# axiomatic definition of the real numbers

## Axiomatic definition of the real numbers

The real numbers consist of a set $\mathbbmss{R}$ together with mappings $+\colon\mathbbmss{R}\times\mathbbmss{R}\to\mathbbmss{R}$ and $\cdot\colon\mathbbmss{R}\times\mathbbmss{R}\to\mathbbmss{R}$ and a relation $<\,\subseteq\mathbbmss{R}\times\mathbbmss{R}$ satisfying the following conditions:

1. 1.

$(\mathbbmss{R},+)$ is an Abelian group:

1. (a)

For $a,b,c\in\mathbbmss{R}$, we have

 $\displaystyle a+b$ $\displaystyle=$ $\displaystyle b+a,$ $\displaystyle(a+b)+c$ $\displaystyle=$ $\displaystyle a+(b+c),$
2. (b)

there exists an element $0\in\mathbbmss{R}$ such that $a+0=a$ for all $a\in\mathbbmss{R}$,

3. (c)

every $a\in\mathbbmss{R}$ has an inverse $(-a)\in\mathbbmss{R}$ such that $a+(-a)=0$.

2. 2.

$(\mathbbmss{R}\setminus\{0\},\cdot)$ is an Abelian group:

1. (a)

For $a,b,c\in\mathbbmss{R}$, we have

 $\displaystyle a\cdot b$ $\displaystyle=$ $\displaystyle b\cdot a,$ $\displaystyle(a\cdot b)\cdot c$ $\displaystyle=$ $\displaystyle a\cdot(b\cdot c),$
2. (b)

there exists an element $1\in\mathbbmss{R}\setminus\{0\}$ such that $a\cdot 1=a$ for all $a\in\mathbbmss{R}$,

3. (c)

every $a\in\mathbbmss{R}\setminus\{0\}$ has an inverse $a^{-1}\in\mathbbmss{R}$ such that $a^{-1}\cdot a=1$.

3. 3.

The operation $\cdot$ is distributive over $+$: If $a,b,c\in\mathbbmss{R}$, then

 $\displaystyle a\cdot(b+c)$ $\displaystyle=a\cdot b+a\cdot c,$ $\displaystyle(b+c)\cdot a$ $\displaystyle=b\cdot a+c\cdot a.$
4. 4.

$(\mathbbmss{R},<)$ is a total order:

1. (a)

(transitivity) if $c\in\mathbbmss{R}$, $a, and $b, then $a,

2. (b)

(trichotomy) precisely one of the below alternatives hold:

 $a

For convenience we make the following notational definitions: $a>b$ means $b, $a\leq b$ means either $a or $a=b$, and $a\geq b$ means either $b or $a=b$.

5. 5.

The operations $+$ and $\cdot$ are compatible with the order $<$:

1. (a)

If $a$, $b$, $c\in\mathbbmss{R}$ and $a, then $a+c.

2. (b)

If $a$, $b$, $c\in\mathbbmss{R}$ with $a and $0, then $ac.

6. 6.

$\mathbbmss{R}$ has the least upper bound property: If $A\subset\mathbbmss{R}$, then an element $M\in\mathbbmss{R}$ is an for $A$ if

 $a

If $A$ is non-empty, we then say that $A$ is bounded from above. That $\mathbbmss{R}$ has the least upper bound property means that if $A\subset\mathbbmss{R}$ is bounded from above, it has a least upper bound $m\in\mathbbmss{R}$. That is, $A$ has an upper bound $m$ such that if $M$ is any upper bound from $M$, then $m\leq M$.

Here it should be emphasized that from the above we can not deduce that a set $\mathbbmss{R}$ with operations $+,\cdot,<$ exists. To settle this question such a set has to be explicitly constructed. However, this can be done in various ways, as discussed on this page (http://planetmath.org/RealNumber). One can also show the above conditions uniquely determine the real numbers (up to an isomorphism). The proof of this can be found on this page (http://planetmath.org/EveryOrderedFieldWithTheLeastUpperBoundPropertyIsIsomorphicToTheRealNumbers).

## Basic properties

In condensed form, the above conditions state that $\mathbbmss{R}$ is an ordered field with the least upper bound property. In particular $(\mathbbmss{R},+,\cdot)$ is a ring, and $(\mathbbmss{R}\setminus\{0\},\cdot)$ is a group, and we have the following basic properties:

###### Lemma 1.

Suppose $a,b\in\mathbbmss{R}$.

1. 1.

$(-a)$ is unique (proof) (http://planetmath.org/UniquenessOfAdditiveIdentityInARing).

2. 2.

$0$ is unique (proof) (http://planetmath.org/UniquenessOfAdditiveIdentityInARing2).

3. 3.

$(-1)\cdot a=(-a)$ (proof) (http://planetmath.org/1cdotAA).

4. 4.

$(-a)\cdot(-b)=a\cdot b$ (proof) (http://planetmath.org/XcdotYXcdotY).

5. 5.

$0\cdot a=0$ (proof) (http://planetmath.org/0cdotA0)

6. 6.

$a^{-1}$ is unique (proof) (http://planetmath.org/UniquenessOfInverseForGroups).

7. 7.

If $a,b$ are non-zero, then $(ab)^{-1}=b^{-1}a^{-1}$ (proof) (http://planetmath.org/InverseOfAProduct).

In view of property 2, we can write simply $-a$ instead of $(-1)\cdot a$ and $(-a)$.

Because of the additive inverse of a real number is unique (by property 1 above), and $(-a)+a=a+(-a)=0$, we see that the additive inverse of $-a$ is $a$, or that $-(-a)=a$. Similarly, if $a\neq 0$, then $a^{-1}\neq 0$ (or we’ll end up with $1=aa^{-1}=a0=0$), and therefore by Property 6 above, $a^{-1}$ has a unique multiplicative inverse. Since $aa^{-1}=a^{-1}a=1$, we see that $a$ is the multiplicative inverse of $a^{-1}$. In other words, $(a^{-1})^{-1}=a$.

For $a,b\in\mathbbmss{R}$ let us also define $a-b=a+(-b)$, which is called the difference of $a$ and $b$. By commutativity, $a-b=-b+a$.  It is also common to leave out the multiplication symbol and simply write $ab=a\cdot b$.  Suppose  $a\in\mathbbmss{R}$  and  $b\in\mathbbmss{R}$  is non-zero.  Then $b$ divided (http://planetmath.org/Division) by $a$ is defined as

 $\frac{a}{b}=ab^{-1}.$

In consequence, if  $a,\,b,\,c,\,d\in\mathbbmss{R}$  and $b,\,c,\,d$ are non-zero, then

• $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{bd}{ac}$,

• $\frac{ab}{b}=a$.

For example,

 $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ab^{-1}}{cd^{-1}}=ab^{-1}(cd^{-1})^{-1}=% ab^{-1}dc^{-1}=\frac{ad}{bc}.$
Title axiomatic definition of the real numbers AxiomaticDefinitionOfTheRealNumbers 2013-03-22 15:39:29 2013-03-22 15:39:29 matte (1858) matte (1858) 17 matte (1858) Definition msc 54C30 msc 26-00 msc 12D99 RealNumber