MT 2123: Introduction to Calculus

This course is

offered by

School of Informatics & Applied Mathematics

Universiti Malaysia Terengganu

By

Shalela Mohd Mahali

September 2014

Contents

1 The Real Number System 1

1.1 Types of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Properties of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Mathematical problems involving inequality . . . . . . . . . . . . . . 4

1.3.1 Solving a Linear Inequality . . . . . . . . . . . . . . . . . . . . 4

1.3.2 Solving a Two–sided Inequality . . . . . . . . . . . . . . . . . 5

1.3.3 Solving an Inequality Involving a Fraction . . . . . . . . . . . 5

1.3.4 Solving a Quadratic Inequality . . . . . . . . . . . . . . . . . . 6

1.3.5 Solving an Inequality Containing an Absolute Value . . . . . . 8

1.4 Number in Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

iii

iv

List of Tables

v

vi

List of Figures

1 Subsets representing real number system . . . . . . . . . . . . . . . . 3

2 An open interval (−∞, 1) . . . . . . . . . . . . . . . . . . . . . . . . 5

3 An open interval (−2, 3

2) . . . . . . . . . . . . . . . . . . . . . . . . . 6

4 Number lines to represent each term in the fraction . . . . . . . . . . 6

5 An open interval (−∞,−2) ∪ [1,∞) . . . . . . . . . . . . . . . . . . . 7

6 Number lines to represent each term of the quadratic expression . . . 7

7 An open interval (−∞,−3) ∪ (2,∞) . . . . . . . . . . . . . . . . . . . 7

8 The distance between a and b . . . . . . . . . . . . . . . . . . . . . . 8

9 |x− 3| < 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

10 Points A, B, C and D plotted on a cartesian coordinate . . . . . . . 10

11 Sketching lines on a cartesian coordinate . . . . . . . . . . . . . . . . 12

12 Shading the region for inequalities . . . . . . . . . . . . . . . . . . . . 13

vii

Chapter 1

The Real Number System

1.1 Types of Real Numbers

Natural numbers

• Natural numbers are also known as ‘counting numbers’.

• The numbers are 1, 2, 3, 4, 5, 6, · · · .

• The set of all natural numbers is denoted as N.

• Natural numbers together with ‘zero’ are calledwhole numbers (i.e 0, 1, 2, 3, 4, 5, 6, · · · )

Integers

• Integers are the positive and negative whole numbers.

• The numbers are · · · ,−3,−2,−1, 0, 1, 2, 3, · · · .

• The set of all integers is denoted as Z

Rational numbers

• Rational numbers are quotients of integers (i.e All numbers of the form a

bwith

a and b are integers and b 6= 0).

1

1.1. TYPES OF REAL NUMBERS

• The set of all rational numbers is denoted as Q.

• All integers are rational numbers because for any integer i ∈ Z, i can be

written as i

1.

• Real numbers with terminating decimal or non-terminating decimals that re-

peat are also rational number.

• Real numbers with non-terminating decimals that do not repeat are irra-

tional numbers. π = 3.14159265 · · · and√2 = 1.41421356 · · · are irrational

numbers.

Real numbers

• Real numbers are the set of all decimals, both terminating and non-terminating.

• The set of all real numbers is denoted as R.

• The above mentioned set of numbers (i.e Natural numbers, Integers, Rational

and Irrational numbers) are subsets of real numbers.

• The relationships of the sets consist in real numbers can be illustrated in Figure

1

2

THE REAL NUMBER SYSTEM

Figure 1: Subsets representing real number system

Exercises 1.1

1 Tick the correct type for the following numbers. Each number may fall into

more than one type.

Number Natural Whole Integer Rational Irrational Real

2√ √ √ √ √

0

0.25

-5√2

8

1.342234223422...

1.234567

−4

5

2 Tick the correct type of the following decimal number. Then, determined

whether the number is a rational or irrational number. If it is a rational

number, rewrite the decimal number in the form of integer fraction, a

b.

3

1.2. PROPERTIES OF REAL NUMBERS

Number Terminating Non–terminating decimal Rational\ Fraction

decimal Repeat Not Repeat Irrational form

3.16792

10.121212...

4.275191919191...

3.14159265...

3.41287548754875...

1.2 Properties of Real Numbers

If a and b are real numbers and a < b, then

i For any real number c, a+ c < b+ c.

ii For any real numbers c and d, if c < d, then a+ c < b+ d.

iii For any real number c > 0, a · c < b · c.

iv For any real number c < 0, a · c > b · c

1.3 Mathematical problems involving inequality

1.3.1 Solving a Linear Inequality

Problem: Solve the linear inequality 5x+ 1 < 6.

Solution: Substract 1 from both sides. Thus we have

(5x+ 1)− 1 < 6− 1

5x < 5

Then divide the resulting inequality by 5

5x

5< 5

5

4

THE REAL NUMBER SYSTEM

Finally we have x < 1 or also can be written as an interval (−∞, 1)

Figure 2: An open interval (−∞, 1)

1.3.2 Solving a Two–sided Inequality

Problem: Solve the two–sided inequality 2 < 8− 4x < 16.

Solution: We work with both inequalities simultaneously. First, substract 8 from

each term. Thus we have

2− 8 < (8− 4x)− 8 < 16− 8

−6 < −4x < 8

Then divide the resulting inequalities by -4. Note that, since −4 < 0, our inequali-

ties are reversed.

