# Real numbers/Ordering axioms/Archimedes/Intervals/Introduction/2/Section

## Definition

A field ${}K$ is called an ordered field if there is a relation ${}>$ (larger than) between the elements of ${}K$ fulfilling the following properties (${}a\geq b$ means ${}a>b$ or ${}a=b$ )

1. For two elements ${}a,b\in K$ we have either ${}a>b$ or ${}a=b$ or ${}b>a$ .
2. From ${}a\geq b$ and ${}b\geq c$ one may deduce ${}a\geq c$ (for any ${}a,b,c\in K$ ).
3. ${}a\geq b$ implies ${}a+c\geq b+c$ (for any ${}a,b,c\in K$ ).
4. From ${}a\geq 0$ and ${}b\geq 0$ one may deduce ${}ab\geq 0$ (for any ${}a,b\in K$ ).

## Lemma

In an ordered field the following properties hold.

1. ${}1\geq 0$ .
2. ${}a\geq 0$ holds if and only if ${}-a\leq 0$ holds.
3. ${}a\geq b$ holds if and only if ${}a-b\geq 0$ holds.
4. ${}a\geq b$ holds if and only if ${}-a\leq -b$ holds.
5. ${}a\geq b$ and ${}c\geq d$ imply ${}a+c\geq b+d$ .
6. ${}a\geq b$ and ${}c\geq 0$ imply ${}ac\geq bc$ .
7. ${}a\geq b$ and ${}c\leq 0$ imply ${}ac\leq bc$ .
8. ${}a\geq b\geq 0$ and ${}c\geq d\geq 0$ imply ${}ac\geq bd$ .
9. ${}a\geq 0$ and ${}b\leq 0$ imply ${}ab\leq 0$ .
10. ${}a\leq 0$ and ${}b\leq 0$ imply ${}ab\geq 0$ .

### Proof

See Exercise.
$\Box$ ## Lemma

In an ordered field the following properties holds.

1. From ${}x>0$ one can deduce ${}x^{-1}>0$ .
2. From ${}x<0$ one can deduce ${}x^{-1}<0$ .
3. For ${}x>0$ we have ${}x\geq 1$ if and only if ${}x^{-1}\leq 1$ .
4. From ${}x\geq y>0$ one can deduce ${}x^{-1}\leq y^{-1}$ .
5. For positive elements ${}x,y$ the relation ${}x\geq y$ is equivalent with ${}{\frac {x}{y}}\geq 1$ .

### Proof

See Exercise.
$\Box$ ## Definition

Let ${}K$ be an ordered field. ${}K$ is called Archimedean, if the following Archimedean axiom holds, i.e. if for every ${}x\in K$ there exists a natural number ${}n$ such that

${}n\geq x\,.$ ## Lemma

1. For ${}x,y\in \mathbb {R}$ with ${}x>0$ there exists ${}n\in \mathbb {N}$ such that ${}nx\geq y$ .
2. For ${}x>0$ there exists a natural number ${}n$ such that ${}{\frac {1}{n}} .
3. For two real numbers ${}x there exists a rational number ${}n/k$ (with ${}n\in \mathbb {Z}$ , ${}k\in \mathbb {N} _{+}$ ) such that
${}x<{\frac {n}{k}} ### Proof

(1). We consider ${}y/x$ . Because of the Archimedean axiom there exists some natural number ${}n$ with ${}n\geq y/x$ . Since ${}x$ is positive, due to fact also ${}nx\geq y$ holds. For (2) and (3) see exercise.

$\Box$ ## Definition

For real numbers ${}a,b$ , ${}a\leq b$ , we call

1. ${}[a,b]={\left\{x\in \mathbb {R} \mid x\geq a{\text{ and }}x\leq b\right\}}$ the closed interval.
2. ${}]a,b[={\left\{x\in \mathbb {R} \mid x>a{\text{ and }}x the open interval.
3. ${}]a,b]={\left\{x\in \mathbb {R} \mid x>a{\text{ and }}x\leq b\right\}}$ the half-open interval (closed on the right).
4. ${}[a,b[={\left\{x\in \mathbb {R} \mid x\geq a{\text{ and }}x the half-open interval (closed on the left).