One of the early objectives of almost any university
mathematics course is to teach people to stop thinking of
the real numbers as infinite decimals and to regard them
instead as elements of the unique complete ordered field,
which can be shown to exist by means of Dedekind cuts,
or Cauchy sequences of rationals. I would like to argue
here that there is * nothing * wrong with thinking of
them as infinite decimals: indeed, many of the traditional
arguments of analysis become more intuitive when one does,
even if they are less neat. Neatness is of course a
great advantage, and I do not wish to suggest that
universities should change the way they teach the real
numbers. However, it is good to see how the conventional
treatment is connected to, and grows out of, more `naive'
ideas. I shall illustrate this later with a discussion of
the square root of two and the intermediate value theorem.

Recall that to `construct' the real numbers means to give an example of a complete ordered field. For this purpose one is allowed to assume the existence of the rational numbers (themselves constructed from the positive integers, which can, if you want, be constructed from sets) and certain notions, such as that of an infinite sequence, which are not entirely unproblematic (for reasons discussed here ) but which are in a sense more elementary.

Here is an indication of how infinite decimals give a perfectly good, and very easily understood, construction of the real numbers. Because of irritating difficulties such as the need to carry digits and to identify 0.999999.... with 1, and because there is something unnatural about the number 10 (or any other number one might choose, though 2 might be an improvement) this construction is less aesthetically pleasing than some. However, it has the advantage, for beginners, of being very close to the picture of real numbers they already have.

To begin with, one defines an infinite decimal in the obvious way, as a finite sequence of elements of the set {0,1,2,3,4,5,6,7,8,9} followed by a decimal point followed by an infinite sequence of elements of the set {0,1,2,3,4,5,6,7,8,9}. This isn't quite the whole definition since one must point out that some of these objects are equal: for example, 0124.383478... is the same number as 124.383478... (assuming of course that the sequences continue in the same way) and 1.9999999... is the same number as 2. (About this last example, by the way, there can be no argument, since I am giving a That defines the set I am constructing. To make it a
complete ordered field, I must now specify the ordering,
explain how to add and multiply infinite decimals, and
prove that the field axioms, order axioms and completeness
axiom are all satisfied. First, then, how does one add
two infinite decimals? The answer is simple, but to
explain it let me introduce some notation. Given an
infinite decimal x, I shall write x(n) for the * finite
* decimal obtained by truncating x at the n^{th}
place after the decimal point. For example, if x is the
square root of two, then x(1)=1.4, x(2)=1.41 and x(3)=1.414.
It doesn't matter too much, but let us say that if x can
be written either as a decimal ending in an infinite string
of nines or as one ending in an infinite string of zeros,
then we will go for the nines - just to remove the ambiguity
from the definition of x(n) above.

To add x and y, the first step is to consider the sequence
of finite decimals x(1)+y(1), x(2)+y(2), x(3)+y(3) and so on.
Let us have a look at this when x=y=pi=3.141592653589793....
Then the sequence begins 6.2, 6.28, 6.282, 6.2830, 6.28318,
6.283184, 6.2831852, 6.28318530, 6.283185306, 6.2831853070,
6.28318530716, 6.283185307178, 6.2831853071794, 6.28318530717958,
6.283185307179586, ... . Now, given a term in this sequence,
you cannot always get the next one by simply putting a new
digit on the end, because sometimes you have to modify one or
more of the earlier digits. For example, after 6.282 came
not 6.282t for some t but 6.2830. However, it is an easy
exercise to show that * no digit is ever modified more
than once *. Therefore, the rule for determining the
infinite decimal for x+y is the following. The n^{th}
digit of x+y is the n^{th} digit of x(n)+y(n), unless
that is later modified, in which case it is the n^{th}
digit of x(m)+y(m), where m > n is the first (and only) time
that the n^{th} digit changes.

We can simplify the above description by defining a notion
of limit appropriate for this context. First, let us note that
it is easy to put infinite decimals in order: x is less than
y if x(n) is less than y(n) for some n. (Here it is important
that I define x(n) unambiguously.) Now suppose we have a
non-decreasing sequence (x_{1}, x_{2},
x_{3}, ...) of infinite decimals (if they are finite
we can make them infinite by putting zeros on the end). Then
for any n the sequence (x_{1}(n), x_{2}(n),
x_{3}(n), ...) is also non-decreasing. (This is the
same sequence as before but with everything truncated at
the n^{th} decimal place.) Either this sequence
changes infinitely often, in which case it is obvious
that the sequence (x_{1}(n), x_{2}(n),
x_{3}(n), ...) is unbounded, and hence so is the
sequence (x_{1}, x_{2}, x_{3}, ...),
or it changes only finitely many times, in which case it
eventually stops changing and becomes some fixed decimal
that terminates at the n^{th} place. Hence, for
any bounded non-decreasing sequence (x_{1}, x_{2},
x_{3}, ...) all the sequences (x_{1}(n), x_{2}(n),
x_{3}(n), ...) are eventually constant. If we define
y(n) to be the finite decimal that (x_{1}(n), x_{2}(n),
x_{3}(n), ...) eventually settles down at, then
whenever m > n the sequence y(m) begins with the sequence
y(n). Therefore, the finite decimals y(1), y(2), ...
are the truncations of an infinite decimal y. This y we call
the limit of (x_{1}, x_{2}, x_{3}, ...).

