Number and Algebraic Patterns

This section discusses binary operations, sets and binomial expressions.

The part on binary operations, discusses the concept and properties such as closure, associativity, commutativity, distributivity, identity elements and inverse. Binary operations are used extensively in algebra, trigonometry, and calculus, and they form the backbone of numerous calculations.

Sets are essential for organising and classifying information and are used in areas like probability and statistics. The section on sets will focus on operations on three-set problems and set algebra. The Binomial theorem provides a powerful formula to expand expressions of the form (a + b)ⁿ, where n is a non-negative integer. It allows us to calculate the coefficients and exponents of the expanded expression. Binomial expansions are crucial for solving various problems in algebra and beyond.

The concepts in this section are essential for science, algebra, trigonometry, and calculus, forming the backbone of numerous calculations in finance to model compound interest, etc.

To understand this section, you will have to rely on your previous knowledge on algebraic expressions, change of subject and substitution at the JHS level. Apart from the four basic operations in arithmetic, other symbols such as ∆, *, ◊, etc., can also be used as operations as you find in this statement, a ∆ b = a + b + (ab). You can note how the symbol ∆ is interpreted by the use of the two basic operations in arithmetic, (×) and (+).

At the end of this section, you should be able to:

1. Recognise binary operations, and apply the knowledge in solving related problems

2. Describe and interpret the characteristics of commutative, associative, distributive and closure properties of binary operations.

3. Determine the identity element and use it to find the inverse of a given element.

4. Establish the properties of operations on set, including commutative, associative, and distributive, sets algebra and apply them to solve problems.

5. Expand Binomial expressions for positive integer indices using Pascal’s triangle

6. Use the combination approach and other approaches to determine the coefficient and exponent of a given term in an expansion.

Key Ideas

• A binary operation is the combination of two elements at a time under a given rule to produce another element.

• Elements in binary operations could be the two given variables, terms or numbers which are connected with the operational symbols. For example in a ∆ b = a + b, a and b are taken as the elements. Similarly, if we have m * n = mn + 2, then the elements are m and n.

• The four basic operations in arithmetic are addition (+), subtraction (–), multiplications (×) and division (÷).

• A set is a well-defined collection of objects. Well-defined means they have the same features or properties. Examples are a set of footballers, a set of cutleries, and a set of counting numbers.

• A set is usually represented by capital letters and written in a curly or closed { }.

• Elements are the individual members in the given set. For example, in a given set A = {0,1,2,3,4,5,6,7,8,9,10}, the elements in the set A are 0,1,2,3,4,5,6,7,8,9,10.

• A binomial is an expression, which involves two terms and can be written in the form (x + y)ⁿ. For example (x + y)²is a square of a binomial. For small values of n , it is relatively easy to write the expansion by using multiplication. For example, the (x + y)²

A binary operation is a rule that combines elements from a non-empty set ℝ to produce another element. Binary operations form the foundation of digital logic circuits.

Binary operations are used in computer programming, electrical engineering, and number theory. In computer programming, they are used for digital logic and computer arithmetic, while in electrical engineering, they are used in coding theory. In number theory, addition and multiplication are binary operations that define properties of numbers like divisibility and prime factorization. Binary operations are also used to interpret true or false statements in logic.

Binary operations can be defined using symbols such as ⊗, ⊞, ⊛, * and ⋄.

A binary operation is the combination of two elements at a time under a rule to produce another element. For example, 4 + 7 = 11 is a binary operation, 4 is the first element and 7 is the second element, whereas the (+) is the operation sign and 11 is the third element.

Similarly, p ∆ q = p + q + pq is a binary operation where the first element is p, q is the second element and the result p + q + pq is the third element, and the operation is ∆.

Furthermore, “An operation is such that the result is the difference of the product of two numbers and the sum of the two numbers”. This statement can be written as a mathematical rule thus:

a * b = ab − (a + b), where a and b are the two numbers and * is the symbol that represents the operation

Activity 1.1

1. To simplify (15 × 5) + 6

2. First solve the terms in the bracket

3. Add the result to 6

4. Write down the final answer Did you get 81? Congratulations!

Note that when you multiplied the two terms in the bracket, you combined the terms based on an operation (multiplication) to get a resulting element (75) as an answer thus you performed a binary operation. Likewise the 75 element was added to 6 to get 81(final resulting element).

Let us further work out the following examples on the concept of binary operations.

In pairs or small groups, work through the examples, discussing them as you go.

Worked Examples 1.1

1. If p ∆ q= p + q + pq, evaluate

i. 3 ∆ 4

ii. 4 ∆ 3

Solution

In this example, looking at the given definition of the operation ∆, and what we are to evaluate, p ∆ q = 3 ∆ 4, where p = 3 and q = 4. We therefore substitute p = 3 and q = 4 into the operation p ∆ q = p + q + pq for (i),

i. 3 ∆ 4 , from p = 3, q = 4 3 ∆ 4 = 3 + 4 + 3(4) = 3 + 4 + 12 = 19

ii. Again, p ∆ q = 4 ∆ 3 Hence, substituting in the operation p∆q= p + q + pq, We have, 4 ∆ 3 = 4 + 3 + 4(3) = 4 + 3 + 12 = 19

2. The operation ǝ on the set R of real numbers is defined by x ǝ y = 4x + 5y + 2xy__________ 3y + 2x for all x, y ∈ R. Evaluate,

i. –2 ǝ 5

ii. 4 ǝ 7

Solution

i. Given, x ǝ y = 4x + 5y + 2xy__________ 3y + 2x Then, –2 ǝ 5 = 4( − 2) + 5(5) + 2( − 2)(5)____________________ 3(5) + 2( − 2) = − 8 + 25 − 20/15 − 4 = − 3/11

ii. 4 ǝ 7 = 4(4) + 5(7) + 2(5)(7)_______________ 3(7) + 2(4) = 16 + 35 + 70/21 + 8 = 121/29 = 4 5/9

3. A binary operation * is defined under the set of real numbers R, by a * b = a²− ab + 4 . Evaluate each of the following equations.

i. –1 * 2

ii. 3 * 4

iii. 3*–2

Solution

i. Given a * b = a²− ab + 4 − 1 * 2 = ( − 1)²—( − 1)(2) + 4 = 1 + 2 + 4 = 7

ii. 3 * 4 = 3²− 3(4) + 4 = 9 – 12 + 4 = 1

iii. 3 * − 2 = 3²− 3(− 2) + 4 = 9 + 6 + 4 = 19

4. Given that m * n = m − n/n , if 3*q = 1/2 find the value of q.

Solution

Given m * n = m − n/n , then 3 * q = m − n/n = 1/2 Hence, 1/2 = 3 − q/q q = 2(3 – q) q = 6 – 2q q + 2q = 6 3q = 6 q = 2

The properties of binary operations include closure, commutativity, associativity and distributive properties. We will investigate these properties using the four basic operations (×, ÷, − , and +) on the set of real numbers (ℚ, ℤ, W, ℕ) and predefined binary operation on a given set using algebraic manipulations and binary tables.

Verifying Properties of Binary Operations

a) Closure A binary operation, Δ is closed under a set, A if for all x and y which are elements of A, xΔy is also in A. For example, the sum of two natural numbers is a natural number hence it can be said that addition is closed under the set of Natural numbers, ℕ.

The set of natural numbers is also closed under multiplication and addition but not closed under subtraction and division. ie when we add or multiply two natural numbers, the result will always be a natural number, however, when you subtract or divide two natural numbers, your result will not always be a natural number.

Activity 1.2

1.

a) Choose any two even numbers

b) Add them up

c) Is the result an even number?

2. Repeat the steps above with different even numbers

3. Is the sum of two even numbers always an even number?

Yes! The sum of two even numbers is always an even number. This demonstrates the closure property of addition with even numbers.

Activity 1.3

1. Perform the operations on the set of natural numbers

a) 2 + 4 =

b) 2 × 4 =

c) 2 – 4

d) 2 ÷4

2. Write down the operations whose results were also natural numbers.

3. Which operations are closed on the set of natural numbers?

Note: If m and n are integers, m + n will be an integer, m × n will also be an integer, likewise m − n . However, m/n will not be integer at all times. Therefore, addition, multiplication and subtraction are said to be closed under the set of integers but division is not.

Can you input real numbers into this example to show that this is true?

Work Example 1.2

An operation Ʌ is defined on the set S = {0, 2, 3, 4} by pɅq = p + q − 2pq.

Determine whether or not the set S is closed under the operation Ʌ.

Solution

Pick two numbers at a time for all members in the set S, and evaluate them with the operation Ʌ.

Given pɅq = p + q − 2pq 0Ʌ0 = 0 + 2 – 2(0)(0) = 0 0Ʌ3 = 0 + 3 – 2(0)(3) = 3 0Ʌ4 = 0 + 4 – 2(0)(4) = 4 2Ʌ0 = 2 + 0 – 2(2)(0) = 2 3Ʌ3 = 3 + 3 – 2(3)(3) = –12 4Ʌ4 = 4 + 4 – 2(4)(4) = –24 2Ʌ3 = 2 + 3 – 2(2)(3) = –7 2Ʌ4 = 2 + 4 – 2(2)(4) = –10 3Ʌ4 = 3 + 4 – 2(3)(4) = –17 From the solution, the numbers 0, –7, –10, –12, –17, and –24 are not in the set S;

therefore the set S is not closed undre the operation Ʌ.

Worked Example 1.3

The operation * is defined on the set M = { 1,3,5,7} by x * y = x + y/2

i. draw a table for * on the set M

ii. determine whether or not the operation * is closed under M

Solution

i. Combining two elements at a time on the given operation x * y = x + y/2 , 1 * 1 = 1 + 1/2 = 1 1 * 3 = 1 + 3/2 = 2 1 * 5 = 1 + 5/2 = 3 1 * 7 = 1 + 7/2 = 4 Combine the other pairs to obtain the table below * 1 3 5 7 1 1 2 3 4 3 2 3 4 5 5 3 4 5 6 7 4 5 6 7

ii. The operation * is not closed because the numbers {2,4, 6} are not members in the set M.

Worked Example 1.4

The binary operation * is defined on the Real numbers R, by x * y = y - 4xy. Show whether * is closed under R in each of the following;

i. 3 * 4

ii. 4 * 5

iii. 5 * –6

iv. –5 * 7

Solution

i. 3 * 4 = 4 – 4(3)(4) = – 44

ii. 4 * 5 = 4 – 4(4)(5) = –75

iii. 5 * –6 = –6 – 4(5)(–6) = 114

iv. –5 * 7 = 7 – 4(–5)(7) = 147 The operation * is closed under the set of R because the results for (i) to (iv) are also real numbers.

b) Commutativity Given that ⋄ is a binary operation defined on a set, S which contains a and b , if a ⋄ b = b ⋄ a, for all a and b in S, then ⋄ is said to be commutative.

In a binary table, the binary operation is only commutative if the results of the operation are symmetrical about the leading diagonal.

Figure 1: Binary table Addition and multiplication are commutative. i.e., 4 + 3 = 3 + 4 = 7 and 4 × 5 = 5 × 4= 20. Subtraction and division are not commutative. i.e., 4 − 3 ≠ 3 − 4, likewise 10/2 ≠ 2/10.

To determine whether the operation ∗ defined by a * b = a + b + 2ab is commutative, we need to check if a * b = b * a for all real numbers a and b.

Let us choose any two different numbers, say 3 and 4, to verify whether the operation ∗ is commutative. Compute 3∗4 and 4∗3 separately:

3∗4 = 4 + 2(3)(4) = 4 + 24 = 26 4∗3 = 4 + 2(4)(3) = 4 + 24 = 26 For the operation to be commutative, 3∗4 must equal 4∗3. Since this is true for all real numbers a and b, the operation ∗ is commutative.

Worked Examples 1.5

1. An operation * is defined on the set of real numbers by a * b = a + 2ab, evaluate whether * is commutative or not, using the following expressions:

i. 5 * 7

ii. 7 * 5

Solution

Given, a * b = a + 2ab ,

i. 5 * 7 = 5 + 2(5)(7) = 5 + 70 = 75

ii. 7 * 5 = 7 + 2(7)(5) = 7 + 70 = 77 Since 5 * 7 ≠ 7 * 5, the operation * is not commutative.

2. The operation ® on the set of real numbers is defined by m® n= m + n − 10mn, evaluate

i. 2 ® 3

ii. 3 ® 2

iii. What can you say about the result in i) and ii)?

Solution

i. 2 ® 3 = 2 + 3 – 10(2)(3) = 5–60 = –55

ii. 3® 2 = 3 + 2 – 10(3)(2) = 5–60 = –55

iii. since 2 ® 3 = 3 ® 2, the operation ® is commutative.

Find out and discuss if this is always going to be commutative no matter what numbers are chosen for m and n.

3. Given that a binary operation, * is defined on a set, S = {− 2, 3, 5} as a * b = ab − (a + b), show whether or not * is commutative.

Solution

Given that a * b = ab − (a + b), and S = {− 2, 3, 5} Substituting any two members of S in the equation, · Using –2 and 3:

− 2 * 3 = − 2(3) − (− 2 + 3) = − 7 3*(− 2) = 3(− 2) − (3 + (− 2)) = − 7 · Using –2 and 5:

− 2 * 5 = − 2(5) − (− 2 + 5) = − 13 5*(− 2) = 5(− 2) − (5 + (− 2)) = − 13 · Using 3 and 5:

3 * 5 = 3(5) − (3 + 5) = 7 5 * 3 = 5(3) − (5 + 3) = 7 Since a * b = b * a for all a and b in S , * is commutative

4. Given that x * y = x + y + xy, show that the operation * is commutative.

Solution

For the operation * to be commutative, x * y must be equal to y * x LHS must be equal to RHS x * y = x + y + xy…………. (1) LHS RHS y * x = y + x + yx…………. (2) RHS Comparing equation (1) to equation (2), LHS = RHS, Hence, the operation * is commutative.

5. Given that a * b = a − 2b , determine whether or not the operation * is commutative.

Solution

For the operation * to be commutative, a * b = b * a LHS a * b = a − 2b …………. (1) RHS b * a = b − 2a…………. (2) Comparing equation (1) to equation (2), LHS ≠ RHS, Hence, the * is not commutative (unless a = b).

6. A binary operation ⨀ is defined on the set P = {x, y, z} by the table below.

Determine whether the operation ⨀ is commutative.

⨀ x y z x x y z y y z x z z x y

Solution

By inspection, we identify that x⨀y = y⨀x = y; x⨀z = z⨀x = z and y⨀z =z⨀y = x ∴ ⨀ is commutative.

Alternatively, since the table is symmetric along the principal diagonal, we conclude that the operation ⨀ is commutative.

⨀ x y z x x y z y y z x z z x y

Worked Example 1.5

The operation * is defined over the set M ={0, 2, 6, 8} as a * b = a + b + 8 .

a) Construct a table of values

b) Show whether the operation * is commutative or not.

Solution

0 * 2 = 0 + 2 + 8 = 10 2 * 0 = 2 + 0 + 8 = 10 2 * 6 = 2 + 6 + 8 = 16 6 * 2 = 6 + 2 + 8 = 16 Continue with the rest of the pairs to obtain the table below.

* 0 2 6 8 0 8 10 14 16 2 10 12 16 18 6 14 16 20 22 8 16 18 22 24

b) the operation * is commutative because for any two numbers picked in the

table, the result for a * b and b * a are equal.

c) Associativity Given that ⋄ is a binary operation defined on a set, S which contains a, b and c, if a ⋄ (b ⋄ c) = (a ⋄ b) ⋄ c, for all, a, b and c in S then ⋄ is said to be associative.

Addition and multiplication are associative.

For example, 2 + (3 + 4) = (2 + 3) + 4 Likewise 2 × (3 × 4) = (2 × 3) × 4 Now, let us go through the following activities or steps to prove or otherwise of the associative property using numbers.

To determine whether the operation ∗ defined by a ∗ b = a + b is associative, we need to check if (a ∗ b) ∗ c = a ∗ (b ∗ c) for all real numbers a, b, and c. In this case, let us check with the specific values 2, 3, and 5 using the following steps:

1. Define the operation:

The operation is a∗b = a + b

2. Compute (a∗b)∗c:

First, find a∗b, substituting the real numbers (e.g. 2 and 3):

a ∗ b = 2 ∗ 3 2 ∗ 3 = 2 + 3 = 5 Next, use this result to compute (a ∗ b) ∗ c:

(a ∗ b) ∗ c = 5 ∗ 5 = 5 + 5 = 10

3. Compute a ∗ (b ∗ c) First, find b∗c: 3∗5 = 3 + 5 = 8 Next, use this result to compute 2 ∗ (3 ∗ 5):

2∗(3∗5) = 2 ∗ 8 = 2 + 8 = 10

4. Compare (a∗b)∗c and a∗(b∗c) (2 ∗ 3) ∗ 5 = 10 2 ∗ (3∗5) = 10 Since (2 ∗ 3) ∗ 5 = 2 ∗ (3 ∗ 5)=10, the operation ∗ is associative for the values 2,3, and 5.

Worked Example 1.6

A binary operation, ∆ is defined on the set, R of real numbers as xωy = x + y − 2xy, where x, y ∈ R . Show whether or not ω is commutative.

Solution

If ω is commutative, it implies xωy= yωx LHS: xωy = x + y − 2xy and RHS: yωx = y + x − 2yx Rearranging, we have RHS: xωy = x + y − 2xy Since xωy = yωx, we conclude that ω is commutative.

Let us narrow it down using the set {− 1, 0, 1, 2}. All possible combinations are shown in the table below.

ω − 1 0 1 2 − 1 − 4 − 1 2 5 0 − 1 0 1 2 1 2 1 0 − 1 2 5 2 − 1 − 4 Since the table is symmetrical about the principal diagonal, it confirms the operation, ω is commutative.

Note that we could have used any set containing real numbers.

Worked Examples 1.7

To determine whether the operation ∗ defined by a ∗ b = a + 2ab is associative, using variables (letters), we need to check if (a ∗ b) ∗ c = a ∗ (b ∗ c) for all real numbers a, b, and c.

Here are the steps:

1. Define the operation:

Given a ∗ b = a + 2ab

2. Compute (a ∗ b) ∗ c:

· First, find a ∗ b: = a + 2ab · Let a * b = m Now, m = a + 2ab · Next, use this result to compute m∗c:

m ∗ c = m ∗ c m ∗ c = m + 2mc …………….. equation (1) substitute m = a + 2ab, into equation (1) = a + 2ab + 2(a + 2ab)c (a * b )*c = a + 2ab + 2ac + 4abc

3. Compute a ∗(b ∗ c):

· First, find b ∗ c:

b ∗ c = b + 2bc · Next, use this result to compute a ∗ (b ∗ c) · Let n = b + 2bc = a ∗ n · Substitute n into the operation:

a ∗ n = a + 2an ………………..equation (2) substitute n = b + 2bc into equation (2) a * (b * c ) = a + 2a(b + 2bc) a * (b * c ) = a + 2ab + 4abc o Compare the two expressions: a + 2ab + 2ac + 4abc and a + 2ab + 4abc o Since (a∗b)∗c ≠ a∗(b∗c), due to the extra 2ac term in (a∗b)∗c, the operation ∗ is not associative.

Now solve the following example either in groups, individually, or with the assistance of your teacher.

Worked Example 1.8

A binary operation * defined on the set R of real numbers by m * n = 2m + 3n-mn where m, n ∈R. Determine whether or not * is associative.

