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 rule a * b: the first number replaces a, the second replaces b
Order usually matters: a ∗ b a * b a ∗ b need not equal b ∗ a b * a b ∗ a .
Worked example · WAEC 2022
WAEC 2022 · Paper 2 · Q1
A binary operation ∗ * ∗ is defined on the set T = { − 2 , − 1 , 1 , 2 } T = \{-2, -1, 1, 2\} T = { − 2 , − 1 , 1 , 2 } by p ∗ q = p 2 + 2 p q − q 2 p * q = p^2 + 2pq - q^2 p ∗ q = p 2 + 2 pq − q 2 , where p , q ∈ T p, q \in T p , q ∈ T .
Copy and complete the table.
∗ * ∗
− 2 -2 − 2
− 1 -1 − 1
1 1 1
2 2 2
− 2 -2 − 2
7 7 7
− 8 -8 − 8
− 1 -1 − 1
2 2 2
− 2 -2 − 2
1 1 1
− 7 -7 − 7
1 1 1
2 2 2
− 1 -1 − 1
Using the table in (a), find the value of p p p such that ( − 2 ∗ p ) ∗ 2 = − 7 (-2 * p) * 2 = -7 ( − 2 ∗ p ) ∗ 2 = − 7 .
Some entries
( − 2 ) ∗ ( − 2 ) = 4 + 8 − 4 = 8 {(-2) * (-2) = 4 + 8 - 4 = 8} ( − 2 ) ∗ ( − 2 ) = 4 + 8 − 4 = 8 and ( − 2 ) ∗ 1 = 4 − 4 − 1 = − 1 {(-2) * 1 = 4 - 4 - 1 = -1} ( − 2 ) ∗ 1 = 4 − 4 − 1 = − 1 .
( − 1 ) ∗ ( − 2 ) = 1 + 4 − 4 = 1 {(-1) * (-2) = 1 + 4 - 4 = 1} ( − 1 ) ∗ ( − 2 ) = 1 + 4 − 4 = 1 and ( − 1 ) ∗ 2 = 1 − 4 − 4 = − 7 {(-1) * 2 = 1 - 4 - 4 = -7} ( − 1 ) ∗ 2 = 1 − 4 − 4 = − 7 .
2 ∗ 1 = 4 + 4 − 1 = 7 {2 * 1 = 4 + 4 - 1 = 7} 2 ∗ 1 = 4 + 4 − 1 = 7 and 2 ∗ 2 = 4 + 8 − 4 = 8 {2 * 2 = 4 + 8 - 4 = 8} 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 {-7} − 7 is in the row of − 1 {-1} − 1 : so x = − 1 {x = -1} x = − 1 .
In the row of − 2 {-2} − 2 , − 1 {-1} − 1 is under 1 {1} 1 : so − 2 ∗ p = − 1 {-2 * p = -1} − 2 ∗ p = − 1 gives p = 1 {p = 1} p = 1 .
Think first. Let x = −2 * p. Which x gives x * 2 = −7?
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More: working out an operation
Is it associative
An operation is associative if moving the brackets never changes the answer, for every a a a , b b b and c c c :
( a ∗ b ) ∗ c = a ∗ ( b ∗ c ) (a * b) * c = a * (b * c) ( a ∗ b ) ∗ c = a ∗ ( b ∗ c )
a b c a * b (a * b) * c = a b c b * c a * (b * c) Two ways to bracket Left: a * b first, then with c. Right: b * c first, then a with it. Associative means they always agree
In ( a ∗ b ) ∗ c (a * b) * c ( a ∗ b ) ∗ c , the answer to a ∗ b a * b a ∗ b is the first number of the second step.
In a ∗ ( b ∗ c ) a * (b * c) a ∗ ( b ∗ c ) , the answer to b ∗ c b * c 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 + 3 a b a * b = a + b + 3ab a ∗ b = a + b + 3 ab . Show that
∗ * ∗ is associative.
Brackets on the left
Rule with a ∗ b a * b a ∗ b first: ( a ∗ b ) + c + 3 ( a ∗ b ) c {(a * b) + c + 3(a * b)c} ( a ∗ b ) + c + 3 ( a ∗ b ) c .
Replace a ∗ b a * b a ∗ b : ( a + b + 3 a b ) + c {(a + b + 3ab) + c} ( a + b + 3 ab ) + c + 3 c ( a + b + 3 a b ) {{} + 3c(a + b + 3ab)} + 3 c ( a + b + 3 ab ) .
Expand: a + b + c + 3 a b + 3 a c + 3 b c + 9 a b c {a + b + c + 3ab + 3ac + 3bc + 9abc} a + b + c + 3 ab + 3 a c + 3 b c + 9 ab c .
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 b * c b ∗ c second: a + ( b ∗ c ) + 3 a ( b ∗ c ) {a + (b * c) + 3a(b * c)} a + ( b ∗ c ) + 3 a ( b ∗ c ) .
Replace b ∗ c b * c b ∗ c : a + ( b + c + 3 b c ) + 3 a ( b + c + 3 b c ) {a + (b + c + 3bc) + 3a(b + c + 3bc)} a + ( b + c + 3 b c ) + 3 a ( b + c + 3 b c ) .
Expand: a + b + c + 3 a b + 3 a c + 3 b c + 9 a b c {a + b + c + 3ab + 3ac + 3bc + 9abc} a + b + c + 3 ab + 3 a c + 3 b c + 9 ab c .
