Subtraction on Numbers is Anticommutative

Theorem
The operation of subtraction on the numbers is anticommutative.

That is:
 * $a - b = b - a \iff a = b$

Proof
Let $a, b$ be elements of one of the standard number sets: $\Z, \Q, \R, \C$.

Necessary Condition
Let $a = b$.

Then $a - b = 0 = b - a$.

Sufficient Condition
Let $a - b = b - a$.

Then:

We have that:
 * $a - b = b - a$

So from the above:
 * $b - a = - \left({b - a}\right)$

That is:
 * $b - a = 0$

and so:
 * $a = b$

Subtraction as defined on the natural numbers is different.

$a - b$ is defined on $\N$ only if $a \ge b$.

If $a > b$, then although $a - b$ is defined, $b - a$ is not.

So for $a - b = b - a$ it is necessary for both to be defined.

This happens only when $a = b$.

Hence the result.