math 100 doc

THE KENYA METHODIST UNIVERSITY

DEPARTMENT OF PURE AND APPLIED SCIENCES

MATH 100: MATHEMATICS

COURSE INSTRUCTOR

ALICE LUNANI MULAMA (MSC. APPLIED MATHS)

DEPARTMENT OF PURE AND APPLIED SCIENCES

P.O BOX 267

MERU

Introduction

The course is designed to provide under graduate students with the basic mathematical

language, skills and logical reasoning. It will cover topic on sets of real numbers,

functions, series, matrices and systems of Equations and introduce them to the new

concept of complex numbers. Concepts that they need to apply in daily life.

OBJECTIVES

During the course, the students will be expected to:

1. Explain the meaning of real numbers

2. Carry out operation with real numbers

3. Solve problems using mathematical logics.

4. Solve linear Equations.

5. Solve Quadratic Equations

6. Solve inequalities.

7. Solve system of equations using matrices.

8. Apply the mathematical knowledge to solve problems in their specific areas and

in real life.

BROAD AREAS TO BE COVERED

1.Elementary algebra

Sets of real numbers, operations with real numbers, intervals and absolute values.

2. Equations and inequalities

Quadratic equations, inequalities, applied problems.

3. Relations, functions and transformations graphics of equations, inverse, graphs of

functions including Transcendental functions, transformations.

4. Exponential and logarithms

Logarithms and laws of logarithms common logarithms and their characteristics

logarithmic equations.

5. Combinatorial algebra

Permutations and combinations.

6. Sequences, series and mathematical logic.

Sequences and series, arithmetic sequence and series, geometric sequence and

services.

7. Matrices and systems of equations

The algebra of matrices, solving linear equations simultaneously by substitution,

elimination, matrix inverse method and Row reduction method.

TEACHING METHODS

Lecture method

Stating and illustrating definition, stating and proving theorems and problems

solve.

Question and Answer

Tutorials take away assignments and solutions discussed.

STUDENTS ASSESSMENT

The lecturer will provide essential facilitation while students are expected to take

responsibility for the learning process for the course objectives to be achieved.

The student performance will be assessed on the following.

Assignment

Course examinations

Individual student’s marks of these assessments will contribute to the final examination

mark and grade for the course.

COURSE OUTLINE FOR THE TRIMESTER

Week 1 topics

Sets of real numbers

Operations with real numbers

Intervals and absolute values

Week 2 topics

Solving Quadratic equation

Solving inequalities

Formulas and applied problems

Week 3 topics

Partial fractions

Graphs of equations

Inverses

Graphs of functions

Week 4 topics

Transcendental functions

Transformations

Week 5 topic

Common logarithms and their characteristics

Week 6 topic

Laws of logarithms

Logarithmic equations

Week 7 topics

Permutations

Combinations

Week 8 topics

Seque4nces

Series

Week 9 topics

Arithmetic

Week 10 topics

Geometric series

Week 11 topics

The algebra of matrices

Week 12 topics

Solving simultaneous equations by substitution, elimination and matrix inverse methods.

Week 13 topics

Matrix reductions

Solving simultaneous equations by matrix reduction method.

Week 14 and Week 15

Assignments

Course examination

LEARNING RESOURCES/REFERENCES

Swokowski, E.W.. Precalculus: “Functions and Graphs”; PWS publishers, 1987.

Larson, R.E. and Hostetlerk, R.P. A graphic Approach; D.C Health and Company,

1993.

Margaret, L.L., Hornsby, J. and miller, C.D. Introductory Algebra, Harper Collins

College Publishers, 1995.

SPECIFIC WEEKLY OBJECTIVES

Week 1

CHAPTER ONE

1.0 INTRODUCTION

1.1 CHAPTER OBJECTIVES

By the end of this chapter the student should be able to:

1. Explain the meaning of a set

2. Differentiate between the sets of real numbers.

3. Add, subtract, multiply and divide positive and negative real numbers.

4. State and explain the properties of real numbers.

5. Make truth tables for statements in mathematical Logics

6. Solve problems involving mathematical logics

1.2 CHAPTER CONTENT

1.2.1 MATHEMETICAL LOGIC

Mathematical Logic deals with statements or propositions.

STATEMENTS

A statement (proposition) is a declarative sentence that is either true or false but not

both

For example

i. All women are Mothers –false

ii. All Africans are black –false

iii.

iv. Electronic mail provide a means of communication –True

When two or more Statements are logically combined, they form of compound

statement.Statements are logically combined by two logical connectives, “and” and “or”.

The compound statement formed by “and” is called a conjunction while that formed by

“or” is called a disjunction.

The connectives are symbolized as for “and” and for “or”.

ACTIVITY

State whether the following statements a conjunction or a disjunction.

i. “violets are blue and roses are red”

ii. “Mary is a kind and humble girl”

iii. “Peter is intelligent or he studies very hard”

iv. “You should be in the field playing or singing in the hall”

Compound Statements connected with “and” are only true when both the first and second

statements are true while those with “or” are true if either the first or the second statement

is true.

BASIC LOGICAL OPERATION

Let p represent the first statement and q the second statement.

1) A conjunction of two statements p and q is written as

Definition 1. If p and q are true then is true, otherwise is false.

2) A disjunction is written as . The truth value of depends only on the truth of

p or q.

Definition 2 If p and q are both false, their is false otherwise is true.

3) Negation

Given a proposition p, the negation is formed by writing “it is not the case that … or “it is

false that …” before P or if possible, by inserting in p the word “not”.

