This web page lists a number of important properties of the real numbers. Most of this is background information that you already know in some form or other and you can think of it as an information sheet setting out the basic properties of the reals that we have been using up to now. You know them all and are familiar with them and with using them, but may not have seen them listed in this way before.

A number system satisfying all the properties listed here is
called an Archimedean Ordered Field. Thus the set of
real numbers forms a Archimedean Ordered Field. So does
the set of rational numbers. The final property describing the
reals and distinguishing it from the rationals and other
Archimedean ordered fields is called *completeness* and
will be described later.

We start here by seeting out the axioms and basic properties of a field. You have seen most of these ideas in the first half of the module, and this might be good revision material. I won't expect you to learn or directly use the axioms of an ordered field in the sequences and series section of the module, and I won't set exam questions on this material, but you might be asked questions about this (or similar) things in reference to the first half of the module.

A field

$F$ is a system of numbers with an
addition operation and a multiplication
operation, satisfying some basic axioms. These axioms say
that $(F,+)$ is an
abelian group
with identity element $0$
and $(F\setminus \left\{0\right\},\cdot )$ is an abelian
group with identity element $1$. (In particular, this means
that $F\setminus \left\{0\right\}$ must be nonempty and hence
a field always contains an element other than $0$.)
Furthermore the distibutivity law holds.

When writing these axioms, it is convenient to
introduce, as well as $0,1$,
notation for the *additive inverse $-x$ of $x$
*
and the *mutiplicative inverse ${y}^{-1}$ of $y\ne 0$
*.

**Definition 2.1**** (of a field)**

A field is a set $F$ with operations of addition $+:{F}^{2}\to F$, multiplication $\cdot :{F}^{2}\to F$, additive inverse $-:F\to F$, and multiplicative inverse ${}^{-1}:F\to F$, and with elements $0,1\in F$, such that

**(1) $(F,+)$ is an abelian group:**

Associativity: $\forall x\in F\forall y\in F\forall z\in F\left((x+y)+z=x+(y+z)\right)$

Commutativity: $\forall x\in F\forall y\in F\left(x+y=y+x\right)$

Identity: $\forall x\in F\left(x+0=x\right)$

Inverses: $\forall x\in F\left(x+(-x)=0\right)$

**(2) $(F\setminus \left\{0\right\},\cdot )$ is an abelian group:**

Non-trivial: $0\ne 1$

Associativity: $\forall x\in F\forall y\in F\forall z\in F\left((x\cdot y)\cdot z=x\cdot (y\cdot z)\right)$

Commutativity: $\forall x\in F\forall y\in F\left(x\cdot y=y\cdot x\right)$

Identity: $\forall x\in F\left(x\ne 0\Rightarrow x\cdot 1=x\right)$

Inverses: $\forall x\in F\left(x\ne 0\Rightarrow x\cdot \left({x}^{-1}\right)=1\right)$

**(3) multiplication and addition satisfy the distributivity law**

Distributivity: $\forall x\in F\forall y\in F\forall z\in F\left(x\cdot (y+z)=(x\cdot y)+(x\cdot z)\right)$

Note that from the way the definition is stated, a field $F$
is *automatically* closed under $+,\cdot ,-$ and
${}^{-1}$. But in an example, you might have to check that
these operations are well-defined on $F$, including the closure
properties.

You already know that the identity and inverses are unique in a group. We can state this formally as a very useful proposition.

**Proposition 2.1**** (uniqueness of identity and inverses)**

Let $F$ be a field. Then

(a) Suppose $x\in F$ satisfies $\forall y\in Fx+y=y$ then $x=0$.

(b) Suppose $x,y\in F$ satisfy $x+y=0$ then $x=-y$ and $y=-x$.

(c) Suppose $x\in F$ satisfies $\forall y\in F\left(y\ne 0\Rightarrow x\cdot y=y\right)$ then $x=1$.

