Contents
0.1
Linear Functions and Linear Inequalities . . . . . . . . . . . . . .
1
0.2
The Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . .
5
0.3
Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . .
7
0.4
Maximum and Minimum . . . . . . . . . . . . . . . . . . . . . . .
8
0.4.1
The Plain Definition . . . . . . . . . . . . . . . . . . . . .
8
0.4.2
The Precise Definition . . . . . . . . . . . . . . . . . . . .
10
0.4.3
Negation of Propositions . . . . . . . . . . . . . . . . . . .
11
0.5
Functions and their Natural Domains
. . . . . . . . . . . . . . .
12
0.6
Appendix: Set Representation . . . . . . . . . . . . . . . . . . . .
12
Chapter 0 Functions and
Inequalities
Functions and inequalities play very important roles in the module Calculus
and Analysis. They will be explored in this preliminary chapter. From the
study of this chapter, we will not only review what we have learnt in middle
school, but have a primary idea of studying math like a mathmatician: stating
and proving mathematical propositions in a rigorous way.
0.1
Linear Functions and Linear Inequalities
We start with the linear case.
1
2
CONTENTS
Definition 1. We assume that k ̸= 0. An inequality of the form
kx + b > c, kx + b ≥ c, kx + b < c or kx + b ≤ c,
is said to be a linear inequality. A number x ∈ R satisfying the inequality is
said to be a solution to the inequality.
We will focus on how to solve such linear inequalities. When we say “solve
an inequality”, we expect to find all solutions satisfying the inequality. Here are
a few examples.
Example 1. Solve
3x + 2 > 5
Solution: By subtracting 2 on both sides, we have 3x > 3. Dividing 3 on both
sides yields x > 1. So the solution is x > 1.
Example 2. Solve
−2x + 4 ≤ 6
Solution: By subtracting 4 on both sides, we have −2x ≤ 2. Dividing −2 on
both sides yields x ≥ −1. So the solution is x ≥ −1.
Remark 1. Be reminded to change direction when you divide a negative number
on both sides of an inequality!
Here is another way to solve Example 1: Since 3x + 2 > 5, 5 > 2, then
3x + 2 > 2. By subtracting 2 on both sides, we have 3x > 0, hence x > 0.
What happened?
Remark 2. Make sure you do not devalue the information when solving an
inequality!
Here are a few examples that can be converted to linear inequalities:
Example 3. Solve
3x + 2 > x − 4
Solution: By subtracting x and subtracting 2 on both sides, we have 2x >
−6. This is equivalent to x > −3.
0.1. LINEAR FUNCTIONS AND LINEAR INEQUALITIES
3
In your homework, you do not have to write every word of the process. We
will adopt the following way
3x + 2 > x − 4
⇐⇒ 2x > −6
⇐⇒ x > −3
where ⇐⇒ is read as “if and only if”, meaning that for each step we keep all
the information with no devaluation at all.
Example 4. Solve
3(x + 4) − 6 < 7x − 2
Solution:
3(x + 4) − 6 < 7x − 2
⇐⇒ 3x + 12 < 7x − 2
⇐⇒ − 4x < −14
⇐⇒ x >
7
2
Now let us explore a few examples with paradox.
Example 5. Solve
x+2>x+1
Easy observation will lead us to substract x on both sides and this yeilds
2 > 1. What should be its solution?
Remark 3. When situation like this happens, it is the definition that judges
everything. We emphasize that solve an inequality is to find all solutions. In
this case, x + 2 > x + 1 ⇐⇒ 2 > 1 which is always true, suggesting that no
matter what x is, the inequality is always true. Hence the solution is the set
of all real numbers R.
What about the inequality
x + 1 ≥ x + 1?
4
CONTENTS
Definition 2. A linear function is of the form
f (x) = kx + b
defined for x ∈ R. Here k ̸= 0 is said to be the slope of the linear function.
We have learnt the graphs of a linear function in middle school. When k > 0,
the line goes upwards; when k < 0, the line goes downwards. These are shown
in below:
Figure 1: The line in blue goes downwards, and the line in red goes upwards
As studying in more advanced level, we will try to state and prove propositions more rigorously. Here, we normally apply the following three steps to
finish an argument. To start, we need to clarify the definitions involved. Then
the proposition needs to be stated rigorously. These two steps are to guarantee
that we are discussing the same concept in the same extent. Finally we give a
proof of the proposition to secure the statement.
