Basic function language

This chapter introduces the fundamental concepts underpinning function theory within Algebra and Functions. It meticulously defines sets, ordered pairs, relations, and their associated properties—domain, range, and codomain—laying the essential groundwork for advanced topics. Mastery of these foundational elements is critical for success in CMI examinations, as they constitute the prerequisite knowledge for comprehending subsequent functional analysis.

---

Chapter Contents

| Topic |

|---|-------| | 1 | Sets and subsets | | 2 | Ordered pairs and Cartesian product | | 3 | Relations | | 4 | Domain | | 5 | Range | | 6 | Codomain |

---

We begin with Sets and subsets.

Part 1: Sets and subsets

Sets and Subsets

Overview

Sets are the basic language of modern mathematics. Before studying functions, relations, and mappings, one must be comfortable with membership, subsets, set operations, and the logic of inclusion. In CMI-style questions, this topic is usually tested through careful definitions, counting subsets, power sets, interval/set description, and symbolic manipulation using union, intersection, and complement. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

understand the meaning of a set, an element, and a subset

distinguish between $\in$ and $\subseteq$

work with union, intersection, difference, and complement

count subsets and understand the power set

translate between verbal, roster, and set-builder descriptions

avoid common logical errors involving empty set and singleton sets

---

What is a Set?

📖 Set

A set is a well-defined collection of objects.

If an object $x$ belongs to a set $A$ , we write

$\qquad x \in A$

If $x$ does not belong to $A$ , we write

$\qquad x \notin A$

📐 Ways to Describe a Set

A set can be written in different ways:

Roster form

\qquad A=\{1,2,3,4\}

Set-builder form

\qquad A=\{x \in \mathbb{N} : 1 \le x \le 4\}

Descriptive form

\qquad A=

the set of natural numbers from

1

4

---

Basic Symbols

📐 Membership and Inclusion Symbols

$x \in A$ means $x$ is an element of $A$

$x \notin A$ means $x$ is not an element of $A$

$A \subseteq B$ means every element of $A$ is also an element of $B$

$A \subset B$ means $A$ is a proper subset of $B$

$A = B$ means both sets have exactly the same elements

⚠️ Do Not Mix Up These Symbols

$\in$ compares an element with a set

$\subseteq$ compares a set with a set

For example, if

A=\{1,2,3\}

, then

$1 \in A$ is true

$\{1\} \subseteq A$ is true

$1 \subseteq A$ is meaningless

$\{1\} \in A$ is false

---

Subsets

📖 Subset

A set $A$ is a subset of a set $B$ if every element of $A$ belongs to $B$ .

We write

$\qquad A \subseteq B$

If $A \subseteq B$ and $A \ne B$ , then $A$ is a proper subset of $B$ , written as

$\qquad A \subset B$

❗ Important Facts About Subsets

For every set $A$ :

$\varnothing \subseteq A$

$A \subseteq A$

So every set has at least two subsets:

the empty set

the set itself

---

Empty Set

📖 Empty Set

The empty set is the set with no elements. It is denoted by

$\qquad \varnothing$

or sometimes by $\{\}$ .

⚠️ Very Important Distinction

These are different:

$\varnothing$ = empty set

$\{\varnothing\}$ = a set whose only element is the empty set

So:

$\varnothing$ has $0$ elements

$\{\varnothing\}$ has $1$ element

---

Equality of Sets

📐 When Two Sets Are Equal

Two sets $A$ and $B$ are equal if and only if they contain exactly the same elements.

