Distributivity of Binary Operations

Distributivity of Binary Operations

Recall from the Associativity and Commutativity of Binary Operations page that if $S$ is a set and $* : S \times S \to S$ is a binary operation on $S$ then we say that $*$ is associative if for all $a, b, c \in S$ we have $a * (b * c) = (a * b) * c$ (nonassociative otherwise) and $*$ is said to be commutative if for all $a, b \in S$ we have $a * b = b * a$ (noncommutative otheriwse).

We will now look at another property regarding binary operations, though this time, we will need two binary operations, call then $* : S \times S \to S$ and $+ : S \times S \to S$.

Definition: Let $*$ and $+$ be binary operations on a set $S$. Then $*$ is said to be Left Distributive if for all $a, b, c \in S$ we have that $a * (b + c) = (a * b) + (a * c)$, and $*$ is said to be Right Distributive if for all $a, b, c \in S$ we have that $(a + b) * c = (a * c) + (b * c)$. $*$ is said to be Distributive if $*$ is both left and right distributive.

For example, standard multiplication $\cdot$ with standard addition $+$ on $\mathbb{R}$ is left distributive since for all $a, b, c \in \mathbb{R}$ we have that:

(1)
\begin{align} \quad a \cdot (b + c) = (a \cdot b) + (a \cdot c) \end{align}

These operations are also right distributive since:

(2)
\begin{align} \quad (a + b) \cdot c = (a \cdot c) + (b \cdot c) \end{align}

Therefore standard multiplication and addition on $\mathbb{R}$ is distributive.

For a more uncommon example, consider the operation $* : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ defined for all $a, b \in \mathbb{R}$ by $a * b = (a \cdot b)^2$ and the operation $+ : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ defined by $a + b = \mid a \mid \mid b \mid$.

(3)
\begin{align} \quad a * (b + c) = a * (\mid b \mid \mid c \mid) = (a \cdot (\mid b \mid \mid c \mid))^2 = a^2 \cdot (\mid b \mid^2 \mid c \mid^2) = a^2 \cdot (b^2 \cdot c^2) \end{align}
(4)
\begin{align} \quad a * b + a * c = (a \cdot b)^2 + (a \cdot c)^2 = \mid a^2b^2 \mid \mid a^2 c^2 \mid = a^4 \cdot (b^2 \cdot c^2) \end{align}

We see that $a * (b + c) \neq a*b + a*c$ so $*$ and $+$ are not left distributive.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License