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

Natural Numbers

$a - b$ 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.

Integers, Rationals, Reals, Complex Numbers

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

Each of those systems is an integral domain, and so is closed under the operation of subtraction.

Necessary Condition

Let $a = b$.

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

$\Box$

Sufficient Condition

Let $a - b = b - a$.

Then:

 $\displaystyle a - b$ $=$ $\displaystyle a + \paren {-b}$ Definition of Subtraction $\displaystyle$ $=$ $\displaystyle \paren {-b} + a$ Commutative Law of Addition $\displaystyle$ $=$ $\displaystyle \paren {-b} + \paren {-\paren {-a} }$ $\displaystyle$ $=$ $\displaystyle -\paren {b - a}$ Definition of Subtraction

We have that:

$a - b = b - a$

So from the above:

$b - a = - \paren {b - a}$

That is:

$b - a = 0$

and so:

$a = b$

$\blacksquare$