We were discussing about how a linear function is going upwards or downwards as its slope changes. Now following the steps, let us study this like a
mathmatician.
0.2. THE ABSOLUTE VALUE
5
To start, we give the precise definition of “going upwards” and “going downwards”:
Definition 3. Let x1 , x2 be two real numbers and we assume x1 < x2 .
(1) If f (x1 ) ≤ f (x2 ), then we say f is an increasing function on R;
(2) If f (x1 ) ≥ f (x2 ), then we say f is an decreasing function on R.
Then, let us state the proposition:
Theorem 1. Let f (x) = kx + b be a linear function. Then
(1) If k > 0, the function is increasing;
(2) If k < 0, the function is decreasing.
Finally, we give a proof:
Proof. Let x1 , x2 be two real numbers and we assume x1 < x2 .
(1) Let k > 0. Then
f (x1 ) − f (x2 ) = (kx1 + b) − (kx2 + b) = k(x1 − x2 ) + b − b = k(x1 − x2 ) < 0
(2) Let k < 0. Then
f (x1 ) − f (x2 ) = (kx1 + b) − (kx2 + b) = k(x1 − x2 ) + b − b = k(x1 − x2 ) > 0
Remark 4. In the definition, if hoping f to be increasing, we require f (x1 ) ≤
f (x2 ). However in the previous proof of (1), we have only obtained f (x1 ) <
f (x2 ).Why can we draw the conclusion directly?
0.2
The Absolute Value
Definition 4. The absolute value or modulus of a real number x is denoted by
|x|, and it is defined as follows:
|x| =

 x if x ⩾ 0

−x if x < 0
6
CONTENTS
Theorem 2. ∀c ∈ R and δ > 0, we have:
|x − c| < δ
⇔
c−δ <x<c+δ
Proof. (⇒) Suppose that |x − c| < δ.
Then x − c ⩽ |x − c| < δ, and
−(x − c) ⩽ |x − c| < δ. So −δ < x − c < δ, that is c − δ < x < c + δ.
(⇐) If c − δ < x < c + δ, then x − c < δ and −(x − c) = c − x < δ. Thus
|x − c| < δ, because |x − c| is either x − c or −(x − c), and in both cases the
numbers are less than δ.
The previous theorem is worth memorying because it will arise over and over
again in future study, especially for the case c = 0, when the theorem becomes
|x| < δ
⇔
−δ < x < δ
(1)
Similarly we can have
|x| = δ
⇔
x = δ or x = −δ
(2)
|x| > δ
⇔
x > δ or x < −δ
(3)
and
With (1), (2) and (3), we can solve inequalities with absolute value. Here
are an example:
Example 6. Solve
| − 2x + 5| ≤ 7
Solution:
| − 2x + 5| ≤ 7
⇐⇒ − 7 ≤ −2x + 5 ≤ 7
⇐⇒ − 12 ≤ −2x ≤ 2
⇐⇒ 6 ≥ x ≥ −1
Hence the solution is −1 ≤ x ≤ 6.
The following inequality is vitally important in future analysis study.
0.3. TRIGONOMETRIC FUNCTIONS
7
Theorem 3 (Triangular Inequality). If x, y are real numbers, then
|x + y| ⩽ |x| + |y|
Proof. For the inequality, we notice that
−|x| ≤ x ≤ |x|
and − |y| ≤ y ≤ |y|
hold for x, y ∈ R. By adding these two inequalities, we have
−(|x| + |y|) ≤ x + y ≤ |x| + |y|
From (1) we have the desired result.
0.3
Trigonometric Functions
You have probably learnt the definitions of trigonometric functions defined in a
right triangle. However, that definition only works for angle between 0 degree
and 90 degrees. In order to extend the definition to real numbers, we seek the
help from the unit circle:
α
sin α
α
cos α
Now we give the definition via the gragh:
1
8
CONTENTS
Definition 5. Consider the unit circle, a circle with radius 1, in the coordinate
system. Then,
(1) We define its perimeter as 2π;
(2) We define the angle as a real number α if the length of its corresponding arc
is α;
(3) We define the sin and cos of α as the length of the line segments showed in
the graph.
From the definition and the graph, we can easily have the following results:
Proposition 1. (1) π > 2;
(2) | sin x| ≤ 1 and | cos x| ≤ 1 hold for x ∈ R;
(3) sin(−x) = − sin x and cos(−x) = cos x hold for x ∈ R.