A common proof method is:

show $A \subseteq B$

show $B \subseteq A$

Then conclude

\qquad A=B

---

Set Operations

📐 Union, Intersection, Difference

For sets $A$ and $B$ :

Union:

\qquad A \cup B = \{x : x \in A \text{ or } x \in B\}

Intersection:

\qquad A \cap B = \{x : x \in A \text{ and } x \in B\}

Difference:

\qquad A \setminus B = \{x : x \in A \text{ and } x \notin B\}

Complement:

U

is the universal set, then

\qquad A^c = U \setminus A

📐 Basic Identities

$A \cup \varnothing = A$

$A \cap \varnothing = \varnothing$

$A \cup A = A$

$A \cap A = A$

$A \cup U = U$

$A \cap U = A$

$A \cup A^c = U$

$A \cap A^c = \varnothing$

---

De Morgan's Laws

📐 De Morgan's Laws

With respect to a universal set $U$ ,

$(A \cup B)^c = A^c \cap B^c$

$(A \cap B)^c = A^c \cup B^c$

These are very important in logic, set algebra, and function language.

---

Power Set and Number of Subsets

📖 Power Set

The power set of a set $A$ , denoted by $\mathcal{P}(A)$ , is the set of all subsets of $A$ .

📐 Counting Subsets

If a finite set $A$ has $n$ elements, then:

number of subsets of $A$ is $\qquad 2^n$

number of proper subsets of $A$ is $\qquad 2^n - 1$

number of non-empty subsets of $A$ is $\qquad 2^n - 1$

number of non-empty proper subsets of $A$ is $\qquad 2^n - 2$

💡 Why

2^n

For each element, there are exactly two choices:

include it

do not include it

So for

n

independent choices, the total number of subsets is

\qquad 2 \cdot 2 \cdot \ldots \cdot 2 = 2^n

---

Intervals as Sets

📐 Useful Interval Language

Real-number sets are often written in interval notation:

$(a,b)=\{x \in \mathbb{R}: a<x<b\}$

$[a,b]=\{x \in \mathbb{R}: a\le x\le b\}$

$(a,b]=\{x \in \mathbb{R}: a<x\le b\}$

$[a,b)=\{x \in \mathbb{R}: a\le x<b\}$

This language becomes important later in domains and ranges of functions.

---

Minimal Worked Examples

Example 1 Let

\qquad A=\{1,2,3\}

List all subsets of

A

. The subsets are

\qquad \varnothing,\ \{1\},\ \{2\},\ \{3\},\ \{1,2\},\ \{1,3\},\ \{2,3\},\ \{1,2,3\}

So there are

\qquad 2^3=8

subsets. --- Example 2 Let

\qquad A=\{1,2,3,4\}, \quad B=\{3,4,5\}

Then

\qquad A \cap B = \{3,4\}

\qquad A \cup B = \{1,2,3,4,5\}

\qquad A \setminus B = \{1,2\}

\qquad B \setminus A = \{5\}

---

Common Logical Traps

⚠️ Avoid These Errors

❌ thinking $\varnothing \in A$ for every set $A$

❌ confusing $\varnothing$ with $\{\varnothing\}$

❌ confusing $x \in A$ with $\{x\} \subseteq A$

❌ forgetting that order does not matter in sets

❌ repeating an element while counting set size

❌ assuming $A \setminus B = B \setminus A$

---

Strategy for CMI-Type Questions

💡 CMI Strategy

Read symbols very carefully. Many questions are really about notation.

When proving equality, use double inclusion.

When counting subsets, think in binary choice form.

For symbolic set algebra, translate union as “or” and intersection as “and”.

Keep empty set and singleton-set distinctions explicit.

In interval problems, rewrite in set-builder form if needed.

---

Practice Questions

:::question type="MCQ" question="If

A=\{1,2,3\}

, which of the following is true?" options=["

1 \subseteq A

","

\\{1\\} \in A

","

\\{1\\} \subseteq A

","

\varnothing \in A

"] answer="C" hint="Distinguish carefully between membership and subset." solution="We check each statement.

1 \subseteq A

is meaningless because

1

is an element, not a set.

