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:
\(\ds a - b\) | \(=\) | \(\ds a + \paren {-b}\) | 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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.1$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 28$. Associativity and commutativity: Definition $2$