The symbol read as “not p”denotes the negation of p. the truth value of ~p depend

on the truth value of p.

Definition 3.If p is true, then ~p is false and if p is false, then ~p is true.

Examples

Statements Truth value

a) Mt. Kenya is in Meru (Statement) F

Mt. Kenya is not in Meru (Negation) T

b) The sun sets in the west (Statement) T

The sun does not set in the West (Negation) F

c) F

≠ 24 T

QUANTIFIERS

The words “all”, “each”, “every” and “none” are called universal quantifiers. While

words or phrases such as “some” “there exists” and “at least one” are called existential

quantifiers. Quantifiers are used to indicate how many cases of a particular situation

exist.

NEGATION OF STATEMENT WITH QUANTIFIERS

Let us consider the following groups of students

GP1: Mary, James, Mary, Steve, Mary, John

GP2: Mary, Jones, Beth, Patrick, Alice, Smith.

GP3: Donna, Gideon, Pauline, Johnson, Sera, Seth.

In GP1: All girls are named Mary

GP2: Some girls are named Mary.

GP3: No girl is named Mary.

If we say;

“All girls in the group are named Mary”. Most people would write the negation as:

“ No girls in the group are named Mary” OR “All girls in the group are not named

Mary ”

None of the two negation statements are correct.

Consider the table below remembering that “some” mean at least one (possibility all)

Truth value as applied to

GROUP

1

GROUP

2

GROUP

3

1. All girls in the group are named Mary T F F

2. No girls in the group is named Mary F F T

3. All girls in group are not named Mary F F T

4. Some girls in the group are not named

Mary

F T T

The negation of statement 1 must have opposite truth values in all cases. Statement (2)

and (3) don’t satisfy this condition for group 2 but statement 4 does. It may be concluded

that the correct negation for the given statement is “Some girls in the group are not

named Mary”

Other ways of stating the negation are

“Not all girls in the group are named Mary”

“It is not the case that all girls in the group are named Mary”

“At least one girl in the group is not named Mary”

The following table can be used to generalize the method of finding the negation of a

statement involving quantifiers.

Statement Negation

1. All do Some do not (not all do)

2. Some do None do (all do not)

The negation of the negation statement is simply the statement itself.

Examples

1) P: Some cats have fleas

Since some means “at least one” the statement is really the same as “At least one

cat has fleas”

~ P: No cat has fleas

2) P: Some cats do not have fleas meaning at least one cat does not have flies.

~P: All cats have fleas

3) P: No cats have fleas

~ P: Some cats have fleas.

Exercise

TRUTH TABLES & EQUIVALENTS STATEMENTS.

The truth functional structure of a statement is determined by use of a truth table.

CONJUNCTION

is true if and only if both p and q are true. AND operator is commutative, that is

The statements “Monday immediately follows Sunday and March immediately follows

February” are true since each component statement is true. While “Monday immediately

follows Sunday and March immediately follows January” is false, even though the 1st part

of the statement is true.

Truth table for

P Q

T T T

T F F

F T F

F F F

Example

Let P represent and q represent find the truth value of .

Solution

P is true and Q is false. From the truth table P is true but Q is false so is false.

DISJUNCTIONS

PVQ is true as long as one of the statements is true. For example the statement in a

college course catalogue, “students must take a statistics course or a logic course to

graduate” would seem to imply that a student meets the requirements if she/he takes a

statistics course or a logic course or takes both statistics and logic.

is false if and only if both p and q are false.

TRUTH TABLE FOR P Q

P Q

T T T

T F T

F T T

F F F

Negation (NOT)

Not (~) transforms a statement into its opposite truth value. That is ~ p is true whenever P

is false and ~ p is false whenever p is true. For example “George Washington was born in

1732” then ~ p is the statement “George Washington was not born in 1732”.

Truth table for negation

P ~ P

T F

F T

Constructing Truth Tables for compound Statement

EXAMPLE 1

Construct a truth table for

Solution

List down all possible combinations of truth values for p and q, then list the truth

values for ~ p

Find ~p^q, ~q and

~ p q ~ q (~ p q) ~ q

~ p

T T F F F F

T F F F T T

F T T T F T

F F T F T T

b) Use the above table to find both values of suppose that both p and q are

true

Solution

If both p and q are true

(2) Construct the truth table for

~p ~q

~p v~ q p (~pv~q)

T T F F F F

T F F T T T

F T T F T F

F F T T T F

COMPOUND STATE WITH VARIABLES

If a compound statement involves three compound statements p q and r, we will use the

std format below for setting up the truth table

r

T T T

T T F

T F T

T F F

F T F

F F T

F F F

EXAMPLE

1) Suppose that p is false, q is true and r is false. What is the truth value of the compound

statement ?

Solution

Truth table

r ~ p ~ r

q v ~ r ~ p

۸ (q v ~ q)

T T T F F T F

T T F F T T F

T F T F F F F

T F F F T T F

F T T T F T T

F T F T T T T

F F T T F F F

F F F T T T T

From the table

~p

(2)(a) Construct a truth table for

Solution

Truth table

r ~p ~q

~p r

~q ~p (~p^r)

T T T F F F F F

T T F F F F F F

T F T F T F F F

T F F F T F F F

F T T T F T F T

F T F T F F F F

F F T T T T T T

F F F T T F T T

b). suppose p is true q is false and r is true find the truth value of

Solution

Exercise

1) Let p represent the statement , q represent and r represent . Decide

whether the following statements are true or false.

a)

b)

c)

EQUIVALENT STATEMENTS

One application of truth tables is to show that 2 statements are equivalent. Two

statements are equivalent if they have the same truth value in every possible situation.

