Trichotomy Law for Real Numbers

Theorem
The real numbers obey the Trichotomy Law.

That is, $\forall a, b \in \R$, exactly one of the following holds:


 * $(1): \quad a > b$ ($a$ is greater than $b$)
 * $(2): \quad a = b$ ($a$ is equal to $b$)
 * $(3): \quad a < b$ ($a$ is less than $b$).

Note that $a > b \iff b < a$.

We also use the following notation:


 * $(a): \quad a \le b \iff a < b \lor a = b$ ($a$ is less than or equal to $b$)
 * $(b): \quad a \ge b \iff a > b \lor a = b$ ($a$ is greater than or equal to $b$).

The following also holds:


 * $\forall a, b, c \in \R: a < b \land b < c \implies a < c$

That is, that $<$ is transitive.

Proof 1
This follows directly from the fact that the real numbers form a totally ordered field.

Proof 2
$\le$ is a total ordering on $\R$.

The trichotomy follows directly from Trichotomy Law.

That $<$ is transitive follows from the definition of a strict total ordering.