Finally let us use proof by cases to prove an important inequality that will
be applicable in the future chapters.
Theorem 4.
| sin x| ≤ |x|,
x∈R
Proof. Case 1: When x ∈ (0, 21 ), then based on the graph above, we have
sin x < x. With x > 0 and sin x > 0, we would have |sin x| ≤ |x|
Case 2: When x ≥
π
2,
then we have x ≥
π
2
> 1 ≥ | sin x|
Therefore, with Case 1 and Case 2, we can obtain when x > 0, |x| ≥ | sin x|.
Now we consider Case 3.
Case 3: When x < 0, then we have −x > 0. Based on our discussion in Case 1
and Case 2, | − sin x| = | sin (−x)| ≤ | − x|. Then, we would have | sin x| ≤ |x|
as desired.
0.4
Maximum and Minimum
In this section we will discuss the maximum and minimum of a set.
0.4.1
The Plain Definition
In most of the cases, their names have explained themselves. Let S be a set of
real numbers. We denote by
max S
0.4. MAXIMUM AND MINIMUM
9
the largest number of S, and
min S
the smallest number of S.
Example 7. Let S = {−1, 4}. Then max S = 4 and min S = −1.
It seems very simple that we do not need any further examples.
Now we turn to inequalities involved maximum and minimum. Firstly let
us consider
max{a, b} ≤ 2
what can we say about a and b?
What if
min{a, b} ≤ 2?
Slightly out of our expectation, the maximum and minimum have secret
connections to the logic words “and ” and “or”. Applying this observation, we
can solve the following inequalities:
Example 8.
max{2x + 4, | − 3x + 6|} ≤ 8
Solution: The original inequality is equivalent to
2x + 4 ≤ 8
and
| − 3x + 6| ≤ 8
The solution to the first one is {x : x ≤ 2}; the second one is equivalent to
− 8 ≤ −3x + 6 ≤ 8
⇐⇒ − 14 ≤ −3x ≤ 2
⇐⇒ −
2
14
≤x≤
3
3
Because we need both inequalities, the solution to the original inequality
10
CONTENTS
should be the intersection of the two sets
{x : x ≤ 2} ∩ {x : −
2
14
2
≤x≤
} = {x : − ≤ x ≤ 2}
3
3
3
Remark 5. Let S ⊂ R. From the previous observation, we can conclude that:
(1) max S ≤ α if and only if any number in the set is ≤ α;
(2) max S ≥ α if and only if there exists a number in the set ≥ α;
(3) min S ≤ α if and only if there exists a number in the set ≤ α;
(4) min S ≥ α if and only if any number in the set is ≥ α.
Now any and there exists have arised a decent number of times to merit
them notations. We reverse the capital letter A, the initial letter of the word
any, as ∀, read as “for any” or “for all”; the capital letter E, the initial letter of
the word exist, as ∃, read as “there exist(s)”. With these notations, the previous
remark can be rewritten as elegant as:
(1) max S ≤ α
⇐⇒
∀s ∈ S, s ≤ α;
(2) max S ≥ α
⇐⇒
∃s ∈ S, s ≥ α.
Please finish the other two about min S to help get used to these notations.
0.4.2
The Precise Definition
The concepts of maximum and minimum make sense when we talk them among
finitely many numbers. However, when we consider the set of infinitely many
numbers, such as the set of all positive numbers S = {x : x > 0}, what is its
maximum? What is its minimum?
Or does every set have to have a maximum or minimum?
As long as there is an argument like this, we need more precise definition.
With the help of the notations ∀ and ∃ we have just learnt, the precise definition
looks very compact:
Definition 6 (Maximum and Minimum). Let S ⊂ R be a set of real numbers.
We say M ∈ R is the maximum in S, if
(1) M ∈ S;
(2) ∀s ∈ S, s ≤ M .
0.4. MAXIMUM AND MINIMUM
11
We say m ∈ R is the minimum in S, if
(1) m ∈ S;
(2) ∀s ∈ S, s ≥ m.
In the definition we emphasize that the maximum or minimum must be an
element in the set.
By applying the precise definition, we can now answer the question raised
previously:
Theorem 5. There is no minimum in the set of positive numbers. In other
words, if S = {x : x > 0}, then min S does not exist.
Proof. We argue this by contradiction. Assume min S exists, let us say, min S =
α. Then according to the definition,
(1) α ∈ S. In this case, this means α > 0.