−6

−4>

−4x

−4>

8

−43

2> x > −2

Rearrange the inequalities, finally we have −2 < x < 3

2or also can be written as an

interval (−2, 3

2)

1.3.3 Solving an Inequality Involving a Fraction

Problem: Solve the inequality x−1

x+2≥ 0

Solution: In order to visualize the function of the fraction, we draw separate number

5

1.3. MATHEMATICAL PROBLEMS INVOLVING INEQUALITY

Figure 3: An open interval (−2, 3

2)

lines for the numerator and the denominator

Figure 4: Number lines to represent each term in the fraction

From the above number lines, we may conclude that the fraction is satisfying the

inequality (i.e nonnegative in this case) whenever x < −2 or x ≥ 0. The solution

also can be written in interval notation as (−∞,−2) ∪ [1,∞)

1.3.4 Solving a Quadratic Inequality

Problem: Solve the quadratic inequality x2 + x− 6 > 0

Solution: By factorising the quadratic term, we have the following equivalent in-

equality to the original problem:

6

THE REAL NUMBER SYSTEM

Figure 5: An open interval (−∞,−2) ∪ [1,∞)

(x+ 3)(x− 2) > 0 (1)

From here, we may draw the number lines to represent the two terms: (x+ 3) and

(x− 2), and finally the combination of both.

Figure 6: Number lines to represent each term of the quadratic expression

The number lines show that the product of the two terms is positive whenever

x < −3 or x > 2. In interval notation, this can be written as (−∞,−3) ∪ (2,∞).

Figure 7: An open interval (−∞,−3) ∪ (2,∞)

7

1.3. MATHEMATICAL PROBLEMS INVOLVING INEQUALITY

Alternatively, you may sketch the quadratic graph and find the interval

whenever the quadratic graph is positive. Try this method for the same

problem above!

1.3.5 Solving an Inequality Containing an Absolute Value

Definition of Absolute Value

The absolute value of a real number x is |x| =

x, if x ≥ 0

−x if x < 0

.

For any real numbers a and b,

1. |a · b| = |a| · |b|.

2. |a+ b| 6= |a|+ |b| in general.

3. |a+ b| ≤ |a|+ |b| (the triangle inequality).

4. |a− b| is referred as the distance between a and b (Figure 8)

Figure 8: The distance between a and b

Problem: Solve the inequality |x− 3| < 5

Solution: The LHS of the inequality refers to the distance between 3 and point x.

Considering the value 5 on the RHS, the inequality shows that the distance between

x and 3 must be less than 5. We may visualise this inequality using the following

figure: Obviously, the solution for x is −2 < x < 8 or in interval notation:(−2, 8).

8

THE REAL NUMBER SYSTEM

Figure 9: |x− 3| < 5

Problem: Solve the inequality |x+ 4| ≤ 7

Solution:

(Hint: |x+ 4| = |x− (−4)|)

There is an alternative method for solving inequalities involv-

ing absolute values.

|x− a| < d

also can be written as the two-sided inequality

−d < x− a < d

. Try this method to solve the previous problems!

1.4 Number in Plane

Our previous discussions only considered real numbers as laid out on a single

line (1 dimension coordinate). Now, we extend our discussion to coordinate in

2-dimensions. We locate points in the plane by using two coordinate lines: the hor-

izontal real line is usually called the x −−axis and the vertical real line is usually

called the y − −axis. Instead of just one real number, the point now is represent

9

1.4. NUMBER IN PLANE

by an ordered pair (x, y) of real numbers, called coordinates of the point. See the

following example to understand how to locate a point based on the given coordinate.

Example: Plotting points

Plot the points A = (3,−2), B = (−3, 6), C = (2, 4) and D = (−1,−2).

Ans:Figure 10

Figure 10: Points A, B, C and D plotted on a cartesian coordinate

Example: Drawing lines

Sketch the following lines:

i x = 1

ii y = −4

iii 4x+ 3y = 24

10

THE REAL NUMBER SYSTEM

Ans:Figure 11

Line x = 1 and y = −4 are easy to skecth.

The third line, 4x + 3y = 24 can be skecthed by finding 2 points on the line and

connecting those 2 points to have a straight line. The easiest 2 points that can be

considered are points when x = 0 and when y = 0.

• Point 1: When x = 0,

4(0) + 3y = 24

3y = 24

y = 8

Hence, the first point is (0, 8).

• Point 2: When y = 0,

4x+ 3(0) = 24

4x = 24

x = 6

Hence, the second point is (6, 0)

11

1.4. NUMBER IN PLANE

Figure 11: Sketching lines on a cartesian coordinate

Example: Graphing Inequalities

Shade the region which contains the points that satisfy the following inequalities:

i y ≤ −4

ii 4x+ 3y > 24

Ans:Figure 12. In order to graph the region for an inequality, we have to determine

the edge of the region first. This can be done by sketching a line representing the

given equation (replace the inequality sign (<,>,≤ or ≥) to an equal sign (=)). In

this example, both lines y = −4 and 4x + 3y > 24 are already sketched in Figure

11. We use solid line to represent edge for the first inequality because of the sign

≤. However, dashed line is used for the second inequality because of the sign > (i.e

does not include equal sign). The reqion for the first inequality is obvious. However,

for the second inequality, we have to choose which side to be shaded by testing one

point at each side.

• Side 1: Point to test is (0,0):

12

THE REAL NUMBER SYSTEM

4(0) + 3(0) = 0 < 24–does not satisfies the inequality.

• Side 2: Point to test is (5,5):

4(5) + 3(5) = 35 > 24–satisfies the inequality. So we choose this side to be

shaded.

Figure 12: Shading the region for inequalities

Your Tasks

Find the formulae and examples for the following topics:

1. The distance and midpoint between 2 points.

2. Equation of a line.

3. Equation of a circle

13

1.4. NUMBER IN PLANE

14

Bibliography

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BIBLIOGRAPHY

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