Example

Are the following statements

~ p and ~ equivalent?

Solution

Make a truth table for each.

Statement

~p ~q

~q ~p

T T F F F

T F F T F

F T T F F

F F T T T

p

v q

~ (q v p)

T T T F

T F T F

F T T F

F F F T

Clearly the truth columns for ~ p (p v q) and ~ are the same ~p ~ q is equivalent

to ~(p v q).

The symbol for equivalent is , hence ~ p ~ q

Example two

Determine whether ~p v~ q

Solution

~ p ~ q

~ p ~ q

T T F F F

T F F T T

F T T F T

F F T T T

p

q

~ (p q)

T T T F

T F F T

F T F T

F F F T

Clearly, ~ v ~ q

These two statements constitute the De Morgan’s law which states that:

For any statement p and q

~ P

and

~

Examples

1) Find the negation of each statement by applying De Morgan’s laws.

a) I got an A or I got a B.

Solution

Let P: I got an A

Q: I got a B

I got an A or I got a B =

By De Morgan’s law, the negation of is ~ .

~

I didn’t get an A and I didn’t get a B.

b) She won’t try and he will succeed

Solution

Let P: She won’t try

Q: He will succeed.

She won’t try and he will succeed

Negation is ~

By De Morgan’s law

~

She will try or he will not succeed.

EXERCISE

A

1) If Q is false, what is the Truth value of

2) If is true, and p is true, then what is Q

3) If ~ is true, what must be the truth value of the component statements P&Q.

B) Let and . Find the truth value of the following compound statements.

1)

2)

3)

4)

5)

6)

C) Let P represent , Q and R . Find the truth

values of the following compound statements.

1)

2) ~

3)

4)

D) Give the number of rows in the truth tables for each of the following compound

statements

1)

2)

3)

E) Construct a truth table for each of the compound statements.

1)

2)

3)

4)

5)

F) Use De Morgan’s laws to write the negation of the statement

1) P: you can pay me now or you can pay me later

2) R: it is summer and there is no show

3) Q: I said yes and she said no

T: 5-1= 4 and 9 +12 7

TAUTOLOGIES AND CONTRADICTIONS

1) A Tautology is statement that contains the truth (T) only in the last column of

their truth tables, that is, it is true for any values of their variables.

~ p

p ~ p

T F T

F T T

Therefore is a tautology.

2) A contradiction is a statement that contains only false (F) in the last columns of its

truth table, that is, it is false for any values of its variables.

For example

~ p

p ~ p

T F F

T F F

Therefore is a contradiction

Note that the negation of a tautology is a contradiction and vice versa. Let

be a tautology and and be any proposition. Since

doesn’t depend on the particular values of its variables we can substitute for

for q in the tautology and still have a tautology

THE0REM 1: Principle of substitution

If is a tautology then is a tautology for any propositions

Laws of the Algebra of statements

Statements satisfy laws which are listed in the table below

1) Idempotent laws

a)

b)Error! Objects cannot be created from editing field codes.

2) Association laws

a)

b)

3) Commutative law

a)

b)

4) Distributive law

a)

b)

5) Identify laws

a)

b)

c)

d)

6) Complement laws

a)

b)

c)

d)

7) Involution laws

a)

8) De Morgan’s laws

a)

b)

CONDITIONAL STATEMENT’S

These are statements in a compound statement that uses the connective “if …….then”

“If p then Q” is written as

Example

“If I am elected, then taxes will go down.”

In , the statement P is the antecedent and Q is the consequent.

The conditional connective may not always be explicitly stated.

For example

1) “ Big girls don’t cry” can be written in the “if---------then” form as if you are a

big girl , then you don’t cry”

2) “ It is difficult to study when you are distracted” can be written as

“If you are distracted, then it is difficult to study,

The conditional is false only when the first part P is true and the 2nd Q is

false.

When P is false, is true regardless of the truth value of Q.

Let’s consider the statement by a politician.

“If I am elected, then taxes will go down” There are 4 possibilities; it is

helpful if we think in terms of the following.

“Did the politician lie?” if she had, the conditional statement is false, if she did

not he, then the conditional statement is true.

Elected Taxes go down

Yes Yes P-T, Q-T

Yes No P-T, Q-F

No Yes P-F, Q-T

No No P-F, Q-F

There are four possibilities

1. The1st case assumes that the politician was elected and taxes went down. The

politician told the truth T. in the 1st row of the truth table (we don’t claim

that taxes went down because she was elected it is possible that she had

nothing to do with it at all).

2. The 2nd case assumes that the senator was elected and taxes didn’t go down,

the politician had therefore lied

3. In the 3rd case assume that politician was defeated but taxes went down

anyway (P-F, Q-T). The politician didn’t lie; she only promised a tax

reduction if she was elected.

4. Last case assumes that politician was defeated and taxes didn’t go down

(P-F, Q-F).

We can’t blame her since she only promised to reduce taxes if elected.

P Q

T T T

T F F

F T T

F F T

Special Characteristics of Conditional Statements.

1. is F only when the antecedent is true and consequent is false.

2. If the antecedent is false, then is automatically true.

3. If the consequent is true, then is automatically true.

Examples

1. construct the truth table for

a)

Solution

P Q ~P ~ Q

T T F F T

T F F T T

F F T T T

P Q ~P

T T F T

T F F F

F T T T

F F T F

T F F

T F F

F T T

T F F

b)

Solution

P Q ~P

T T T F T T

T F F F F T

F T T T T T

F F T T T T

Negation of a conditional statement.