Solution

For (m∗n)∗p First, compute m∗n:

m∗n = 2m + 3n−mn Now, compute (m∗n)∗p:

Let k = m∗n = 2m + 3n−mn. Then we need to compute k∗p:

k∗p = (2m + 3n−mn)∗p = 2(2m + 3n−mn)+ 3p−(2m + 3n−mn)p = 4m + 6n−2mn + 3p−2mp−3np + mnp For m∗(n∗p):

First, compute n∗p:

n∗p = 2n + 3p−np Now, compute m∗(n∗p):

Let r = n∗p = 2n + 3p−np.

Then we need to compute m∗r:

m∗r = m∗(2n + 3p−np) = 2m + 3(2n + 3p−np)−m(2n + 3p−np) = 2m + 6n + 9p−3np−2mn−3mp + mnp From the computations above, we have:

(m∗n)∗p = 4m + 6n−2mn + 3p−2mp−3np + mnp m∗(n∗p) = 2m + 6n + 9p−3np−2mn−3mp + mnp Since the expressions (m∗n)∗p and m∗(n∗p)m are not equal, the operation ∗ defined by m∗n = 2m + 3n−mn is not associative.

Worked Example 1.9

Determine whether or not the operation * is associative given that x * y = x + y + xy.

Solution

For the operation * to be associative, x * (y * z) = (x * y)*z LHS:

x * (y * z) (y * z) = y + z + yz x * (y + z + yz) = x + y + z + yz + x(y + z + yz) x * (y * z) = x + y + z + xy + yz + xz+ xyz ………..(1) RHS:

(x * y) * z (x * y) = x + y + xy (x + y + xy)*z = (x + y + xy) + z + (x + y + xy)z x * (y * z) = x + y + z + xy + xz + yz+ xyz ………..(2) LHS = RHS, Hence, the operation * is associative.

d) Distributive Property Given that * and ⊗ are binary operations defined on the set, S = {a, b, c} then * is distributive over⊗ifa *(b ⊗ c)=(a * b)⊗(a * c). Example. 2 × (3 + 4) = (2 × 3) + (2 ×4).

The distributive property is applied when we are opening brackets in mathematics.

For example, 2(x + y) = 2x + 2y.

Worked Example 1.10

If two binary operations * and ∆ are defined as x * y = 2xy and x ∆ y = 2x + 3y on the set R of real numbers. Show whether or not

a) ∆ is distributive over *

b) * is distributive of ∆

Solution

a) If ∆ is distributive of * , then for all x, y, z ∈R We should have x ∆ (y * z) = (x ∆ y)*(x ∆ z) LHS: x ∆ (y * z) = x ∆ (2yz) = 2x + 3(2yz) = 2x + 6yz RHS: (x ∆ y)*(x ∆ z) = (2x + 3y)*(2x + 3z) = 2(2x + 3y)(2x + 3z) = (4x + 6y)(2x + 3z) = 8 x²+ 12xz + 12xy + 18yz Since LHS≠ RHS, the operation ∆ is not distributive over * .

b) If * is distributive of ∆ , then for all x, y, z ∈R We should have x*(y ∆ z) = (x * y) ∆ (x * z) LHS: x*(y ∆ z) = x*(2y + 3z) = 2x(2y + 3z) = 4xy + 6xz RHS: (x * y) ∆ (x * z) = (2xy) ∆ (2xz) = 2(2xy) + 3(2xz) = 4xy + 6xz Since LHS = RHS, the operation * is distributive over ∆.

Worked Example 1.11

Given that Δ and + are binary operations defined on the set, S = { x, y, z} as x ∆ y = x + 2xy determine whether ∆ is distributive over + for x ∆ ( y + z) = (x ∆ y) + (x ∆ z)

Solution

LHS:

x ∆ ( y + z) = x + 2(x)(y + z) = x + 2xy + 2xz ……….(1) RHS:

(x∆y) + (x∆z) = x + 2xy + x + 2xz = 2x + 2xy + 2yz……………….. (2) Since x ∆ ( y + z) ≠ (x ∆ y) + (x ∆ z) , Δ is not distributive over +

Worked Example 1.12

Two binary operations * and ∆ are defined as a * b = 2ab and a ∆ b = 2a + 3b + 2 for all a, b ⋲ R.

Evaluate:

i. 2 * (3 ∆ 4)

ii. (2 * 3) ∆ (2 * 4)

iii. What can you say about (i) and (ii)?

Solution

i. 3 ∆ 4 = 2(3) + 3(4) + 2 = 20 2 * 20 = 2(2)(20) = 80

ii. 2 * 3 = 2(2)(3) = 12 2 * 4 = 2(2)(4) = 16 12 ∆ 16 = 2(12) + 3(16) + 2 = 74

iii. Comparing (i) and (ii):

2 * (3 ∆ 4) = 80 and (2 * 3) ∆ (2 * 4) = 74 These two expressions are not equal. This means that the operation * is not distributive over ∆ Since 2 * (3 ∆ 4) ≠ (2 * 3) ∆ (2 * 4)

Identity / Neutral Element The identity element, e of a set, S under an operation, ∆ on S exists if there is an element e ∈ S such that a ∆ e = e ∆ a = a. If a binary operation is not commutative, it cannot have an identity element. Under the operation of addition, (+), the identity element is 0 ie 6 + 0 = 6, 14 + 0 = 14. Also, under the operation of multiplication, the identity element is 1, i.e., 3 × 1 = 3, 100 × 1 = 100.

Worked Example 1.13

A binary operation ∇ is defined on the set R or real numbers by p∇q = p + q− pq where p and q ∈ ℝ. Find the identity element under the operation ∇.

Solution

Check for commutativity:

p∇q = p + q + pq q∇p = q + p + qp Since p∇q = q∇p, ∇ is commutative and hence could have an identity element.

Let e be the identity element, where e ∈ ℝ, then by definition, p∇e = p But p∇e= p + e + ep ⟹ p + e + ep = p p + e(1 + p) = p e(1 + p) = p − p e(1 + p) = 0 e = 0

Worked Example 1.14

A binary operation ∇ is defined on the set R of real numbers by m ∇ n = m + 2mn where m and n ∈ ℝ. Find the identity element under the operation ∇.

Solution

Check for commutativity:

m∇n= m + 2mn n∇m = n + 2nm Since m∇n ≠ n∇m, ∇ is not commutative and hence does not have an identity element.

Worked Example 1.15

A binary operation ◊ is defined on the set R or real numbers by a ◊ b = a − 2b + ab where a and b ∈ ℝ. Find the identity element under the operation ◊

Solution

Check for commutativity:

a ◊ b= a − 2b + ab b ◊ a = b − 2a + ba Since a ◊ b ≠ b ◊ a, ◊is not commutative and hence would not have an identity element.

Worked Example 1.16

The binary operation ∋ is defined on the set ϕ = {u, v, w, x, y} by the table below.

∋ u v w x y u v w y u x v w y x v u w y x u w v x u v w x y y x u v y w Find the identity element.

Solution

If eϵϕ is the identity element, then for all R ∈ ϕ, R ∋ e = e ∋ R = R.

By inspection, we have u ∋ x = x ∋ u = u v ∋ x = x ∋ v = v w ∋ x = x ∋ w = w x ∋ x = x ∋ x = x y ∋ x = x ∋ y = y Therefore, the identity element is x.

The Inverse of an Element

The inverse of an element a of a set S under an operation ∆ on S is an element a−1 ∈ S such that a ∆ a−1 = a−1 ∆ a = e, where e is the identity element of S under the operation, Δ. A binary operation with no identity has no inverse for the general elements. Under the operation of addition, 2 + (–2) = 0, therefore the additive inverse of 2 is –2.

Under the operation of multiplication, 2 × 1/2 = 1 , therefore the multiplicative inverse of 2 is 1/2. Remember that if a binary operation is not commutative it cannot have an identity element.

Worked Example 1.17

A binary operation ∇ is defined on the set R or real numbers by p∇q = p + q− pq where p and q ∈ R. Find the inverse element under the operation ∇.

Solution

Let p−1∈ R be the inverse element, then p∇p−1 = p + p−1 + p p−1 = e Check for commutativity:

p∇q= p + q + pq q∇p = q + p + qp Since p∇q = q∇p, ∇ is commutative and hence could have an identity element Let e be the identity element, where e ∈ ℝ, then by definition, p∇e = p But p∇e= p + e + ep ⟹ p + e + ep = p p + e(1 + p) = p e(1 + p) = p − p e(1 + p) = 0 e = 0 ⇒p + p−1+ p p−1 = 0 ⟹p−1 = – p/1 + p The inverse element, p−1 of p under the operation ∇ exist and it is − p/1 + p

Worked Example 1.18

A binary operation ∇ is defined on the set R or real numbers by a∇b = a + b + 2ab where a and b ∈ R.

i. Find the inverse element under the operation ∇.

ii. Hence determine the inverse of 4.

Solution

Let a−1∈ R be the inverse element, then a∇a−1 = a + a−1+ 2a a−1 = e Check for commutativity:

a∇b= a + b + 2ab b∇a = b + a + 2ba Since a∇b = b∇a, ∇ is commutative and hence could have an identity element Let e be the identity element, where e ∈ ℝ, then by definition, a∇e = a But a∇e= a + e + 2ea ⟹ a + e + 2ea = a a + e(1 + 2a) = a e(1 + 2a) = a − a e(1 + 2a) = 0 e = 0 a + a−1+ 2a a−1 = e ⇒a + a−1 + 2a a−1 = 0 a−1 + 2a a−1 = − a a−1(1 + 2a) = − a ⟹a−1 = –a/1 + 2a The inverse element, p−1 of p under the operation ∇ exist and it is − a/1 + 2a The inverse of 4 is − 4/1 + 2(4) = − 4/1 + 8 = − 4/9

Worked Example 1.19

1. The operation * is defined on the set {1, 3, 5} as shown in the table below * 1 3 5 1 3 1 5 3 1 3 5 5 5 5 5

a) Evaluate (3 * 1)*(5 * 3)

b) State the identity element for *

c) Which of the elements has no inverse

Solution

a) From the table 3 * 1 = 1 5 * 3 = 5 1 * 5 = 5

b) For identity element, e, a * e = a From the table, the identity element for * is 3

c) 5 has no inverse under *

Worked Example 1.16

2. The operation * is defined over the set of real numbers by a * b = a + b + 3ab

a) Show that *

i. is commutative

ii. is associative

b) Find the identity element for the operation *

c) Find the inverse and the value for which a has no inverse solution.

d) Determine whether or not a * (b + c) = (a * b ) + (a * c )

Solution

2a.

i. for the operation * to be commutative, a * b = b * a LHS a * b = a + b + 3ab …………(1) RHS b * a = b + a + 3ba…….…(2) a * b = b * a , therefore the operation * is commutative.

ii. For the operation * to be associative, a * (b * c ) = (a * b )*c a * [b + c + 3bc] a + b + 3bc + 3[(a)(b + c + 3bc] a + b + 3bc + 3[ab + ac + 3abc] a + b + 3bc + 3ab + 3ac + 9abc a * b = a + b + 3ab (a * b )*c = a + b + 3ab + c + 3[(a + b + 3ab)c] a + b + 3ab + 3[ac + bc + 3abc] a + b + 3ab + 3ac + 3bc + 9abc Since LHS = RHS , the operation * is associative.

b) Let e be the identity element a * e = a a + e + 3ae = a e + 3ae = a-a e(1 + 3a) = 0 e = 0/1 + 3a e = 0

c) Let a⁻¹be the inverse a * a⁻¹= e a + a⁻¹+ 3aa⁻¹= 0 a⁻¹+ 3aa⁻¹= -a a⁻¹(1 + 3a)= -a a⁻¹= − a/1 + 3a For a to have an inverse, 1 + 3a ≠ 0 1 + 3a = 0 3a = –1 a = − 1/3 , hence a ≠ –1/3 in order for a to have an inverse

d) a * (b + c) = (a * b ) + (a * c ) LHS:

a * (b + c) = a + b + c + 3[(a)(b + c)] a + b + c + 3[ab + ac)] a + b + c + 3ab + 3ac (a * b ) + (a * c ) = a + b + 3ab + a + c + 3ac 2a + c + 3ab + 3ac Therefore a * (b + c) ≠ (a * b ) + (a * c )

Worked Example 1.20

The operation * is defined over the set of real numbers by m * n = m + n + 10 Find

i. the identity element

ii. the inverse of –3 and 7 under *

Solution

i. Check for commutativity:

m∇n= m + n + 10 n∇m = n + m + 10 Since m∇n = n∇m, ∇ is commutative and hence could have an identity element.

Let e be the identity element, where e ∈ ℝ , Then by definition, m∇e = m But m∇e= m + e + 10 ⟹ m + e + 10 = m e = m − m − 10 e = − 10

ii. Let a⁻¹be the inverse:

m * m⁻¹= e m + m⁻¹+ 10 = –10 m⁻¹= –10 – 10 – m m⁻¹= –20 – m The inverse of –3 under *:

m⁻¹= –20 – (–3) = –17 The inverse of 7 under *:

m⁻¹= –20 – (7) = –27

In your class, consider the following scenarios:

i. A girl using a cutlery set for the first time in boarding school

ii. The football team of your school

iii. Natural numbers less than 10 From the above scenarios, the following can be deduced;

· Members of the cutlery sets are fork, spoon and knife which can conveniently be used for eating.

· There are eleven members in a football team playing different roles of defending, midfield and attacking to win a game · The members of natural numbers less than 10 are 1,2,3,4,5,6,7,8 and 9.

· All these groups belong to common features called a set.

Therefore a set is a well–defined collection of objects. Sets are usually denoted by capital letters such as A, B, C, D…etc

Sets are commonly represented by any of the following methods:

1. Statement form (word description).

2. Tabular or Roster form (Listing).

3. Rule or set builder notation form.

Suppose you want to describe the following situations in statement form:

1. Set of trees in the forest that are taller than 20 feet.

2. Set of birds in your region that move north to south in harmattan.

3. Positive integers less than 10.

From the above, you can deduce the following as statement form of the sets:

1. {trees in the forest that are taller than 20 feet}

2. {b : b is a bird in your region that moves north to south in harmattan}

3. {postive integers less than 10}

1. The Statement Form (Word Description)

The statement form is a method used to describe a set using natural language. We use words to convey the characteristics or elements of the set. For example, the set of even numbers less than 8 is written as: {even numbers less than 8}

2. The Tabular or Roster Form

The roster notation involves listing all the elements of a set. We enclose the elements within curly brackets { } and separate them with commas.

For example, let N denote the set of the first five natural numbers. Therefore, N = {1, 2, 3, 4, 5}. Similarly, if M denotes the set of the months beginning with the letter “J”, then M = {January, June, July}.

3. The Rule or Set Builder Form

In this method, the elements of the set are described by using the variable “x” or any other variable, followed by a colon. The symbol “:” or” “/” as used in this case is read “such that”.

Thereafter, the property possessed by the elements is written in brackets (called “set of all”). For example, P is the set of counting numbers greater than 12. In set builder notation or form, P is expressed as P = {x: x is a counting number greater than 12}. This is read as “P is the set of elements of x such that x is a counting number greater than 12. The set builder notation also involves the use of inequality symbols to describe the range of data.

Likewise, if A is the set of even numbers between 6 and 14, then A is written in set builder notation as A = {x / x is even, 6 < x < 14} or A = {x: x in P, 6 < x < 14, P is an even number} Also, if K = {4, 5, 6, 7} then set builder form of K is: K = {x: x is a natural number and 3 < x < 8} Also, if K = {4, 5, 6, 7} then set builder form of K is: K = {x: x is a natural number and 3 < x < 8}.

In pairs or small groups, read and discuss the following, explaining the concepts to one another.

1. The Null Set or Empty Set

Consider the following statements:

· A set of students in your school who are 100 years old.

· A set of dogs that can fly.

The two statements above would contain nothing. Thus a set which does not contain any elements is called a null set. It is denoted by ∅ (read as phi). The null set can also be denoted by { }.

It can be concluded that the number of elements in an empty set is 0. An empty set is therefore an example of a finite set (see below). For example, the set of whole numbers less than 0. Since there is no whole number less than 0, the set is said to be a null set.

Likewise M = {x: x is a composite number less than 4}. M is an empty set because there is no composite number (remember that a composite number is a number which has more than two factors) less than 4.

Note: ∅ ≠ {0} because {0} is a set which has one element 0.

2. The Unit Set or Singleton Set

A Unit set is a set which contains only one element. For example, the set of Headmasters in your school could be a unit set. Why?

Take R = {x:x is neither prime nor composite}, then R = {1} Other examples are:

· S = {x:x is an even prime} = {2}, n(S) = 1 · T = {x:x is a natural number and x²= 4} = {2}, n(T) = 1

3. Finite Set

This is a set which contains a definite number of elements. A finite set can contain all its elements. For example, A = {set of days of the week} B = {2, 3, 5, 7….19} C = {x:x is a natural number and x < 7}

4. Infinite Set

It is a set whose elements cannot be listed completely. In other words it is a set containing never–ending elements. For example, A = {x : x N, x > 1} B = {Set of all prime numbers} Cardinality of Set The total number of elements in a set P, is called the cardinal number of P, denoted by n(P). For example,

1. A = {x:x is a natural number, x < 5} A = {1, 2, 3, 4} n(A) = 4

2. B = {set of elements in the word {“MATHEMATICS”} n(B) = 11

Operations on sets which will be discussed in this section are Union, Intersection and Complements of sets. Read the information below so you can investigate them as a class and see how they are represented on Venn diagrams.

1. Intersection of Sets

Activity 1.4

Consider three students, Kojo (K), Amina (A) and Joseph (J) who selected their favourite fruits as follows:

K ={ orange, mango, pawpaw} A={orange, strawberries, pawpaw} J={mango, pawpaw, banana, pineapple, orange} Is there any fruit(s) that were selected by Kojo, Amina and Joseph?

Yes, all of them selected pawpaw.

We therefore say that the intersection of K, A and J is pawpaw. This is written mathematically as K ∩ A ∩ J = {pawpaw}.

Now, describe intersection of sets in your own words.

The intersection of sets A, B and C is all the elements that are common for both A, B and C. Symbolically, we denote the intersection of sections A, B and C as A ∩ B ∩ C.

For example the intersection for the sets, A ={2, 4, 6, 8}, B ={4, 8, 12, 16, 20}, and C ={4, 8, 12, 16, 20, 24, 28}is A ∩ B ∩ C = {4, 8}

Worked Example 1.21

1. Three learners were asked to select their four best subjects (not in any particular order) by their class teacher. The responses were as follows:

A = {Science, Computing, Mathematics and English} B = {Mathematics, Computing, French and English} C = {Ghanaian Language, Computing, Agriculture and Mathematics} If the teacher wanted to find out the subject that learners like most in his school, then we can use A ∩ B ∩ C = {Mathematics, Computing}

2. If U, the universal set, = {positive integers less than 20} M = {factors of 18} N = {multiples of 3} Q = {prime numbers} M, N and Q are subsets of the universal set U. Find M ∩ N ∩ Q

Solution

U = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19} M = {1,2,3,6,9,18} N = {3,6,9,12,15,18} Q = {2,3,5,7,11,13,17,19} M ∩ N ∩ Q = {3}

2. Union of Sets

If a set is the union set of two or more sets, then it consists of all elements that are in the individual sets and those that are common to the sets. Note that the common elements to the set are written once.

