Pointwise Addition is Commutative
Theorem
Let $\mathbb F$ be one of the standard number sets: $\Z, \Q, \R$ or $\C$.
Let $f, g: S \to \mathbb F$ be functions.
Let $f + g: S \to \mathbb F$ denote the pointwise sum of $f$ and $g$.
Then:
- $f + g = g + f$
That is, pointwise addition is commutative.
Proof
\(\ds \forall x \in S: \, \) | \(\ds \map {\paren {f + g} } x\) | \(=\) | \(\ds \map f x + \map g x\) | Definition of Pointwise Addition | ||||||||||
\(\ds \) | \(=\) | \(\ds \map g x + \map f x\) | Commutative Law of Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {g + f} } x\) | Definition of Pointwise Addition |
$\blacksquare$
Specific Contexts
This result can be applied and proved in the context of the various standard number sets:
Pointwise Addition on Integer-Valued Functions is Commutative
Let $f, g: S \to \Z$ be integer-valued functions.
Let $f + g: S \to \Z$ denote the pointwise sum of $f$ and $g$.
Then:
- $f + g = g + f$
Pointwise Addition on Rational-Valued Functions is Commutative
Let $f, g: S \to \Q$ be rational-valued functions.
Let $f + g: S \to \Q$ denote the pointwise sum of $f$ and $g$.
Then:
- $f + g = g + f$
Pointwise Addition on Real-Valued Functions is Commutative
Let $f, g: S \to \R$ be real-valued functions.
Let $f + g: S \to \R$ denote the pointwise sum of $f$ and $g$.
Then:
- $f + g = g + f$
Pointwise Addition on Complex-Valued Functions is Commutative
Let $f, g: S \to \C$ be complex-valued functions.
Let $f + g: S \to \C$ denote the pointwise sum of $f$ and $g$.
Then:
- $f + g = g + f$