Consider the statement.

“If it values then l take my umbrella”

When will the person have lied to you? When it rains and the person doesn’t take the

umbrella .

Let P “it rain” and “I take my umbrella”.

Assume is the negation of , that is

Proof

P Q ~Q

T T F T F F

T F T F T T

F T F T F F

F F T T T F

Clearly

Conditional as disjoint

Since , negating each expression we have

Applying De Morgan’s laws

That is statement (1) indicates that a conditional statement may be written as a disjoint

Examples

1) Write the negation of each statement.

a) If you go, he will come.

Solution

The above statement is

Negation is

“You go and he will not come”

b) “all dogs have fleas”

Solutions

The statement must be restated.

In “if …………..then” form

“If it is a dog; then it has fleas”

Negation

“It is a dog and it does not have fleas”

2) Write each conditional statement as an equivalent statement with using “if ……

then”

a) If the Harambee Stars win the game, then couch Mulei will be happy”

Solution

Harambee star win the game or couch Mulei will not be happy.

b) If it’s Delmont’s, its good to be good.

Example

It’s not Delmont or it’s got to be good.

CONVERSE, INVERSE & CONTRAPOSITIVE

Any conditional statement is made up of an antecedent and a consequent. If they are

interchanged we form the converse.

For example, in the statement: “If you stay, then I go”

Interchanging the antecedent “you stay” and the consequent “I go” we obtain the

converse of the given statement that is “If I go, then you stay”.

By negating both the antecedent and the consequent, we obtain the inverse of the

given statement that is “If you do not stay, then I do not go”

If the antecedent and the consequent are both interchanged and negated, the

Contra positive of the given statement is formed. This is “If I do not go, then you do

not stay”

SYMBOLICALLY

Direct statement

Converse Error! Objects cannot be created from editing field

codes.

Inverse Error! Objects cannot be created from editing field codes.

Contra positive

Example

1. Write the converse, inverse and the contra positive of the statements

“if I live in Miami, then I lice in Florida”

a) “ If I live in Florida, then I live in Miami

b) “If do not live in Miami, then I do not live in Florida.

c) “If I don’t live in Miami”

P Q

T T T T T T

T F F T T F

F T T F F T

F F T T T T

Note that the direct statement and its contra positive always have the same truth

values making it possible to replace any statement with its contra positive without

affecting the logical meaning.

BI-CONDITIONAL STATEMENTS

The compound statement P is, if and only if Q is, abbreviated as P iff Q is known as a

Bi-Conditional statement. It is symbolized as and is interpreted as the

conjunction of 2 conditional statements.

and =

So by definition

Truth tables

P Q

T T T

T F F

F T F

F F T

Exercise

1) Verify that the statement is a tautology.

2) Show that the statements. and are logically equivalent.

3) Show that

4) Rewrite the following statements without using the “ if ……then ….”

a) If it is cold, he wears a hat.

b) If productivity increases, then wages rise.

5) Determine the contra positive of the following statements

a) If John is a poet, then he is poor.

b) Only if mark studies will he pass the test.

ANALYSING ARGUMENTS WITH TRUTH TABLES.

Consider the argument

First premise (P1 ):“If the floor is dirty, then I must mop it”

Second premise (P2): The floor is dirty

Conclusion :I must mop it.

To test the validity of this argument, we 1st identify the component statement.

Let P: the floor is dirty’

Q: I must mop it.

Rewrite the 2 premises and conclusion using symbols.

To decide if the argument is valid, we must determine whether the conjunction of

both premises implies the conclusion for all possible cases of the truth values of P and

Q

Finally construct the truth table for this conditional statement

P Q

T T T T T

T F F F T

F T T F T

F F T F T

Since the final column indicates that the conditional statement that represents the

argument is true for all possible truth values of P and Q, it is a tautology thus the

argument is valid.

The pattern of the argument in the above example.

Is called Modus ponens or the law of attachment.

Steps in testing validity.

1. Assign a letter to represent each component statement in the argument.

2. Express each premise and the conclusion symbolically.

3. Form the symbolic statement of the entire argument by writing the conjunction of

all the premises as the antecedent of the conditional statement and the conclusion

of the argument as the consequent.

4. Compete the truth table for the conditional statement formed above if it is a

tautology, then the argument is valid, otherwise it’s invalid.

EXAMPLE

1) Determine the validity of the arguments below.

2) If my Cheque arrives on time I will register for the fail semester. I’ve registered

for the fail semester my Cheque arrived in time.

Using symbols

P Q

T T T T T

T F F F T

F T T T F

F F T T T

It’s not a tautology hence is not valid

b) If a man could be in two places at one time, I would be with is I am not with you.

P: “a man could be in 2 places at one time”

Q: I would be with you.

P Q ~P ~Q

T T T F T F T

T F F T F F T

F T T F T F T

F F T F T T T

This is a tautology hence the argument is valid

EXERCISE

1) Determine whether the argument is valid or invalid.

a) I will buy a car of I will take a vacation

I won’t buy a car.

I will take a vacation.

Solution

P: I will buy a car

Q: I will take a vacation.

P Q ~P

T T T F F T

T F T F F T

F T T T T T

F F F T F T

The statement is a tautology hence argument is valid.

2) Determine whether the argument is valid /invalid.

a) If I study, then I will not fail mathematics

If I do not play basketball, then

I will study.

But I failed mathematics

I must have played basketball.

b) If Roy plays, the opponent gets shut out.

The opponent does not get shut out

Roy does not play.

c) If it rains, Erick will be sick

Erick was not sick.

