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: $\Z, \Q, \R, \C$.


 * First, suppose that $a = b$.

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


 * Next, suppose that $a - b = b - a$.

We have by definition of subtraction:

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.