MAT01A1 0.5cm Numbers, Inequalities and Absolute Values ......Solving absolute value inequalities: I...

MAT01A1

Numbers, Inequalities and Absolute Values

(Appendix A)

Dr Craig

5/6 February 2020

Introduction

Who:

Dr Craig

What:

Lecturer & course coordinator for MAT01A1

Where:

C-Ring 508 acraig@uj.ac.za

Web:

http://andrewcraigmaths.wordpress.com

Important information

Module code: MAT01A1

NOT: MAT1A1E, MAT1A3E, MATE0A1,

MAEB0A1, MAA00A1, MAT00A1,

MAFT0A1

Learning Guide: available on Blackboard.

Please check Blackboard twice a week.

Student email: check this email account

twice per week or set up forwarding to an

address that you check frequently.

Important information

Textbook: the textbook for this module is

Calculus: Early Transcendentals

(International Metric Edition)

James Stewart

8th edition

Please wait until the end of this week before

buying the e-book (or hardcopy). It is

possible that it will be provided by the

University. If you want to buy a hard copy,

older editions are fine.

Other information

I This module is taught on both APK and

DFC. Some announcements will only

apply to one campus or the other.I Need help? Visit the Maths Learning

Centre in C-Ring 512.

Mon 10h30–15h25

Tue 08h00–15h25

Wed 08h00–15h25

Thu Closed for BSc students

Fri 08h00–15h25

Lecturers’ Consultation Hours (APK)

Tuesday:

11h20–12h55 Dr Robinson (C-508)

Tuesday:

14h40–15h25 Dr Craig (C-514)

Wednesday:

12h10–12h55 Dr Craig (C-508)

Thursday:

08h50–10h25 Dr Robinson (C-514)

Thursday:

14h40–16h15 Dr Craig (C-508)

Appendix A:

I Number systems

I Set notation

I Inequalities

I Absolute value

Number systems

The integers are the set of all positive and

negative whole numbers, and zero:

. . . ,−3,−2,−1, 0, 1, 2, 3, 4, . . .

The set of integers is denoted by Z.

From the integers we can construct the

rational numbers. These are all the ratios

of integers. That is, any rational number r

can be written as

r =m

nwhere m,n ∈ Z, n 6= 0.

The following are examples of rational

numbers (elements of Q):

1

3

22

7

−67

3 =24

8=

3

1

1.72 =172

100=

43

251 =

1

1

Number systems

Some numbers cannot be written as mn for

m,n ∈ Z. These are the irrational numbers.√2

3√9 π e log10 2

Combining the rational and irrational

numbers gives us the set of real numbers,

denoted R. Every x ∈ R has a decimal

expansion. For rationals, the decimal will

start to repeat at some point. For example:

1

3= 0.33333 . . . = 0.3

1

7= 0.142857

The real numbers

Q: Why the name?

A: To distinguish them from imaginarynumbers (explained next week).

Fun fact: there are as many integers as

there are positive whole numbers. In fact,

there are as many rational numbers as there

are positive whole numbers. However, there

are more real numbers than rationals.

Read more: “The Pea and the Sun” by

Leonard Wapner

The real numbers

The real numbers are totally ordered. We

can compare any two real numbers and say

whether the first one is bigger than the

second one, whether the second is bigger

than the first, or whether they are equal.

The following are examples of true

inequalities:

7 < 7.4 < 7.5 − π < −3√2 < 2

√2 6 2 2 6 2 − 10 <

√100

Important: there is a big difference between

6 and <, and also between > and >.

In order to score a distinction for MAT01A1

(or any module at UJ), you must have a final

mark > 75%.

You will not get exam entrance if your

semester mark is < 40%.

Set notation

A set is a collection of objects. If S is a set,

we write a ∈ S to say that a is an element

of S. We can also write a /∈ S to mean that

a is not an element of S.

Example: 3 ∈ Z but π /∈ Z.

Example of set-builder notation:

A = {1, 2, 3, 4, 5, 6}= {x ∈ Z | 0 < x < 7 }

Another example of set notation

{−2,−1, 0, 1, 2, 3} = {x ∈ Z | −2 6 x 6 3}= {x ∈ Z | −3 < x 6 3}= {x ∈ Z | −3 < x < 4}= {x ∈ Z | −2 6 x < 4}

Intervals For a, b ∈ R,

(a, b) = {x ∈ R | a < x < b }

whereas

[a, b] = {x ∈ R | a 6 x 6 b }.

Now let us look at Table 1 on page A4 of the

textbook. This shows how different intervals

can be written using interval notation,

set-builder notation, and how they can be

drawn on the real number line.

