Subtraction on Numbers is Anticommutative/Integral Domains

From ProofWiki
Jump to navigation Jump to search


The operation of subtraction on the numbers is anticommutative.

That is:

$a - b = b - a \iff a = b$


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


Sufficient Condition

Let $a - b = b - a$.


\(\ds a - b\) \(=\) \(\ds a + \paren {-b}\) 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:

$a = b$


Also see