Binary Operations

Lessons

Senior 4 · Mathematics

Binary Operations

4 lessons

Lesson 2 of 4

Associativity, identity and inverse elements, distributivity

25 min

Brackets, "do nothing" and "undo"

To add $2 + 3 + 4$ nobody asks where the brackets go: $(2 + 3) + 4 = 2 + (3 + 4)$ . Yet in the last lesson the operation $a * b = 2 a - b + 3$ gave $(2 * 5) * 1 = 6$ but $2 * (5 * 1) = - 5$ . Ordinary arithmetic also has a number that changes nothing ( $a + 0 = a$ ) and, for each number, a partner that undoes it ( $5 + (- 5) = 0$ ). This lesson turns those three everyday facts into properties that any binary operation may or may not have, and adds a fourth that links two operations: distributivity.

Definition

Associative operation: a binary operation $*$ on $E$ is associative if

(a * b) * c = a * (b * c) \quad \text{for all } a, b, c \in E .

Then $a * b * c$ can be written without brackets. Examples: $+$ , $\times$ , $\cup$ , $\cap$ and composition of functions are associative; subtraction is not, since $(5 - 3) - 1 = 1$ but $5 - (3 - 1) = 3$ .

Worked example 1: proving associativity

Show that $a * b = a + b - 3 a b$ is associative on $\mathbb{R}$ .

Solution Expand both sides for arbitrary real $a, b, c$ .

(a * b) * c = (a + b - 3 a b) + c - 3 (a + b - 3 a b) c = a + b + c - 3 a b - 3 a c - 3 b c + 9 a b c .

a * (b * c) = a + (b + c - 3 b c) - 3 a (b + c - 3 b c) = a + b + c - 3 a b - 3 a c - 3 b c + 9 a b c .

The two expressions are identical, so $*$ is associative.

Definition

Identity (neutral) element: an element $e \in E$ such that

e * a = a * e = a \quad \text{for all } a \in E .

Inverse (symmetric) element: if $(E, *)$ has identity $e$ , an element $a' \in E$ is an inverse of $a$ when

a * a^{'} = a^{'} * a = e .

Examples: for $+$ on $\mathbb{Z}$ the identity is 0 and the inverse of $a$ is $- a$ . For $\times$ on $\mathbb{Q}^*$ the identity is 1 and the inverse of $a$ is $\frac{1}{a}$ . For $\cup$ on the subsets of $S$ , the identity is $\varnothing$ ; for $\cap$ it is $S$ itself.

Important

Theorem 1. A binary operation has at most one identity element.

Proof. Suppose $e$ and $f$ are both identities. Then $e * f = f$ (because $e$ is an identity) and $e * f = e$ (because $f$ is an identity). So $e = f$ . $\blacksquare$

Theorem 2. If $*$ is associative with identity $e$ , an element has at most one inverse.

Proof. Let $a^{'}$ and $a^{''}$ both be inverses of $a$ . Then $a^{'} = a^{'} * e = a^{'} * (a * a^{''}) = (a^{'} * a) * a^{''} = e * a^{''} = a^{''}$ . $\blacksquare$ (Notice where associativity is used.)

Worked example 2: finding the identity and the inverses

For $a * b = a + b - 3 a b$ on $\mathbb{R}$ , find the identity element and the inverse of each element that has one.

Solution Identity: we need $a * e = a$ for every $a$ : $a + e - 3 a e = a$ , so $e (1 - 3 a) = 0$ for all $a$ . That forces $e = 0$ . Since $*$ is commutative, $0 * a = a$ too. So $e = 0$ .

Inverse: we need $a * a^{'} = 0$ : $a + a^{'} - 3 a a^{'} = 0$ , so $a^{'} (1 - 3 a) = - a$ and

a' = \frac{a}{3a - 1}, \qquad a \ne \frac{1}{3} .

The element $\frac{1}{3}$ has no inverse: $\frac{1}{3} * b = \frac{1}{3} + b - b = \frac{1}{3}$ for every $b$ , never 0. Check with $a = 1$ : $a' = \frac{1}{2}$ , and $1 * \frac{1}{2} = 1 + \frac{1}{2} - \frac{3}{2} = 0$ . ✓

Worked example 3: identity and inverses on $\mathbb{Z}$

On $\mathbb{Z}$ , $a \circ b = a + b + 1$ . Find the identity, the inverse of a general element, and the inverse of 4.

Solution $a \circ e = a$ gives $a + e + 1 = a$ , so $e = - 1$ (and $\circ$ is commutative). Then $a \circ a' = -1$ gives $a + a^{'} + 1 = - 1$ , so $a^{'} = - a - 2$ , which is an integer for every $a$ . The inverse of 4 is $- 6$ . Check: $4 \circ (-6) = 4 - 6 + 1 = -1 = e$ . ✓

Warning

An identity must work on both sides, for every element. For $a * b = a + 2 b$ on $\mathbb{R}$ , $a * 0 = a$ for all $a$ , so 0 is a right identity. But $0 * a = 2a \ne a$ when $a \ne 0$ , so 0 is not an identity, and no other number works either. Without an identity, the word "inverse" has no meaning.

Definition

Distributive operation: let $*$ and $\circ$ be binary operations on $E$ . The operation $*$ is distributive over $\circ$ if, for all $a, b, c \in E$ ,

a * (b \circ c) = (a * b) \circ (a * c) \quad \text{and} \quad (b \circ c) * a = (b * a) \circ (c * a) .

