MT2002 Analysis

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Axioms for the Real numbers

We saw before that the Real numbers R have some rather unexpected properties. In fact, there are many things which it is difficult to prove rigorously.

Examples

How do we know that sqrt2 exists? In other words how can we be sure that there is some real number whose square is 2?
It is easy to convince yourself that (say) 2 + 3 = 3 + 2. How about sqrt2 + sqrt3 = sqrt3 + sqrt2 or e +p = p + e?

One's intuition about what should be true works pretty well for N or Z or even for Q. Things don't get hard until we are forced (like the Pythagoreans) to admit the existence of irrationals.

There are constructive methods for making the full set R from Q and hence starting with N. The first rigorous construction was given by Richard Dedekind (1831 to 1916) in 1872.

You can see more about Dedekind's construction.

However, for the moment we will simply give a set of axioms for the Reals and leave it to intuition that there is something that satisfies these axioms.

We start with a set, which we'll call R and a pair + . of binary operations.

The Axioms

These are divided into three groups.

I The algebraic axioms

R is a field under + and .
This means that (R, +) and (R, .) are both abelian groups and the distributive law (a + b)c = ab + ac holds.

II The order axioms

There is a relation > on R.
(That is, given any pair a, b then a > b is either true or false).

It satisfies:

a) Trichotomy: For any a belongs R exactly one of a > 0, a = 0, 0 > a is true.

b) If a, b > 0 then a + b > 0 and a.b > 0

c) If a > b then a + c > b + c for any c

Something satisfying axioms I and II is called an ordered field.

Examples

  1. The field Q of rationals is an ordered field.

    Proof
    Define a/b > c/d provided that b, d > 0 and ad > bc in Z. One may easily verify the axioms.


  2. The field C of complex numbers is not an ordered field under any ordering.

    Proof
    Suppose i > 0. Then -1 = i2 > 0 and adding 1 to both sides gives 0 > 1.
    But squaring both sides gives (-1)2 = 1 > 0 and so we get a contradiction.
    A similar argument starting with i < 0 also gives a contradiction.


The above two groups of axioms can be used to deduce any algebraic or order properties of R.

Example

The ordering > on R is transitive.
That is, if a > b and b > c then a > c.

Proof
a > b if and only if a - b > b - b = 0 by Axiom II c)
a > c if and only if a - c > c - c = 0
Hence (a - b) + (a - c) > 0 and so a - c > 0 and we have a > c.


The thing which distinguishes R from Q (and from other subfields) is the Completeness Axiom.

Definitions

An upper bound of a non-empty subset A of R is an element b belongs R with b gte a for all a belongs A.
An element M belongs R is a least upper bound or supremum of A if
M is an upper bound of A and if b is an upper bound of A then b gte M.
That is, if M is a lub of A then (thereexistsb belongs R)(forallx belongs A)(b gte x) implies b gte M
A lower bound of a non-empty subset A of R is an element d belongs R with d lte a for all a belongs A.
An element m belongs R is a greatest lower bound or infimum of A if
m is a lower bound of A and if d is an upper bound of A then m gte d.

We can now state:

III The Completeness Axiom

If a non-empty set A has an upper bound, it has a least upper bound.

Something which satisfies Axioms I, II and III is called a complete ordered field.

Remark

In fact one can prove that up to "isomorphism of ordered fields", R is the only complete ordered field.

Note that the ordered field Q is not complete
For example, the set {q belongs Q | q2 < 2} is bounded but does not have a least upper bound in Q. We will see why in a little while.

Some consequences of the completeness axiom.

  1. A subset A which has a lower bound has a greatest lower bound.

    Proof
    Let B = {x belongs R | -x belongs A}. Then B is bounded above by -(the lower bound of A) and so has a least upper bound b say. It is then easy to check that -b is a greatest lower bound of A.


  2. The Archimedean property of the Reals

    If a > 0 in R, then for some n belongs N we have 1/n < a.
    Equivalently: Given any x belongs R, for some n belongs N we have n > x.

    Proof
    This last statement is equivalent to saying that N is not bounded above. This seems like a very obvious fact, but we will prove it rigorously from the axioms.
    Suppose N were bounded above. Then it would have a least upper bound, M say. But then M - 1 is not an upper bound and so there is an integer n > M - 1. But then n + 1 > M contradicting the fact that M is an upper bound.

    Remark

    This result has been attributed to the great Greek mathematician (born in Syracuse in Sicily) Archimedes (287BC to 212BC) and appears in Book V of The Elements of Euclid (325BC to 265BC).

    From this we can deduce :

  3. Between any two real numbers is an rational number.

    Proof
    Let a noteq b be real numbers with (say) a < b. Choose n so that 1/n < b - a. Then look at multiples of 1/n. Since these are unbounded, we may choose the first such multiple with m/n > a.
    We claim that m/n < b. If not, then since (m-1)/n < a and m/n > b we would have 1/n > b - a.


    Remark

    A set A with the property that an element of A lies in every interval (a, b) of R is called dense in R.
    We have just proved that the rationals Q are dense in R. In fact, the irrationals are also dense in R.

    We can now prove the result we stated earlier.

  4. The real number sqrt2 exists.

    Proof
    We will get sqrt2 as the least upper bound of the set A = {q belongs Q | q2 < 2 }. We know that A is bounded above (by 2 say) and so its least upper bound b exists by Axiom III.
    We now prove that b2 < 2 and b2 > 2 both lead to contradictions and so we must have b2 = 2 (by the Trichotomy rule).
    So suppose that b2 > 2. Look at (b - 1 /n )2 = b2 - 2b /n + 1 /n2 > b2 - 2b/n.
    When is this > 2 ?
    Answer: When b2 - 2b/n > 2 which happens if and only if b2 - 2 > 2b/n or 1/n < (b2 -2)/2b and we can choose such an n by the Archmedean property. Thus b - 1/n is an upper bound, contradicting the assumption that b was the least upper bound.
    Similarly, if b2 < 2 then (b + 1/n)2 = b2 + 2b/n + 1/n2 > b2 + 2b/n.
    Can this be < 2 ?
    Answer: Yes, when b2 + 2b/n < 2 which happens if and only if 2 - b2 > 2 b/n or 1/n < (2 - b2)/2b and we can choose an n satisfying this, leading to the conclusion that b would not be an upper bound.


  5. Real numbers can be defined by decimal expansions.

    Proof
    Given the decimal expansion (say) 0.a1a2a3... consider the set (of rationals) {0.a1 , 0.a1a2 , 0.a1a2a3 , ...} = { a1/10 , (10 a1 + a2)/100 , (100 a1 + 10 a2 + a3)/1000 , ... }.
    This is bounded above -- say by (a1+ 1)/10 or by (10 a1+a2+ 1)/100 etc. and so it has a least upper bound.
    This is the real number defined by the decimal expansion.



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JOC September 2002