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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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