It did not rain.

Week 2

CHAPTER TWO


1. Solve simple equations using the addition and subtraction.

2. solve quadratic equations by factoring and using the principle of zero

products.

3. Solve quadratic quadratic equations by completing the square method.

4. Derive the Quadratic formula.

5. Solve quadratic equations using the formula.

6. Formulate Quadratic equations and solve using an appropriate method.

7. Solve simple inequalities.

Week3

CHAPTER THREE


1. Write fractional algebraic expression into partial fractions

2 Determine the domain and range of a relation.

3.Draw graphs of functions

5. Determine the symmetry, evenness and oddness of a function.

6. Sketch the graph of the universe of a given function

Week 4

CHAPTER FOUR


1. Given an equation of a relation, describe the effect on the graph of replacing y

by. (y – a) and x by (x – b).

2. Given an equation y = f(x), describe the effect on the graph of multiplying f(x)

by a constant c and x everywhere it occurs by a constant d.

3. Given a graph y= f(x), sketch a graph of

y = a + f(x), y = f (x + b), c f (x) and y = f (x + d) where a, b, c, d, are

Constants.

Week 5

CHAPTER FIVE


1. Use tables and scientific calculators to find logarithms and antilogarithms.

2. Find products, quotients, powers, roots and combinations of these using

logarithms.

3. Write logarithmic equations into exponential form and exponential equation

into logarithmic form.

4. Solve exponential and logarithmic equations.

5. Change logarithms from one base to another.

6. Solve applied problems involving exponential and logarithmic equations.

CHAPTER SIX


1. Explain the meaning of complex numbers

2. Carryout mathematical operations with complex numbers

3. Solve Quadratic equations whose roots are complex

4. Represent Complex numbers on a Cartesian plane

COMPLEX NUMBER: IMAGINARY NUMBERS

Given any number k and then the number is real.In real numbers has no solution because and negative number have no roots.To solve such equations we need a set of numbers whose squares are negative eg

This set of numbers are called imaginary numbersGenerally is an imaginary number where n is real

Imaginary numbers can be added, subtracted, multiplied and divided just like the Real numbers

Example ;

The number or quotient of two imaginary numbers is always a real number Powers of imaginary numbersNote that (-1) is always replaced by and (1) by or vice versa.

Imaginary numbers helps to solve quadratic equation whose

COMPLEX NUMBERSA complex number is an ordered pair of numbers / expression of the form where

And are real numbers It is an ordered pair because is not the same as Complex numbers are usually denoted as = Where

The Real part of

The Imaginary part of

When That is Imaginary numbers

When

That is real numbers.The field of complex numbers includes the real number set and the imaginary number set.

OPERATIONS ON COMPLEX NUMBERSThe fundamental rules of algebra which are used in the manipulation of real numbers

Complex numbers are a logical extension of the concept of real numbers.

ADDITION AND SUBSTRACTIONReal terms and imaginary terms are compounded into two separate groupAdditionEXAMPLE S1)

2) If and

Find

Generally if

SubtractionEXAMPLES

1) a)

b)

2) a)

b)

3)

Generally , given that

MULTIPLICATION

The multiplication of complex numbers is Distributive.ExamplesMultiply the following

The product is a real number. and are said to be

Conjugates each of the other.Given and

If the conjugate of is

and

If

The product of conjugate complex numbers is always a real number.

ExamplesEvaluate1)2)3)

DIVISION Complex numbers can not be divided directly because the denominator is made up of two independent terms. This difficulty is overcome by making the denominator real, a process known as realization of the denominator.Recall: The product of conjugate complex numbers is always a real numbers.

EXAMPLES

1)

2)

EXERCISE1) If

Find ,

The Zero complex numberA complex number is zero if and only if the real term and imaginary term are each zero. iff and

Equal complex numbersConsider

Two complex numbers are equal if the real terms and the imaginary terms are separately equal.

A complex number equation is therefore equivalent to two separate equations hence alternatively we can divide.Division 1

Solve and simultaneous and

2) Show that

Square root of complex numbers.The method of equating real and imaginary parts of a complex equation can be used to determine the square root of complex numbers.

EXAMPLE 1Find

Let

SolutionFrom equation 2

Substitute in equation 1

Let

Or

4 -1-2 1

Hence

EXAMPLE 2Find the square root of Solution

Let

COMPLEX ROOTS OF QUADRATIC EQUATIONS1) Consider the equation

2) Form the equation whose roots are

3) Solve

Week 7 AND 8

CHAPTER SEVEN

7.0 INTRODUCTON

7.1 CHAPTER OBJECTIVES


1. Express a factorial in expanded form and evaluate.

2. Apply the formula nPn = n(n -1 ) (n – 2 ) – 3.2:1

= n!

to find the total number of permutation

3. Use the formula to find the number of permutations of n objects

taken r at a time without repetition.

4. Given a set of n objects in which there are n1 of first kind, n2 of second kind. ..nk

of the kth kind, apply the expression n! to find the number of n1. n2 … nk

Distinguishable permutations

7. Apply the expression (n-1)! To find the number of circular arrangements of n

objects.

8. Apply the expression nr to ding the number of distinct arrangements of n

objects taken r at a time, with repetition.

9. Apply the formula

nCr = n! = n = nPr r! (n – r)! r r!

to find the number of combinations taken r at a time of a set of n objects.

10. Solve applied problems involving combinations.

PERMUTATIONS AND COMBINATIONS

Topic involves finding number of different arrangements or selections which are possible from a set of given objects.

