Binary operations · Binary operations

This lesson builds on binary operations.

Working out an operation

A binary operation $*$ combines two numbers by a rule. Put the first number in for the first letter and the second for the second:

a * b = 2a + b

5 * 3 = 2(5) + 3 = 13

Following the rulea * b: the first number replaces a, the second replaces b

Order usually matters: $a * b$ need not equal $b * a$ .

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q1

A binary operation $*$ is defined on the set $T = \{-2, -1, 1, 2\}$ by $p * q = p^2 + 2pq - q^2$ , where $p, q \in T$ .

Copy and complete the table.

$*$	$-2$	$-1$	$1$	$2$
$-2$		$7$		$-8$
$-1$		$2$	$-2$
$1$	$-7$			$1$
$2$		$-1$

Using the table in (a), find the value of $p$ such that $(-2 * p) * 2 = -7$ .

Some entries
- ${(-2) * (-2) = 4 + 8 - 4 = 8}$ and ${(-2) * 1 = 4 - 4 - 1 = -1}$ .
- ${(-1) * (-2) = 1 + 4 - 4 = 1}$ and ${(-1) * 2 = 1 - 4 - 4 = -7}$ .
- ${2 * 1 = 4 + 4 - 1 = 7}$ and ${2 * 2 = 4 + 8 - 4 = 8}$ ; the rest follow the same way.
Think first. p is the row heading, q the column heading.
Solve with the table
- In the column under 2, ${-7}$ is in the row of ${-1}$ : so ${x = -1}$ .
- In the row of ${-2}$ , ${-1}$ is under ${1}$ : so ${-2 * p = -1}$ gives ${p = 1}$ .
Think first. Let x = −2 * p. Which x gives x * 2 = −7?

More: working out an operation

Is it associative

An operation is associative if moving the brackets never changes the answer, for every $a$ , $b$ and $c$ :

(a * b) * c = a * (b * c)

Two ways to bracketLeft: a * b first, then with c. Right: b * c first, then a with it. Associative means they always agree

In $(a * b) * c$ , the answer to $a * b$ is the first number of the second step.
In $a * (b * c)$ , the answer to $b * c$ is the second number.
To show it is associative, expand both sides with letters and show they are the same expression.
To show it is not, one set of numbers that gives two different answers is enough.

Worked example

A binary operation is defined on the real numbers by

a * b = a + b + 3ab

. Show that

*

is associative.

Brackets on the left
- Rule with $a * b$ first: ${(a * b) + c + 3(a * b)c}$ .
- Replace $a * b$ : ${(a + b + 3ab) + c + 3c(a + b + 3ab)}$ .
- Expand: ${a + b + c + 3ab + 3ac + 3bc + 9abc}$ .
Think first. a * b is now the first number. Put it in for a in the rule.
Brackets on the right
- Rule with $b * c$ second: ${a + (b * c) + 3a(b * c)}$ .
- Replace $b * c$ : ${a + (b + c + 3bc) + 3a(b + c + 3bc)}$ .
- Expand: ${a + b + c + 3ab + 3ac + 3bc + 9abc}$ .
Think first. b * c is now the second number. Put it in for b in the rule.
Compare
- The two expansions match term by term.
- So $(a * b) * c = a * (b * c)$ for all real $a$ , $b$ , $c$ : $*$ is associative.
Think first. Are the two expressions the same for every a, b and c?

More: is it associative

WAEC 2014 · Paper 2 · Q2A binary operation $*$ is defined on the set of rational numbers by $m * n = \dfrac{m^2 - n^2}{2mn}$ , $n \ne 0$ and $m \ne 0$ .

The identity element

The identity $e$ leaves every number unchanged: $a * e = a$ (and $e * a = a$ ). Solve $a * e = a$ for $e$ ; the answer must not depend on $a$ .

⊗	1	2	3	4
1	1	2	3	4
2	2	4	1	3
3	3	1	4	2
4	4	3	2	1

The identity in a tableIts row and column repeat the headings

Worked example · WAEC 2016

WAEC 2016 · Paper 2 · Q1

A binary operation $*$ is defined on the set $\mathbb R$ of real numbers by $m * n = m + n + 2$ . Find the:

identity element under the operation $*$ ;

inverse of $n$ under the operation $*$ .

The identity
- ${m + e + 2 = m}$ , so ${e = -2}$ .
Think first. Solve m * e = m.
The inverse
- ${n + n^{-1} + 2 = -2}$ .
- So ${n^{-1} = -4 - n}$ .
Think first. n * n⁻¹ must give the identity, −2.

More: the identity element

WAEC 2023 · Paper 1 · Q32The table shows the operation $*$ on the set $\{x, y, z, w\}$ .

Inverses

The inverse of $a$ is the number $a^{-1}$ with $a * a^{-1} = e$ . Solve for $a^{-1}$ ; some values of $a$ may have no inverse.

Identity and inversePick an operation, set its constant and a

−3identity e−8inverse of 2

c = 3a = 2

a * b = a + b + 3. Identity: a + e + 3 = a, so e = −3. Inverse: a + a⁻¹ + 3 = −3, so a⁻¹ = −a − 6. For a = 2: a⁻¹ = −8. Check: 2 * (−8) = −3 = e ✓.

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q1

A binary operation $\Delta$ is defined on the set of real numbers, $\mathbb R$ , by $p\,\Delta\,q = pq + p + q$ for $p, q \in \mathbb R$ .

Find the: (i) identity element; (ii) inverse element.

Given that $m\,\Delta\,8 = 35$ , find the value of $m$ .

The identity
- ${pe + p + e = p}$ , so ${e(p + 1) = 0}$ for every $p$ : ${e = 0}$ .
The inverse
- ${pp^{-1} + p + p^{-1} = 0}$ , so ${p^{-1}(p + 1) = -p}$ .
- ${p^{-1} = -\frac{p}{p + 1}}$ , for ${p \ne -1}$ .
Think first. p Δ p⁻¹ = 0.
Solve m Δ 8 = 35
- ${8m + m + 8 = 35}$ , so ${9m = 27}$ and ${m = 3}$ .

More: binary operations

Your turn

WAEC 2019 · Paper 2 · Q3 (a)

(a)
A binary operation $*$ is defined on the set of real numbers, $\mathbb R$ , by $x * y = x + y - 3xy$ , where $x, y \in \mathbb R$ . Find the identity element in $\mathbb R$ under the operation $*$ .

Worked solution (try it first)

(a)

The identity $e$ satisfies $x * e = x$ : $x + e - 3xe = x$ .
Take $x$ from both sides: $e - 3xe = 0$ , so $e(1 - 3x) = 0$ .
This must hold for every $x$ , so $e = 0$ .
Check: $x * 0 = x + 0 - 0 = x$ ✓.

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