axiomatic definition of the real numbers

Axiomatic definition of the real numbers

The real numbers consist of a set $ℝ$ together with mappings$+:ℝ\times ℝ\to ℝ$ and $\cdot :ℝ\times ℝ\to ℝ$ and arelation satisfyingthe following conditions:

1. $(ℝ,+)$ is an Abelian group:

   1. For $a,b,c\in ℝ$, we have

      $a+b$$=$$b+a,$

      $(a+b)+c$$=$$a+(b+c),$

   2. there exists an element $0\in ℝ$ such that $a+0=a$ for all $a\in ℝ$,

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

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

   1. For $a,b,c\in ℝ$, we have

      $a\cdot b$$=$$b\cdot a,$

      $(a\cdot b)\cdot c$$=$$a\cdot (b\cdot c),$

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

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

3. The operation $\cdot$ is distributive over $+$: If $a,b,c\in ℝ$, then

   $a\cdot (b+c)$$=a\cdot b+a\cdot c,$

   $(b+c)\cdot a$$=b\cdot a+c\cdot a.$

4. is a total order:

   1. (transitivity) if $c\in ℝ$, , and , then ,

   2. (trichotomy) precisely one of the below alternativeshold:

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

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

   1. If $a$, $b$, $c\in ℝ$ and , then .

   2. If $a$, $b$, $c\in ℝ$ with and , then .

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

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

Here it should be emphasized that from the above we can not deduce thata set $ℝ$ with operations exists. To settle this question sucha set has to be explicitly constructed. However, this can be done in various ways, asdiscussed 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 onthis page (http://planetmath.org/EveryOrderedFieldWithTheLeastUpperBoundPropertyIsIsomorphicToTheRealNumbers).

Basic properties

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

Lemma 1.

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

1. The additive inverse $(-a)$ is unique (proof) (http://planetmath.org/UniquenessOfAdditiveIdentityInARing).

2. The additive identity $0$ is unique (proof) (http://planetmath.org/UniquenessOfAdditiveIdentityInARing2).

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

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

5. $0\cdot a=0$ (proof) (http://planetmath.org/0cdotA0)

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

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 $a\ne 0$, then $a^{-1}\ne 0$ (or we’ll end up with $1=a{a}^{-1}=a0=0$), and therefore by Property 6 above, $a^{-1}$ has a unique multiplicative inverse. Since $a{a}^{-1}=a^{-1}a=1$, we see that $a$ is the multiplicative inverse of $a^{-1}$. In other words, ${(a^{-1})}^{-1}=a$.

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

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

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

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

• $\frac{ab}{b}=a$.

For example,

$$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a{b}^{-1}}{c{d}^{-1}}=a{b}^{-1}{(c{d}^{-1})}^{-1}=a{b}^{-1}d{c}^{-1}=\frac{ad}{bc}.$$