If $*$ is commutative, the first equality is enough. Example: on $\mathbb{R}$ , $\times$ is distributive over $+$ : $a (b + c) = a b + a c$ . But $+$ is not distributive over $\times$ : $2 + 3 \times 4 = 14$ , while $(2 + 3) (2 + 4) = 30$ .

Distributivity with sets

For subsets of a set $S$ , intersection distributes over union, and union distributes over intersection:

A \cap (B \cup C) = (A \cap B) \cup (A \cap C), \qquad A \cup (B \cap C) = (A \cup B) \cap (A \cup C) .

In the diagram, the shaded region is the part of $A$ that lies in $B$ or in $C$ . Read it as " $A$ and ( $B$ or $C$ )" and as "( $A$ and $B$ ) or ( $A$ and $C$ )": both describe the same region.

Worked example 4: testing distributivity

On $\mathbb{Z}$ , define $a \oplus b = a + b - 1$ and $a \otimes b = a + b - ab$ . (a) Is $\otimes$ distributive over $\oplus$ ? (b) Is $\oplus$ distributive over $\otimes$ ?

Solution (a) Both operations are commutative, so one side is enough.

a \otimes (b \oplus c) = a + (b + c - 1) - a(b + c - 1) = 2a + b + c - ab - ac - 1,

(a \otimes b) \oplus (a \otimes c) = (a + b - ab) + (a + c - ac) - 1 = 2a + b + c - ab - ac - 1 .

They are equal for all integers, so $\otimes$ is distributive over $\oplus$ .

(b) Try $a = b = c = 0$ : $a \oplus (b \otimes c) = 0 \oplus 0 = -1$ , but $(a \oplus b) \otimes (a \oplus c) = (-1) \otimes (-1) = -1 - 1 - 1 = -3$ . One counterexample: $\oplus$ is not distributive over $\otimes$ .

Tip

A checklist for any new operation: closed? commutative? associative? identity? inverses? Find the identity first, by solving $a * e = a$ for $e$ and checking that the answer does not depend on $a$ ; only then solve $a * a^{'} = e$ for the inverse.

Understanding Mathematics Academy: Binary operations, finding the identity and inverse elements. Watch for the two equations: a * e = a gives the identity, then a * a′ = e gives the inverse.

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Practice

Show your working; give a counterexample for every "no".

Is $a \circ b = a + b + ab$ associative on $\mathbb{Z}$ ?
For $a \circ b = a + b + ab$ on $\mathbb{R}$ , find the identity and the inverse of $a$ . Which element has no inverse?
On $\mathbb{Q}^*$ , $a * b = \frac{ab}{2}$ . Find the identity and the inverse of $a$ .
On $\mathbb{R}$ , $a * b = a + b - 5$ . Find the identity, the inverse of $a$ , and the inverse of 3.
Show that $a * b = a + 2 b$ on $\mathbb{R}$ is not associative.
Is $\times$ distributive over $a * b = a + b - 5$ on $\mathbb{R}$ ?
Use a Venn diagram, described in words, to show that $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ .
On $\mathbb{Z}$ , $a * b = a + b - 2$ and $a \circ b = ab - 2a - 2b + 6$ . Check that $a \circ (b * c) = (a \circ b) * (a \circ c)$ when $a = 3, b = 4, c = 5$ , then prove it for all integers.

Show answersHide answers

Yes: both $(a \circ b) \circ c$ and $a \circ (b \circ c)$ expand to $a + b + c + a b + a c + b c + a b c$ .
$a + e + a e = a$ for all $a$ gives $e = 0$ . $a + a^{'} + a a^{'} = 0$ gives $a' = \frac{-a}{1 + a}$ , $a \ne -1$ . The element $- 1$ has no inverse.
$\frac{ae}{2} = a$ gives $e = 2$ ; $\frac{aa'}{2} = 2$ gives $a' = \frac{4}{a}$ .
$e = 5$ ; $a + a^{'} - 5 = 5$ gives $a^{'} = 10 - a$ ; the inverse of 3 is 7.
$(0 * 0) * 1 = 0 * 1 = 2$ but $0 * (0 * 1) = 0 * 2 = 4$ .
No. $a (b * c) = a b + a c - 5 a$ but $(a b) * (a c) = a b + a c - 5$ ; they differ when $a \ne 1$ , e.g. $a = 2, b = c = 0$ : $-10 \ne -5$ .
Both sides are the region that is in $A$ , or in both $B$ and $C$ : the whole of circle $A$ together with the lens shared by $B$ and $C$ .
$b * c = 7$ , $3 \circ 7 = 21 - 6 - 14 + 6 = 7$ ; $3 \circ 4 = 4$ , $3 \circ 5 = 5$ , $4 * 5 = 7$ . ✓ In general both sides equal $a b + a c - 4 a - 2 b - 2 c + 10$ .

Quick check

On ℤ, a ∘ b = a + b + 1. What is the identity element?

Key points

Associative: $(a * b) * c = a * (b * c)$ for all $a, b, c$ .
Identity: $e * a = a * e = a$ for all $a$ ; it is unique.
Inverse: $a * a^{'} = a^{'} * a = e$ ; unique when $*$ is associative.
$*$ distributes over $\circ$ when $a * (b \circ c) = (a * b) \circ (a * c)$ , and similarly on the right.

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples, the practice items and the questions.
Video: Binary operations: finding the identity & inverse elements, Understanding Mathematics Academy, embedded from YouTube.

1Brackets, "do nothing" and "undo"

2Distributivity with sets

Brackets, "do nothing" and "undo"

Distributivity with sets