For example, if set P = {1,2,3,6,12}, Q={2,4, 6,8,10} and R = {5, 10,15,20} then P ∪Q ∪ R = {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20} Again, if P = {a, b, c, d, e}, Q ={a, e, i, o, u} and R = {g,i,r,l} Then P∪Q ∪ R= {a,b,c,d,e,g,i,l,o,r,u}

Worked Example 1.22

A, B and C are integers and A = {− 6, − 5, 0, 3, 4}, B = {b : b is an odd number, 3 ≤ b < 12} and C = {positive prime numbers less than 8.} Find A ∪ B ∪ C

Solution

A = {− 6, − 5, 0, 3, 4} B = {3, 5, 7, 9, 11} C = {2, 3, 5, 7} A ∪ B ∪ C = {− 6, − 5, 0, 2, 3, 4, 5, 7, 9, 11}

3. Complements of Sets

The complement of a set is the set of all elements in a universal set that are not in the given set. Given a universal set U and a set A, the complement of A, denoted as A′ or _ A, includes all elements in U that are not in A.

Here are a few examples to illustrate the concept of the complement of a set:

1. Universal Set U={1,2,3,4,5}, Set A={1,2,3} Complement of A: A′={4,5}

2. Universal Set: U={a,b,c,d,e,f}, Set B={a,c,e} Complement of B: B′={b,d,f}

3. Universal Set: U = {apple,orange,banana,grape,mango}, Set C = {apple,mango} Complement of C: C ′={orange,banana, grape}

4. Universal Set: U={red,blue,green,yellow,purple}, Set D={red,yellow} Complement of D: D′={blue,green,purple}

Activity

Go through the following activities to identify various regions in a Venn diagram

1. Use three different writing materials of different colours (crayons, markers, pens) to draw the three intersecting circles in a Venn diagram and label them A, B and C.

2. Identify the various regions that have;

a) only one of the colours

b) two different colours meeting

c) all three colours meeting

d) only two colours meeting Now, study the diagrams below in groups or individually and verify the shaded regions.

Regions of a Three-Set Venn Diagram

The various operations on sets can be represented in a Venn diagram as depicted below.

P ∩ Q P ∩ Q ∩ R P ∩ Q ∩ R′ (P and Q only) P ∩ (Q ∪ R)′ (P only)

1. Study the shaded portions of the three set Venn diagram

(a) (b) A ∩ B ∩ C Aʹ ∩ B ∩ C′

(c) (d) Aʹ ∩ B′∩ C A ∩ B′∩ C′

(e) (f) A ∩ B A ∩ (B ∪ C)

(g) (h) (A ∪ B) ∩ C Aʹ ∩ B ∩ C

(i) (j) (A ∪ B) ∩ C′ A ∪ B ∪ C (k) (A ∪ B ∪ C)′ Let us go through the following examples to enable us illustrate three-sets problems in a Venn diagram.

Worked Example 1.23

The universal set, U = {1,2,3,4,5,6,7,8,9,10,11}, A, B and C are subsets of U such that A={factors of 6}, B={multiples of 3}, C= {odd numbers greater than 1} Find

a) i. (A ∩ B)

ii. (B ∩ C)

iii. (A ∩ C)

iv. A ∩ B ∩ C

b) Illustrate the above information on a Venn diagram.

Solution

A = {1,2,3,6}, B ={3, 6, 9}, C = {3, 5, 7, 9, 11}

a) i. A ∩ B = {3, 6}

ii. B ∩ C = {3,9}

iii. A ∩ C ={3}

iv. A ∩ B ∩ C = {3}

b) U={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

Worked Example 1.24

A group of students were asked to say which type of fruits they like. Some students like more than one fruit and some did not like any of the three types of fruits.

Explain what each of the numbers in the diagram represents;

4 students like pawpaw only.

3 students like pawpaw and melon but do not like mango.

2 students like melon only.

2 students like pawpaw and mango but do not like melon.

5 students like all three fruits.

1 student likes melon and mango but does not like pawpaw.

3 students like mango only.

10 students do not like any of the three fruits.

Find the number of students who like;

i. Melon

ii. Only one of the three fruits

Solution

i. 3 + 5 + 2 + 1 = 11

ii. 4 + 3 + 2 = 9

Worked Example 1.25

In a class of 55 students, some students study at least one of the following subjects: Manufacturing, Robotics and Aviation. 20 students study none of them.

The following gives details of the subjects:

Manufacturing only = 4 Robotics only = 5 Aviation only = 7 All three subjects = 3 Manufacture and Aviation = 7 Robotics and Aviation = 8

a) Illustrate the above data in a Venn diagram

b) Find the number of students who study

i. Manufacturing or Aviation or both, but not Robotics

ii. Robotics

Solution

Let Manufacturing be denoted by M, Robotics by R and Aviation by A and the class of students by U n(M only) = 4, n(R only) = 5, n (A only) = 7, n(M ∩ R ∩ A) = 3, n(M ∩ A ) = 7, n(R ∩ A) = 8, n(U) = 55

a) In a Venn diagram the information can be represented as shown, letting x represent those who study M and R only.

Given n (U) = 55 4 + 4 + 3 + x + 5 + 5 + 7 + 20 = 55 48 + x = 55 x = 55 – 48 x = 7 b)

i. Manufacturing or Aviation or both but not Robotics refer to all elements in the circle of both M and A but outside R ∴ required number = 4 + 4 + 7 = 15

ii. Number of students who study Robotics are in all the elements in the circle of R ∴ required number = 5 + 5 + 3 + 7 = 20

Worked Example 1.26

The Ghana Tourism Authority conducted a survey to find out the number of students who had visited various tourism centres in Ghana. The result of the survey revealed the following information: 48 had visited Kakum National Park, 50 had visited Cape Coast Castle, and 59 had visited Mole National Park. 33 had visited Kakum National Park and Cape Coast Castle 29 had visited Cape Coast Castle and Mole National Park, 36 Kakum and Mole National parks. 24 visited all the places. They also observed that 2 students had not visited any of the three places.

a) Illustrate the above information on a Venn diagram.

b) How many students

i. were surveyed?

ii. had visited Mole National Park, but not Cape Coast nor Kakum?

iii. had visited Cape Coast or Mole National Park?

iv. had visited exactly one of the tourism sites?

v. had visited exactly two of the tourism sites?

vi. had visited at least two tourism sites?

Solution

Let K = {Kakum National Park}, C = {Cape Coast Castle}, M = {Mole National Park} n(U) = y, n(K) = 48, n(C) = 50, n(M) =59, n(K ∩ C ) = 33, n(C ∩ M ) =29, n(K ∩ M ) =36, n(K ∩ C ∩ M ) = 24, n(C ∪ P ∪ M )ʹ = 2 n(K ) = 48 a + (33–24) + 24 +(36–24) = 48 a + 9 + 24 + 12 = 48 a = 48– (9 + 24 + 12) a = 3 n(C) = 50 c + (33–24) + 24 +(29–24) = 50 c + 9 + 24 + 5 = 50 c = 50 – (9 + 24 + 5) c = 12 n(M ) = 59 b + (36–24) + 24 +(29–24) = 59 b + 12 + 24 + 5 = 59 b = 59 – (12 + 24 + 5) b = 18

i. n(U)=3 + 9 + 12 + 5 + 24 + 12 + 18 + 2 = 85 85 students were surveyed

ii. 18 students

iii. 9 + 12 + 5 + 24 + 12 + 18 = 80 students

iv. 18 + 12 + 3 =33 students

v. 36 – 24 + 33 – 24 + 29 – 24 = 26 students

vi. 36 – 24 + 33 – 24 + 29 – 24 + 24 = 50 students Properties of Operation on Sets

1. Commutative Property

i. A ∪ B = B ∪ A Hence, Union is commutative.

ii. A ∩ B = B ∩ A Thus, intersection is commutative.

Verification If A = {1, 2, 3, 4, 5, 6, 8}, and B = {2, 3, 5, 7, 8, 9, 10}

i. A ∪ B= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B ∪ A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Therefore A ∪ B = B∪A

ii. A ∩ B= {2, 3, 5, 8} and B ∩ A = {2, 3,5, 8} Therefore A ∩ B = B ∩ A

2. Associative property

i. (A ∩ B) ∩ C = A ∩ (B ∩ C) Hence, intersection is associative.

ii. A ∪ (B ∪ C) = (A ∪ B) ∪ C Thus, Union is associative.

3. Distributive property

i. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C), Intersection distributes over union.

ii. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) Union distributes over intersection.

Worked Example 1.27

Now, in pairs, do the same to verify these properties, the solution is below.

If A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {2, 4, 8} are subsets of the universal set U = {1, 2, 3. . . 10}, list the elements of the sets

a) i) (A ∩ B) ∩ C

ii. A ∩ (B ∩ C)

iii. What can you say about your result in a. i) and ii)?

b) i) A ∪ (B ∪ C)

ii. (A ∪ B) ∪ C

iii. What can you say about your result in b i) and ii)

Solution

a) i) (A ∩ B) ∩ C = {2,4}

ii. (A ∩ B) ∩ C = {2,4},

iii. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) , showing associative property

b) i) A∪(B∪C) = {1,2,3,4,6,8}

ii. (A∪B)∪C ={1,2,3,4,6,8},

iii. A ∪ (B ∪ C) = (A ∪ B) ∪ C, showing associative property

Worked Example 1.28

Let U = {1, 2, 3, …, 12} and U ∈ ℤ A = {x ∈ U : x is a prime number} B = {x ∈ U : x is an even number} C = {x ∈ U: x is divisible by 3}.

Find the following sets:

a) A ∪ B

b) A ∩ C

c) (A ∩ C) ∪ (B ∩ C)

d) A ∩ B

Solution

A = {2, 3, 5, 7, 11} B = {2, 4, 6, 8, 10, 12} C = {3, 6, 9, 12}

a) A ∪ B = {2, 3, 4, 5, 6, 7, 8, 10, 11, 12}

b) A ∩ C = {3}

c) (A ∩ C) ∪ (B ∩ C) = {3} ∪ {6, 12} = {3, 6, 12}

d) A ∩ B = {2, 3, 5, 7, 11} ∩ {2, 4, 6, 8, 10, 12} = {2}

A binomial is a simplified polynomial with two terms i.e., of the form (p + q) . (2 x²+ 3), (y − 2x) and (− 5 + 0.2x) are all examples of binomials since they have exactly two terms only but −1/4 y³, (2 x²+ 3y − 4) and (y + 3 − 0.1x) are not binomials since in their simplest forms, do not have exactly two terms.

In certain situations, in mathematics, it is necessary to write (p + q)ⁿas the sum of its terms. Because p + q is a binomial, this process is called expanding the binomial. For small values of n , it is relatively easy to write the expansion by using multiplication. For example, (p + q)²can be written as (p + q)(p + q) and expanded to obtain p²+ 2pq + q²and (p + q)³, written as (p + q)²(p + q) and so on. We could continue to build on previous expansions and eventually have quite a comprehensive list of binomial expansions. Instead, we could look for a theorem that will enable us to expand directly. This theorem is the binomial theorem.

One great French mathematician and philosopher, Blaise Pascal who lived from 1623 to 1662 helped to develop a triangular array of numbers and it was named after him, hence Pascal’s triangle. The Pascal triangle depicts the coefficients of binomial expansions based on the given power.

Let us now go through the following activities to help us generate the coefficient of the terms by Pascal’s triangle.

The Coefficient of Terms

Suppose you buy a number of snacks at the school canteen, biscuits(b) and popcorn(p).

How will you write out the combination of these snacks mathematically?

Did you get (b + p)? Good!

Given a flat square tray, with dimension (b + p), how many snacks can be arranged on the tray. Did you get (b + p)²? Congratulations!

How many snacks can you pack into a cube-shaped box of dimension (b + p)?

Is it (a + b)³? Your guess is as good as mine!

In Algebra, it must be recalled that the letters of the alphabet are used to represent items, objects, things or numbers to make problem-solving easier. Also, the number in front of a letter is called its coefficient.

Terms that are like terms can be grouped together or added. For example, adding five mangoes to three mangoes will result in eight mangoes. However, one mango and four oranges will remain the same because they are not alike.

Worked Example 1.29

Expand and simplify the two brackets (2x + 1)(3x − 4).

If your answer is 6x²− 5x − 4 , then you are correct.

If not this is one way of doing it:

(2x + 1)(3x − 4) = (2x + 1)(3x − 4) = 2x(3x − 4) + 1(3x − 4) = 6x²− 8x + 3x − 4 = 6x²− 5x − 4 The coefficient of x²is 6, that of x is –5 and –4 is the constant.

Similarly, in the expression, x³+ 7x y²− 9 y³, the coefficient of x³is 1, that of xy²is 7, and it is –9 for y³.

Worked Example 1.30

Given the expression x²+ 10xy + 24y², find the coefficient of i) x²ii) y².

Solution

(i) Coefficient of x²is 1

(ii) Coefficient of y²is 24

Now, let us consider the expansion of the following consecutive powers of a binomial, The expression a + b is actually a shortened version for 1a + 1b. the next power of a + b is (a + b) (a + b), which is also written as (a + b)².

Using Pascal’s triangle, this can be expanded as follows:

(a + b)²= (a + b)(a + b) = a(a + b) + b(a + b) = a²+ ab + ab + b²= a²+ 2ab + b²Similarly, (a + b)³can be expanded as follows:

(a + b)³= (a + b) (a + b)²= (a + b)(a²+ 2ab + b²) (by substitution) =a (a²+ 2ab + b²) + b(a²+ 2ab + b²) = a³+ 2 a²b + ab²+ a²b + 2a b²+ b³= a³+ 3 a²b + 3 ab²+ b³(a + b)⁴= (a + b) (a + b)³= (a + b)(a³+ 3 a²b + 3a b²+ b³) =a( a³+ 3 a²b + 3 ab²+ b³) + b( a³+ 3 a²b + 3 ab²+ b³) = a⁴+ 3 a³b + 3 a²b²+ ab³+ a³b + 3 a²b²+ 3a b³+ b⁴= a⁴+ 4 a³b + 6 a²b²+ 4 ab³+ b⁴Example 1.31

1. Use the pattern to obtain the coefficient for the following expression:

i.. (a + b)⁵You may have noticed a pattern in the coefficients of the expansions. Writing the coefficients of the expansions of (a + b)², (a + b)³(a + b)⁴, we get n coefficients 1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1 5 1 5 10 10 5 1 You may have observed that

1. Every subsequent row is obtained by starting with 1 and ending with 1.

2. Each interior member is the sum of the two elements directly above it. For

example 1 + 4 = 5, 4 + 6 = 10, 6 + 4 = 10, 4 + 1 = 5 For completeness, it may be observed that (a + b)⁰=1 and (a + b)¹= 1a + 1b 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 When an expression is written as a series of terms, it is said to be expanded, and the series is its expansion.

This pattern created by the coefficients is referred to as Pascal’s triangle.

Pascal’s triangle therefore is the triangular arrangement of numbers that represent the coefficients in the expansion of any binomial expression such as (p + q)ⁿ. The numbers are so arranged that they reflect as a triangle.

The numbers in Pascal’s triangle are placed in such a way that each number is the sum of two numbers just above it.

Take note of the following characteristics with respect to the expansion of (a + b)ⁿ, where n is a positive integer.

i. Reading from either end of each row, the coefficients are the same.

ii. There are (n + 1) terms.

iii. Each term is of degree (sum of the powers of the two terms) n.

iv. The coefficients are obtained from the row in Pascal triangle.

v. The exponent on a decreases by 1 for each successive term.

vi. The exponent on b increases by 1 for each successive term.

Go through the steps below to expand a binomial expression

Step 1: Write down the binomial coefficients:

Step 2: Write the expression with the binomial coefficients:

Step 3: Simplify the expression:

Step 4: Simplify further:

Worked Example 1.32

Expand (a + b)⁶in descending powers of a.

Solution

Write down the binomial coefficients 1, 6, 15, 20, 15, 6, 1 Write the expression with the binomial coefficients:

1(a)⁶(b)⁰+ 6(a)⁶⁻¹(b)0 + 1 + 15(a)⁶⁻²(b)0 + 2 + 20(a)⁶⁻³(b)0 + 3 + 15(a)⁶⁻⁴(b)0 + 4 + 6(a)⁶⁻⁵(b)0 + 5 + 1(a)⁶⁻⁶(b)0 + 6 Simplify the expression 1(a⁶) + 6(a⁵)(b¹) + 15(a)⁴(b)²+ 20(a)³(b)³+ 15(a)²(b)⁴+ 6(a)(b)⁵+ 1(a)⁰(b)⁶Therefore, the expansion of (a + b)⁶in descending powers of a is a⁶+ 6a⁵b + 15 a⁴b²+ 20 a³b³+ 15 a²b⁴+ 6a b⁵+ b⁶Worked Example 1.33 Expand (x + 3y)³in descending powers of x.

Solution

Here, a = x, b = 3y and there will be four terms involving (3 + 1 = 4) x³, (x²)(3y), (x) (3y)², (3y)³ Their coefficients obtained from Pascal’s triangle are respectively 1, 3, 3, 1 Therefore, the expansion of (x + 3y)³in descending powers of x is x³+ 3(x²)(3y) + 3(x) (3y)²+ (3y)³Which simplifies to x³+ 9x²y + 27x y²+ 27y³Worked Example 1.34 Expand (2x + 3y)⁵in descending powers of x

Solution

(2x)⁵+ 5(2x)⁴(3y) + 10 (2x)³(3y)²+ 10 (2x)²(3y)³+ 5 (2x)(3y)⁴+ (3y)⁵32x⁵+ 5( 16x⁴)(3y) + 10 (8x³) (9y²)+ 10 (4x²) (27y³)+5 (2x)(81y⁴)+ (243y⁵) 32x⁵+ 240x⁴y + 720x³y²+ 1080x²y³+810xy⁴+ 243y⁵Worked Example 1.35 Expand (x + 1/x)⁴Solution

Step 1: Write down the binomial coefficients:

1, 4, 6, 4, 1 Write the expression with the binomial coefficients:

(x + 1/x)⁴=1(x)⁴(¹__ x)⁰+ 4 (x)⁴−1 (¹__ x)⁰⁺¹+ 6 (x)⁴−2 (¹__ x)⁰⁺²+ 4(x)⁴−3 (¹__ x)⁰⁺³+1(x)⁴−4 (¹__ x)0/(+4) Simplify the expression (x)⁴(¹_ x)⁰+ 4 (x)³(¹_ x)¹+ 6 (x)²(¹_ x)²+ 4 (x)¹(¹_ x)³+ (x)⁰(¹_ x)⁴= x⁴+ 4 x²+ 6 + 4/x²+ 1/x ⁴Simplify further:

= x⁴+ 4 x²+ 6 + 4x−2 + x−4

Worked Example 1.36

a) Expand (¹__ x − √x)⁵b) Expand (¹__ x + 1)⁵Solution

a) From the expansion of (a + b)⁵in the previous example and by comparison i.e., a = 1/x and b = √x, (¹__ x − √x)⁵= (1_ x)⁵+ 5 (1_ x)⁴(−√x) + 10 (1_ x)³(−√x)²+ 10 (1_ x)²(−√x)³+ 5(1_

x) (−√x)⁴+ (−√x) ⁵= 1/x⁵− 5 √x/x⁴+ 10/x²− 10 √x/x²+ 5x − √x⁵= x−5 − 5 x−⁷_ ₂ + 10 x−2 − 10 x−¹_ ₂ + 5x − x⁵_ ₂

b) From the expansion of (a + b)⁵in the previous example and by comparison i.e., a = ¹_ₓ and b = 1 , (¹__ x + 1)⁵= (¹__ x)⁵+ 5 (¹__ x)⁴(1) + 10 (¹__ x)³(1)²+ 10 (¹__ x)²(1)³+ 5(¹__

x) (1)⁴+ (1)⁵= 1_ x⁵+ 5_ x⁴+ 10_ x³+ 10_ x²+ 5_ x + 1 = x−5 + 5 x−4 + 10 x−3 + 10 x−2 + 5 x−1 + 1

Worked Example 1.37

Follow the steps below to complete the tasks use the binomial expansion: find the value of

a) (1.01)³b) (2.1)⁴without using a calculator.

