# Ordering/Examples/Example Ordering on Integers

## Example of Ordering

Let $\preccurlyeq$ denote the relation on the set of integers $\Z$ defined as:

$a \preccurlyeq b$ if and only if $0 \le a \le b \text { or } b \le a < 0 \text { or } a < 0 \le b$

where $\le$ is the usual ordering on $\Z$.

Then $\preccurlyeq$ is an ordering on $\Z$.

## Proof

### Reflexivity

Let $0 \le a$.

Then $0 \le a \le a$ and so $a \preccurlyeq a$.

Let $a < 0$.

Then $a \le a < 0$ and so $a \preccurlyeq a$.

Thus in both cases $a \preccurlyeq a$.

So $\preccurlyeq$ has been shown to be reflexive.

$\Box$

### Transitivity

Let $a, b, c \in \R$ such that:

$a \preccurlyeq b \text { and } b \preccurlyeq c$

Let $0 \le a \le b$.

Then:

$0 \le b \le c$

and so:

$0 \le a \le c$

That is: $a \preccurlyeq c$

Let $b \le a < 0$.

Then there are two cases for $c$:

$(1): \quad c \le b < 0$

Then we have:

$c \le a < 0$

That is: $a \preccurlyeq c$

$(2): \quad b < 0 \le c$

Then we have:

$a < 0 \le c$

That is: $a \preccurlyeq c$

In both cases:

$a \preccurlyeq c$

Let $a < 0 \le b$.

Then:

$0 \le b \le c$

and so:

$0 < a \le c$

That is: $a \preccurlyeq c$

In all cases: $a \preccurlyeq c$

So $\preccurlyeq$ has been shown to be transitive.

$\Box$

### Antisymmetry

Let $a, b \in \R$ such that:

$a \preccurlyeq b \text { and } b \preccurlyeq a$

Let $0 \le a \le b$.

Then as $b \ge 0$:

$0 \le b \le a$

Thus:

$a \le b \text { and } b \le a$

and as $\le$ is the usual ordering, and so antisymmetric:

$a = b$

Let $b \le a < 0$.

Then as $b < 0$:

$a \le b < 0$

Thus:

$a \le b \text { and } b \le a$

and as $\le$ is the usual ordering, and so antisymmetric:

$a = b$

Let $a < 0 \le b$.

But then because $b \le a$:

$0 \le b \le a$

which contradicts $a < 0$.

Hence if both $a \le b$ and $b \le a$, it cannot happen that $0 \le b \le a$.

All cases are accounted for, and it has been shown that:

$a \preccurlyeq b \text { and } b \preccurlyeq a \implies a = b$

So $\preccurlyeq$ has been shown to be antisymmetric.

$\Box$

$\preccurlyeq$ has been shown to be reflexive, transitive and antisymmetric.

Hence by definition it is an ordering.

It is seen that $\preccurlyeq$ looks like:

$-1 \preccurlyeq -2 \preccurlyeq -3 \preccurlyeq \cdots \preccurlyeq 0 \preccurlyeq 1 \preccurlyeq 2 \preccurlyeq \cdots$

$\blacksquare$