FACTORIAL NOTATIONConsider the following

5 x 4 x 3 x 4 x 1 = 120 = 5!

10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x1 = 10!

32 x 3! x 30 x 29 x 28 x 27 x 26 x 25 x ………x 1 = 10!

Writing the product of large consecutive numbers in expanded form and evaluating can be cumbersome and tedious, hence the necessity for a shortened notation.The product of consecutive integers is usually denoted as n!When the product of consecutive integers is expressed as n!, it is said to be in factorial notation.

In general n! represents n x (n - 1) x (n - 2) x …. x2 x1 it n means the product of all the integers from 1 to is inclusive and is called n factorial.

Examples13! = 13 x 12 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1. 1! = 1 0! = 1Evaluate1. (a) 15! = 15 x 14 x 13 x 12 x 11 x ……x 2 x1

12! 12 x 11 x ……. x 2 x 1

= 15 x 14 x 13

(b) 20! = 20 x 19 x 18 x 17 x …….. x 2 x 1 17! 3! 17 x 16 x…… x 2 x 3 x 3 x 2 x1

= 20 x 19 x 18 3 x 2 x 1

(c) (3!) = 3! 3! 2! 4! 2! 4!

= 3 x 2 x 1 x 3 x 2 x1 2 x 1 x 4 x 3 x 2 x1

= 3 4

2. Express (a) 10! as a multiple of 7! 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

= 720 x 7!(b) n! as a multiple of (n-1)!

n! = n x (n-1) x (n-2) x …… x 2 x 1= n (n-1)!

3. Write in factorial form(a) 11x 10 = 11!

9!

(b) (n+1) n (n-1) = (n+1) ! (n-2)!

(c) 8 x 7 x 6 = 8! 3!

6 x 5 x 4 6! 5!

4. Factorize(a) 8! + 9! = 8! + 9. 8!

= 8! + (1+9)

= 10. 8!

(b) 3 (10!) + 4 (8!) = 3 (10 x 9.8!) + 4 (8!)

= 8! (270 + 4)

= 274.8!

(c) (n + 1)! – (n-1)! = (n + 1) n (n-1)! – (n-1)!

= (n-1)! [(n+1) n-1]

= (n2 + n-1) (n-1) !

(d) 7! + 6 ! = 7! 4! + 6! 3!3! 4! 3! 4!

(e) 7! + 6! = 4 ! 7.6! + 6! 3! 3! 4! 3! 4!

= 6! 3! (4x7) + 1 3! 4!

= 2.9 (6!) 4!

(f) 2 (n+1) ! - 3n! = (r + 7) ! 2 (n + 1) n! - 3n!. r! r! ( r + 1) (r! ) (r + 1)!

= n! r! [2(n+1) (r+1) - 3](r + 1)!

= n! (2{nr + n + r + 1}-3)(r + 1)!

= n! (2nr + 2n + 2r - 1)(r + 1)!

Exercise1. Write in factorial form

(n + r)! (n + r – 3)!

(a) (n + r) (n + r - 1) (n – r - 2);

(b) 14 x 13 ; 3 x 3 x 1

(c) (n-2) (n-3) (n-4) ; 4 x 3 x 2 x 1

2. Factorize(a) 10! + 9! + 8! ; 8! 2! 7! 2! 6! 2!

(b) n! + (n-1)! ; r! (r+1)!

(c) 2 (n+1)! - 3n! ;r! (r+1)!

(d) n! + 2 (n+1)! ; r! (n-r)! r! (n-r +1)!

(e) n! r 2 - 2 (n-1)! ; (n-r)! r! (r-1)! (n-r)!

PERMUTATIONS1. Given the letters P, Q, R in how many ways can these letters be arranged

14! 12! 3!

(n - 2)! 4! (n -5)! 3)!

109 (8!) 28(6! 2!) 3)!

(n - 1)! (nr + n + 1 ) (r +1)!

n! (2nr + 2r +2n -1) (r+1)!

n! (3n-r + 3) r! (n - r +1)!

(nr -2) (n-1)! (n – r)! (r – 1)!

PQR QPR RPQ 6 different ways

PRQ QRP RQP

Ordered arrangements of objects known as permutations and each arrangement is called a permutation, Hence in the case above, there are 6 permutations and each permutation is made up of 3 letters.Definition: A permutation of a set of n objects

2. Suppose the letters are 4 ; P, Q, R, S

PQRS QPRS RPQS

PQSR QPSR RPSQ

PRQS QRPS RQPS

PRSQ QRSP RQSP

PSQR QSPR RSPQ

PSRQ QSRP RSQP

There are 24 permutations each made up of 4 letters.This exercise of listing is tedious and difficult without listing them as followsPermutations can be calculated without listing. Consider the above 2 cases

a) In how many ways can 1st letter be chosen ; 3 2nd letter 2

3rd letter 1The total number of different arrangements is 3 x 2 x 1 = 3! =6

b) In how many ways can the:

1st letter can be chosen 4

2nd 3

3rd 2

4th 1

No of permutations = 4 x 3 x 2 x 1 = 4! = 24

In general, the total number of permutations for n unlike objects taken all at a time is

n!

nPn = n x (n-1) x (n-2) ……….. x 3 x 2 x 1 = n!

Exercise

1. In how many ways can five different pens be arranged?

Solution

nPn = 5!

= 120

2. In how many ways can six different pens be arranged?

Solution

nPn = 6!

= 720

3. A club wants to make a flag by joining 5 different coloured strips of cloths together, find the number of possible ways in which they can be joined horizontally on top of each other.

