Repeating Decimals

Date: 5/31/96 at 16:49:42
From: Anonymous
Subject: Repeating Decimals

Hi, 

I am a high school teacher who is having a running argument with some 
students about repeating decimals. I claim that 0.9 repeating is equal 
to 1. One of my proofs involves the lack of numbers between these two 
numbers, much like any other two identical numbers. Another good proof 
is the pattern that comes from 1/9 = 0.1 repeating, 2/9 = 0.2 
repeating, ... 8/9 = 0.8 repeating, and therefore 9/9 = 0.9 repeating 
which of course = 1 (9/9). 

They still refuse to believe me, and have asked that I consult a 
higher authority. Can you help?

Thanks!

Date: 6/3/96 at 10:46:14
From: Doctor Tom
Subject: Re: Repeating Decimals

Okay, I'm a "higher authority," and I pronounce that you're right!

Is that enough?

If not, there are other ways to think of it:

For example, if they believe that 1/3 = .3333..., then
clearly 1/3 + 1/3 + 1/3 = 1 = .99999...

Or, if they believe that .00000... is zero, then if .99999... is 
different from 1, it must clearly be less. So what is 1 - .99999...?  
If it's not zero, its decimal expansion must have some non-zero term.  
Wherever that term is, it is easy to show that the difference between
1 and .99999... is less than that.

Or, if they believe that

.99999... = 9/10 + 9/100 + 9/1000 + 9/10000 + ...,

then just sum up the geometric series, where the first term is 9/10, 
and the factor is 1/10.

If S = a + ar + ar^2 + ar^3 + ..., then

S = a/(1-r).  a = 9/10, (1-r) = 9/10, so S = 1.

I hope this helps.

-Doctor Tom,  The Math Forum

One more variation to consider that has been suggested before is the 
following proof sent in by koops@magnet.at (Karsten Koops):

     Suppose                           X = 0,99. periodical
           Multiply by 10, so you get       10X=9,99.
           Subtract 1X (0,99.)               9X=9
           Divide by 9 and you get            X=1

-Doctor Steve   The Math Forum
 Check out our Web site:  http://mathforum.org/dr.math/