Solution

a) Step 1: write 1.01 as a binomial and since 1.01 is closer to 1, write it as 1 + 0.01

Step 2: Expand the expression:

(1 + 0.01)³= 1³+ 3(1²)(0.01) + 3(1)(0.01²) + 1(1)⁰(0.01³)

Step 3: Simplify each term:

1 + 3(1)(0.01) + 3(1)(0.0001) + 1(1)(0.000001) 1 + 0.03 + 0.0003 + 0.000001 1.030301

Step 4: Write the final answer:

1.0303 (4 decimal places)

b) go through the steps in (a) to solve the question (b) and verify your answer with the calculator.

Congratulations on obtaining 19.4481 as your answer. You are good to move on.

Worked Example 1.38

Use Pascal’s triangle of binomial expansion to obtain the value of (3.004)⁴, correct to six decimal places

Solution:

(3.004)⁴= (3 + 0.004)⁴= 3⁴+ 4(3³)(0.004) + 6(3²)(0.004²) + 4(3)(0.004³) + (0.004⁴) = 81 + 4(27)(0.004) + 6(9)(0.000016) + 4(3)(0.000000064) + (0.000000000256) = 81 + 0.432 + 0.0000864 + 0.000000768 + 0.000000000256 = 81.43208717 ≈ 81.432087(6 dp)

Combination is the selection in which the order of selection is not important.

Combination is used in lottery in calculating the probability of winning numbers.

It also used in choosing committee members from a large group of members. It is not practical to use Pascal’s triangle to expand binomial expressions with larger positive integer values as exponents since the method is recursive. It requires that to obtain the coefficients of the expansion of (p + q)ⁿ, we would need to find the coefficients of the expansion of (p + q)ⁿ−1.

Thus, to find the 100th row of the Pascal’s triangle, we must first find the preceding 99 rows. The coefficients in the Pascal’s triangle however, can be generated using the combination approach.

Activity 1.5

1. Look for n_(C)_(Γ) on your calculator.

2. Now use n_(C)_(Γ) on your calculator to evaluate i) 3_(C)₀

ii) 3_(C)₁

iii) 3_(C)₂

iv) 3_(C)₃ Did you obtain? 1 3 3 1.

You are right, congratulations!

What do these numbers represent? They represent the coefficients of n = 3 as obtained in the Pascal triangle. This means there are other ways to obtain the coefficients in binomial expansions.

Using n_(C)_(Γ) to obtain the coefficients in binomial expansion is the combination method.

If n is a positive integer, the expansion (a + b)ⁿis given by:

(a + b)ⁿ= [ ⁿC₀ aⁿb⁰+ ⁿC₁ aⁿ−1 b¹+ ⁿC₂ aⁿ−2 b²+ …+ⁿCₙ aⁿ−n bⁿ] The combination of n items taking r at a time is given by:

(n r ) = n_(C)_(Γ) = n!_______ (n − r)!r !

Where n! is read as n factorial and is explained as n!= n(n − 1)(n − 2)(n − 3)…2 × 1 Worked Examples 1.40

1. 5 ! = 5 (5 − 1 )(5 − 2 )(5 − 3 )(5 − 4 ) = 5 × 4 × 3 × 2 × 1 = 120

2. 7 ! = (7 − 1 )(7 − 2 )(7 − 3 )(7 − 4 )(7 − 5 )(7 − 6 ) = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Example 1.41

Study the computation of the following factorials to appreciate the derivation of the Binomial Theorem:

i. 6_(C)₃ = 20

ii. 4_(C)₂ = 6

iii. 7_(C)₄ = 35 Confirm your answers with a calculator.

Note: It must be noted that n_(C)₀ = 1 and 0!=1 Binomial Expansion of Positive Integers Using the Combination Method Using the combination method, the binomial (a + b)ⁿ, n being a positive integer, is expanded as:

(a + b)ⁿ=(a + b)(a + b)… …(a + b) to n factors.

(a) Choosing an a from each bracket we obtain aⁿ(b) The term aⁿ−1 is obtained by choosing one b from one bracket and ‘a’s from the other n–1. This can be done in n_(C)₁ ways, giving n_(C)₁ ₐ ₙ−1 b.

(c) The term aⁿ−2 is obtained by choosing one b from two brackets and ‘a’s from the other n–2 . This can be done in n_(C)₂ ways, giving n_(C)₂ ₐₙ−2 b2

(d) The term in aⁿ−r is obtained by choosing one b from r brackets and ‘a’s from other n-r. This can be done in n_(C)ᵣ ways, giving n_(C)ᵣ ₐₙ−r br.

(e) Choosing a b from each bracket we obtain bⁿ.

Worked Example 1.42

Use the binomial theorem to write out the first four terms of the binomial expansion and simplify (2x+ y) ¹⁰(3x + y)¹⁰= x¹⁰+ (10/1 ) (2x)⁹(y)¹+ (10/2 ) (2x)⁸(y)²+ (10/3 ) (2x)⁷(y)³Technology Tip (use of calculator evaluation) (10/1 ) = 10/1 = 10 (10/2 ) = 10 . 9/1 . 2 = 45 (10 3 ) = 10 . 9 . 8_ 1 . 2 . 3 = 120 (2x+ y)¹⁰= (2 x)¹⁰+ (10/1 ) (2x)⁹(y)¹+ (10/2 ) (2x)⁸(y)²+ (10/3 ) (2x)⁷(y)³(2x+ y)¹⁰= (2 x)¹⁰+ (10/1 ) (2x)⁹(y)¹+ (10/2 ) (2x)⁸(y)²+ (10/3 ) (2x)⁷(y)³+ (10/4 )(2 x)⁶(y)⁴w

Worked Example 1.43

By using the combination approach, expand completely the expression (2 + x)⁵.

Solution

(2 + x)⁵= 5_(C)₀ 2⁵+ 5_(C)₁ × 2⁴× x + 5_(C)₂ × 2³× x²+ 5_(C)₃ × 2²× x³+ 5_(C)₄ × 2 × x⁴+ x⁵= 1 × 2⁵+ 5 × 2⁴× x + 10 × 2³× x²+ 10 × 2²× x³+ 5 × 2 × x⁴+ x⁵= 32 + 80x + 80x²+ 40x³+ 10x⁴+ x⁵Worked Example 1.44 Expand (x–2y)⁶using the combination method, and hence find the value of (1.06)⁶, correct to 5 decimal places.

Solution

(a + b)ⁿ= (a + b)ⁿ= [ ⁿC₀ aⁿb⁰+ ⁿC₁ aⁿ−1 b¹+ ⁿC₂ aⁿ−2 b²+ …+ⁿCₙ aⁿ−n bⁿ] (x–2y)⁶= (6/0) x⁶( − 2y)⁰+ (6/1) (x)⁶−1 (–2y) + (6/2) (x)⁶−2 × ( − 2y)²+ (6/3) (x)⁶−3 × ( − 2y)³+ (6/4) × (x)⁶−4 × ( − 2y)⁴+ (6/5) × (x)⁶−5 × ( − 2y)⁵+ (6/6) ( − 2y)⁶= (1) x⁶(1) + 6 (x)⁵(–2y) + 15 (x)⁴× 4y²+ 20 (x)³× − 8y³+ 15 × (x)²× 16y⁴+ 6 ×(x)¹× − 32y⁵+ 1 × 64y⁶= x⁶− 12x⁵y + 60x⁴y²− 160x³y³+ 240 x²y⁴− 192x y⁵+ 64 y⁶(1.06)⁶= (1 + 0.06)⁶Comparing (x–2y)⁶= (1 + 0.06)⁶x = 1, –2y = 0.06 y= –0.03 (1.06)⁶= (1)⁶− 12(1)⁵( − 0.03) + 60(1)⁴( − 0.03)²− 160(1)³(− 0.03)³+ 240 (1)²( − 0.03)⁴− 192(1) ( − 0.03)⁵+ 64( − 0.03)⁶= 1 + 0.36 + 0.054 + 0.00432 + 0.0001944 + 0.0000046656 + 0.000000046656 (1.06)⁶= 1.41851911 = 1.41852(5d.p)

Worked Example 1.45

The coefficient of x³in the expansion of (k + 3x)⁴is 54. Find the value of k.

Solution

4_(C)₃ k¹(3x)³= 54 x³4k . 27 x³= 54x³∴ 4k . 27 = 54 k = 54/4 × 27 k = 1/2

Worked Example 1.46

Find the numerical coefficient of x¹⁷in the expansion of (x + y)²⁰Solution 20C₃ x¹⁷y³= 1140 x¹⁷y³Therefore, the numerical coefficient is 1140.

Worked Example 1.47

Without using tables or calculator, evaluate (√2 + 1)⁴− (√2 − 1)⁴________________ 2

Solution

(√2 + 1)⁴= (√_ 2)⁴+ 4 . (√_ 2)³1¹+ 6 (√_ 2)²1²+ 4 (√_ 2)¹1³+ 1⁴= 4 + 8√_ 2 + 12 + 4√_ 2 + 1 = 17 + 12√_ 2 (√2 – 1)⁴= (√_ 2)⁴− 4 . (√2)³1¹+ 6 (√2)²1²− 4 (√_ 2)¹1³+ 1⁴= 4 - 8√_ 2 + 12 –4√_ 2 + 1 = 17–12√_ 2 (√2 + 1)4 − (√2 − 1)⁴________________ 2 = 17 + 12 √_ 2 − (17 − 12√_ 2)___________________ 2 = 24 √2/2 = 12 √2

Worked Example 1.48

a) Write down the binomial expansion of (2 + x)⁴.

b) Use your expansion to evaluate (1.97)⁴, correct to two decimal places.

Solution

a) (2 + x)⁴= 2⁴+ 4 × 2³× x¹+ 6 × 2²× x²+ 4 × 2¹× x³+ x⁴= 16 + 32x + 24 x²+ 8x³+ x⁴b) Comparing (2 + x)⁴≡ (1.97)⁴, 2 + x = 1.97 x = 1.97 − 2 = –0.03 ∴ (1.97)⁴= 16 + 32(− 0.03) + 24 (− 0.03)²+ 8(− 0.03)³+ (− 0.03)⁴= 16 – 0.96 + 0.0216 – 0.000216 + 0.00000081 = 15.06138481 = 15.06 ( 2 d.p)

Review Question 1

1. State with reason whether the following statements are true or false:

a) The set of odd integers are closed under multiplication.

b) The set of rational numbers are closed under addition.

c) The set of whole numbers are closed under subtraction.

2. A binary operation ∇ is defined under the set of real numbers R = {2, 3, 4, 5} by m ∇ n = m + n + mn.

Use this definition to copy and complete the table below.

∇ 2 3 4 5 2 11 14 17 3 11 15 19 4 19 5 17 29 35 From the table, by giving reason(s) determine whether

a) the operation ∇ is commutative

b) the operation is closed under the set of real numbers

3. Study the table below under * and answer the questions that follow:

* p q r s P q p s r q p q r s r s r q p s r s p q By giving reason(s) determine whether the operation *

i. is closed ii) is commutative

iii. has an identity element

iv. has identity element, and find the inverses of p, q, r and s.

4. The operation ° is defined under the set of real numbers R by p ° q= p + q/3 .

Evaluate

i. 3 ° 2

ii. 2 ° 3

iii. What can be said about the results in i) and ii)?

iv. if p ° 4= − 1, find the value of p.

5. A baker produces two types of bread: wheat and potato. The wheat bread has a production cost of GH¢ 250.00, and potato bread has a production cost of GH¢ 300.00 . The baker sells the wheat bread to retailers for GH¢ 6.00 each and the potato bread for GH¢ 7.50 each. Last month, the baker produced 500 loaves of wheat bread and 300 pieces of potato bread.

a) Explain how you will use binary operation to find the total revenue generated from the sale of bread last month and determine the overall profit or loss.

b) Calculate the total revenue generated from the sale of bread last month and determine the overall profit or loss.

6. You have six shirts, two trousers and two pairs of shoes. Explain how you will use a binary operation to determine how many ways a shirt, a trouser and a pair of shoes can be worn.

7. A set P = {1, 4, 7, 8, 10, 11}. The operation ® is defined as follows: if a and b are any elements of P, then a ® b denotes the remainder when the results of adding a to b is divided by 5; For example, 4 * 10 = 4

a) Draw a table for the above operation ®

b) Determine if the binary operation:

i. is closed

ii. is commutative

iii. has an identity element

c) Solve the equation 11 * x = 4

8. The operation * is defined on the set of real numbers such that d * e = d²+ 2de + 5e. If d = 2x + 1 and e = x – 1, evaluate d * e in terms of x.

9. Two students were asked if a, b ∈ R, such that a * b = 2a – 3b, to find the identity element.

Ama’s solution was:

a *e = a 2a – 3e = a

–3e = – a e = a/3 Her study partner, Kwame solved it as a * b = b * a a * b = 2a – 3b b * a = 2b – 3a a * b ≠ b * a, therefore, the identity element did not exist.

State with reasons who is correct.

10. The operation * is defined over the set of real numbers by p * q= p + 2q/3 Investigate if *

a) i. is commutative

ii. is associative

b) Determine if the operations has an identity and an inverse.

c) determine whether or not p * (q + r) = (p * q ) + (p * r )

11. An operations ʘ is defined on the set of real numbers by 1/2 m + 1/3 n + 5mn, evaluate

a) (4 ʘ 5)

b) 3 ʘ (4 ʘ 5)

12. A mathematical set costs GH¢ 30.50 and a calculator costs GH¢ 120.00.

How much will Ms Ekua Bonney pay for 5 mathematical sets and 7 calculators she procured for her bookshop.

13. m, n ∈R, the operation *is defined by m * n = 2m + n______ 4n − m, evaluate

i. –8 * 5

ii. 4*(5 * 6)

14. A binary operation * is defined on the set of real numbers R, by p * q = pq/p + q, evaluate

i. 5/7 *2/3

ii. If (x + 1) *3 = –5, find the value of x.

15. Kofi had 20 GH¢ 100.00 notes and 25 GH¢ 200.00 notes in his wallet.

How much is in the wallet?

16. A binary operation * is defined on the set of real numbers R, by a * b = a__ 2b + b__ 7a, find 3*–5

17. The relation for paying for total cost of a number of books(x) and a number of pencils(y) was x * y = 15x + 5y + 10. If the total payment made was GH¢ 300 and 5 books were bought, how much was a pencil?

18. An operation * is defined on the T = {1, 2, 3, 4, 6} by a * b = a + 2b + 1.

a) Copy and complete the table below * 1 2 3 4 6 1 6 14 2 3 10 4 7 6 11

i. Is the operation * closed?

ii. Is the operation * commutative?

19. An operation * is defined on the set of real numbers by p * q = 2p + q/p , determine whether or not, the operation * is commutative.

20. An operation ɵ is defined on the set of real numbers by x ɵy= x + y +1__ 2xy, determine whether or not the operation ɵ is associative Review Question 2

1. P and Q are subsets of U such that; U = {12, 13, 14,… 21} P={X : 12 ≤ X < 16} Q = {multiples of 3 ≤18}, find

a) P ∩ Q

b) n(P ∩ Q¹)

c) Q ∪ (P¹∩ Q¹)

2. Given that U = {a, b, c, d, e, f, g, h}, A = {a, b, d, g} and B ={b, d, e, h}

a) Illustrate U, A and B in a Venn diagram

b) Write down the elements of the following sets: (i) A ∩ B (ii) A ∪ B

3. In the diagram n(A) = 2n(B). Find x

4. Consider the sets:

A = {red, green, blue}, B = {red, yellow, orange}, C = {red, orange, yellow, green, blue, purple} D = {yellow, white} and the universal set, U = A ∪ B ∪ C ∪ D

a) Find A′∩C

5. In the Venn diagram below. The numbers indicate the number of elements in the region Determine the following

a) n(A) b) n(C)

c) n(B) d) n(A ∩ C)

e) n(B ∪ C) f) n((B ∩ C) ∪ A)

6. If A = {x:x is a factor of 72}, B = {x:x < 20}, C= {multiples of 3}, and A,B,C ⊂ U, where U = {integers} List the elements of A ∩ B ∩C

7. Given μ = {1, 2, 3, 4,…,10}, A = {1, 4, 9}, B = {1, 3, 5, 7, 9} and C = {1, 3, 6,10}, show that:

a) (A∪B)∪C = A∪(B∪C ) b) A∪(B∩C ) = (A∪B)∩(A∪C )

c) (A∩B)∩C = A∩(B∩C ) d) A∩(B∪C ) = (A∩B)∪(A∩C )

8. Write the following in roster form:

a) The set of whole numbers less than 20 and divisible by 3.

b) The set of integers greater than –2 and less than 4.

c) The set of integers between - 4 and 4, including both –4 and 4.

d) The squares of the first four natural numbers.

e) The set of prime factors of 36.

f) The set of counting numbers between 15 and 35, each of which is divisible by 6.

9. Write each of the following in set builder form:

a) A = {16, 25, 36, 39, 64}

b) B = {2, 4, 6, 8, 10}

c) C = {2, 3, 5, 7, 11}

d) D = {1, 3, 5, 7, 9}

e) E = {a, e, i, o, u}

10. The universal set, U = {1,2,3,4,5,6,10,20,50,100}, P, Q and R are subsets of U, P={1,2,5,10,20} Q={2,5,10,20,50} and R= {5,10,20,50,100} Find

i. n(P ∩ Q) ii) P ∪ (Q∩R)

iii. (P ∪ R) ∩ Q

11. U = {m,h,e,a,t,c,s,i} A= {m,e,s,t,i,c} B = {s,e,m,t,c,} C= {m,e,a,t}, A,B,C are subsets of U. List the elements of

a) A ∪ (B ∩ C) b) (A ∩ B') ∪ C

c) (A ∪ B) ∩ (B ∪ C)′

12. U = {x:1≤x≤20}, where x is an integer and P, Q and R are subsets of U such that P={x:x is multiple of 2}, Q={x:x a multiple 3} R = {x:x is multiple of 9}

i. What is the relationship between Q and R?

ii. Show the relationship between P, Q and R in Venn diagram, listing the elements of each region.

iii. List the elements of

a) PÇQÇR¹b) P¹ÇQÇR¹13. Given U = {1, 2, 3, 4,…,15} A = {1, 4, 9, 15} B = {1, 3, 5, 7, 9, 11, 13} C = {1, 3, 6, 10, 15} Show that:

a) (A ∪ B) ∪ C = A ∪ (B ∪ C)

b) A ∪(B∩C) = (A ∪ B) ∩ (A ∪ C)

c) (A ∩ B) ∩ C = A ∩ (B ∩ C)

d) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

e) (A ∪ B) ∩ (B ∪ C) = (C ∪ B) ∩ (B ∪ A)

14. Let P = {0, 2, 4, 6, 8, 10}, Q = {0, 1, 2, 3, 4, 5, 6}, and R = {4, 5, 6, 7, 8, 9, 10}.

Find

a) P ∩ Q ∩ R. b) P ∪ Q ∪ R.

c) (P ∪ Q) ∩ R. d) (P ∩ Q) ∪ R.