This definition more or less contains a proof of the completeness axiom for infinite decimals, in the monotone-sequences form. (What I have not done is show that this notion of limit agrees with the usual one. That is, however, not very difficult.) And now I can simply define x+y to be the limit of the bounded non-decreasing sequence x(1)+y(1), x(2)+y(2), x(3)+y(3), ... . (Actually, it may sometimes decrease if one of x and y is negative. For negative numbers one could add two integers to make them positive, add those numbers together and then subtract the two integers again. See below for a different convention.)

How might one now prove that addition is commutative and
associative? It is very easy: for example, the sequences
(x(1)+y(1), x(2)+y(2), x(3)+y(3), ...) and
(y(1)+x(1), y(2)+x(2), y(3)+x(3), ...) are equal and therefore
have the same limit. A similar remark does associativity. It is
clear that 0 is an identity. Inverses are a little bit more complicated
because of the problems with negative numbers experienced
above. One could define limits for more general sequences
than increasing ones. I prefer to define infinite decimals
as follows: start with an integer k (positive or negative)
and follow it with a decimal point and then an infinite
string s of numbers from 0 to 9. This represents the number
k + 0.s. For example, by (-2).386... I mean the number that
would normally be written -1.613... . Then the inverse of
n.s_{1}s_{2}s_{3}... is
(-(n+1)).t_{1}t_{2}t_{3}... where
each t_{i} is 9-s_{i}. It is not hard
to check that these two numbers add up to
(-1).999999... which equals zero.

It is not hard to prove that multiplication
thus defined is commutative and associative and that 1
is an identity element. Once again, inverses present
more of a problem, though not much more. Again, let
us imagine trying to do the calculations for a specific
example - say pi again. One natural way to do it would
be to take the reciprocals of pi(n), which would be
infinite decimals, and watch them gradually converge.
One would obtain a decreasing sequence rather than an
increasing one, but it would be easy to extend the
definition of limit to accommodate this. Alternatively,
one could define pi[n] to be the smallest decimal
which is larger than pi and terminates at the n^{th}
place so that the sequence 1/pi[n] would now be increasing.
Yet another possibility would be to define r(n) to be
the largest decimal that terminates at the n^{th}
place and satisfies the inequality r(n)pi < 1. Any
one of these will work, though a little bit of effort is
needed to show that the resulting number really does
give 1 when multiplied by pi.

I hope, even if the above discussion falls short of a complete proof, that I have given enough detail to convince you that there are natural definitions of addition and multiplication and ordering of infinite decimals, and that with these definitions they form a complete ordered field. Now let us look at a `naive' approach to the existence of the square root of two (about which further discussion can be found here ).

One reason it seems obvious that the square root of
two exists is that one can calculate it as an infinite
decimal. If you have worked out the first few digits,
then the next one will be the largest number between
0 and 9 such that the resulting number squares to less
than 2. For example, 1.4^{2}=1.96 < 2 and
1.5^{2}=2.25 > 2, so the first digit after the
decimal point is 4. Then, since 1.41^{2} < 2 and
1.42^{2} > 2, we find that the next digit is 1,
and so on.

I shall now show that this idea forms the basis for
a perfectly good and correct proof of the existence of the
square root of two. First of all, if we have defined real
numbers as infinite decimals, then the procedure just outlined
really does unambiguously define a real number. In fact,
it is the unique infinite decimal x such that, for every n,
x(n)^{2} < 2 and (x(n)+10^{-n})^{2} > 2.
This doesn't quite prove that 2 has a square root though, since
it might be that x, though well defined, doesn't actually
square to 2.

Why does this suggestion seem ridiculous? Is it not
obvious that x squares to 2? Actually, it is not quite
obvious, though it is not too hard either. Here is how
an argument would go. If you look back to the definition
of multiplication for infinite decimals, you will see that
what we have to show is that, as n increases, x(n)^{2}
is a decimal consisting of 1 followed by a decimal point
followed by a string of 9s whose length increases to
infinity as n increases to infinity. That way, the square
of x will be (by definition) 1.999999999.... which (by
definition) equals 2.