Solution

nPn = 5!

= 120

4. A student has 14 books to put on a bookshelf. There are 5 History books, 4 Mathematics, 3 language and 2 Chemistry. He arranges them so that all the books declining with the same subject are together on shelf. He also puts Maths books first followed by History, chemistry and language. How many different arrangements are possible?

Solution

Total permutations = 4! x 5! x 2! x 3!

= 34560

b) If student wants the books with same subject together without any other restrictions, find the number of arrangements that are possible.

SolutionTotal permutations = 4! (5! x 4! x 3! x 2!)

= 4! x 34560

= 24 x 34560

= 829440

Permutations of r objects from n unlike objects: This is an arrangement of r objects in a specific order.

1. Example

How many three digit numbers can be made from the integers 2, 3,4,5,6, if

a) Each integer is used once

Permutation = 5 x 4 x 3 =

= 60

b) There are no restrictions on number of times an integer is used

Permutation = 5 x 5 x 5

= 125

2. How many ways can 4 digit number be formed from 1, 2, 3,4,5,6,7,8,9 if each is used only, once

P= 9 x 8 x 7 x 6 = 9! 5!

= 3024

Hence in general

Order of choice 1st 2nd 3rd …(r-1)th rth

Number of possible n n-1 n-2… (n- (r + 1-1) (n- (r-1))

Finding number of permutation for r objects from n unlike object

Total permutations= n x (n-1) x (n-2) ……….. x (n-r + 2) x (n –r +1)

= n (n-1) (n-2) (n-r +1) (n-r) x……… x 2 x 1 (n-r) x ……. x 2 x 1

= n! (n-r) x ……. 2 x 2

The total number of permutation for r objects from n unlike object is denoted as nPn = nPr = P (n, r)

Thus nPr = nPr = n! (n-r)!

Provided

Examples1. How many different 4 letter words can be made from the alphabet of each letter

can be used only on a 26P4 = 26!

(26-4)!

= 26! =26 x 25 x 24 23 22!

=358800

2. How many ways can a team of 11 players be taken from 30 players

30P11 = 30! (30-11) ! = 30! You may leave the answer in factorial form. 19!

3. How many different 3-digit numbers can be formed from the digits 1,2,3,4,5 if

a) There is no repetition5P3 = 5! = 5!

(5-3)! 2!

= 5 x 4 x 3 = 60

b) Repetition is allowed5P3 with repetition = 5 x 5 x 5 = 53 = 125

In general, the number of permutations of in different objects taken r at a time with repetition is nr.

nPr = nr.EXERCISE1. How many 5 letter code symbols can be formed with the letters A,B,C,D, if

repetitions are allowed.

2. How many letter code symbols can be formed by repeated use of letters of alphabet?

Selecting officers

Example 1From a committee of 8 people, in how many ways can we choose a chair and a vice chair assuming one person can’t hold more than one position?

Solution We are asking for number of permutations of 8 objects taken 2 at a time.

8P2 = 8! (8-2)!

= 8! = 8 x 7. 6!6! 6!

= 56

2. From a committee of 10 people, in how many ways can we choose a chair, vice-chair and secretary if one person can’t hold more than one position?

Solution

10P3 = 10! = 10! (10-3)! 7!

= 10.9.8

= 720

CIRCULAR ARRANGEMENTS

Theorem The number of distinct circular arrangements of n unlike objects is (n - 1)!

Example1. How many ways can 9 different foods be arranged around a lazy Susan?

P = (9-1)! = 8!

= 40320

2. How many ways can 7 men be arranged around a round table.

Solution P = (7-1)! = 6!

= 720

3. In how many ways can the numbers on a clock face be arranged?

Solution P= (12-1)! = 11!

= .39916800

(4 )The representatives of five countries attend a conference. In how many ways can they be seated at a round table?Solution:As all chairs are identical and there are no specific features that influence choice of position and they must always be state.

In such problems fix one person and arrange the others.

Number of permutations = 4!= 4 x 3 x 2 x 1= 24

EXERCISE(1) In how many ways can 10 people sit at a round table?

(2) In how many ways can five beads chosen from eight different beads be threaded on a ring?

Permutation of identical objects

Consider a set of n objects in which n1 are of one kind, n2 are of a second kind ………. nk are of kth kind.Total number of permutation of the set is n! Let P be number of distinguishable permutations. There are n1! Permutations of objects of 1st kind.

n2! of 2nd kind and so on

Total number of actual = Pn1 ! n2! n3!... nk! = n!Permutation

P = n! n! n2!...nk

In the general the number of permutation of n objects of which n1 are alike n2 ... nk are alike is n!

n! n2!...nk!

Examples

1.How many distinguishable ways can the letters of the word CINCINNATI be arranged?Solution P= 10! 2! 3! 3! 1! 1! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

2 x 1 x 3 x 2 x 1 x 3 x 2 x 1 x 1 x 1 = 504002. In how many ways can 5 blue beads, 4 green beads, 2 red beads and 1 white bead be arranged in a row if beads of the same colour are indistinguished. Total beads = 12

Number of Permutations = 12! 5! 4! 2!

= 12 x 11 x 10 x 9 x 8 x 7 x 6 4 x 3 x 2 x 2 x 1

= 220 x 9 x 42

= 83160.