15. Write an expression for the shaded region.

16. Write an expression for the shaded regions in the diagrams below:

a) b)

17. A survey asked people what alternative transportation modes they use.

30% use the bus; 20% ride a bicycle, 25% walk; 5% use the bus and ride a bicycle 10% ride a bicycle and walk; 12% use the bus and walk; 2% use all three. Use the data to complete a Venn diagram, then determine:

a) What percent of people only ride the bus

b) How many people don’t use any alternate transportation

18. A survey asks: “Which online social media have you used in the last month:

WhatsApp, Facebook, Both”. The results show 42% of those surveyed have used WhatsApp, 70% have used Facebook, and 20% have used both.

How many people have used neither WhatsApp nor Facebook?

19. In a survey of 115 pet owners, 26 said they own a dog, and 64 said they own a cat. 5 said they own both a dog and a cat. Use a Venn diagram to determine how many of the pet owners surveyed owned neither a cat nor a dog?

20. A pharmaceutical company is considering manufacturing new toothpaste.

They are considering two charcoal flavours, strawberry and mint. In a sample of 74 people, it was found that 45 liked strawberries, 37 liked mint and 21 liked both types

a) Create a Venn diagram to model the information.

b) How many liked only strawberry?

c) How many liked only mint?

d) How many liked exactly one of the two (that is they liked one but not the other)?

Review Questions 3

1. Indicate which of the following statements is true and which is false.

a) Pascal’s triangle is a triangular array of numbers in which the first and the last numbers in each row are 1.

b) In Pascal’s triangle, each number is the two numbers above it added together.

2. Use the combination formula to expand the following:

a) (2a + 3b)³b) (3x − 5y)⁴3. Obtain the expansion of (2x + 1/2) 4 in descending powers of x.

4. Use Pascal’s triangle to expand the following:

a) (2a + 3b)²b) (4x + 1/2) 4

5. Obtain the expansion of (2x − 1/2) 4 in descending powers of x.

6. Find the coefficient of the term (x)¹⁰in the binomial expansion of the expression: (1 + x)²⁵7. Find the coefficient of the term x⁹y⁵in the binomial expansion of the expression:

(2x + 3y)¹⁴.

8. Find the x³term in the expansion of (x + 3)¹².

9. Find the expression for the 4th term of the expression ( 2__ 3x − √x) 6 .

10. i) Using the binomial theorem, expand (1 + 2x)⁵, simplifying all terms.

ii. Use your expansion to calculate the value if 1.05⁵, correct to six decimal places.

11. If the first three terms of the expansion (1 + px)ⁿin ascending powers of x are 1 + 20x + 160x², find the values of n and p.

12. Kwame was granted a loan of GH¢ 5000.00 at a rate of 5% for 3 years from a bank at compound interest (CI). The bank used the relation CI= P(1 + r%)ⁿ, where P is the principal, r is the rate and n is the number of years. Find the total amount he had paid at the end of third year using the binomial theorem.

13. The population of a senior high school in Ghana increases according to the formula N = P(1 + 0.025)ᵗ, where t is the number of years, P is the current population, and N is the total population. If the current population for that school is 1400 which is almost full capacity of the school, what steps should government take for the school in 5 years’ time.

14. Use the Pascal triangle to expand and simplify all terms in (2 – 3y)⁴, hence find the value of (1.97)⁴.

15. Expand

i. (2x – 4)⁴ii. (3y + 1)⁵, using Pascal triangle.

16. Use binomial theorem to expand (1–2x)⁶and simplify all coefficients.

17. Find the numerical coefficient of x⁴in the expansion (2–x)¹⁰.

18. Find the x⁵term in the expansion (2x-y)⁷.

19. The first three terms of the binomial expansion of (1 + bx)ⁿare 1 + 2x + 5/3 x².

a) Using this information, work out the values of b and n.

Surds will provide a foundation for understanding more complex mathematical concepts. Surds being a square root of Non-perfect Squares, introduce learners to irrational numbers. These numbers cannot be expressed as simple fractions and have decimal expansions that neither terminate nor repeat. Familiarity with surds helps learners to grasp the concept of irrationality. The use of surds enables the engineers to calculate the dimensions and angles for bridges and buildings, ensuring structural stability and load–bearing capacity.

At the end of this section, you will be able to:

· Investigate the properties of surds and perform basic arithmetic operations on surds · Rationalise surds with binomial denominators · Recollect the initial laws of indices and establish other laws for negative powers and roots · Recognise the relationship between surds and indices and apply laws of indices to simplify expressions · Pose and solve simple equations involving indices · Establish the relationship between indices and logarithms and use the properties of logarithms to solve related problems in one base

Key Ideas:

· Surds are values expressed in the square root, that cannot be further simplified into whole numbers or integers. Examples include √_ 2, √_ 3, √_ 5.

· Surds have decimal representations that go on forever without repeating.

For example, √_ 2 is approximately 1.4142135………….,√_ 3 is approximately 1.7320508…………etc. These decimals are considered irrational.

· Values or numbers in the form √__ 16 = 4, √_ 4 = 2, ³√8 = 2 are not surds as their respective answers are whole numbers or exact figures and not repeating decimals.

· In general, expressions of the form aᵐare called indices, where a is called the base and m is called the index or exponent or power and it is read as “a exponent m” or “a to the power of m”.

· On the other hand, if we have the equation we say that y = aᵐ, the exponent m is the logarithm of y to base a. it is written logₐy = m.

Surds were discovered and defined by a European mathematician, Gherardo of Cremona, in 1150 BC (Joseph, 2010). He used the Pythagoras’ theorem to find the diagonal of a square and the value of the first surd. He termed this value ‘voiceless’ because the root value had no meaning at that time. Surds are numbers with roots that cannot be simplified to whole numbers. They are square roots, or other roots, that cannot be written as a simple fraction. Surds or radical expressions contain roots (like √_

2) that are not whole numbers. For example, √_ 2 = 1.4142135624 √3 = 1.7320508076, √35 = 5.9160797831 etc. However, √25 = 5 and √81 = 9, both have whole number solutions and, thus, are not surds.

The behaviour of numbers can lead to some classifications or groupings.

Activity 2.1

Now, let’s go through the following activities either in small groups or individually.

1. Use your calculator to compute the following numbers.

i. √2, √3, √5,

ii. √4, √36, √49,

2. What was your observation?

You will observe that the numbers in (1) did not have exact values and so were approximated. However, the numbers in (2) had exact values.

Expressions of the form √2, 3 √5, 2 + √7 and so on that involve or contain square roots of positive integers which are not perfect squares are called Surds. When expressed in decimal form, surds are non-terminating and non-repeating.

Though a calculator with a square root key can give values of surds, it must be noted that, these values they provide are only approximations.

A surd is of the form a ± √_ b or √_ a ± √_ b where a and b are natural numbers, not perfect squares, is often referred to as a binomial surd. The numbers√2, √3 , √5, √7, √8, √10, etc, are all surds. On the other hand, the numbers where a certain number multiplied itself to give the number is a perfect square. Examples of perfect squares are, 1, 4, 9, 16, 25.

Have a go at writing more perfect squared numbers and verify them with the use of calculator.

Types of Surds

1. Pure Surds: A surd having only a single irrational number is called a pure surd.

Example, √7, √3, √5, √2

2. Mixed Surds: A surd having a mix of a rational number, and an irrational number is called a mixed surd.

Example, 5 √3, 12 √5, 7 √11,

3. Compound Surds: A surd composed of two surds, or a surd and a rational number is called a compound surd.

Example, √3 + √10, 3 + √7,

4. Binomial Surd: When two surds give rise to one single surd, the resultant surd is known as a binomial surd.

Example, √30 = √15 × 2 Rules/Properties of surds Surds have rules which influence the way they behave in algebra. Note the following carefully.

Activity 2.2

In a small group, or individually, use your calculator to verify the following properties of surds using any numbers of your choice.

Multiplication Rules

1. √a × √a = ( √a )²= a, if a ≥ 0 Examples:

i. √3 × √3 = ( √3 )²= 3

ii. √7 × √7 = ( √7 )²= 7 · The square of a surd is equal to its radical.

2. √a × √b = √(ab) Examples:

i. √2 × √3 = √(2 × 3)= √6

ii. √5 × √11 = √(5 × 11)= √55 · The product of two surds is equal to the square roots of their products.

3. m √a × √a = m( √a )²= m × a = ma Examples:

i. 5 √3 × √3 = 5( √3 )²= 5 × 3 = 15

ii. 3 √7 × √7 = 3( √7 )²= 3 × 7 = 21

4. m √a × n √a = m × n( √a )²= mn × a = mna Examples:

i. 4 √5 × 3 √5 = 4 × 3( √5 )²= 12 × 5 = 60

ii. 2 √3 × 5 √3 = 2 × 5( √3 )²= 10 × 3 = 30

5. m √a × √b = m √a × b= m √ab Examples:

i. 5 √2 × √3 = 5 √6

ii. 3 √7 × √5 = 3 √7 × 5 = 3 √35

6. m √a × n √b = (m × n) √a × b = mn √ab Examples:

i. 5 √2 × 4 √3 = (5 × 4) √2 × 3 = 20 √6

ii. − 3 √5 × 2 √6 = (− 3 × 2) √5 × 6 = − 6 √30 Division Rules

1. √a ÷ √a= √a_ √a = √a_ a = √1 = 1, where a ≠ 0 Examples:

i. √5 ÷ √5= √5_ √5 = √5_ 5 = √1 = 1

ii. √3 ÷ √3= √3_ √3 = √3_ 3 = √1 = 1 · The quotient of two identical surds is equal to 1.

2. √a ÷ √b= √a_ √b = √a_ b , where b ≠ 0 Examples:

i. √2 ÷ √3= √2_ √3 = √2_ 3

ii. √11 ÷ √7= √11_ √7 = √11_ 7 · The quotient of two surds is equal to the square root of the quotient.

Addition Rule

1. m √a+ n √a = (m + n) √a Examples:

i. 4 √3+ 5 √3 = (4 + 5) √3 = 9 √3

ii. − 3 √2+ 7 √2 = (− 3 + 7) √2 = 4 √2

2. m √a+ n √b = m √a+ n √b Examples:

i. 6 √3+ 4 √5 = 6 √3+ 4 √5

ii. 3√2+ 5 √3 = 3 √2+ 5 √3 · Like surds can be added together.

· Unlike Surds cannot be added together.

Subtraction Rule

1. m √a− n √a = (m − n) √a Examples:

i. 7 √3− 5 √3 = (7 − 5) √3 = 2 √3

ii. 3 √2− 7 √2 = (3 − 7) √2 = − 4 √2

2. m √a− n √b = m √a− n √b Examples:

i. 6 √3 − 4 √5 = 6 √3 − 4 √5

ii. 3 √2 − 5 √3 = 3 √2 − 5 √3 · Like surds can be subtracted.

· Unlike Surds cannot be subtracted.

Simplification of surds In simplifying surds, you find two factors of the number such that one is a perfect square. This can be done by dividing the number by the prime numbers 2, 3, 5, 7, etc in turn as exemplified in the table until you are left with a perfect square.

Or you can divide by a perfect square (4, 9, 16 etc) until you have no remainder.

Number Factors

8 4 × 2 12 4 × 3 32 16 × 2 45 9 × 5 68 4 × 17 567 81× 7 75 25 × 3 Let us look at √8, the number is a product of a perfect square and a prime number.

Can you guess the two numbers?

The surd √8 = √4 × 2 √4 × 2 = √4 × √2 2 × √2 2 √2 So, to simplify a surd into basic form, look for two products of which one must be a perfect square.

Now let us simplify each of the following surds in small groups or individually.

Example 1

Simplify the following

i. √12

ii. √___ R18

iii. √___ R45

iv. √___ 108

Solution

i. Finding two numbers that we can multiply to get 12, of which one must be a perfect square, the two numbers will be 4 and 3 √12 = √4 × 3 √4 × 3 = √4 × √3 = 2 × √3 = 2 √3

ii. √18 =√9 × 2 √9 × 2 =√9 × √2 = 3 × √2 = 3 √2

iii. √45 = √9 × 5 √9 × 5 = √9 × √5 = 3 × √5 = 3 √5

iv. √108 = √36 × 3 √36 × 3 = √36 × √3 = 6 × √3 = 6 √3

Example 2

A square has an area of 50cm². What is the exact length in a simplified form of one side of the square?

Area of a square is (x²) For one side √50 = √25 × 2 = 5 √2 cm Addition and Subtraction of Surds Surds are added and subtracted in the same way as algebraic expressions. Just as we add or subtract algebraic expressions by grouping like terms, we do the same for surds i.e. we add and subtract surds by grouping like terms. For example, if we have 2√2 + 5 √2, they are alike because they have the same root, we add to get (2 + 5)√2 = 7 √2, likewise, 2√2 − 5 √2, = (2 − 5) √2 = − 3 √2. We can also simplify surds by writing or breaking some surds down into basic form before adding or subtracting.

Now simplify the following expressions

Example 1

√24 + √96 −√600

Solution

= √4 × 6 + √16 × 6 −√100 × 6 [write surd as product of perfect square and another number] = √4 ×√6 + √16 ×√6 −√100 ×√6 = 2√6 + 4√6 − 10 √6 [ simplify perfect squares] = (2 + 4 − 10) √6 [ add surds] = − 4 √6

Example 2

√147 + √75 − √9

Solution

√49 × 3 + √25 × 3 − 3 = √49 ×√3 + √25 ×√3 − 3 = 7 √3 + 5 √3 − 3 = (7 + 5) √3 – 3 = 12 √3 − 3

Example 3

Simplify 4 √2 + 5 √2 − 7 √2

Solution

(4 + 5 − 7) √2 = 2 √2

Example 4

Simplify 10 √2 + 12 √3 − 7 √2 + 13 √3

Solution

10 √2 − 7 √2 + 12 √3 + 13 √3 = 3 √2 + 25 √3

Example 5

Simplify 10 √12 + 12 √32 − 7 √27 + 13 √50

Solution

10 √4 × 3 + 12 √16 × 2 − 7 √9 × 3 + 13 √25 × 2 = 10 √4× √3 + 12 √16 × √2 − 7 √9 × √3 + 13 √25 × √2 = 10 × 2 × √3 + 12 × 4 × √2 − 7 × 3 × √3 + 13 × 5 × √2 = 20 √3 + 48 √2 − 21 √3 + 65 √2 = 20 √3 − 21 √3 + 48 √2 + 65 √2 = −√3 + 113 √2 Multiplication of surds We multiply surds in same way as algebra including the distributive property.

For example, in algebra, 2x × 4y = 2 × 4 × x × y = 8xy. If we exchange x and y for surds, this expression can be written in surd form as 2 √2 × 4 √3 = 2 × 4 × √2 × √3 = 8 √6

Example 1

Simplify the following

i. 5 √3 × 13 √3

ii. 10 √11(3 √2 − 2 √11)

iii. (2 + 3 √3)(4 − 5 √2)

iv. (− 6 + 2 √5)(7 − 5 √5)

v. (3 − 4 √3)²vi. 4 √x(5 √x − 10 √y)

vii. ( 4 + 2 √3)²− (4 − 2 √3)²Solution

i. 5 × 13 √3 × 3 = 5 × 13 × 3 = 195

ii. 10 √11(3 √2 − 2 √11) = 10 √11 × 3 √2 − 10 √11 × 2 √11) = 10 × 3 × √11 × 2 − 10 × 2 √11 × 11 = 30 √22 − 10 × 2 × 11 = 30 √22 − 10 × 2 × 11 = 30 √22 − 220

iii. (2 + 3 √3)(4 − 5 √2) = 2(4 − 5 √2) + 3 √3(4 − 5 √2) = 8 − 10 √2 + 12 √3 − 15 √6

iv. (− 6 + 2 √5)(7 − 5 √5) = − 6(7 − 5 √5) + 2 √5(7 − 5 √5) = − 42 + 30 √5 + 14 √5 − 50 = − 42 − 50 + 30 √5 + 14 √5 = − 92 + 44 √5

v. (3 − 4 √3)²= (3 − 4 √3)(3 − 4 √3) = 3(3 − 4 √3) − 4 √3(3 − 4 √3) = 9 − 12 √3 − 12 √3 + 16 × 3 = 9 + 48 − 12 √3 − 12 √3 = 57 − 24 √3

vi. 4 √x(5 √x − 10 √y) = 4 × 5 × x − 4 × 10 × √x ×√y = 20 √x − 40 √xy

vii. ( 4 + 2 √3)²− (4 − 2 √3)²We can apply the difference between two squares a ²− b²= (a + b)(a − b) to solve this question.

= (4 + 2 √3 + 4 − 2 √3)[(4 + 2 √3 − (4 − 2 √3)] = (4 + 4)(4 − 4 + 2 √3 + 2 √3) = (8)(4 √3) = 32 √3 Alternatively, we can expand the brackets and simplify.

= (4 + 2 √3)(4 + 2 √3) − [(4 − 2 √3)(4 − 2 √3)] = (16 + 8 √3 + 8 √3 + 12) − [(16 − 8 √3 − 8 √3 + 12)] = (28 + 16 √3) − [(28 − 16 √3)] = 28 + 16 √3 − 28 + 16 √3 = 32 √3 Conjugate of surds Conjugate surds are like mirror images in the world of surds. They are formed by simply changing the sign of the radical part in a surd expression. For example, the conjugate of √2 + 3 is √2 − 3. When you multiply a surd by its conjugate, the result is a rational expression due to the difference of squares property, making them helpful for simplifying expressions containing surds.

The table below lists various surds and their conjugates.

Surd Conjugate Example

Surd Conjugate

√a √a √2 √2 a √b √b 2 √3 √3 a + √b a − √b 2 + √3 2 − √3 a − √b a + √b 2 − √3 2 + √3 a + b √c a − b √c 2 + 3 √5 2 − 3 √5 a − b √c a + b √c 2 − 3 √5 2 + 3 √5 √a + √b √a − √b √2 + √3 √2 − √3 √a − √b √a + √b √2 − √3 √2 + √3

Example 1

Write down the conjugate surd of the following

i. √11

ii. 2 + 3 √3

iii. √3 + 4 √2

iv. √6 − √5

Solution

i. √11

ii. 2 − 3 √3

iii. √3 − 4 √2

iv. √6 + √5 It must be noted that the product of a surd (irrational) and its conjugate is a rational number or expression, i.e. the expression is now ‘rationalised’ as it no longer contains a surd.

Rationalising the denominator of surds means making the denominator of a fraction a rational number by multiplying it by its conjugate surd. If the denominator is a single surd, the conjugate is the same surd. If the denominator is a binomial expression with a surd, the conjugate is the same expression with the opposite sign in the middle. The numerator of the fraction is also multiplied by the same conjugate. The answer is then simplified.

Given the expression 2___ √5 , simplifying this expression would make it easier to work with if the denominator was a rational number. Using the idea of equivalent fractions, we can multiply both the numerator and denominator by the conjugate denominator to obtain a ‘rationalised denominator’ fraction.