\{1\} \in A

is false because the elements of

A

are

1,2,3

, not the set

\{1\}

\{1\} \subseteq A

is true because every element of

\{1\}

belongs to

A

\varnothing \in A

is false because the empty set is not listed as an element of

A

Hence the correct option is

\boxed{C}

." ::: :::question type="NAT" question="A set has

4

elements. How many non-empty proper subsets does it have?" answer="14" hint="Use the total number of subsets first." solution="If a set has

4

elements, then the total number of subsets is

\qquad 2^4 = 16

Among these:

one subset is the empty set

one subset is the set itself

So the number of non-empty proper subsets is

\qquad 16 - 2 = 14

Therefore the answer is

\boxed{14}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["

\varnothing \subseteq A

for every set

A

","If

A \subseteq B

and

B \subseteq A

, then

A=B

","

\\{\varnothing\\}

has no elements","If a set has

n

elements, then it has

2^n

subsets"] answer="A,B,D" hint="Recall the role of the empty set and the power set." solution="1. True. The empty set is a subset of every set.

True. This is the double inclusion criterion for equality of sets.

False. The set

\{\varnothing\}

has exactly one element, namely

\varnothing

True. A set with

n

elements has

2^n

subsets.

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="Let

A=\{1,2,3,4\}

and

B=\{2,4,6\}

. Find

A \cap B

A \cup B

, and

A \setminus B

." answer="

A \cap B=\\{2,4\\},\\ A \cup B=\\{1,2,3,4,6\\},\\ A \setminus B=\\{1,3\\}

" hint="Intersection means common elements, union means all elements, and difference means elements of

A

not in

B

." solution="We compare the elements of the two sets. The elements common to both

A

and

B

are

2

and

4

, so

\qquad A \cap B = \{2,4\}

The union contains all distinct elements from both sets, so

\qquad A \cup B = \{1,2,3,4,6\}

The difference

A \setminus B

means elements of

A

that are not in

B

, so

\qquad A \setminus B = \{1,3\}

Therefore,

\boxed{A \cap B=\{2,4\},\ A \cup B=\{1,2,3,4,6\},\ A \setminus B=\{1,3\}}

." ::: ---

Summary

❗ Key Takeaways for CMI

$\in$ is for element-to-set comparison, while $\subseteq$ is for set-to-set comparison.

The empty set is a subset of every set.

A set with $n$ elements has $2^n$ subsets.

$\varnothing$ and $\{\varnothing\}$ are completely different objects.

Union means “or”, intersection means “and”, and set difference means “in one but not the other”.

Sets are the language in which relations and functions are naturally expressed.

---

💡 Next Up

Proceeding to Ordered pairs and Cartesian product.

---

Part 2: Ordered pairs and Cartesian product

Ordered Pairs and Cartesian Product

Overview

Ordered pairs and Cartesian products form the language in which relations and functions are written. This topic looks elementary, but in CMI-style questions it is often tested through counting, set construction, domain-codomain reasoning, and logical precision. The key point is that order matters in ordered pairs, and Cartesian products create the universe from which relations and functions are built. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

define and interpret an ordered pair correctly.

distinguish between $(a,b)$ and $(b,a)$ .

construct Cartesian products of finite sets.

count the number of elements in a Cartesian product.

describe relations and functions as subsets of Cartesian products.

---

Ordered Pair

📖 Ordered Pair

An ordered pair is an expression of the form

$\qquad (a,b)$

where:

$a$ is called the first component

$b$ is called the second component

Two ordered pairs are equal if and only if their corresponding components are equal:

\qquad (a,b) = (c,d) \iff a=c \text{ and } b=d

❗ Most Important Fact

In general,

$\qquad (a,b) \ne (b,a)$

unless

$\qquad a=b$

So order matters.

---

Basic Examples

$(2,5)$ and $(5,2)$ are different ordered pairs

$(3,3)$ and $(3,3)$ are equal

if $(x+1,2x)=(5,8)$ , then

\qquad x+1=5

and

\qquad 2x=8

x=4

---

Cartesian Product

📖 Cartesian Product of Two Sets

If $A$ and $B$ are sets, then the Cartesian product of $A$ and $B$ is the set

$\qquad A \times B = \{(a,b) : a \in A,\ b \in B\}$

That is, $A \times B$ is the set of all ordered pairs whose first component comes from $A$ and second component comes from $B$ .