Intersections and Unions

Intersection of two intervals:

(a, b) ∩ (c, d) =

{x ∈ R | x ∈ (a, b) AND x ∈ (c, d)}= {x ∈ R | a < x < b AND c < x < d}

Union of two intervals:

(a, b) ∪ (c, d) =

{x ∈ R | x ∈ (a, b) OR x ∈ (c, d)}= {x ∈ R | a < x < b OR c < x < d}

Examples: unions & intersections

Simplify and give your answer in interval

notation.

I (1, 3) ∪ (2, 4]

I [−4,√2) ∩ [0, 1)

I [√3, 5) ∩ (1.8, 7)

I [0, 4) ∩((−2, 1) ∪ [3, 7)

)Recall,

√2 ≈ 1.41 . . . and

√3 ≈ 1.73 . . ..

Examples: unions & intersection

Solutions:

I (1, 3) ∪ (2, 4] = (1, 4]

I [−4,√2) ∩ [0, 1) = [0, 1)

I [√3, 5) ∩ (1.8, 7) = (1.8, 5)

I [0, 4) ∩((−2, 1) ∪ [3, 7)

)= [0, 1) ∪ [3, 4)

Understanding inequalities graphically

Consider the following functions:

f (x) = x2 − 1

g(x) = (x− 1)2

and the inequality g(x) < f (x).

Also, notice the difference between solving

for x in yesterday’s tut question:

x2 + 3x− 18 = 0

andx2 + 3x− 18 > 0

Inequalities

Let a, b, c ∈ R.

1. If a < b, then a + c < b + c.

2. If a < b and c < d, then a + c < b + d.

3. If a < b and c > 0, then ac < bc.

4. If a < b and c < 0, then ac > bc.

5. If 0 < a < b, then 1a >

1b .

Very important: note that for (3) and (4)

we must know the sign of c. We cannot

multiply or divide by a term if we do not

know its sign!

Solving inequalities

We will often make use of a number line or a

table to determine the sign of the function

on particular intervals. We use critical values

(where a function is 0 or undefined) to

determine the intervals.

Examples:

1. Solve for x: 1 + x < 7x + 5

2. Solve for x: x2 < 2x

Solving x2 < 2x

Solving inequalities continued

Find solutions to the following inequalities

and write the solutions in interval notation.

1. 4 6 3x− 2 < 13

2. x2 − 5x + 6 6 0

3. x3 + 3x2 > 4x (Don’t divide by x!)

4.x2 − x− 6

x2 + x− 66 0

5.x2 − 6

x6 −1 (Don’t multiply by x!)

Solution to (4)

Solution to (5)

Absolute value

The absolute value of a number a,

denoted by |a| is the distance from a to 0

along the real line. A distance is always

positive or equal to 0 so we have

|a| > 0 for all a ∈ R.

Examples

|3| = 3 | − 3| = 3 |0| = 0

|2−√3| = 2−

√3 |3− π| = π − 3

In general we have

|a| = a if a > 0

|a| = −a if a < 0

We can write the absolute value function as

a piecewise defined function.

|x| =

{x if x > 0

−x if x < 0

We can also replace the x above with

something more complicated.

Sketching y = |x|:

If f (x) = |x|, calculate f (−5), f (4) and

f (0).

I f (−5) = −(−5) = 5

I f (4) = 4

I f (0) = 0

Note: for any a ∈ R, |a| =√a2

Example

Write |3x− 2| without using the absolute

value symbol.

|3x− 2| =

{3x− 2 if 3x− 2 > 0

−(3x− 2) if 3x− 2 < 0

Hence

|3x− 2| =

{3x− 2 if x > 2

3

2− 3x if x < 23

Below is a sketch of y = |3x− 2|. Note how

the function changes at x = 23.

Properties of Absolute Values

Suppose a, b ∈ R and n ∈ Z. Then

1. |ab| = |a||b|2. |ab | =

|a||b| (b 6= 0)

3. |an| = |a|n

Let a > 0. Then

4. |x| = a if and only if x = a or x = −a.

5. |x| < a if and only if −a < x < a.

6. |x| > a if and only if x > a or x < −a.

Example: Solve |2x− 5| = 3.

Solving absolute value inequalities:

I Solve: |x− 5| < 2.

I Solve: |3x + 2| > 4.

To solve the first inequality above we must

solve

−2 < x− 5 < 2

For the second inequality we have

3x + 2 > 4 OR 3x + 2 6 −4

Solving |x− 5| < 2 gives x ∈ (3, 7).

Solving |3x + 2| > 4 gives

x ∈ (−∞,−2] ∪[23,∞

)

The triangle inequality

If a, b ∈ R, then |a + b| 6 |a| + |b|.

For an example of when this inequality is

strict, consider a = 1 and b = −2.

An example application:

If |x− 4| < 0.1 and |y − 7| < 0.2, use the

Triangle Inequality to estimate |(x+ y)− 11|.