2___ √5 × √5___ √5 = 2 √5/5 , We observe from our discussion on conjugate surds that the conjugate of √5 is √5 To rationalise the denominator of an expression, we multiply both the numerator and the denominator by the conjugate of the surd.

Example 1

a_ √b = a___ √_ b × √b_ √b = a √b_ b Also, 1/3 − √5 = 1/3− √_ 5 × 3 + √5/3 + √5 = 3 + √__ R5_______________ (3 − √__ R5)(3 + √__ R5) = 3 + √__ R5/9 − 5 = 3 + √__ R5/4

Example 2

Rationalise each of the following surds.

a. 1___ √5 b. 3___ √__ 11 c. √5___ √13 d. ³√2_ √3

Solution

a. 1___ √5 = 1___ √_ 5 × √5_ √5 = √_ 5/5 = ¹_ 5 √5

b. 3___ √__ 11 = 3___ √11 × √__ 11____ √__ 11 = 3 √__ 11/11

c. √5___ √13 = √5_ √13 × √13_ √13 = √65_ 13

d. 3 √_ 2____ √_ 3 = 3 √_ 2____ √_ 3 × √_ 3___ √_ 3 = 3 √_ 6/3 = √_ 6

Example 3

A carpenter was to cut 90m of wood into 2√3 m, how many pieces will she have? [leave your answer in surd form]

Solution

90_ 2 √3 = 90 × √3_ 2 √3 × √3 = 90 × √3_ 2 × 3 = 90 × √3_ 6 = 15 √3

Example 4

Rationalise the following

a. 1 + √_ 3/2 + √_ 3 b. 4 + √2/3 + 2 √3 c. 3 − √5/6 − √3 d. 2 − 3 √3/5 − 5 √6

Solution

a. 1 + √3/2 + √3 = (1 + √3)(2 − √3)_____________ (2 + √3)(2 − √3) [multiplying both numerator and denominator by conjugate surd] = (2 − √3 + 2 √3 − 3)_______________ ( 2)²− ( √3)²[Open the bracket ] = − 1 + √3/4 − 3 [simplify] = − 1 + √3

b. 4 + √2/3 + 2 √3 = (4 + √2)(3 − 2 √3)_____________ (3 + 2 √3)(3 − 2 √3) = 12 + 3 √2− 8 √3 − 2 √6)_______________ ( 3)²− (2 √3)²= 12 + 3 √2 − 8 √3 − 2 √6/9 − 12 = 12 + 3 √2 − 8 √3 − 2 √6_______________ − 3 = 12___ − 3 + 3√__ R2____ − 3 − 8√__ R3____ − 3 − 2√__ R6____ − 3 = − 4 − √2 8 √3_ 3 + 2√__ R6/3

c. 3 − √5/6 − √3 = (3 − √5)(6 + √3)___________ (6 − √3)(6 + √3) = 18 + 3 √3 − 6 √5 − √15_______________ ( 6)²− ( √3)²= 18 + 3 √3 − 6 √5 − √15/36 − 3 = 18 + 3 √3 − 6 √5 − √15/33 = 18_ 33 + 3√_ 3/33 − √15_ 33 = 6_ 11 + √3_ 11 − √15_ 33

d. 2 − 3 √3/5 − 5 √6 = (2 – 3√__ R3)(5 + 5√__ R6)________________ (5 – 5√__ R6)(5 + 5√__ R6) = 10 + 10 √6 − 15 √3 − 15 √18)___________________ ( 5)²− ( 5 √6)²= 10 + 10 √6 − 15 √3 − 15 √9 × 2/25 − 150 = 10 + 10 √6 − 15 √3 − 45 √2__________________ − 125 = 10_ − 125 + 10 √_ 6/125 + 15 √_ 3/125 + 45 √_ 2/125 = − 2/25 + 2 √_ 6/25 + 3 √_ 3/25 + 9 √_ 2/25 Extended Activities Please have a look at finding the square root of a surd if you have time to spare this week. It is more fun maths!

Finding the square root of a surd When two or more expressions involving surds are separated by an equal sign, they are said to be equal. Given two surds a + √b and c + √d, if a + √b= c + √d then a = c , b = d.

Note: ( √a + √b)²= a + b + 2 √ab and ( √a − √b)²= a + b − 2 √ab We use this concept of equality of two surds to help us determine the square root of surds expressions. Let us with the support of our small group members or teacher go through some examples.

Example 1

Simplify √(16 + 2√__ 55)

Solution

Let √______________________ 16 + 2√__ 55 = ±(√_ a + √_ b) Squaring both sides (√(16 + 2√__ 55))²= (√a + √_ b)²16 + 2√__ 55 = a + 2 √(ab) + b 16 + 2√__ 55 = a + b + 2√___ (ab) Comparing corresponding coefficients 16 = a + b (1) 2√55 = 2√___ (ab) 55 = ab (2) From equation (1) a = 16 − b (3) Substitute equation (3) into equation (2) (16 – b) b = 55 16b-b²= 55 b²− 16b + 55 = 0 (b − 5)(b − 11) = 0 b = 5 or b = 11 when b = 5, a = 16-5 = 11 when b = 11, a = 16-11= 5 ∴ √___ (16 + 2√__ 55)= (√5 + √__ 11) It will be necessary to square the results to see if we will get the initial surd.

(√5 + √__ 11)²= (√_ 5)²+ 2 √5 . √11 + (√__ 11)²= 5 + 2 √55 + 11 = 16 + 2√__ 55 It is now verified.

Example 2

Calculate the square root of 18 − 12√_ 2

Solution

Let √__________ (18 − 12√_

2) = ±(√_ a − √_ b) Squaring both sides (√__________ (18 − 12√_ 2))²= (√a − √_ b)²18 − 12 √2 = a − 2(√__ ab) + b 18 − 12 √2 = a + b − 2√___ (ab) Comparing corresponding coefficients 18 = a + b (1) − 12 √2 = − 2√___ (ab) 6√_ 2 = √___ (ab) 72 = ab (2) From equation (1) a = 18 − b (3) Substitute equation (3) into equation (2) (18 − b)b = 72 18b − b²= 72 b²− 18b + 72 = 0 b²− 12b − 6b + 72 = 0 b(b − 12) − 6(b − 12) = 0 (b − 6)(b − 12) = 0 ∴ b = 6 or b = 12 When b = 6, a = 18 – 6 = 12 When b = 12, a = 18 –12 = 6 ∴ √__________ (18 − 12 √2) = √12 − √6 = 2 √3 − √_ 6

Example 3

1. Find the positive square root of 11 − 4 √6

Solution

The square root of 11 − 4 √6 , must be of the form √a − √b , where a ≥ b , as the root must be non-negative.

√11 − 4 √6 = √a − √b 11 − 4 √6 = ( √a − √b)²11 − 4 √6 = (a + b) − 2 √ab 11 − 2 √4 × 6 = (a + b) − 2 √ab By comparing the left-hand side and the right-hand side 11 = (a + b)………………. 1 24 = ab……………………2 From equation 1 b = 11 − a, substituting b = 11 − a into equation 2, we have a(11 − a) = 24 11a − a²= 24 a²− 11a + 24 = 0 (a − 8)(a − 3) = 0 a = 8 or a = 3 Hence, the square root of 11 − 4 √6, is √8 − √3 = 2 √2 − √3 .

Example 4

Find the square root of 23 + 4 √15

Solution

23 + 4 √15, the square root of 23 + 4 √15 must be of the form √a + √b √23 + 4 √15 = √a + √b 23 + 4 √15 = ( √a + √b)²23 + 2 √4 × 15 = a + b + 2 √ab 23 + 2 √60 = (a + b) − 2 √ab Comparing the left-hand side and the right-hand side of the equation 23 = (a + b)………………. 1 60 = ab……………………2 From equation 1 b = 23 − a, substituting b = 23 − a into equation 2 a(23 − a) = 60 23a − a 2 = 60 a²− 23a + 60 = 0 (a − 3)(a − 20) = 0 a = 3 or a = 20 a = 3 and b = 20 or a = 2− and b = 3 Hence, the square root of 23 + 4 √15, is √3 + √20 = √3 + 2 √5 .

We have reached a stage where we must know the difference between the expressions 4a and a⁴.

4a = a + a + a + a a⁴= a × a × a × a For a⁴which is read as a exponent 4 or a to the power of 4, a is the base and 4 is the power, or exponent.

Index is the power or exponent of a number or a variable. The plural for index is indices. For instance, in the expression, 3⁴, 3 is called the base and 4 is called the index, power or exponent and it is read as “three exponent four” or “three to the power of 4”. The concept of indices is widely applied in simplifying mathematical expressions, algebraic manipulation, writing scientific notation, differentiation and integration, geometry and trigonometry, compound interest calculation, etc.

From our previous lessons, we learnt that 2 × 2 × 2 × 2 × 2 × 2 = 2⁶. Likewise a × a × a × a = a⁴. In the example 2 × 2 × 2 × 2 × 2 × 2 = 2⁶, the 2 is called the base, (the number multiplying itself) and the 6 is the index or exponents (the number of times the base is multiplying).

Example 1

Write the following single numbers as exponents in their simplest forms.

i. 16

ii. 27

iii. 100

Solution

i. Least prime factor of 16 = {2} 16 = 2 × 2 × 2 × 2 = 2⁴ii. Least prime factor of 27 = {3} 27 = 3 × 3 × 3 = 3³iii. Prime factors of 100 = {2, 5} 100 = 10 × 10 = 2 × 5 × 2 × 5 = 2²× 5²

Example 2

1. Identify the base and the exponent in following

a. 5⁸b. (2/3) −3

c. xᵐd. 16¹_ 2

Solution

Base Exponent

a. 5 8

b. 2_ 3 -3

c. x m

d. 16 1_ 2

Example 3

1. Write the following base and exponent form.

a. 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4

b. 3/5 × 3/5 × 3/5 × 3/5 × 3/5

c. y²× y²× y²× y²× y²× y²× y²Solution

a. 4⁸b. (³_ 5)⁵c. (y²)⁷Verification of rules of indices Consider 3⁴× 3⁵, if we expand each of them and simplify, we will have 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3. If we write this in index form, the result will be 3⁹. Is there any other way to get the result without the expansion? Yes, we can add the exponent since they are of the same base, thus, 3⁴× 3⁵= 34 + 5 = 3⁹.

This helps us to establish a rule that, when multiplying the powers of the same base, we simply add the indices (power or exponent) aᵐ× aⁿ= aᵐ+ n and by extension once the base (a) is equal aᵐ× aⁿ× aᵖ× …= aᵐ+ n + p + … We can apply our knowledge on this rule to verify the other rules that follow.

ii) aᵐ÷ aⁿ= aᵐ−n

Example:

3⁸÷ 3⁶= 3⁸⁻⁶= 3²iii) (aᵐ)ⁿ= aᵐⁿ, and by extension, (aᵐ× bᵐ× …)ⁿ= aᵐⁿ× bᵐⁿ× … (5²)⁷= 52×7 = 5¹⁴vi) (ᵃ__ b)ᵐ= aᵐ__ bᵐ(3_ 4) 4 = 3⁴_ 4⁴v) (a × b × …________ c × d × …) m = aᵐ× bᵐ× …__________ cᵐ× dᵐ× … (2 × 3 × 4_ 5 × 6 × 7) 3 = 2³× 3³× 4³_ 5³× 6³× 7³vi) (ab)ᵐ= aᵐbᵐ(3 × 4)⁵= 3⁵× 4⁵vii) (aᵐ___ bᵐ) n = aᵐⁿ___ bᵐⁿ(2³_ 3³) 4 = 23×4 _ 33×4 = 2¹²_ 3¹²viii) (a ᵐ× bᵐ× …__________ cᵐ× dᵐ× … ) n = aᵐⁿ× bᵐⁿ× …___________ cᵐⁿ× dᵐⁿ× …

ix) a ¹_ n = ⁿ√a 3¹_ 2 = ²√3 y ¹_ 2 = √y Other rules are:

1. Negative exponent Rule, that is a−m = ¹_ aᵐby extension (ᵃ_ b)−n = (ᵇ_ a)ⁿ2. Zero power rule, that is a⁰= 1 , where a ≠ 0

Example 4

Simplify 8²ˣ_ 3

Solution:

8²ˣ_ 3 = (³√8)²ˣ= 2²ˣ= 4 x Or 2³(²ˣ_ 3 ) = 2²ˣ= (2²)ˣ= 4ˣExample 5 Simplify 27¹_ 3 × 1/81

Solution

(3³)¹_ 3 × 81⁻¹= (3³)¹_ 3 × (3⁴)⁻¹= 3 × 3⁻⁴= 3⁻³Example 6 Simplify the following

a. (64/27) −²_ 3

b. (0.0125)¹_ 4

c. √y × ³√y_ y³d. ¹²¹_ 3 × 6¹_ 3_ 81¹_ 6

e. (t + 1)¹_ 3 × (t + 1)³_ 4_ (t + 1)−¹_ 2

Solution

a. (64/27) −²_ 3 = (27_ 64) 2_ 3 = (3³_ 4³) 2_ 3 = 33ײ_ 3 _ 43ײ_ 3 = 3²_ 4²= 9_ 16

b. (0.0625)¹_ 4 = (625 × 10⁻⁴)¹_ 4 = (5⁴× 10⁻⁴)¹_ 4 = 5⁴× ¹_ 4 × 10⁻⁴×¹_ 4 = 5 × 10⁻¹= 5_ 10 = 1_ 2

c. √y × ³√y/y³= y ¹_ 2 × y ¹_ 3 _ y ³= y ¹_ 2 × y ¹_ 3 × y ⁻³= y ¹_ 2 + ¹_ 3 + (−3) = y −¹³_ 6

d. 12¹_ 3 × 6¹_ 3 ______ 81¹_ 6 = (3 × 4)¹_ 3 × (2 × 3)¹_ 3 ____________ (3⁴)¹_ 6 = 3¹_ 3 × 4¹_ 3 × 2¹_ 3 × 3¹_ 3 ___________ (3⁴)¹_ 6 = 3¹_ 3 × (2²)¹_ 3 × 2¹_ 3 × 3¹_ 3 _____________ (3⁴)¹_ 6 = 3¹_ 3 × 3¹_ 3 ×3−⁴_ 6 × 2²_ 3 × 2¹_ 3 = 3¹_ 3+ ¹_ 3−⁴_ 6 × 2²_ 3 + ¹_ 3 = 3¹_ 3 + ¹_ 3 −⁴_ 6 × 2²_ 3 + ¹_ 3 3⁰× 2¹= 1 × 2 = 2

e. (t + 1)¹_ 3 × (t + 1)³_ 4_ (t + 1)−¹_ 2 = (t + 1)¹_ 3+ ³_ 4 _ (t + 1)−¹_ 2 = (t + 1)¹_ 3+ ³_ 4 × (t + 1)¹_ 2 = (t + 1)¹_ 3+ ³_ 4+ ¹_ 2 = (t + 1)¹_ 3+ ³_ 4+ ¹_ 2 = (t + 1)¹⁹_ 12

As already explained, surds are the root values that cannot be written as whole numbers. More so, indices are the exponents of a value. Thus, given 2⁵, 5 is the index while 2 is the base. Taking a square root is the inverse process of squaring.

Solving indicial problems involving surds

Example 7

If a = 3 − √3, show that a²+ 36/a²= 24

Solution

Substituting a = 3 − √3 into a²+ 36/a²⇒ a²+ 36/a²= (3 − √3)²+ 36_______ (3 − √_ 3)²= 9 − 6 √3 + 3 + 36/9 − 6 √_ 3 + 3 =12 − 6 √3 + 36/12 − 6 √_ 3 =12 − 6 √3 + 6/2 − √_ 3 × 2 + √_ 3/2 + √_ 3 =12 − 6 √3 + 1 12 + 6 √3_ 4 − 3 = 24

Example 8

If m= 2 + √2 and n = 2 − √2 , find the value of ¹_ m²+ n²Solution 1_ ( 2 + √2)²+ (2 − √2)²= 1_____________ (2 + √2)(2 + √2) + (2 − √2)(2 − √2) = 1___________ (4 + 4 √2 + 2) + (4 − 4 √2 + 2) = 1_ (6 + 4 √2) + (2 − 4 √2) = 1(6 − 4 √2)_____________ (6 + 4 √2)(6 − 4 √2) + (2 − 4 √2) = 1(6 − 4 √2)_____________ (6 + 4 √2)(6 − 4 √2) + (2 − 4 √2) = 1(6 − 4 √2)_ (36 − 32) + (2 − 4 √2) = 1(6 − 4 √2)_ 4 + ((2 − 4 √2)_ 1 = 6 − 4 √2 + 4(2 − 4 √2)______________ 4 = 6 − 4 √2 + 8 − 16 √2)______________ 4 = 14 − 20 √2_ 4 = 7 − 10 √2_ 2

Areas under indicial equations to be explored include;

1. Solving simple indicial or exponential equations

2. Solving simultaneous equations involving exponents or indices

3. Application of exponential indicial equations (such as growths and/or decays)

Example 9

Solve the equation 2ˣ= 16

Solution

We know 2⁴= 16 [write both sides with the same base] 2ˣ= 2⁴[equate their exponents] x = 4

Example 10

Find the value of 25ˣ⁻¹= 5ˣ⁺²Solution (5²)ˣ⁻¹= 5ˣ⁺² 5²(x−1) = 5ˣ⁺²2x − 2 = x + 2 2x − x = 2 + 2 x = 4

Example 11

If (5x + 4)³= 8 , find the value of x.

Solution

(5x + 4)³= 2³(as exponents are equal we can equate the bases) 5x + 4 = 2 5x = 2 – 4 5x = –2 x = –2/5

Example 12

If 5ˣ_ 3 = 0.04 , find the value of x 5ˣ_ 3 = 4/100 5ˣ_ 3 = 1/25 5ˣ_ 3 = 1/5²5ˣ_ 3 = 5⁻²x_ 3 = − 2 x = − 6

Example 13

Given that y = 2x and 3ˣ+y = 27 , Find x

Solution

Substitute y = 2x into 3ˣ+y = 27 Giving us 3ˣ⁺²ˣ= 3³, equating exponents, x + 2x = 3 3x = 3 ∴ x = 1

Example 14

Find the values of x and y in the equations 2ˣ−y = 8 and 2³ˣ−y = 128

Solution

2ˣ−y = 2³2³ˣ−y = 2⁷x − y = 3 equation 1 3x − y = 7. equation 2 2x = 4. equation 2 − equation 1 x = 2 2 − y = 3. substitute x = 2 into equation 1 to find y y = − 1

Example 15

Solve for x and y given that 3(x+1) = 27 and 4(y−2) = 16

Solution

3(x+1) = 3³x + 1 = 3 x = 3 − 1 x = 2 4(y−2) = 16 2²(y−2) = 2⁴2(y − 2) = 4 2y − 4 = 4 2y = 8 y = 4 Therefore x = 2, y = 4

Example 16

Solve the simultaneous equations 9²ᵃ+b= 1/729 3ᵃ(9ᵇ) = 27

Solution

9²ᵃ+b = 1/729 3²(2a+b) = 1/3⁶3⁴ᵃ⁺²ᵇ= 3⁻⁶4a + 2b= − 6 ……………..(eqn1) 3ᵃ(9ᵇ) = 27 3ᵃ(3²(b)) = 3³3ᵃ× 3²ᵇ= 3³3ᵃ⁺²ᵇ= 3³a + 2b = 3 ………………….(eqn2) From eqn2, a = 3 − 2b Substituting a = 3 − 2b into (eqn1) ⇒ 4(3 − 2b) + 2b= − 6 12 − 8b + 2b= − 6 − 6b= − 18 dividing both sides by − 6 b = 3 From a = 3 − 2b From a = 3 − 2(3) = 3 − 6 = − 3 Thus, a = − 3, b = 3

Example 17

A bacteria culture doubles every hour. If there are initially 100 bacteria, how many bacteria will be present after 5 hours?

