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.