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axiomatic definition of the real numbers (Definition)

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. $ (\mathbbmss{R},+)$ is an Abelian group:
    1. 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. there exists an element $ 0\in \mathbbmss{R}$ such that $ a+0=a$ for all $ a\in \mathbbmss{R}$,
    3. every $ a\in \mathbbmss{R}$ has an inverse $ (-a)\in\mathbbmss{R}$ such that $ a+(-a)=0$.
  2. $ (\mathbbmss{R}\setminus\{0\}, \cdot)$ is an Abelian group:
    1. 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. there exists an element $ 1\in \mathbbmss{R}\setminus\{0\}$ such that $ a\cdot 1=a$ for all $ a\in \mathbbmss{R}$,
    3. every $ a\in \mathbbmss{R}\setminus\{0\}$ has an inverse $ a^{-1}\in\mathbbmss{R}$ such that $ a^{-1}\cdot a=1$.
  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. $ (\mathbbmss{R},<)$ is a total order:
    1. (transitivity) if $ c\in\mathbbmss{R}$, $ a<b$, and $ b<c$, then $ a<c$,
    2. (trichotomy) precisely one of the below alternatives hold:
      $\displaystyle 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$, $ c\in\mathbbmss{R}$ and $ a<b$, then $ a+c<b+c$.
    2. If $ a$, $ b$, $ c\in\mathbbmss{R}$ with $ a<b$ and $ 0<c$, then $ ac<bc$.
  6. $ \mathbbmss{R}$ has the least upper bound property: If $ A\subset \mathbbmss{R}$, then an element $ M\in \mathbbmss{R}$ is an upper bound for $ A$ if
    $\displaystyle a<M,$    for all $\displaystyle \ a\in 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\le 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. 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

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. 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 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=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 by $ a$ is defined as

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



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topic entry on real numbers (Topic) by matte
rational numbers are real numbers (Result) by matte
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Cross-references: consequence, multiplication, commutativity, difference, multiplicative inverse, identity, additive, properties, group, ring, ordered field, isomorphism, upper bound, least upper bound, bounded from above, least upper bound property, order, compatible, definitions, trichotomy, transitivity, total order, distributive, operation, inverse, abelian group, relation, mappings, real numbers
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This is version 14 of axiomatic definition of the real numbers, born on 2006-02-05, modified 2006-03-13.
Object id is 7591, canonical name is AxiomaticDefinitionOfTheRealNumbers.
Accessed 2990 times total.

Classification:
AMS MSC: 12D99 (Field theory and polynomials :: Real and complex fields :: Miscellaneous)
 26-00 (Real functions :: General reference works )
 54C30 (General topology :: Maps and general types of spaces defined by maps :: Real-valued functions)
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