Difference between revisions of "real number"

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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, n²/2 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,