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