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 ) (a * b) * c = a * (b * c) ( a ∗ b ) ∗ c = a ∗ ( b ∗ c ) for all real a a a , b b b , c c c : ∗ * ∗ is associative.
Think first. Are the two expressions the same for every a, b and c?
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Common mistake
Trying one set of numbers, getting equal answers, and calling the operation associative. With a ∗ b = a 2 − b a * b = a^2 - b a ∗ b = a 2 − b , both ( 2 ∗ 1 ) ∗ 3 (2 * 1) * 3 ( 2 ∗ 1 ) ∗ 3 and 2 ∗ ( 1 ∗ 3 ) 2 * (1 * 3) 2 ∗ ( 1 ∗ 3 ) give 6, yet the first check shows it is not associative.
Equal answers for some numbers prove nothing. To show it is associative, expand with letters. To show it is not , give one set of numbers with different answers.
The identity element
The identity e e e leaves every number unchanged: a ∗ e = a a * e = a a ∗ e = a (and e ∗ a = a e * a = a e ∗ a = a ). Solve a ∗ e = a a * e = a a ∗ e = a for e e e ; the answer must not depend on a a a .
The identity in a table Its row and column repeat the headings
Worked example · WAEC 2016
WAEC 2016 · Paper 2 · Q1
A binary operation ∗ * ∗ is defined on the set R \mathbb R R of real numbers by m ∗ n = m + n + 2 m * n = m + n + 2 m ∗ n = m + n + 2 . Find the:
identity element under the operation ∗ * ∗ ;
inverse of n n n under the operation ∗ * ∗ .
The identity
m + e + 2 = m {m + e + 2 = m} m + e + 2 = m , so e = − 2 {e = -2} e = − 2 .
Think first. Solve m * e = m.
The inverse
n + n − 1 + 2 = − 2 {n + n^{-1} + 2 = -2} n + n − 1 + 2 = − 2 .
So n − 1 = − 4 − n {n^{-1} = -4 - n} n − 1 = − 4 − n .
Think first. n * n⁻¹ must give the identity, −2.
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More: the identity element
Inverses
The inverse of a a a is the number a − 1 a^{-1} a − 1 with a ∗ a − 1 = e a * a^{-1} = e a ∗ a − 1 = e . Solve for a − 1 a^{-1} a − 1 ; some values of a a a may have no inverse.
Identity and inverse Pick an operation, set its constant and a
a * b = a + b + 3
−3 identity e −8 inverse of 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 Δ \Delta Δ is defined on the set of real numbers, R \mathbb R R , by p Δ q = p q + p + q p\,\Delta\,q = pq + p + q p Δ q = pq + p + q for p , q ∈ R p, q \in \mathbb R p , q ∈ R .
Find the: (i) identity element; (ii) inverse element.
Given that m Δ 8 = 35 m\,\Delta\,8 = 35 m Δ 8 = 35 , find the value of m m m .
The identity
p e + p + e = p {pe + p + e = p} p e + p + e = p , so e ( p + 1 ) = 0 {e(p + 1) = 0} e ( p + 1 ) = 0 for every p p p : e = 0 {e = 0} e = 0 .
The inverse
p p − 1 + p + p − 1 = 0 {pp^{-1} + p + p^{-1} = 0} p p − 1 + p + p − 1 = 0 , so p − 1 ( p + 1 ) = − p {p^{-1}(p + 1) = -p} p − 1 ( p + 1 ) = − p .
p − 1 = − p p + 1 {p^{-1} = -\frac{p}{p + 1}} p − 1 = − p + 1 p , for p ≠ − 1 {p \ne -1} p = − 1 .
Think first. p Δ p⁻¹ = 0.
Solve m Δ 8 = 35
8 m + m + 8 = 35 {8m + m + 8 = 35} 8 m + m + 8 = 35 , so 9 m = 27 {9m = 27} 9 m = 27 and m = 3 {m = 3} m = 3 .
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Your turn
(a) A binary operation ∗ * ∗ is defined on the set of real numbers, R \mathbb R R , by x ∗ y = x + y − 3 x y x * y = x + y - 3xy x ∗ y = x + y − 3 x y , where x , y ∈ R x, y \in \mathbb R x , y ∈ R . Find the identity element in R \mathbb R R under the operation ∗ * ∗ .
Worked solution (try it first) (a) The identity
e e e satisfies
x ∗ e = x x * e = x x ∗ e = x :
x + e − 3 x e = x x + e - 3xe = x x + e − 3 x e = x .
Take
x x x from both sides:
e − 3 x e = 0 e - 3xe = 0 e − 3 x e = 0 , so
e ( 1 − 3 x ) = 0 e(1 - 3x) = 0 e ( 1 − 3 x ) = 0 .
This must hold for every
x x x , so
e = 0 e = 0 e = 0 .
Check:
x ∗ 0 = x + 0 − 0 = x x * 0 = x + 0 - 0 = x x ∗ 0 = x + 0 − 0 = x ✓.
Watch out
In (a), the identity must work for every x x x , so e = 0 e = 0 e = 0 ; x = 1 3 x = \frac13 x = 3 1 is not the answer. In (b), multiplying by − 1 -1 − 1 turns ≤ \le ≤ into ≥ \ge ≥ . Then test a value to confirm between or outside. Report a problem with this question