Solution

Since the culture doubles, the growth factor is 2 (each bacteria becomes 2 after an hour).

We know the initial quantity (100) and want to find the final quantity (let it be Q) after 5 hours (represented by exponent t). The general formula for exponential growth is:

Q = A (growth factor)ᵗQ = 100 × (2)⁵Q = 100 * 32 = 3200 Therefore, there will be 3200 bacteria after 5 hours.

Example 18

Suppose that a culture initially contains 1000 bacteria and that this number doubles each hour. Write a general formula for the number of bacteria N present after t hours

Solution

After one hour, there are 1000 × 2 bacteria After two hours, there are 1000 × 2 × 2 = 1000 ×2²bacteria After three hours, there are 1000 × 2²× 2 = 1000 × 2³bacteria and so on.

Following the pattern, if there are bacteria after t hours, then N = 1000 × 2ᵗbacteria

Definition: If N = aˣ, in indices, we observed that a is the base, x is the index or power, and N is the result. For example, 100 = 10²The logarithmic function is defined as logₐ(N) = x, (logₐN is read “logarithm of N to the base a) where N is the number such that N = aˣwith N>0 and a being positive constant other than 1 Relationship Between Indices and Logarithms to Solve Problems To find the logarithm of a number a to the base b, that is, log_(b)(a), we ask the question, ‘What power do I raise b , to obtain a?

Taking a logarithm is the inverse process of taking a power. Generally, if a > 0 and x > 0, then a ^(log)ax = x logₐaˣ= x Extended Activities Let’s go through the following activities to use the relationship between indices and logarithms applying the laws of logarithm

Example 19

Solve the equation: log₃81

Solution

The logarithm (log) to base 3 of 81 ((log₃81) means what is the exponent to which we have to raise 3 to get 81.

81 = 3ˣ3⁴= 3ˣx = 4

Example 20

Evaluate log₁₂₅25

Solution

By writing, log₁₂₅25 = x; we have 125ˣ= 25 which gives 5³ˣ= 5²;

Equating indices, 3x = 2, so x = ²_ 3

Example 21

Solve for y if log₂32 = y + 1

Solution

32 = 2ʸ⁺¹2⁵= 2ʸ⁺¹5 = y + 1 y = 5 − 1 y = 4

Example 22

Evaluate

a) log₂64 b) log₁₀1000

c) log₅125 d) log_(0.1)10

Solution

a) Let log₂64 =x 64 = 2ˣ2⁶= 2ˣ∴ x = 6

b) Let log₁₀1000 = x 1000 = 10ˣ10³= 10ˣ x = 3

c) Let log₅125 = x 125 = 5ˣ5³= 5ˣx = 3

d) Let log_(0.1)10 = x 10 = 0.1ˣ10¹= 10−x x = − 1 The Laws of Logarithm

1. 1ˢᵗLaw of Logarithm

logₐxy = logₐx + logₐy

2. 2ⁿᵈLaw of Logarithm

logₐ(ˣ_ y) = logₐx − logₐy

3. 3ʳᵈLaw of Logarithm

logₐxⁿ= n logₐx NB: the laws are numbered for convenience, and it is not that they should necessarily be in a certain order. Also, log with no given base can be assumed to be log to the base 10.

Example 23

Let us use the laws of logarithm to express in terms of loga, logb and logc each of the following;

i. loga/c ii. log1/b

iii. log a²b³_ 2 iv. log ¹_ 100 b²Solution

i. loga/c = loga − logc

ii. log¹_ b = log1 − logb = − logb

iii. log a²b³_ 2 = log a²+ log b³_ 2 = 2loga + ³_ 2 logb

iv. log ¹_ 100 b²= log1 − log100 b²= 0 − log100 − logb²= − log 10²− 2logb = − 2log10 − 2logb Note, log with no given base can be taken to be log to the base 10, ie log₁₀ = log = − 2 − 2logb Extended Activities

1. In our small groups, or individually, let us also use the laws of logarithm to express each of the following as a single logarithm;

i. log2 + log3

ii. log18 − log9

iii. 3log2 + 2log3 − 2log6

iv. 2 + 3loga

Solution

i. log2 + log3 = log2 × 3 = log6

ii. log18 − log9 = log¹⁸_ 9 = log2

iii. 3log2 + 2log3 − 2log6 = log 2³+ log 3²− log 6²= log8 + log9 − log36 = log8 × 9/36 = log2

iv. 2 + 3loga = 2log10 + 3loga = log 10²+ loga³= log100 + loga³= log100 a³ We are now going to consider exponential equations where the bases are different.

2. Solve the following equations;

i. 2ˣ= 5

ii. 3ˣ= 2

iii. 3⁴ˣ= 4

iv. 2ˣ× 2ˣ⁺¹= 10

Solution

i. 2ˣ= 5 Taking logarithm to base 10 on both sides log₁₀2ˣ= log₁₀5 x log₁₀2 = log₁₀5 x = ^(log105)_ log₁₀2 x = 0.69897/0.30103 using calculator x= 2.3219

ii. 3ˣ= 2 Taking logarithm to base 10 on both sides log₁₀3ˣ= log₁₀2 xlog₁₀3 = log₁₀2 x = ^(log102)_ log₁₀3 x = 0.30103/0.47712 using calculator x = 0.6309

iii. 3⁴ˣ= 4 Taking logarithm to base 10 on both sides log₁₀3⁴ˣ= log₁₀4 4xlog₁₀3 = log₁₀4 x = ^(log104)_ 4log₁₀3 x = 0.60206/1.90849 using calculator x = 0.3157

iv. 2ˣ× 2ˣ⁺¹= 10 2ˣ+x+1 = 10 2²ˣ⁺¹= 10 Taking logarithm to base 10 on both sides log₁₀2²ˣ⁺¹= log₁₀10 (2x + 1)log₁₀2 = 1 2xlog₁₀2 + log₁₀2 = 1 2xlog₁₀2 = 1 − log₁₀2 x = ¹− log₁₀2_ 2 log₁₀2 x = 0.69870/0.60206 using full calculator display x= 1.1610 Change of base of logarithm We can change the base of any logarithm to any base. We cannot directly calculate log₂(7) or log₂(10) without a calculator.

In practice, you would use a calculator with a log function and approximate the answer. To help us calculate logₐ(b) without using calculator, go through the

activity that follow immediately.

Extended Activities

1. With the assistance of your teacher or fellow learner, go through the following work which illustrates how to change base.

Let us consider y = logₐb ⇒ b = aʸ[ converting logarithm to indices] Taking logarithm to base c on both sides log_(c)b = log_(c)aʸlog_(c)b = ylog_(c)a ∴ y = ^(logcb)_ log_(c)a Hence logₐ = ^(logcb)_ log_(c)a

2. Let us use this relationship to change the base of the following to log₁₀;

i. log₄15

ii. log_(2.5)8

iii. log₂25 × log₅8

Solution

i. log₄15 = ^(log1015)_ log₁₀4 = 1.17609/0.60206 using calculator =1.9534

ii. log_(2.5)8 = ^(log108)_ log₁₀2.5 = 0.90309/0.39794 using calculator = 2.2694

iii. log₂25 × log₅8 = ^(log1025)_ log₁₀2 × ^(log108)_ log₁₀5 = ^(log1052)_ log₁₀2 × ^(log1023)_ log₁₀5 = ^(2log105)_ log₁₀2 × ^(3log102)_ log₁₀5 = 2 × 3 = 6

Review Questions 2.1

1. Put the following surds in their simplest form where possible. For those that cannot be further simplified, state the reasons why.

i. √5 + √7

ii. 3 √2 + 5 √2

iii. √7 − √5

iv. 3 √2 − 5 √2

v. 3 √2 × 5 √2

vi. √15 ÷ √5

2. Simplify the following surd expression:

√12 + √27

3. Solve for x:

2 √3x + 5 = 4 √x + 1

4. Given that √a + √b = 7 and √a − √b = 1, find the value of a and b.

5. Rationalise the denominator of the fraction: 1___ √5

6. Simplify and rationalise the expression: 2 + √6_____ √2

7. Rationalise the denominator of the expression: ( √3 + √7)________ ( √7 + 3)

8. Calculate √0.9

9. Which of the following statements is/are true?

i. 2√3 >3 √2,

ii. 4√2 >2 √8

10. What is the conjugate of

i. 1 + √3

ii. 8 √5 + 6

11. What is the square root of (10 + √25)(12 − √49)?

12. If (3 + 2 √5)²= 29 + k √5, find the value of k

13. Find the value of y, if √64 − 3 √64 = − 4 √y, where y > 0,

14. Simplify √48 + 2 √27_________ √12 , given your answer as an integer

15. Express 1 − 5 √5/3+ √5 in the form m – n√5 ; m, n∈Z

16. Show that √75 + √27________ √3 is an integer and find its value

17. Show that x − 25_____ √x + 5 = √x − 5

18. Rationalise the denominator of 8/1 + 2 √3

19. Simplify 8/1 + 2 √3 × 8/1− 2 √_ 3

Review Questions 2.2

1. Simplify and write the answer with positive indices: (x³)⁴____ (x ⁵)²2. Simplify and write the answer with positive indices: (a²b⁴)_____ (a ⁵b³)

3. Simplify 2²× 4−4 ÷ 16−3

4. Find the value of x in the equation 2³× 3⁴× 72= 6ˣ5. If 2ⁿ= 32 find the value of n

6. Solve 3³−x = 27ˣ– 1.

7. Show that

i. 32−²_ 5 = ¹_ 4

ii. (2x−²_ 5 )⁵= 32/x²8. Find the value of x given 6 25^(0.17)× 625^(0.08)= 25ˣ× 25−³_ 2

9. If (3/5) x = ( 81/625), then what is the value of xˣ10. Given that (7/5) 4x × (7/5) 3x−1 = (7/5) 8 , find the value of x that satisfies this equation.

11. Find the value of a if 5³ᵃ−1 ×125 = 2 5²ᵃ−1

12. Given that y = 5x and 3ˣ+y = 81 , Find x

13. Solve the simultaneous equations 9²ᵃ+b = 2187 and 3ᵃ× 9ᵇ= 3

14. Simplify 8²_ 3

15. Simplify log_(b)x²+ log_(b)x³− log_(b)x⁴16. Calculate log₇8 to four decimal places

17. If √5ˣ= 25 , find the value of x

18. It is given that x = √3 and y= √12.

Find in the simplest form, the value of

i. xy

ii. y_ x

iii. (x + y)²19. Given log₇2= α, log₇3 = β and log₇5 = γ, express in terms of α, β , and γ;

i. log₇6

ii. log₇ 15/2

A matrix is a rectangular arrangement of numbers, symbols, or expressions enclosed within brackets ( ) or [ ]. Each element within the matrix has a specific location identified by its row (horizontal position) and column (vertical position).

We describe a matrix by its dimensions, specifying the number of rows and columns. For example, a matrix with 2 rows and 3 columns is called a (2 × 3) matrix. When a matrix has the same number of rows and columns for example (3 × 3), it is called a square matrix. A square matrix with non-zero entries only on its main diagonal (from top left to bottom right) is called a diagonal matrix. A special square matrix with ones (1s) on its main diagonal and zeros (0s) elsewhere is called an identity matrix. It plays a crucial role in solving matrix equations. Matrices have extensive applications in various fields: They are used to organise and analyse large datasets in statistics, health, economics, and social sciences. In computer graphics, matrices are essential for representing 3D objects, transformations, and lighting effects in computer graphics. Matrices are also used to analyse electrical circuits and solve complex problems related to currents and voltages. They are applied in physics to represent physical systems like forces and motion, simplifying calculations and analysis.

At the end of this section, you will be able to:

· Recognise a matrix including types of matrices and state its order · Find the determinant of a (2 × 2) matrix · Add and subtract matrices (2 × 2) matrix · Multiply a matrix by a scalar and a matrix by a matrix (2 × 2) matrices Key Idea:

A matrix (plural matrices) is a rectangular array of numbers, symbols, or expressions arranged in rows and columns.

A matrix (plural: matrices) is a rectangular array of numbers, symbols, or expressions, organised in rows and columns. Each entry in a matrix is called an element or an entry, and it is identified by its row and column indices.

4 columns 2 rows → → [ ↓ ↓ ↓ ↓ 2 7 3 6 − 2 1 3 5] Everyday situations that exemplify the concept of matrices include

a. Classroom seating arrangement

b. Provision items in a shop

c. A pack of bottled water, among others.

Example:

Suppose that we wish to express the information of possession of pens and pencils by Afiba and his two friends Enyonam and Nana, which is as follows:

Afiba has 2 pens and 7 pencils, Enyonam has 1 pen and 5 pencils, Nana has 3 pens and 2 pencils.

Now, this could be arranged in tabular form as the individual items in the matrix are called the elements or entries.

Describing a Matrix

Individuals Pens Pencils

Afiba 2 7 Enyonam 1 5

Nana 3 2 Which could be expressed in matrix form as;

A = ( 2 7 1 5 3 2) or A = [ 2 7 1 5 3 2] The horizontal arrays are called rows, and vertical arrays are called columns ROWSCOLUMNS d b c a Describing a Matrix (Order or dimension of a matrix) A matrix is described by stating the dimensions. For example, [a b] is a (1 × 2) (read one–by–two)n matrix, [a/b] is a (2 × 1) (read two–by–one), [ a b c d] is a (2 × 2) (read two–by–two) and [a b c x y z] is (2 × 3) (read two–by–three) all because of their respective number of rows and columns.

An (m × n ) matrix has m rows (horizontal) and n columns (vertical). Each element of a matrix is denoted by a variable with subscripts. For example, a₂₃,represents the element in the 2ⁿᵈ(second) row and 3ʳᵈ(third) column of the matrix.

1 2 … n 1 a₁₁ a₁₂ ⋯ a₁ₙ 2 a₂₁ a₂₂ ⋯ a₂ₙ For example, the matrix A= 3 a₃₁ a₃₂ ⋯ a₃ₙ ⋮ ⋮ ⋮ ⋮ ⋮ m aₘ₁ aₘ₂ ⋯ aₘₙ is an (m × n ) dimensional matrix having m number of rows and n number of columns.

The order of the matrix M = (− 3 4 − 1 5 2 0 ) is 2 × 3

Note: Commas are not used in matrices. Gaps are left between the columns.

Types of Matrices

Square matrix: A matrix that has the same number of rows and the same number of columns, it is called a square matrix.

For example:

[a b c d] [ p q r s t u v w x] 2 × 2 matrix 3 × 3 matrix Rectangular matrix: A matrix in which the number of rows is not equal to the number of columns.

For example, [1 2 3 0 1 5] (2× 3) or [ a b c d e f ] (3× 2) The Zero Matrix A zero matrix is a matrix whose entries are all zeros or are equivalent to zero.

A = [0 0 0 0], B = [0 0 0 0 0 0] (0 0), ( 0 0

0) and ( 0 − i + i 0 0 0 0 0 0 x − x) are all zero matrices

Example:

Which of the following matrices is/are zero matrix/matrices?

a. A = (x − x 0 0 0 − b + b 0)

b. B = ( 1 − 1 0 0 0 0 )

c. C = (m − m 0 0 0)

Solution:

a. Simplifying the entries in the matrix A gives (0 0 0 0 0 0), thus A is a zero matrix

b. Matrix B has two (2) non-zero entries, thus B is not a zero matrix

c. Matrix C can be simplified to (0 0 0 0) making it a zero matrix The Unit or Identity Matrix A unit matrix or identity matrix is a square matrix having every non-main- diagonal element equal to zero and every main diagonal element equal to one.

For example, [1 0 0 1] [ 1 0 0 0 1 0 0 0 1 ] Equality of Matrix Two matrices are said to be equal if their corresponding elements or entries are the same (equivalent).

For example, if [a b c d] = [ e f g h], then a = e, b = f, c = g, d = h .

Also, if [a b c d ] = [2 3 4 5 ] , then a = 2, b = 3, c = 4, d = 5 For example, M = ( 3 5 − 2 ³_ 4 4 ¹_ 2 0 −²_

5) and N= ( 3 5 − 2 ³_ 4 4 ¹_ 2 0 −²_

5) are two equal matrices since they have the same dimension and their corresponding elements are the same.

Example

If P = ( 3_ 2 x y − x − 3) and Q = ( x 1.5 3.5 − 3) are two equal matrices, find the values of x and y .

Solution

Since P = Q , 3/2 = x and y − x = 3.5 y − 3/2 = 3.5 y = 5 DETERMINANTS OF (2 × 2) MATRICES Imagine a woven basket from a village, but instead of holding your favourite fruits, it holds numbers arranged in a neat grid, like two rows of cowrie shells. This grid is what mathematicians call a (2 × 2) matrix. In Ghana, we have a word for unlocking secrets – “Odomankoma” (key). Determinants act as the Odomankoma for these matrices, revealing a unique property based on how the numbers are arranged.

Here’s a breakdown for (2 × 2) matrices, like our basket of numbers:

Basic Structure: A (2 × 2) matrix looks like this:

(a b c d) (where a represents the element in the first row, first column; b represents the element in the first row, second column, and so on) Determinant Formula: There’s a special formula to calculate the (determinant, det ) of a (2 × 2) matrix:

det = (a × d) − (b × c) where a, b, c and d represent the elements (oduas or twigs) of the matrix as shown above.

Visualizing the Determinant: Imagine drawing a diagonal line across the basket, starting from the top left corner (a) and reaching the bottom right corner (d). Now, draw another diagonal line starting from the top right corner (b) and reaching the bottom left corner (c). The determinant captures the difference between the product of the elements along one diagonal (a × d), the ′m ain diagona l′, and the product of elements along the other diagonal (b × c).

Example:

Consider this (2 × 2) matrix:

A = (2 3 1 4) Using the formula, the determinant (det) would be:

det(A) = (2 × 4)− (3 × 1) = 8 − 3 = 5 Importance:

The determinant of a (2 x 2) matrix has various applications, including:

a. Solving systems of linear equations: Determinants play a crucial role in finding solutions to systems of linear equations with two variables.

b. Invertibility of matrices: A non-zero determinant indicates that the matrix is invertible, meaning it has an inverse matrix.

c. Area calculation: In specific contexts, the determinant can be used to calculate the area enclosed by a parallelogram defined by the matrix’s row vectors. Imagine a farmer needs to calculate the area of a rectangular plot of land represented by a (2 × 2) matrix, where each element represents the length and width of the plot in meters. The determinant, in this case, can be used to calculate the area (note, this application has limitations for general area calculation)

Example 1

Evaluate the determinants of the following matrices

a. [21 4 17 9]

b. [ 2 − 8 − 3 6 ]

c. [a + 3 7 − a a 7 ]

Solution

Given a 2 × 2 matrix A = [a b c d], determinant of A, det(A) = ad − bc

a. det|²¹⁴17 9| = (21 × 9) − (4 × 17) = 121

b. det| ²− 8 − 3 6 | = ((2 × 6) − (− 8 × − 3)) = − 12

c. det|ᵃ+ 3 7 − a a 7 | = (7(a + 3) − a(7 − a)) = 7a + 21 − 7a + a ²= a ²+ 21

Example 2

If the matrix A = (2 1 3 4) and B = (3 − 2 6 5 ), find the determinants of A and B.

