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Are All Infinitely Long Repeating Numbers Even?


Date: 06/06/2000 at 17:53:50
From: Richard Huggins
Subject: All infinitely long numbers of repeating form are even?

Take any infinitely long repeating series:

     x = 1234123412341234...

This one repeats after a section of length 4.

Multiply the number by 10^(section length), in this case 10^4. Thus we 
have:

     10000x = 12341234123412341234...
and
          x =     1234123412341234......

Subtract the two:

      9999x = 1234000000000000000000000000...

Clearly the right-hand side is even, because it ends in zero, i.e. 
"9999x" is even. But since 9999 is odd, therefore x must be even.

Thus all infinitely long numbers of a repeating form can be shown to 
be even.

Does this make sense? Is it correct? How could a series consisting 
entirely of odd numbers (ex. 1357135713571357....) be even?

Thanks in advance for any response.

Richard Huggins


Date: 06/07/2000 at 18:03:38
From: Doctor Rick
Subject: Re: All infinitely long numbers of repeating form are even?

Hi, Richard. Interesting question!

If you do the subtraction with any finite number of terms in the sum, 
you will see that the difference does not end in zero; it does not 
have to be even.

       1243124312430000
     -     124312431243
       ----------------
       1242999999998757

Even if you have a huge number of digits, you can't ignore those last 
digits just because they are "infinitely small" compared to the 
number. However small they are, they still determine whether the 
number is even or odd.

What makes your argument interesting is its resemblance to the 
argument that 0.999... (with the digit 9 repeated infinitely) = 1. See 
our Dr. Math FAQ on " 0.9999... = 1" at:

   http://mathforum.org/dr.math/faq/faq.0.9999.html   

The simple form of that argument looks like this:

           x = 0.9999...
         10x = 9.9999...
     10x - x = 9.9999... - 0.9999...
          9x = 9
           x = 1

Very much like yours; however, it's important to recognize what's 
going on behind the scenes here. Students are right to question this 
argument; we need to know more to be sure it's valid.

The number 0.99999, to pick a finite number of 9s, is really a 
shorthand way of writing

     9/10 + 9/100 + 9/1000 + 9/10000 + 9/100000

The number 0.9999... is a shorthand way of writing

     the limit of the sequence 0.9, 0.99, 0.999, 0.9999, ...
     as the number of 9s increases without end

We can write this in more formal mathematical notation as

                n
      Limit  [ Sum 9/10^i ]
     n->inf.   i=1

For any finite number of decimal places (say, n = 5), we have: As n 
increases, the sum approaches a limit, which is 1. By this we mean 
that you can pick any number, however tiny, and I can find a value of 
n such that the difference between the sum (0.999... with n or more 
9's) and 1 is less than the number you chose. That, in essence, is 
what we mean by a limit.

The FAQ gives an equivalent proof to the simple form above, using the 
notation of limits and sums. It hinges on a theorem in the theory of 
limits, that you can interchange a limit and a sum *provided* that the 
limit exists.

We can treat your argument similarly. Your "infinitely long repeating 
series" can be written formally as

                n
      Limit  [ Sum M*10^(Ki) ]
     n->inf.   i=1

where M is the repeating part (like 1234) and K is the number of 
digits in the repeating part.

Your argument breaks down because the limit above does not exist. In 
other words, the series (sum) does not "converge." The nth term 
increases without limit: 1234, 12340000, 123400000000, ... If a series 
does not converge to a limit, you can't do operations on it as you can 
with numbers.

Here's another way of saying it: you are subtracting two "infinite" 
numbers. Infinity is not a number, and you can't do ordinary 
arithmetic on it as if it were. See our Dr. Math FAQs on "Infinity and 
large numbers" and "Dividing by 0" for more discussion of this 
subject. But the better way of saying why your argument fails is that 
you are subtracting two series that do not converge, and this is 
invalid.

I hope I've stimulated some thought about important mathematical 
concepts. Fallacious proofs are a great way to get us thinking, and 
this is a good one!

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Number Theory
High School Sequences, Series

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