3. How many odd numbers greater than 4000 can be formed using the digits 1, 4, 7, 8?Solutions (a)Without repetition Number of Permutations =3 x 2 x 1 x 2 = 12 numbers

b) With repetition Number of Permutations = 3 x 5 x 5 x 3

= 15 x 15 = 225 numbers

Exercise1. In how many ways can the letters of the following words be arranged in a row? (i)TAMAA (ii) TITTER (iii)PILI PILI

4. How many odd members greater than 70,000 can be former using the digits 0, 1, 4, 7, 8, 9 (a) Without repetition (b) With repetition

5. A man dines at the same hotel for 3 consecutive days and the menu each day is one of any four types of goat dish, two types of chicken dish, 2 types of chicken dish,

or one type of vegetarian dish. Has many ways can he arrange his lunches over the three days if he does not have goat dish two days running or no repeat of dish.

6. In how many ways can 3 objects be taken from 8 objects and be arranged in a row.

COMBINATIONSSuppose there are seven pictures labeled a, b, c, d, e, f and g and only three are to be displayed one possible choice is A, B, C. regardless of how they are hanged this group of 3 is called a combination.ABC, BCA, BCA, BAC, CAB, CBA, ACB are different ways in which they can be

arranged – permutation ABC is the same combination as BCA, BAC, CAB, etc.The concern is the combination is what part of the selection is but not how they are

arranged.Definition.A combination is therefore an unordered selection of a number of items from a given

set.Consider the example In how many ways can 3 objects be taken from 8 objects and arranged in a row? 8P3 = 8!

5!

Suppose the problem is not one of arranging but selection, how do we determine the number of selection?

One selection of 3, results into 3 arrangements. Let the number of selection is be C, the total number of permutations will be

C x 3! = 8! 5!

Make C the subject C = 8!

5! 3!

Generally, if we wish to know the number of ways of selecting r objects from n available objects

Total = total permutation of r objects from n Number of permutation among the r objects.

n Pr =

Let c represent number of selections. Each selection gives r arrangement Total arrangement = C. r!

C.r! =

C =

This expression which gives the number of ways of selecting r objects from n available objects is known as in combination r denoted as nCr .

nCr =

Examples 1. Evaluate

10C5 = 10! = 10 x 9 x 8 x 7 x 6 5! 5! 5 x 4 x 3 x 2 x 1

= 4 x 63 = 252.

9C6 = =

=

=84

= 20 x 35 = 700

1 EXERCISE

2. A team of four children is to be selected from a class of twenty children to complete in a quiz game in how many ways can the team be chosen if

(a) Any four can be chosen(b) The four must include the oldest in the class.

3. A shop stocks ten different varieties of pocket soup in how many ways can a shopper buy 3 packets of soup if

(a) Each packet is a different variety (b) Two packets are same variety

INDEPENDENT PERMULATION AND COMBINATIONSConsider the case where the choice of one type of objects doesn’t affects the choice and arrangement of another type BExample There are 30 bottles of wine and 15 cans of soda .find(a)The number of selection of 3 bottles of wine and 10 cans of soda Solution N umber of selections =

=

=

= 4060 x 3003 = 12192180

(b) Number of ways in which the 3 bottles of wine and 10 cans of soda can be arranged in line. Solution

Number of ways = . This is a very large number

Therefore leave it in factorial form.

Generally the number of permutation p1 of objects from one set combined with the number permutation p2 of objects from an independent set is p1 x p2.

This is also true for combination of objects from two independent set and can be extended to cover more than two sets.

ExamplesIn how many ways can a customer at a supermarkets select 4 different types of soap from 30 available types and 10 different packets of biscuits from 12 different types.Solution

= 27405 x 66 = 1808730

Mutually exclusive permutation and combinations

Example

In how many ways can a class of 20 children be split into two groups of 8 and 12 respectively if there are two twins who must not be separated SolutionOnce the group of 8 has been selected that of 12 is formed hence.The twins are either in the group of 8 or out

If included the number of ways is8 C6

If excluded the number of ways is18C8

Total number of ways = 18 + 18 12! 6! 10! 8!

= 18564 + 43758 = 62322 ways.

Exercise(1)Given 3 integers 2, 3, 4, find the number greater than 20 can be made without repetition (2)Five books are to be selected from 20 books of which 8 are paper back and twelve are hard back. How many selections are possible if at least one paper back book has to be included?

Week 9

CHAPTER EIGHT


1. Given a sequence, find the pattern and write the general rule.

2. Differentiate between finite and infinite sequence.

3. Find the series associated to given terms of a sequence.

4. Write a serves in sigma notation like,

4 ∑ (2k + 1)

K=1

Week 10 and 11

CHAPTER NINE


1. Identify the first term and common difference of an Arithmetic

sequence/series

2. Find the nth term by Tn = a1 + (n – 1 )d

3. Find the sum of given (n) terms of an arithmetic series using the formula

Sn = n (a1 + Tn)

Or Sn = n (2a1 + ( n – 1) d) 2

4. Solve problems involving arithmetic series.

5 Find the common ratio (r) of a geometric sequence.

6. Determine the nth term and r using an = a1 r n-1

7. Find the sum of n terms of a geometric series using the formula

Sn = a1 (1 – r n ) 1 – r

8. Solve applied problems involving geometric sequence and series.

Week 12 and 13

CHAPTER TEN


1. State the dimension of a matrix.

2. Add matrix

3. Find the scalar product of a number k and a matrix A

4. Find the product AB

5. Find the determinant of a matrix.

6. Find the inverse of a matrix

7. Write a matrix equation and solve the system.

8. Reduce a square matrix to echelon form.

9 Find the inverse by reduction.

10. Solve a system of equation by matrix reduction method.

Week 14 and 15

1. Students return/hand in the assignment.

2. Take the sit in course examination.

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