mat01a1 0.5cm numbers, inequalities and absolute values ......solving absolute value inequalities: i...
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MAT01A1
Numbers, Inequalities and Absolute Values
(Appendix A)
Dr Craig
5/6 February 2020
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Introduction
Who:
Dr Craig
What:
Lecturer & course coordinator for MAT01A1
Where:
C-Ring 508 [email protected]
Web:
http://andrewcraigmaths.wordpress.com
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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.
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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.
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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
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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)
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Appendix A:
I Number systems
I Set notation
I Inequalities
I Absolute value
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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.
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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
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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
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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
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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
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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%.
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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 }
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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}
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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.
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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}
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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 . . ..
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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)
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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
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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!
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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
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Solving x2 < 2x
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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!)
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Solution to (4)
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Solution to (5)
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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
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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.
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Sketching y = |x|:
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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
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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
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Below is a sketch of y = |3x− 2|. Note how
the function changes at x = 23.
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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.
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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
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Solving |x− 5| < 2 gives x ∈ (3, 7).
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Solving |3x + 2| > 4 gives
x ∈ (−∞,−2] ∪[23,∞
)
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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|.