Solution

A = det|²¹3 4| = (2 × 4) − (3 × 1) = 8 − 3 = 5 B = det|³− 2 6 5 | = (3 × 5) − (6 × − 2) = 15 + 12 = 27

Example 3

If A = ( 4 − 2 3x 5 ),find the value of x if the determinant of A = 32

Solution

detA = (4)(5) − (3x)(− 2) = 20 + 6x detA = 20 + 6x, but detA = 32 ∴ 32 = 20 + 6x 12 = 6x x = 2

Example 4

Given that B = (2 4 6 12), find the determinant of B

Solution

detB = det|²⁴6 12| = 12(2) − 6(4) = 24 − 24 = 0

Example 5

Evaluate the determinant of:

(a) A = (− 4y 3 6 5), if y = − 2

(b) B = ( − 2 − 2 2r + 3 − 5), if r = 4

Solution

detA = det|− 4y 3 6 5| = 5(− 4y) − (6)(3) = − 20y − 18 but y = 2 ∴ detA = − 20(2) − 18 = − 58 detB = det| − 2 − 2 2r + 3 − 5| = − 2(− 5) − (2r + 3)(− 2 ) = 10 − (− 4r − 6) = 10 + 4r + 6 but r = 4 ∴ detB = 10 + 4(4) + 6 = 10 + 16 + 6 = 32

Example 6

Solve the following equations

a. | ˣ+ 1 2 2x − 1 3| = 4 b. |⁵ˣ+ 2 6x − 3 4 3 | = 0

c. |³ˣ− 2 2 1 2x + 1| = 1 d. |²ˣ− 1 3x + 1 x − 1 x + 1 | = 2

Solution

a. | ˣ+ 1 2 2x − 1 3| = 4 ⟹ 3(x + 1) − 2(2x − 1) = 4 3x + 3 − 4x + 2 = 4 5 − x = 4 x = 1

b. |⁵ˣ+ 2 6x − 3 4 3 | = 0 ⟹ 3(5x + 2) − 4(6x − 3) = 0 15x + 6 − 24x + 12 = 0 18 − 9x = 0 x = 2

c. |³ˣ− 2 2 1 2x + 1| = 1 ⟹ (3x − 2)(2x + 1) − 2 = 1 ⟹ 6x²+ 3x − 4x − 2 − 2 = 1 6x²− x − 4 − 1 = 0 6x²− x − 5 = 0 (x − 1)(6x + 5) = 1 x = 1 or x = − 5/6

d. |²ˣ− 1 3x + 1 x − 1 x + 1 | = 2 ⟹ (2x − 1)(x + 1) − (3x + 1)(x − 1) = 2 2x²+ x − 1 − (3 x ²− 2x − 1) = 2 2x²− 3 x²+ x + 2x − 1 + 1 = 2 − x²+ 3x = 2 x²− 3x + 2 = 0 (x − 1)(x − 2) = 0 x = 1 or x = 2

Addition of Matrices

Two matrices can be added only if they have the same size. To add two matrices, add the elements in the corresponding positions in each matrix.

For example, given (2 × 2 matrices):

[a b c d] + [ e f g h] = [ a + e b + f c + g d + h] resulting in another (2× 2) matrix.

Given that A = ( a₁₁ a₁₂ a₂₁ a₂₂) and B = ( b₁₁ b₁₂ b₂₁ b₂₂), A + B = ( a₁₁ a₁₂ a₂₁ a₂₂) + ( b₁₁ b₁₂ b₂₁ b₂₂) = ( a₁₁ + b₁₁ a₁₂ + b₁₂ a₂₁ + b₂₁ a₂₂ + b₂₂)

Example 1

If A = (− 5 3 2 − 1), B = (4 − 3 7 − 5) and C = (− 1 3 5 4), find

a) A + B

b) B + A

c) A + (B + C)

d) (A + B) + C

e) What is the relationship between your answers in a) and b)

f) What is the relationship between your answers in c) and d)

Solution:

a) A + B = (− 5 3 2 − 1) + (4 − 3 7 − 5) = (− 5 + 4 3 − 3 2 + 7 − 1 − 5) = (− 1 0 9 − 6)

b) B + A = (4 − 3 7 − 5) + (− 5 3 2 − 1) = (4 − 5 − 3 + 3 7 + 2 − 5 − 1) = (− 1 0 9 − 6)

c) B + C = (4 − 3 7 − 5) + (− 1 3 5 4) = (4 − 1 − 3 + 3 7 + 5 − 5 + 4) = ( 3 0 12 − 1) A + (B + C) = (− 5 3 2 − 1) + ( 3 0 12 − 1) = (− 5 + 3 3 + 0 2 + 12 − 1 − 1) = (− 2 3 14 − 2)

d) (A + B) + C = (− 1 0 9 − 6) + (− 1 3 5 4) = (− 1 − 1 0 + 3 9 + 5 − 6 + 4) = (− 2 3 14 − 2)

e) A + B = B + A therefore, matrix addition is commutative

f) A + (B + C) = (A + B) + C therefore, matrix addition is associative

Example 2

A shopkeeper has two separate shops which she opens on Mondays and Fridays.

In a particular week, she made the following sales in the two shops. Represent the total sales she made in the week in matrix form.

Shop A Coke Sprite

Monday 9 8 Friday 6 7

Shop B Coke Sprite

Monday 5 4 Friday 2 3

Solution

The sales in each shop can be put in (2 × 2) matrix where the rows are indexed as days of the week and the columns are indexed as types of drinks.

A = ( 9 8 6 7 ), B = ( 5 4 2 3 ) The total sales of the two shops is given by the sum of the matrices T = A + B T = ( 9 8 6 7 )+ ( 5 4 2 3 ) = ( 14 12 8 10)

Example 3

The matrices Q and R are given by Q = ( 2 6 4 8 ) and R = ( 3 − 5 − 7 9 ).

Find Q + R

Solution

Q + R = ( 2 6 4 8 )+ ( 3 − 5 − 7 9 ) = ( 2 + 3 6 + ( − 5) 4 + ( − 7) 8 + 9 ) = ( 5 1 − 3 17 ) Take care with the signs when adding the numbers.

Example 4

The matrices A and B are given by; A = ( 3 − 1 − 8 6 ) and B = ( − 2 5 0 − 4 ).

Find A + B

Solution

A + B = ( 3 − 1 − 8 6 )+ (− 2 5 0 − 4 ) = ( 3 + ( − 2) − 1 + 5 − 8 + 0 6 + ( − 4)) = ( 1 4 − 8 2) Subtraction of Matrices Two matrices can be subtracted only if they have the same size. To subtract two matrices, subtract the elements in the corresponding positions in each matrix. For

example, given:

A − B = ( a₁₁ a₁₂ a₂₁ a₂₂) − ( b₁₁ b₁₂ b₂₁ b₂₂) = ( a₁₁ − b₁₁ a₁₂ − b₁₂ a₂₁ − b₂₁ a₂₂ − b₂₂) [a b c d]− [ e f g h] = [ a − e b − f c − g d − h] resulting in another (2 × 2) matrix

Example 1

If A = (− 5 3 2 − 1), B = (4 − 3 7 − 5) and C = (− 1 3 5 4), Evaluate the following:

a) A − B

b) B − A

c) B − C

d) A − (B − C)

e) (A − B) − C

f) What is the relationship between your answers in a) and b)

g) What is the relationship between your answers in c) and d)

Solution

a) A − B = (− 5 3 2 − 1) − (4 − 3 7 − 5) = ( − 5 − 4 3 − ( − 3) 2 − 7 − 1 − ( − 5)) = (− 9 6 − 5 4)

b) B − A = (4 − 3 7 − 5) + (− 5 3 2 − 1) = ( 4 − ( − 5) − 3 − 3 7 − 2 − 5 − ( − 1)) = (9 − 6 5 − 4)

c) B − C = (4 − 3 7 − 5) − (− 1 3 5 4) = (4 − ( − 1) − 3 − 3 7 − 5 − 5 − 4) = (5 − 6 2 − 9)

d) A − (B − C) = (− 5 3 2 − 1) − (5 − 6 2 − 9) = ( − 5 − 5 3 − ( − 6) 2 − 2 − 1 − ( − 9)) = (− 10 9 0 8)

e) (A − B) − C = (− 9 6 − 5 4) − (− 1 3 5 4) = (− 9 − ( − 1) 6 − 3 − 5 − 5 4 − 4) = ( − 8 3 − 10 0)

f) A − B ≠ B − A therefore, matrix subtraction is not commutative

g) A − (B − C) ≠ (A − B) − C therefore, matrix subtraction is not associative

Example 2

A car dealer sells two types of cars Toyota (T) and Opel (O), and two models for each brand Prius (P) and Corsa (C) the inventory of the cars is shown below.

P C T 4 5 O 2 6 If the dealer makes sales the following month as given by the table below, find the new inventory.

H C T 3 3 O 2 1

Solution

The inventory can be put in a (2× 2) matrix where the row of indexed as brands and the columns are indexed as models.

R = ( 4 5 2 6 ), the sales can also be put in a (2 × 2) matrix as S = (3 3 2 1) The new inventory will be R− S = ( 4 5 2 6 )− (3 3 2 1) = (4 − 3 5 − 3 2 − 2 6 − 1) = (1 2 0 5)

Example 3

The matrices A, B and C are given by:

A = (− 2 4 3 − 1), B = (5 3 0 1) and C = (− 1 − 2 − 3 4 ).

Evaluate

(i) ( A − B )

(ii) ( B – C )

Solution

(i) A – B = (− 2 4 3 − 1) − (5 3 0 1) = (− 2 − 5 4 − 3 3 − 0 − 1 − 1) = (− 7 1 3 − 2)

(ii) B – C = (5 3 0 1) − (− 1 − 2 − 3 4 ) = (5 + 1 3 + 2 0 + 3 1 − 4) = (6 5 3 − 3) Multiplication of Matrices Scalar multiplication of matrices All the entries of a matrix can be multiplied by a common factor called a scale factor, k, in a process called scalar multiplication to obtain a scalar product. That is, if A = (a b c d), then kA = k (a b c d) = (ka kb kc kd).

If k = − 1 , then –A = (− a − b − c − d) is called the negative of A

Example 1

Given that A = ( 5 3

2) and the matrix P = − 2A , write out the matrix P

Solution

P = − 2A = − 2( 5 3

2) = ( − 10 − 6 − 4 )

Example 2

Given that A = ( 6 3 18 9), then A = 3 (2 1 6 3) where k is a scale factor (k = 3).

Example 3

If P = (4 − 1 2 3 ) and Q = ( 12 13 − 3 − 9) , find:

i) − 5P

ii) 1/3 Q

Solution

i) P = (4 − 1 2 3 ) ∴ − 5p = − 5(4 − 1 2 3 ) = (− 5 × 4 − 5 × − 1 − 5 × 2 − 5 × 3 ) = (− 20 5 − 10 − 15)

ii) Q = ( 12 13 − 3 − 9) ∴ 1/3 Q = 1/3( 12 13 − 3 − 9) = ( 12/3 13/3 − 3/3 − 9/3 ) = ( 4 4.3 − 1 − 3)

Example 4

If A = (− 3 0 7 − 4), B = ( 2 − 1 − 7 4 ) and C = ( 1 0 − 2 − 4), find 2A – 3B + 4C.

Solution

2A = 2(− 3 0 7 − 4) = (− 6 0 14 − 8), 3B = 3( 2 − 1 − 7 4 ) = ( 6 − 3 − 21 12 ) 4C = 4( 1 0 − 2 − 4) = ( 4 0 − 8 − 16) ∴ 2A – 3B + 4C = (− 6 0 14 − 8) − ( 6 − 3 − 21 12 ) +( 4 0 − 8 − 16) = (− 8 3 27 − 36) Multiplication of matrices Matrix multiplication, unlike multiplying individual numbers, involves a specific process for combining elements from two matrices to create a new matrix. The two matrices must have compatible dimensions for multiplication. The number of columns in the first matrix Aₘ×n must equal the number of rows in the second matrix Bₙ× q. The resulting product matrix will have dimensions m × q . We don’t directly multiply corresponding elements between the matrices. Instead, to find an element at any row (i) and column (j) of the resulting product matrix, we take the dot product of the row i from the first matrix (A) with column j from the second matrix (B). The dot product involves multiplying corresponding elements between the row and column vectors and summing those products. We repeat this process for each element in the resulting product matrix, considering all possible row-column combinations.

For example, given a matrix, M = (a b c d) and N = ( e f g h), = (a b c d)(e g f

h) Multiplying N by the 1st row of M Which gives the entries for first-row elements of the product as (ae + bg af + bh) = (a b c d)(e g f

h) Multiplying N by the 2nd row of M Giving the entries for first-row elements of the product MN as (ce + dg cf + dh) The result will be MN = ( ae + bg af + bh ce + dg cf + dh) In summary, ⎡ ⎢ ⎣ (a b)( e f g h) (c d)(e f g g) ⎤ ⎥ ⎦ multiply the first row by the second matrix and the second row by the second matrix ⎛ ⎜ ⎝ (a b)( e

g) (a b)( f h) (c d)( e

g) (c d)( f h) ⎞ ⎟ ⎠ Multiply:

· First row in first matrix by first column in second matrix, · Second row in first matrix by the first column in the second matrix · First row in first matrix by first second column in second matrix · Second row in first matrix by the second column in the second matrix You will obtain ( a × e + b × g a × f + b × h c × e + d × g c × f + d × h) ( ae + bg af + bh ce + dg cf + dh)

NOTE: Matrix multiplication is not commutative (AB ≠ BA) in general.

The order of the matrices matters when multiplying them.

Example 1

Given that A = (4 − 2 1 3 ) and B = (− 2 3 − 2 − 7) Evaluate the following:

a) AB

b) BA

c) what is the relationship between your answers in a) and b)

Solution

a) AB = (4 − 2 1 3 )(− 2 3 − 2 − 7) = (4(− 2) + (− 2)(− 2) 4(3) + (− 2)(− 7) 1(− 2) + 3(− 2) 1(3) + 3(− 7) ) = (− 4 26 − 8 − 18)

b) BA = (− 2 3 − 2 − 7)(4 − 2 1 3 ) = ( − 2(4) + 3(1) (− 2)(− 2) + 3(3) (− 2)(4) + (− 7)(1) (− 2)(− 2) + (− 7)(3)) = ( − 5 13 − 15 − 17)

c) AB ≠ BA therefore, matrix multiplication is not commutative

Example 2

If A = (1 2 3 4) and B = (− 1 2 − 3 1), find p and q if AB = ( p − 2 − 6 4 ) + 3( 4 2 − 3 q)

Solution

AB = (1 2 3 4)(− 1 2 − 3 1) = ( − 7 4 − 15 10) AB = ( p − 2 − 6 4 ) + 3( 4 2 − 3 q) = ( − 7 4 − 15 10) ( p − 2 − 6 4 ) + ( 12 6 − 9 3q) = ( − 7 4 − 15 10) ( p + 12 4 − 15 4 + 3q) = ( − 7 4 − 15 10) Equating corresponding entries, it follows that − 7 = p + 12 ∴ p = − 19 10 = 4 + 3q ⟹ 3q = 6 ∴ q = 2

Example 4

Find AV where A = (2 3 1 2) and V = (5 4)

Solution

AV = (2 3 1 2)(5

4) = (2 × 5 + 3 × 4 1 × 5 + 2 × 4) = (10 + 12 5 + 8 ) = (22 13)

Example 5

If P = (2 3) and A = (4 6 5 2), find PA

Solution

PA = (2 3)(4 6 5 2) = ( 2 × 4 + 3 × 5 2 × 6 + 3 × 2) = (23 18)

Example 6

Given the matrices A = ( 5 9 − 2 4 ) and B = (2 3 4 1 ), evaluate

(i) AB

(ii) BA

(iii) A(AB)

Solution

(i) AB = ( 5 9 − 2 4 )(2 3 4 1 ) = ( 5(2) + 9(4) 5(3) + 9(1) − 2(2) + 4(4) − 2(3) + 4(1)) = ( 10 + 36 15 + 9 − 4 + 16 − 6 + 4 ) AB = (46 24 12 − 2 )

(ii) BA = (2 3 4 1 )( 5 9 − 2 4 ) = ( 2(5) + 3( − 2) 2(9) + 3(4) 4(5) + 1( − 2) 4(9) + 1(4)) = (10 − 6 18 + 12 20 − 2 36 + 4 ) BA = (− 4 30 18 40 )

(iii) A(BA) We obtained BA = (− 4 30 18 40 ) A(BA) = ( 5 9 − 2 4)(− 4 30 18 40 ) = ( 5( − 4) + 9(18) 5(30) + 9(40) − 2( − 4) + 4(18) − 2(30) + 4(40)) = (− 20 + 162 150 + 360 8 + 72 − 60 + 160 ) A(BA) = (142 510 80 100 )

Activity 4.1

Use the appropriate steps to solve the following question in pairs or individually and cross-check your answer with the ones provided.

Given that M = ( y + 1 7 1/3 + x q) and N = (− 2 m + 11 − 2 − 8 − q) and MN are equal vectors.

(i) Find the values of x, y, m and q

(ii) Hence find the determinant of M²Expected Answers

(i) y = − 3 , m = − 4, x = −⁷_ 3, q = − 4

(ii) the determinant of M²= 484

Review Questions 4.1

1. The matrices A = ( 3 0 0 4 ) and B = ( a b 0 c ) are such that AB = A + B.

Find the values of a, b and c

2. Given that the following matrices are equal, find the values of x, y and z :

( x + 3 − 1 4 5 ) = ( 6 y z − 3 5 )

3. If ( a − b 3 2 a + b) = ( 2 3 2 6 ) , find the values of a and b

4. Find a (2 × 2) matrix A and B such that A + 2B = ( 1 − 2 0 1 ) and 2A + 3B = ( 2 1 − 1 0).

5. If A = ( 1 2 3 4 ) and B = ( − 1 2 − 3 1 ) , find p and q, if AB = ( p − 2 − 6 4 ) + 3 ( 4 2 − 3 q ).

6. If T = ( 4 − 2 3x 5 ), find the value of x, if the determinant of T = 32

7. A = ( 1 2 3 4 ) and B = ( 5 1 2 0 )

(i) Find the sum (A + B ).

(ii) Find the difference (A − B ).

8. Given that A = [ 2m + 1 − 1 mn + 4 8 − m + 7 − 4 ] and B = [ − 5 5 − 3n − 2 n − 2m 10 − n + 2/3 m] and A = B, find the values of m and n

9. Evaluate the determinant ( − 6 2 5p + 5 − 4 ) , if p = 2

10. Evaluate the determinant ( − 4 − 3 − 4 2n − 4 ) , if n = − 4

11. For the matrices A = ( 2 3 1 4 ) , B = ( 3 0 2 − 1 ) and C = ( 1 1 − 2 5 ), verify that:

(i) (AB)C = A(BC)

(ii) A(B + C) = AB + AC

(iii) (A + B)C = AC + BC

12. Find the ( 2 × 2 ) matrices A and B such that 2A + B = ( 3 1 − 8 6 ) , 3A + 2B = ( 5 0 − 13 10).

13. Consider two matrices:

C = (2 − 1 1 3 ) and D = (4

5) Multiply matrix C by Matrix D