and its meaning in math.

[stuff deleted]

>I appeal to any math gurus out there to comment on the validity of .9(r)
>equals one, and anything else which may increase my understand of the
>concept of infinity and how putting an "infinite" amount of 9's after
>a decimal point can become 1.

>Thanks
>-- 
>					-- Mike

Okay, here is an attempt to enlighten you about numbers
with infinitely repeating digits to the right of the decimal
point.

It kind of depends on the notions of equivalence classes and
convergent sequences.  If these are new ideas (and you have
time to spare), look it up in any text on elementary analysis...


Firstly, to answer your question:    .9(r) = 1.


A Question for motivation:  How do the digital
representation of a number and its actual meaning relate?

As you should know, all numbers that we play with can be
represented in terms of their decimal expansion (base 10),
and that is what we will focus on here.  A number is
"originally" written as the infinite sequence of rational
numbers, which converge to that number.   So
1 = {1, 1, 1, 1, 1, 1, 1, ...}
1/2 = {.5, .5, ...} ; .75 = {.7, .75, ....} ;
pi = {3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ...}

However, we point out that by representing a number
(which now is simply an element in an equivalence class
of sequences of rational numbers), we must be able to
accept the fact that some numbers have nonunique
representations (in fact, infinitely many (countably
infinite) rational numbers have more than one digital
representation).  All numbers which can be written
with a finite digital expansion can also be written
as one with an infinitely repeating expansion (in spirit
of 1 and .9(r).  eg.  1/2 = .5 = .49999...).

A digital number's true meaning is dependent on the
equivalence class in which it resides.  We can determine
if two numbers lie in the same class if the sequence of numbers
maejohns@cobray:~  :cat inf
Okay, here is an attempt to enlighten you about numbers
with infinitely repeating digits to the right of the decimal
point.

It kind of depends on the notions of equivalence classes and
convergent sequences.  If these are new ideas (and you have
time to spare), look it up in any text on elementary analysis...


Firstly, to answer your question:    .9(r) = 1.


A Question for motivation:  How do the digital
representation of a number and its actual meaning relate?

As you should know, all numbers that we play with can be
represented in terms of their decimal expansion (base 10),
and that is what we will focus on here.  A number is
"originally" written as the infinite sequence of rational
numbers, which converge to that number.   So
1 = {1, 1, 1, 1, 1, 1, 1, ...}
1/2 = {.5, .5, ...} ; .75 = {.7, .75, ....} ;
pi = {3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ...}

However, we point out that by representing a number
(which now is simply an element in an equivalence class
of sequences of rational numbers), we must be able to
accept the fact that some numbers have nonunique
representations (in fact, infinitely many (countably
infinite) rational numbers have more than one digital
representation).  All numbers which can be written
with a finite digital expansion can also be written
as one with an infinitely repeating expansion (in spirit
of 1 and .9(r).  eg.  1/2 = .5 = .49999...).

A digital number's true meaning is dependent on the
equivalence class in which it resides.  We can determine
if two numbers lie in the same class if the sequence of numbers
defined by the difference of each element of the numbers' sequence
representation goes to zero...

eg:   are pi and 22/7 the same number?

a = pi = 3.1415926...   = {3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ...}
b = 22/7 = 3.1428571... = {3, 3.1, 3.14, 3.142, 3.1428, 3.14285, ...}

{b-a} = {0, 0, 0, .001, .0013, .00126, ...}

we can continue this indefinitely and we will see that the jth element
of this sequence does NOT go to zero as j->infinity.  Hence
these numbers don't lie in the same equivalence class and
therefore are not equal.


.9(r) = 1  ?

.9(r) = {a} = {.9, .99, .999, .999, .9999, .99999, .999999, ...}
1     = {b} = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, .....}

So, {b-a} = {.1, .01, .001, .0001, .00001, .000001, .0000001, ...}

We see that the lim (j->infinity) of {b-a}_j = 0.  Hence, {b} and
{a} lie in the same equivalence class which implies that
.9(r) = 1.  QED



Hope this helps to some extent.  A more thorough discussion
can be found in many texts on real analysis (and many other
topics)...


Mark E. Johnson
Graduate Student,
Institute for Scientific Computing
   and Applied Mathematics
Newsgroups: sci.math