📐 Finite Set Cardinality

If $A$ and $B$ are finite sets, then

$\qquad |A \times B| = |A|\cdot|B|$

---

Examples of Cartesian Products

\qquad A=\{1,2\},\quad B=\{a,b,c\}

then

\qquad A\times B=\{(1,a),(1,b),(1,c),(2,a),(2,b),(2,c)\}

\qquad |A\times B|=2\cdot 3=6

But

\qquad B\times A=\{(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)\}

and this is a different set from

A\times B

. ---

Important Properties

📐 Basic Properties

If $A=\varnothing$ or $B=\varnothing$ , then

\qquad A\times B=\varnothing

In general,

\qquad A\times B \ne B\times A

If $A \subseteq C$ and $B \subseteq D$ , then

\qquad A\times B \subseteq C\times D

If $A$ and $B$ are finite, then

\qquad |A\times B|=|A||B|

⚠️ Do Not Confuse These

$A\times B$ is a set of ordered pairs

$A\cup B$ is a set of elements

$A\times B$ is not the same thing as $\{a,b\}$

$A\times A$ is not the same as $A$

---

Ordered Pair Equality

📐 Equality Rule

If

$\qquad (p,q)=(r,s)$

then necessarily

$\qquad p=r \quad \text{and} \quad q=s$

This is one of the fastest tools in algebraic questions involving ordered pairs.

Example If

\qquad (2x-1,\ x+3)=(5,\ 7)

then

\qquad 2x-1=5

and

\qquad x+3=7

Both give

\qquad x=3

So the equality is consistent. ---

Cartesian Product with More Than Two Sets

📖 Triple Cartesian Product

For sets $A,B,C$ ,

$\qquad A\times B\times C = \{(a,b,c): a\in A,\ b\in B,\ c\in C\}$

If all three sets are finite, then

$\qquad |A\times B\times C|=|A||B||C|$

Example If

\qquad A=\{1,2\},\ B=\{0,1\},\ C=\{x,y\}

then

\qquad |A\times B\times C|=2\cdot 2\cdot 2=8

---

Relations as Subsets of Cartesian Products

📖 Relation

A relation from a set $A$ to a set $B$ is any subset of

$\qquad A\times B$

So a relation does not need to contain all ordered pairs; it can contain some of them.

Example If

\qquad A=\{1,2\},\ B=\{a,b\}

then

\qquad A\times B=\{(1,a),(1,b),(2,a),(2,b)\}

A possible relation is

\qquad R=\{(1,a),(2,b)\}

since

R \subseteq A\times B

. ---

Functions as Special Relations

📖 Function

A function from $A$ to $B$ is a relation $f \subseteq A\times B$ such that each element of $A$ appears exactly once as first component.

So for every $a\in A$ , there is exactly one $b\in B$ such that

$\qquad (a,b)\in f$

💡 Why Ordered Pairs Matter in Functions

A function is literally described by ordered pairs:

first component = input

second component = output

That is why understanding ordered pairs and Cartesian products is necessary before studying functions properly.

---

Counting Questions

📐 How Many Ordered Pairs?

If $|A|=m$ and $|B|=n$ , then the number of ordered pairs in $A\times B$ is

$\qquad mn$

If order matters and repetition is allowed from the same set $A$ , then the count of ordered pairs from $A\times A$ is

$\qquad |A|^2$

Example If

A=\{1,2,3,4\}

, then

\qquad |A\times A|=4^2=16

---

Common Mistakes

⚠️ Avoid These Errors

❌ Writing $(a,b)=(b,a)$ automatically

✅ They are equal only when

a=b

❌ Thinking $A\times B = B\times A$

✅ These are usually different because order changes.

