Axiomatic definition of the real numbers

1. is an Abelian group:
   1. For , we have \begin{eqnarray*} a+b&=&b+a, \\ (a+b)+c&=&a+(b+c), \end{eqnarray*}
   2. there exists an element such that $a+0=a$ for all ,
   3. every has an inverse such that $a+(-a)=0$ .
2. is an Abelian group:
   1. For , we have \begin{eqnarray*} a\cdot b&=&b\cdot a, \\ (a\cdot b)\cdot c&=&a\cdot (b\cdot c), \end{eqnarray*}
   2. there exists an element such that $a\cdot 1=a$ for all ,
   3. every has an inverse such that $a^{-1}\cdot a=1$ .
3. The operation $\cdot$ is distributive over $+$ : If , then \begin{eqnarray*} a\cdot (b+c)&=a\cdot b + a\cdot c, \\ (b+c)\cdot a&=b\cdot a + c\cdot a. \end{eqnarray*}
4. is a total order:
   1. (transitivity) if , $a<b$ , and $b<c$ , then $a<c$ ,
   2. (trichotomy) precisely one of the below alternatives hold:$$ a<b, \quad a=b, \quad b<a.$$
   For convenience we make the following notational definitions: $a>b$ means $b<a$ , $a\le b$ means either $a<b$ or $a=b$ , and $a\ge b$ means either $b<a$ or $a=b$ .
5. The operations $+$ and $\cdot$ are compatible with the order $<$ :
   1. If $a$ , $b$ , and $a<b$ , then $a+c<b+c$ .
   2. If $a$ , $b$ , with $a<b$ and $0<c$ , then $ac<bc$ .
6. has the least upper bound property: If , then an element is an upper bound for $A$ if$$ a<M, \mbox{ for all }\ a\in A.$$ If $A$ is non-empty, we then say that $A$ is bounded from above. That has the least upper bound property means that if is bounded from above, it has a least upper bound . 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 that a set 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. 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.

Basic properties

Lemma 1   Suppose .

1. The additive inverse $(-a)$ is unique (proof).
2. The additive identity $0$ is unique (proof).
3. $(-1)\cdot a=(-a)$ (proof).
4. $(-a)\cdot(-b)=a\cdot b$ (proof).
5. $0\cdot a=0$ (proof)
6. The multiplicative inverse $a^{-1}$ is unique (proof).
7. If $a,b$ are non-zero, then $(ab)^{-1}=b^{-1} a^{-1}$ (proof).

In view of property , 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 $1=aa^{-1}=a0=0$ ), 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 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 and is non-zero. Then $b$ divided by $a$ is defined as $$ \frac{a}{b} = ab^{-1}. $$ In consequence, if and $b,\,c,\,d$ are non-zero, then

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