# real number

Informally, the "real numbers" are the rational numbers with all the holes plugged.

Before we proceed with any formalism, let's exhibit an example of a
"hole" in the rational numbers. Take, for our "hole", the square root
of 2. Suppose <math>x^2 = 2</math>. If x is a rational number then
we can express it as a fraction, and what's more, we can express it as
a *reduced* fraction. Let **I** = the set of integers. Then:

- (1) <math>x = {n \over m},\ \ n,m \in \mathbb{I}</math>

and

- (2) <math>\not\exists p,q,k \in \mathbb{I} \left (k > 1 \ \and \ n = p \cdot k \ \and \ m = q \cdot k \right )</math>

Then we must have

- (3) <math> {\left ({ n \over m} \right ) }^2 = 2</math>

- (4) <math> {{n^2} \over {m^2}} = 2</math>

- (5) <math> n^2 = 2 \cdot m^2</math>

And so <math>n^2</math> must be *even*. But since the square of any *odd* integer is also *odd*, we see that **n** must *also* be even.

If n is even, then clearly <math>n^2</math> must be divisible by
*4*. In that case, n^{2}/2 must also be even,
and since we have

- (6) <math> {{n^2} \over 2} = m^2</math>

we see that <math>m^2</math> is even also. By the same argument, therefore, m must be even. But if both n and m are even, we must be able to find p and q such that,

- (7) <math>p,q \in \mathbb{I}\ (n = 2 \cdot p \ \and \ m = 2 \cdot q)</math>

But this contradicts (2), and we conclude that <math>\sqrt 2</math> must not be expressible as a rational number.

The real numbers add the *continuum axiom* to the axioms satisfied by
the rational numbers:

- (A.1)
**Any nonempty subset of the real numbers which is bounded above has a least upper bound.**

There are actually several equivalent ways to state the continuum axiom. The statement I've just given, though perhaps not maximally intuitive, is convenient for use in the construction of the real numbers.

We've just shown, above, that the rational numbers
do *not* satisfy the continuum axiom. For consider the set:

- <math>S \equiv \{x \ | \ x^2 < 2\}</math>

This is certainly nonempty: <math>1 \in S</math>. It's certainly bounded above: 3 is larger than any element of S. But the least upper bound of the set is <math>\sqrt 2</math> which is not a rational number.

Very well, so the rationals don't satisfy (A.1) -- but how do we
know *any* set will satisfy it? How do we know the real numbers
exist? We will construct them ... or, rather, we will sketch the
construction of the reals; the details actually fill a slim book (see
references).

We will start with the rational numbers, which we call Q. We first
define a *cut* (also called a "Dedekind cut"). If a nonempty subset
of the rationals, Γ, is a *cut*, then it divides the rational
numbers in two:

- (C.1) <math>x \in \Gamma \ \and \ y < x \Rightarrow y \in \Gamma</math>

- (C.2) <math>x \not\in \Gamma \ \and \ y > x \Rightarrow y \not\in \Gamma</math>

- (C.3) <math>(y \not\in \Gamma \Rightarrow y \geq x) \Rightarrow x \in \Gamma</math>

Property (C.1) says a cut contains a continuous block of numbers, extending to the "left". Property (C.2) says that the numbers that are *not* in the cut are also a continuous block, extending to the *right*.
Property (C.3) says that, if a cut has a rational least upper bound,
then that bound is a member of the cut set.

We observe immediately that each rational number, q, corresponds to a cut:

- <math>\Gamma(q) \equiv \{ x \in Q \ | \ x \leq q \}</math>

From here on, we will use rational numbers interchangeably with the cuts corresponding to them. In particular, we'll use "0", "1", and "-1" to refer to Γ(0), Γ(1), and Γ(-1), as the need arises.

At this point, we observe that the set **S**, defined above, is also a
cut, and it corresponds to <math>\sqrt 2</math>. If we view the
rational numbers as being contained in the set of cuts, via the
correspondence <math>q \leftrightarrow \Gamma(q)</math>, then the
set of cuts must therefore be a *proper* superset of the rational
numbers.

The set of all cuts will be our model of the real numbers. But we're not done yet: we still need to define comparisons, and we need to define multiplication and addition.

Comparisons are easy. It's simpler to define <math>\leq</math> than <math> < </math>, so that's what we'll do. For two cuts, G and H,

- <math>G \leq H \equiv x \in G \ \Rightarrow \ x \in H</math>

Addition is easy, too.

- <math> G + H \equiv \{x+y\ | x \in G\ \and\ y \in H\}</math>

Multiplication is a bit trickier. The problem is that all our cuts
have "tails" extending arbitrarily far to the left, and if we just
multiply all the members of two cuts we'll get something that has a
tail extending arbitrarily far to the *right*. So we need to be
cleverer than that.

We can define the product of two *non-negative* cuts:

- <math>G,H \geq 0 \Rightarrow G \cdot H = \{x\ |\ \exists y \in G, z \in H (y \geq 0 \ \and \ z \geq 0 \ \and \ x \leq yz)\}</math>

Now, before we define the product of two *general* cuts, we need a few
"helper definitions". The first of these is *negation*:

- <math>-G = \{x\ |\ \forall y \in G (x \leq -y) \}</math>

We can define absolute value in terms of negation:

- <math>|G| = \begin{cases} G, & \mbox{if } G \geq 0 \\ -G, & \mbox{if } G < 0 \end{cases}</math>

We'll define one "helper function":

- <math>Sgn(G) \equiv \begin{cases}1, & \mbox{if } G \ \geq 0 \\ -1, & \mbox{if } G < 0 \end{cases} </math>

and

- <math>Sgn(G,H) \equiv Sgn(G) \cdot Sgn(H)</math>

We also need to define multiplication of a cut by *-1*:

- <math>-1 \cdot G \equiv -G</math>

And finally, we can define multiplication of two general cuts:

- <math>G \cdot H \equiv Sgn(G,H) \cdot (|G| \cdot |H|)</math>

This completes the creation of the model. We have defined comparisons, addition, and multiplication. The rationals are embedded in the new model, and the comparison operation is clearly an extension of the comparison operation for the rationals. The definitions for addition and multiplication are pretty clearly the appropriate extensions from the rationals but rigorous proofs of those claims, as well as the claim that the model satisfies the axioms of the reals (the field axioms) would take more work than what I have displayed here.

The continuum axiom, as I stated it here, just describes a cut of the rationals. In our model for the real numbers, the least upper bound of such a set is the cut itself. So the model does indeed satisfy the continuum axiom.

**The Axiom of Choice**

It's worth calling some attention to one of the steps I glossed over. I rather casually said we would form the set of all cuts of the rationals. That's actually an enormous set, and it's formed as a subset of an equally enormous set, which is the power set of the rational numbers. It actually requires an uncountably infinite number of operations to form the set of all cuts. The assertion that we can form that set depends on the axiom of choice. This leads to the real numbers having some rather peculiar properties, and not everyone feels it's entirely legitimate.

It is possible to develop a version of analysis that doesn't depend on the axiom of choice, based on what are called the constructible reals.

References:

See Edmund Landau, Foundations of Analysis, for a thorough exposition of the construction of the real numbers. I've glossed over a lot and left out a great deal on this page. Landau does it "right". If you can find it in the library, that's the way to go -- it's well worth reading once, but it's not a book you'll refer to a lot.

--SAL 03:39, 19 Nov 2004 (UTC)