Subtraction on Numbers is Anticommutative

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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:

\(\ds a - b\) \(=\) \(\ds a + \paren {-b}\) Definition of Subtraction
\(\ds \) \(=\) \(\ds \paren {-b} + a\) Commutative Law of Addition
\(\ds \) \(=\) \(\ds \paren {-b} + \paren {-\paren {-a} }\)
\(\ds \) \(=\) \(\ds -\paren {b - a}\) 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:

$a = b$

$\blacksquare$


Also see


Sources

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