(d) Suppose $x,y\in F$ are not equal to $0$ and satisfy $x\cdot y=1$ then $x={y}^{-1}$ and $y={x}^{-1}$.

Most of the usual algebraic properties of numbers (such as $(-x)\cdot (-y)=xy$) follow from the axioms of a field via the above proposition. I'll do some examples here. My proofs will be correct, but rather abbreviated. You'll have to work quite hard to read them and work out the reasons behind each step. For the remaining propositions in this section, we fix a field $F$.

**Proposition 2.2**

$-1\ne 0$

**Proof**

If $-1=0$ then $0+0=0=1+-1=1+0$ so $1=0$ by cancellation in the additive group.

**Proposition 2.3**

$\forall x\in Fx\cdot 0=0=0\cdot x$

**Proof**

$x\cdot x+0=x\cdot x=x\cdot (x+0)=x\cdot x+x\cdot 0$ so $0=x\cdot 0$ by cancellation. The other direction now follows by commutativity of $\cdot $.

**Proposition 2.4**

$\forall x\in Fx\cdot 1=x=1\cdot x$

**Proof**

Except for when $x=0$ this is an axiom. The case $x=0$ is covered by the previous proposition.

Please read the next few statements carefully. $-x$ is the
additive inverse of $x$. It has nothing
*
a priori*
to do with multiplication by minus one.

**Proposition 2.5**

$\forall x\in F-(-x)=x$

**Proof**

$x+-x=0$, so x is the additive inverse of $-x$ (by uniqueness) so $x=-(-x)$.

**Proposition 2.6**

$\forall x\in F\left(-1\right)\cdot x=-x$

**Proof**

$x+\left(-1\right)\cdot x=1\cdot x+\left(-1\right)\cdot x=(1+-1)\cdot x=0\cdot x=0$ so $\left(-1\right)\cdot x$ is the additive inverse of $x$.

**Proposition 2.7**

$\forall x\in F\forall y\in F(-x)\cdot y=-(x\cdot y)$

**Proof**

$(x\cdot y)+\left(\right(-x)\cdot y)=(x+-x)\cdot y=0\cdot y=0$ so $(-x)\cdot y$ is the additive inverse of $x\cdot y$.

**Proposition 2.8**

$\forall x\in F\forall y\in F(-x)\cdot (-y)=x\cdot y)$

**Proof**

$(-x)\cdot (-y)=-(x\cdot (-y\left)\right)=--(x\cdot y)=x\cdot y$ by previous propositions and commutativity of $\cdot $.

In particular note from the last proposition the real reason why
a negative times a negative is a positive

: it is the only way
to get the distributive law to work out!

I could go on, but this is enough to get us started. Some of the propositions above are not obvious consequences of the axioms. Many other basic properties you are familiar with are now much easier to prove with these propositions done.

A further important feature of the reals that we have been
using all along is the $<$ relation. The material in this
section *is* part of the sequences and series

syllabus, and could be
examined, but should be easy to learn and use.

I choose here to discuss $<$ first and then define $>,\le ,\ge $ from it. There are other (similar) approaches!

The relation $<$ relates two numbers, and is called less than

. We write
$x<y$ to mean $x$ is less than $y$.

**Definition 3.1**** (of an ordered field)**

An ordered field is a field $F$ with binary relation $<$ satisfying

Transitivity: $\forall x\in F\forall y\in F\forall z\in F\left(\right(xy\wedge yz)\Rightarrow xz)$

Irreflexivity: $\forall x\in F\left(\neg xx\right)$

Linearity: $\forall x\in F\forall y\in F(xy\vee x=y\vee yx)$

One is positive: $0<1$.

Addition respects $<$: $\forall x\in F\forall y\in F\forall z\in F(xy\Rightarrow x+zy+z)$

Multiplication respects $<$: $\forall x\in F\forall y\in F\forall z\in F\left(\right(0z\wedge xy)\Rightarrow x\cdot zy\cdot z)$

**Definition 3.2**

We write $x\ge y$ for $\neg (x<y)$ and also write $x>y$ for $y<x$ and $x\ge y$ for $y\le x$.

