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MAT01A1 0.5cm Numbers, Inequalities and Absolute Values ......Solving absolute value inequalities: I Solve: jx 5j 4. To solve the rst inequality above we

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  • 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 [email protected]

    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 andDFC. 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 systemsI Set notationI InequalitiesI Absolute value

  • Number systems

    The integers are the set of all positive andnegative 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 ratiosof 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 decimalexpansion. 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 asthere 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” byLeonard Wapner

  • The real numbers

    The real numbers are totally ordered. Wecan 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 between6 and 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 elementof S. We can also write a /∈ S to mean thata 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 − 1g(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− 6x2 + x− 6

    6 0

    5.x2 − 6x

    6 −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 0along 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) andf (0).

    I f (−5) = −(−5) = 5I f (4) = 4I f (0) = 0

    Note: for any a ∈ R, |a| =√a2

  • Example

    Write |3x− 2| without using the absolutevalue symbol.

    |3x− 2| =

    {3x− 2 if 3x− 2 > 0−(3x− 2) if 3x− 2 < 0

    Hence

    |3x− 2| =

    {3x− 2 if x > 232− 3x if x < 23

  • Below is a sketch of y = |3x− 2|. Note howthe function changes at x = 23.

  • Properties of Absolute Values

    Suppose a, b ∈ R and n ∈ Z. Then1. |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 < 2For 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 givesx ∈ (−∞,−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 theTriangle Inequality to estimate |(x+ y)− 11|.