Is 0.999... (nines repeating forever) equal to 1?

Yes.

In article <3fjels$qi@barracuda.dadd.ti.com>,
Mark Shaw <mns1@.rtc.sc.ti.com> wrote:

>I maintain that 0.999... != 1.

Ok, here's more than you ever wanted to know about equality of real
numbers.

There is more than one way to define what an infinite decimal expansion
means, and for that matter, there is more than one way to define the
real numbers.  Here is an approach that is commonly used.

Let's start with Q, the field of rational numbers, and consider
sequences in Q.  In particular, we are interested in sequences {a_i}
that satisfy the Cauchy property:  given any epsilon > 0, there exists
N > 0 such that |a_i - a_j| < epsilon whenever i, j > N.  It's tempting
to say that the Cauchy sequences are sequences that converge to
something, but that would be putting the cart before the horse.  We
haven't even said what a real number is yet, and there are Cauchy
sequences of rationals that do not converge to any rational number.

It's easy to see, though, that every infinite decimal expansion is
associated with a Cauchy sequence of rationals.  For example, the
decimal expansion

        3.14159...

corresponds to the Cauchy sequence

        3, 31/10, 314/100, 3141/1000, 31415/10000, 314159/100000, ...        (1)

(Obviously that's a sequence of rationals, but why is it a Cauchy
sequence?  That follows from the definition, but be sure you understand
why before continuing.)

On the other hand, there are lots of Cauchy sequences that do not
correspond in a simple way with any decimal expansion, because it's
hard to predict how many terms you might need in order to get n-place
accuracy for a given n.  For example, the partial sums of the sequence

        4 (1 - 1/3 + 1/5 - 1/7 + ...)

which are

        4, 8/3, 52/15, 304/105, ...                                        (2)

would be one such sequence.  You might recognize it as one that
converges to pi, but extremely slowly.  We aren't in a position to
say what pi (or any other real number) is just yet, but somehow we
would like to capture the idea that both sequences (1) and (2) are
somehow representative of pi, even though they look very different from
each other.

The key lies in the concept of an equivalence relation.  We say that
two Cauchy sequences {a_i} and {b_i} are equivalent to each other if
the corresponding terms eventually get arbitrarily close.  In other
words, for every epsilon >0, there exists N > 0 such that |a_i - b_i| <
epsilon whenever i > N.

If {a_i} and {b_i} are Cauchy sequences of rationals, let's write

        {a_i} ~ {b_i}

to mean they are equivalent in the sense just explained.  It's easy to see
that for any Cauchy sequences {a_i}, {b_i}, and {c_i}, the relation "~"
has the following properties:

        (1) {a_i} ~ {a_i}                                (~ is reflexive)
        (2) if {a_i} ~ {b_i}, then {b_i} ~ {a_i}        (~ is symmetric)
        (3) if {a_i} ~ {b_i} and {b_i} ~ {c_i},
            then {a_i} ~ {c_i}                                (~ is transitive)

A relation that is reflexive, symmetric and transitive is called an
"equivalence relation" in mathematics, and therefore ~ is well named
(we said it meant the sequences are equivalent, and now that
terminology has been made precise.)

Given the set of all Cauchy sequences of rationals, we can partition
this huge set into a whole bunch of subsets called "equivalence
classes", having the property that any two members of the same
equivalence class are "equivalent" (in the ~ sense) to each other, and
no two members of different equivalence classes are ~ to each other.

Now, at last, we are in a position to say what a real number is.

        Definition.  A real number is an equivalence class of
        Cauchy sequences of rationals.

It will probably help to see this definition in action at least once
before proceeding.  Given two numbers a and b, what do we mean by the
sum a+b?

That's easy.  Choose a sequence {a_i} that represents a, and another
sequence {b_i} that represents b.  Now form the sequence {a_i+b_i}.
That sequence is a Cauchy sequence (requires proof, but it's easy).
Then we simply define a+b to be the equivalence class corresponding to
the Cauchy sequence just constructed.

The tricky part, which you should think about a little, is what happens
if someone else decides to find the sum a+b by using different Cauchy
sequences to represent the two numbers.  In order to show that the
concept is well defined, it's necessary to show that if

        {a_i} ~ {p_i}
and
        {b_i} ~ {q_i}
then
        {a_i+b_i} ~ {p_i+q_i}.

This all follows from the definitions of equivalent sequences, plus the
commutative and associative laws for addition of rational numbers, and
therefore it makes no difference which representatives you choose in
defining the sum.  Differences, products, and quotients work in much
the same way.  You may choose to use the sequence

        1, 1, 1, ...

to describe the real number 1, but someone else might use

        1/2, 3/4, 7/8, 15/16, ...

and a third person might use

        2/1, 3/2, 4/3, 5/4, 6/5, ...

to describe the same number. It makes no difference, since they all
lead to the same results.

Oh, by the way:  another member of that same class is:

        9/10, 99/100, 999/1000, 9999/10000, ...

which is just the sequence of partial sums corresponding to the decimal
expansion

        0.999...

and therefore we now know very precisely what it means to say that

        0.999... = 1

-- Dave Seaman <ags@seaman.cc.purdue.edu>

In rec.puzzles Darryl Tam <ba625@FreeNet.Carleton.CA> wrote:
> I know that 0.999... is equal to 1 but does anyone have proof?

Be careful of proofs that go:

let x=.999999...
10x = 9.99999...

so 10x-x = 9.9999... - .99999... = 9
9x=9 so x=1

There is a problem... namely that it is not logically correct.

For finite decimals (.234=234/1000) we have just ratios... and one can
prove that moving the decimal point one to the right is multiplying by
10.  Now... PROVE (not just say it because it feels right) that for
infinite decimals!  In fact, one can give a fairly natural meaning to
.999... in a non-standard extension of the reals where ALL the standard
operations on rationals work "right" but where 10*.9999999.... =
9.99999...+inf where inf is an infinitesimal.

One can show that if order is preserved and .9999.... is a real number
(not 1-inf where inf is an infinitesimal, which is not a real number)
then .999... is equal (as a real number) to 1 (without assuming how
multiplication by ten "works" or "defining" .9999... as the limit of a
sequence in the reals).

The trick is that between any two different real numbers there is a
rational number.

Suppose .99999.... < 1 AND .9999... is a real number and order is
preserved.

Then there is a rational, p/q (p,q positive integers) with
.9999...<p/q<1

But then p/q<1 so p<q and p<=q-1 (integers!)

But .9<.9999...<p/q, so p/q>.9=9/10=1-1/10 and p>q-q/10
now we have q-q/10 <p<= q-1 so q>10

also .99<.9999...<p/q and doing it again q>100

and .999<.9999...<p/q so q>1000

So... if .9999...<1 and .9999... is a real number, there would have to
be a positive integer q with q>10 and q>100 and q>1000 and q>10000 etc.

But there is no "infinitely big" integer.  Can't happen.

-- John McGowan <jmcgowan@shell.inch.com>, May 1997