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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)
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.
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
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