(2) ∀s ∈ S, α ≤ s.
But if we consider β =
On the other hand,
0.4.3
α
2
α
2.
On the one hand, since α > 0, β =
α
2
> 0, β ∈ S.
< α. This is a contradiction to (2).
Negation of Propositions
From the last theorem, we can see that arguing by contradiction is very useful
in proving propositions containing “not” or “no”. We normally use ¬ to denote
the negation of a proposition. For example, ¬x ≤ 4 is equivalent to x > 4.
In the last proof, we finally constructed a new number, β =
α
2
α
2,
and it satisfies
< α. Essentially, we have obtained
∃β ∈ S, β < α
and concluded a contradiction. In general, if we denote P (x) is a proposition
involving x, then
¬∀x ∈ S, P (x)
⇐⇒
∃x ∈ S, ¬P (x)
¬∃x ∈ S, P (x)
⇐⇒
∀x ∈ S, ¬P (x)
and
12
CONTENTS
0.5
Functions and their Natural Domains
We have discussed several types of functions so far: linear functions, absolute
functions, trigonometric functions, maximum functions and minimum functions.
Quadratic functions, exponential functions and logarithmic functions were introduced in your middle school. All of these functions contain a formula, or a
representation, looking like f (x) =. Indeed, the concept of function is not so
narrow as this. In this section, we introduce the precise definition of a function
and its natural domain.
Definition 7. Let A, B ⊂ R. We define f : A → B be a function, if ∀a ∈ A,
∃!b ∈ B, such that we have the correspondence f (a) = b. We say A to be the
domain of f , B the codomain of f .
Here ! showing up in the definition means ”unique”.
Example 9. (1) Define a function f : R → R with the correspondence f (x) =
x2 . We can check that this is the a well-defined function from the definition.
(2) How about f : R → [0, +∞) with the correspondence f (x) = x2 ?
(3) How about f : R → [1, +∞) with the correspondence f (x) = x2 ?
Example 10. Define g : R → R with the correspondence g(x) =
1
x.
What is
g(0)?
From the last example, we can realize that the domain of a function can
sometimes be not as large as the set R. This also shows up in other cases, for
√
example f (x) = x can only be defined when x ≥ 0. Hence we define the
natural domain of a function as its largest possible domain as a subset in R.
0.6
Appendix: Set Representation
We take this oppotunity to standardise how to represent a set. The notation
that most of the modern mathematician are adopting is like a sentance with an
attributive clause. In English, we usually say
The real numbers that are between 1 (included) and 2 (not included)
0.6. APPENDIX: SET REPRESENTATION
13
In math, we write the set as
{x ∈ R : 1 ≤ x < 2}
Sometimes we write in short
{x : 1 ≤ x < 2}
when there is no ambiguity from the context.
You may wonder the possibility of writing the set as
{1 ≤ x < 2}
This is not recommended, because in future study we will encounter cases like
{a ≤ x}
But ambituity arises because we can argue which letter is the “main character”.
In other words, from the set can you say whether we are saying those a less than
or equal to x, or those x greater than or equal to a? Because of this possible
ambiguity, we write {a : a ≤ x} or {x : a ≤ x} instead.
When sets involve “supporting character”, like the previous one, we need
also claim the largest possible area (indeed the natural domain) of those letters.
Let us explore the following example.
Example 11. Define tan x =
sin x
cos x .
Write the natural domain of the function
f (x) = tan x.
Because cos x appears in denominator, this is equivalent to find all real
numbers such that cos x ̸= 0. Applying what we have learnt, we can write a
raw answer:
{x ∈ R : cos x ̸= 0}
Nextly we try to solve the equation cos x = 0. From the definition of cos,
combining the following graph:
14
CONTENTS
Figure 2: Graph of cos x
we can see that the zero points appear at x = ± π2 , ± 3π
2 , · · · , so we may write
the natural domain as
π 3π
{x ∈ R : x ̸= ± , ± , · · · }
2
2
This has been fine as an official answer. To write it more elegant, we can try
this way:
{x ∈ R : x ̸= ±
(2k + 1)π
, k ∈ Z}
2
here the supporting character k appears. Besides what we have discussed previously, we also emphasize that we need to give all the possible values for all
those supporting characters. In this case, k is the supporting character, and we
write k ∈ Z to restrict k as an integer.