**Definition 3.3**

Using $<$ we define $\left|x\right|$ as $x$ if $x\ge 0$ or $-x$ is $x<0$. Hence we define the distance function $d(x,y)=\left|x-y\right|$, etc.

We give a few simple but important propositions concerning the order in a field here. For the rest of this section fix an ordered field $F$.

**Proposition 3.1**

$\forall x\in F\forall y\in F(xy\Rightarrow x\ne y)$.

**Proof**

If $x>y=x$ then $x>x$ contradicting irreflexivity.

**Proposition 3.2**

$-1<0$

**Proof**

$0<1$ so $-1=0+-1<1+-1=0$.

**Proposition 3.3**

For all positive $n\in \mathbb{N}$, $1+1+\dots +1>0$ ($n$ 1s).

**Proof**

By induction on $n$, we show that $1+1+\dots +1>0$ ($n$ 1s). For $n=1$ this is an axiom. If $1+1+\dots +1>0$ holds then $1+1+\dots +1+1>0+1=1>0$ so the induction hypothesis is true for $n+1$, by transitivity of $>$ and by $1>0$.

**Proposition 3.4**

For all positive $n\in \mathbb{N}$, $\left(-1\right)+\left(-1\right)+\dots +\left(-1\right)<0$ ($n$ 1s).

**Proof**

Similar.

The last two propositions have the useful consequences that $1+1+\dots +1\ne 0$ and $\left(-1\right)+\left(-1\right)+\dots +\left(-1\right)\ne 0$. These statements concern a field but don't mention the order. However the order relation is needed to prove them. To see this consider the field ${F}_{2}$ of integers modulo 2. In this field we have $1+1=0$.

**Proposition 3.5**

$\forall x\in F\left(\right(x0\Rightarrow -x0)\wedge (x0\Rightarrow -x0\left)\right)$.

**Proof**

If $x>0$ then $0=x+(-x)>0+(-x)=-x$. The other one is similar.

**Proposition 3.6**

$\forall x\in Fx\cdot x\ge 0$ and $\forall x\in F(x\cdot x=0\Rightarrow x=0)$.

**Proof**

By cases. If $x>0$ then $x\cdot x>x\cdot 0=0$. If $x=0$ then $x\cdot x=0\ge 0$. And if $x<0$ then $-x>0$ so $(-x)\cdot (-x)=x\cdot x>0$ by results above.

Note too that the only one of the three cases above where we had $x\cdot x=0$ was the case $x=0$. So the second statement holds too.

**Proposition 3.7**

$\forall x\in F(x0\Rightarrow {x}^{-1}0)$.

**Proof**

Given that $x>0$, suppose ${x}^{-1}\le 0$. Then $1=x\cdot {x}^{-1}\le x\cdot 0=0$ which is impossible.

**Proposition 3.8**

For all positive $n\in \mathbb{N}$, $\forall x\in F(xy\wedge y0\Rightarrow {x}^{n}{y}^{n})$.

**Proof**

Given $x>y>0$, multiplying by $x$ and $y$ we have $x\cdot x>y\cdot x>y\cdot y$, and $x\cdot x\cdot x>y\cdot y\cdot x>y\cdot y\cdot y$, etc.

**Proposition 3.9**

$\forall x\in F\forall y\in F\left(\right(xy\wedge y0)\Rightarrow {x}^{-1}{y}^{-1})$.

**Proof**

Assume $x>y\wedge y>0$. Then $1=x\cdot {x}^{-1}>y\cdot {x}^{-1}$ as ${x}^{-1}>0$ and ${y}^{-1}={y}^{-1}\cdot 1>{y}^{-1}\cdot y\cdot {x}^{-1}={x}^{-1}$ as ${y}^{-1}>0$.

Finally for this section, we note that the proof of the triangle inequality we gave elsewhere works completely in the setting of an arbitrary ordered field, where distance is defined by $d(x,y)=\left|x-y\right|$.

