# Subtraction on Numbers is Anticommutative

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## Contents

## 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:

- $a = b$

$\blacksquare$

## Also see

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 4.2$. Commutative and associative operations: Example $61$