MAT01A1 0.5cm Numbers, Inequalities and Absolute Values · PDF fileMAT01A1 Numbers, Inequalities and Absolute Values (Appendix A) Dr Craig 8 February 2017

MAT01A1

Numbers, Inequalities and Absolute Values

(Appendix A)

Dr Craig

8 February 2017

Leftovers from yesterday:

limn→∞

n∑i=1

3

n

[( in

)2+ 1]= lim

n→∞

3

n

n∑i=1

[ i2n2

+ 1]

= limn→∞

3

n

( n∑i=1

i2

n2+

n∑i=1

1)

= limn→∞

3

n

( 1

n2× n(n + 1)(2n + 1)

6+ n)

= limn→∞

(2n2 + 3n + 2

2n2+ 3)

= limn→∞

(1 +

3

2n+

1

n2+ 3)= 4

Introduction

Who:

Dr Craig

What:

Lecturer & course coordinator for MAT01A1

Where:

C-Ring 508 [email protected]

Web:

http://andrewcraigmaths.wordpress.com

Important information

Course 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

Lecture times: Tuesday 08h50 – 10h25

Wednesdays 17h10 – 18h45

Lecture venues: C-LES 102, C-LES 103

Tutorials: Tuesday afternoons (only!)

13h50 – 15h25: D-LES 104 or D-LES 106

OR

15h30 – 17h05: C-LES 203 or D1 LAB 408

Other announcements

I No tuts for MAT01A1 on Wednesdays. If

you see this on your timetable, it is an

error. (To move your Chem. prac., email

Mr Kgatshe [email protected])

I CSC02A2 students. Email Dr Craig

regarding tutorial clash.

I Maths Learning Centre in C-Ring 512:

10h30 – 14h35 Mondays

08h00 – 15h30 Tuesday to Thursday

08h00 – 12h55 Fridays

Lecturers’ Consultation Hours

Monday:

10h30 – 11h30 Ms Richardson (C-503)

Wednesday:

14h30 – 16h00 Ms Richardson (C-503)

Thursday:

11h00 – 13h00 Dr Craig (C-508)

13h30 – 14h00 Ms Richardson (C-503)

Friday:

11h30 – 13h00 Dr Craig (C-508)

Today’s lecture:

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 rationalnumbers. 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

10=

86

51 =

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 on 15 Feb.)

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 }

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.

Understanding inequalities graphically

Consider the following functions:

f (x) = x2 − 1

g(x) = (x− 1)2

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

Solving for x (yesterday’s tut question):

x2 + 3x− 18 = 0

and

x2 + 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 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!)

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:

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,∞

)

Next time

The triangle inequality

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

Applications:

Example: If |x− 4| < 0.1 and |y − 7| < 0.2,

use the Triangle Inequality to estimate

|(x + y)− 11|.

Exercise: show that if |x + 3| < 12, then

|4x + 13| < 3.