Pointwise Multiplication is Commutative

Theorem

Let $S$ be a non-empty set.

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:

$f \times g = g \times f$