Subtraction on Numbers is Anticommutative/Integral Domains

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

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:

$a = b$

$\blacksquare$


Also see


Sources