Let us see, then, why one can choose n in such a way that
the string of 9s in x(n)^{2} has length at least
100. The rough argument goes like this. If you set
n=101, then x(n)^{2} < 2 while
(x(n)+10^{-101})^{2} > 2. However,
x(n) and x(n)+10^{-101} differ by only 10^{-101},
from which it follows, after a brief calculation, that
x(n)^{2} and (x(n)+10^{-101})^{2}
differ by less than 10^{-100}. From this it follows
that x(n)^{2} > 2-10^{-100}, as we wanted.

The brief calculation is the following: the difference
between (y+c)^{2} and y^{2} is 2cy+c^{2}.
If y=x(n) < 2 and c < 10^{-101}, then this is less
than 10^{-100}. Clearly, a similar calculation shows
that if you want a string of m 9s, then you can achieve it
with x(m+1) (which, just to remind you, is the truncation at
the (m+1)^{st} place of the decimal expansion of the
square root of two.)

One of the usual proofs of the existence of the square root
of two goes as follows. The function f(x)=x^{2} is
continuous, f(1) < 2 and f(2) > 2, so by the intermediate
value theorem there exists x such that f(x)=2. Examining
the proof of the intermediate value theorem one sees that it
specifies a value of x, namely the supremum of the set A of
real numbers y such that y^{2} < 2. Every step of
this usual proof has its counterpart in the argument I have
just given in terms of decimals. Let me spell this out.

1. To define the infinite decimal x, I made its decimal
expansion as big as possible, taking care only that x(n)^{2}
was less than 2. This corresponds in a natural way to picking the
supremum of A above. We can make the correspondence even closer
by redefining the notion of supremum in terms of infinite decimals.
To do this, let B be any set of infinite decimals which is bounded
above, and define a number x=sup B by setting x(n) to be the maximum of
all y(n) such that y is in B. It is not hard to see that x(m) is
an initial segment of x(n) when m < n, as is needed, and that
the resulting infinite decimal x is the smallest possible upper
bound of the set B (obviously x(n) has to at least as big as all
y(n) for x to be an upper bound, and equally obviously if, for every
n, x(n) is as big as all y(n), then x is an upper bound). Thus, it is
easy to prove the least upper bound axiom for our construction. It is
also not hard to see that if we let B be the set of all
infinite decimals with squares less than 2, then sup B will
be exactly the same as the decimal defined earlier.

2. When I showed that we could make the string of 9s as long
as we liked, I used the fact that a change late on in a decimal
expansion of a number x has only a small effect on x^{2}.
This comes to the same thing as saying that the function
f(x)=x^{2} is continuous, and the calculation I did
was the usual calculation needed to show this (at least in
the range 1 < x < 2).

From the point of view of decimals, what was the exact property
of the function f(x)=x^{2} that allowed our argument to work?
It was the following. For any given m there exists
an n such that, in order to work out f(x) to within an accuracy of
10^{-m} it is enough to know x up to the n^{th} place
after the decimal point - that is, to know x(n). This is a simple
reformulation of the definition of continuity in terms of decimals.
(See here for a further discussion
of continuity.)

Using this definition, let us try to prove the intermediate
value theorem in a hands-on way. For this purpose, suppose that f
is a continuous function from the interval [a,b] to R, with f(a) < c
and f(b) > c. How might we try to find x such that f(x)=c?
Why don't we try to do it exactly as we did when f(x) was
x^{2} and c was 2? Thus, we should start by finding
the largest possible x(1) such that f(x(1)) < c, and then
the largest possible x(2) that starts like x(1) and satisfies
f(x(2)) < c and so on. If we continue this process, then
we end up by constructing the largest possible infinite
decimal x with the property that f(x(n)) < c for every
n. (Note that this may not be the supremum of the set
of all y such that f(y) < c. With a slightly different
argument, one can arrive at this x instead.) An easy inductive
argument shows that for every n we have f(x(n)) < c and
f(x(n)+10^{-n})__>__c. Since f is continuous, for
every m we can choose n such that f(x(n)) and
f(x(n)+10^{-n}) are both within 10^{-m}
of f(x), from which it follows that f(x) is within
10^{-m} of c. Since we can do this for every m,
f(x) must equal c.

The above argument is a slightly disguised form of another common proof of the intermediate value theorem, which uses repeated bisection. (We, of course, repeatedly split our intervals into ten parts rather than two.) The fact that it is also pretty similar to the proof via the supremum of {x:f(x) < c} is an indication that the usual proofs of the theorem are all, fundamentally, based on the same simple and intuitive idea.