**Theorem 3.1**** (The triangle inequality)**

For all $x,y,z\in F$ we have $d(x,y)\le d(x,z)+d(z,y)$

The next axiom concerns $\left[x\right]$, the integer part

function.
We have seen that this function plays a useful role in our theory, but it turns
out that not all ordered fields have it defined. A field for which integer part is
defined is called an Archimedean Ordered Field.

Before stating the axiom, it is important to know how the integers
lie in an ordered field. We have seen that such a field contains elements
called 0 and 1. The *additive subgroup* that are generated by these
is a copy of the integers $\mathbb{Z}$.

More formally, given an ordered field $F$, we define a map from natural numbers to $F$ by $n\mapsto 1+1+\dots +1$ ($n$ $1$s). This is extended to the whole of the integers by $0\mapsto 0\in F$ and $-n\mapsto -(1+1+\dots +1)$ ($n$ $1$s).

Fortunately this mapping is one-to-one (i.e., an injection). That is because all the values $(1+1+\dots +1)$ and $\left(\right(-1)+(-1)+\dots +(-1\left)\right)$ are all different in an ordered field. (We proved the special case that they are all non-zero above. This proof can be adapted to the more general result.)

In fact (and there is still something more to prove here) it turns out that this mapping is an embedding of $\mathbb{Z}$ into $F$, and it is usual practice to pretend that the integer $n$ and its copy in $F$ are the same object, thus identifying the two objects and simplifying our discussion so that we can regard $\mathbb{Z}\subseteq F$.

More generally a rational number $\frac{m}{n}\in \mathbb{Q}$ can be identified with $m{n}^{-1}\in F$, since if $n\ne 0$ then the copy of $n$ in $F$ is also nonzero. So we can regard $\mathbb{Z}\subseteq \mathbb{Q}\subseteq F$.

**Definition 4.1**** (of an Archimedean ordered field)**

An Archimedean ordered field is an ordered field $F$ such that

Archimedean Property: $\forall x\in F\exists n\in \mathbb{Z}xn$

If $F$ is an Archimedean ordered field we can define $\left[x\right]$ to be $n-1$ where $n$ is the least $n\in \mathbb{Z}$ such that $x<n$. The Archimedean Property guarantees there is such an $n$. The least number principle for the integers says there is a least such $n$.

Examples of Archimedean ordered fields include the reals $\mathbb{R}$ and the rationals $\mathbb{Q}$. We have already implicity used the Archimedean Property of the reals every time we have used the integer-part function $\left[x\right]$, or its cousin, $\lceil x\rceil $. We will continue to do so. There will also be some proofs later on in the course where the Archimedean Property of the reals will be used explicitly to good effect.

We can add other axioms at this stage to distinguish the reals and the rationals. For eample, we could consider the square root function (which isn't available in $\mathbb{Q}$) via an axiom such as:

Square roots: $\forall x\in F(x0\Rightarrow \exists y\in Fy\cdot y=x)$

Everything we have done in the course up to this point (i.e., up to but not including the monotone convergence theorem) works for any Archimedean ordered field with square roots. Actually, square roots were only required for some examples so the formal theory we have been working through works in slightly less.

However, it turns out that Archimedean ordered field with square roots
need not have such numbers as $e$ and $\pi $, etc. To really characterize
the reals and distinguish it from other fields we need the *completeness axiom*
which is actually a version of a useful theorem called *the monotone convergence theorem*.
There will be more about this elsewhere.

You have seen the axioms for Archimedean ordered fields, the two key examples being the reals and the rationals, and some of the basic consequences.

This web page is available in xhtml, html and pdf. It is copyright and is one of Richard Kaye's Sequences and Series Web Pages. It may be copied under the terms of the Gnu Free Documentation Licence (http://www.gnu.org/copyleft/fdl.html). There is no warranty. Web page design and creation are by GLOSS.