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### Irrational Decimals

Date: 08/04/98 at 08:54:13
From: Scott Bennett
Subject: Decimal representations of rational numbers

I was recently shown a proof stating that 0.9 (repeating) is actually
equal to one. I have come to accept this proof, but I'm wondering from
a representation standpoint whether they are actually considered to be
mathematically equal.

The proof involves using an sum of 9/(10^n) for n = 1 to infinity,
which is an expression for 0.9 repeating. I was actually able to
justify this by pulling the 9 out of the sum and multiplying it by the
sum of 1/(10^n) for n = 1 to infinity.

This opens up a whole new can of worms, specifically that any
irrational number could possibly be expressed as an infinite sum of
rational numbers. And since the sum of two or more rational numbers is
another rational number, there may not be any irrational numbers as the
term is currently defined.

My question is, are the decimal representations of rational numbers and
the rational numbers themselves mathematically equal, or would they
more appropriately be called equivalent?

Date: 08/04/98 at 16:59:58
From: Doctor Rob
Subject: Re: Decimal representations of rational numbers

You are beginning to discover some of the beauties and mysteries of
mathematics!

It is true that the sum of any finite number of rational numbers is
again a rational number, but the sum of an infinite number of rational
numbers may be irrational. After all, any decimal expansion, such as:

Pi = 3.14159265358979323846...
sqrt(2) = 1.414213562...

and so on, are actually infinite sums of rational numbers:

3.14159... = 3 + 1/10 + 4/10^2 + 1/10^3 + 5/10^4 + 9/10^5 + ...

Yes, every irrational number is the sum of an infinite number of
rational numbers. That does not preclude them from existing. The
rational numbers are just those whose decimal expansions are
eventually periodic. All the others are irrational.

rational numbers are mathematically equal. Some rational numbers have
more than one decimal expansion, as you have come to accept:

1 = 1.00000000... = 0.99999999...

for example. That is not an impediment.

- Doctor Rob, The Math Forum
Check out our web site! http://mathforum.org/dr.math/

Date: 08/05/98 at 11:39:15
From: Anonymous
Subject: Re: Decimal representations of rational

How can you say that an infinite sum of rational numbers can be an
irrational number? If you can add two rational numbers and get a
rational number, then by the associative property of addition, you
could add up any number of rational numbers, even an infinite number,
and get a rational number.

It defies logic to say that you can show irrationality by demonstrating
that a number cannot be shown to be the root of Qx+P = 0. There are an
infinite number of integers, and no one can know whether there are
numbers out there for which this method will or will not work. Maybe
our computing power just is not enough yet to find these numbers, but
this does not mean they don't exist, they're just beyond our current
understanding.

To me, you could prove irrationality by showing that there is not a
repeating pattern. But what if the repeating pattern is infinitely
long? And what if the pattern is not necessarily (I don't know the
right word for it) linear? Maybe something like:

1.32435465768798109111012 ...

There is a repeating pattern, but it's not linear, like:

1.35246135246 ...

(like I said, I don't know the right word for it, but linear seems
appropriate).

Thanks, this has been fun and educational, and has definitely made me
think. This revelation has only come to me in the past few days through
some research and just plain thinking about things, and it has shaken
me up a little bit, but I think that's good now and then.

Scott

Date: 08/05/98 at 16:09:16
From: Doctor Rob
Subject: Re: Decimal representations of rational

When you pass from a finite number of terms to an infinite number of
terms, your proof breaks down, and no substitute proof is available.
You may get for your sum either a rational or an irrational number.
See below.

An irrational number is defined as one which cannot be written in the
form a/b for integers a and b with b nonzero. (You might as well assume
that b > 0, since a/[-b] = [-a]/b.) Those are precisely the solutions
of b*x + (-a) = 0, so any number which is not the solution of such a
linear equation with integer coefficients is by the definition an
irrational number.

Every rational number a/b has a finite repeating sequence of digits
in its decimal expansion. The proof goes like this. Use long division
to divide b into a. Look at the sequence of remainders you get.
All of them are smaller than b (if you have done the long division
correctly!). After at most b steps, you will either get a remainder
of 0 (and the decimal expansion terminates) or you will get a nonzero
remainder you have already seen previously. This is an application of
the Pigeonhole Principle. (If you put n+1 things in n boxes, at least
one box must contain two of them.) From that point on, the division
will exactly duplicate the steps following that previously seen point,
including getting that same remainder yet again, after the same number
of steps. If you are getting the same sequence of remainders, you will
be getting the same sequence of quotient digits, so that will repeat
with the same period as the remainder sequence.

The above argument is independent of the size of a and b, so is valid
for every integer choices for those numbers. If b is large, the period
may be long, but it is finite, since its length is at most b-1. (Why

Here is an example:  16/37.

.432432432432
------------------
37 ) 16.0000000000000
14 8
----
1 20
1 11
----
90
74
--
160
148
---
120
111
---
90
74
--
16

Sequence of remainders:  16, 12, 9, 16, 12, 9, 16, 12, 9, 16, ...
---------  ---------  ---------  --------

If two remainders are equal, then when you bring down the 0, the result
will also be the same, and the quotient digit will be the same, and the
new remainder will be the same, so all the steps afterwards will look
like a repetition of earlier steps.

Another example:  93/176.

.528409090909...
---------------------
176 ) 93.000000000000...
88 0
----
5 00
3 52
----
1 480
1 408
-----
720
704
---
160
0
---
1600
1584
----
160
0
---
1600
1584
----
16

Sequence of remainders: 93, 50, 148, 72, 16, 160, 16, 160, 16, ...
-------  -------  -------

There is a certain pattern to the decimal digits, but they do not form
a periodic sequence of digits. All those decimals cannot be numbers of
the form a/b, for any a or b, by the above argument. Thus they are
irrational. Here is another kind:

1/10^1 + 1/10^4 + 1/10^9 + 1/10^16 + 1/10^25 + ... + 1/10^(n^2) + ...
= 0.1001000010000001000000001000000000010000000000001000...

The gaps between the 1's get longer and longer, so it is not a periodic
repeating decimal. That means it must be irrational.

Keep thinking and asking good questions!

- Doctor Rob, The Math Forum
Check out our web site! http://mathforum.org/dr.math/

Associated Topics:
High School Sequences, Series

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