❌ Forgetting that $A\times B$ contains ordered pairs, not ordinary elements

✅ Each element has the form

(a,b)

❌ Adding cardinalities instead of multiplying

✅ For finite sets,

\qquad |A\times B|=|A||B|

❌ Thinking every subset of $A\times B$ is a function

✅ A function is a very special kind of subset.

---

CMI Strategy

💡 How to Think in Exam Questions

First check whether order matters.

If the question asks for all possible input-output combinations, think Cartesian product.

If the question asks for a relation, think subset of a Cartesian product.

If the question asks for a function, check uniqueness of second component for each first component.

In counting problems, multiply sizes, do not add them.

---

Practice Questions

:::question type="MCQ" question="If

A=\{1,2\}

and

B=\{a,b,c\}

, then how many elements are there in

A\times B

?" options=["

3

","

5

","

6

","

8

"] answer="C" hint="Use the formula for the cardinality of a Cartesian product." solution="We have

\qquad |A|=2,\quad |B|=3

\qquad |A\times B|=|A||B|=2\cdot 3=6

Therefore the correct answer is

\boxed{C}

." ::: :::question type="NAT" question="If

(2x+1,\ 3x-2)=(7,\ 7)

, find the value of

x

." answer="3" hint="Equate corresponding components." solution="Equality of ordered pairs gives:

\qquad 2x+1=7

and

\qquad 3x-2=7

From the first,

\qquad 2x=6

\qquad x=3

Check in the second:

\qquad 3(3)-2=9-2=7

So the value of

x

\boxed{3}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If

(a,b)=(c,d)

, then

a=c

and

b=d

.","In general,

A\times B = B\times A

.","A relation from

A

B

is a subset of

A\times B

.","If

A

is a set with

4

elements, then

|A\times A|=8

."] answer="A,C" hint="Check equality, order, and counting carefully." solution="1. True. This is the defining property of equality of ordered pairs.

False. In general,

A\times B

and

B\times A

are different because order matters.

True. By definition, a relation from

A

B

is a subset of

A\times B

False. If

|A|=4

, then

\qquad |A\times A|=4\cdot 4=16

Hence the correct answer is

\boxed{A,C}

." ::: :::question type="SUB" question="Let

A=\{1,2\}

and

B=\{x,y,z\}

. Write

A\times B

explicitly and explain why

A\times B \ne B\times A

." answer="

A\times B=\\{(1,x),(1,y),(1,z),(2,x),(2,y),(2,z)\\}

and order matters, so

A\times B \ne B\times A

." hint="List all ordered pairs with first component from

A

and second from

B

." solution="By definition,

\qquad A\times B=\{(a,b):a\in A,\ b\in B\}

\qquad A\times B=\{(1,x),(1,y),(1,z),(2,x),(2,y),(2,z)\}

Now

\qquad B\times A=\{(x,1),(x,2),(y,1),(y,2),(z,1),(z,2)\}

These are not the same, because for example

\qquad (1,x)\in A\times B

but

\qquad (1,x)\notin B\times A

Hence

\qquad \boxed{A\times B \ne B\times A}

The reason is that ordered pairs depend on the order of components." ::: ---

Summary

❗ Key Takeaways for CMI

An ordered pair $(a,b)$ remembers order.

$(a,b)=(c,d)$ only when both corresponding components are equal.

$A\times B$ is the set of all ordered pairs with first entry from $A$ and second from $B$ .

For finite sets, $|A\times B|=|A||B|$ .

Relations are subsets of Cartesian products.

Functions are special relations with exactly one output for each input.

---

💡 Next Up

Proceeding to Relations.

---

Part 3: Relations

Relations

Overview

Relations are the language used to describe how elements of one set are connected to elements of another set, or to the same set. In CMI-style algebra, this topic is not just about definitions: it tests whether you can move comfortably between set notation, ordered pairs, properties of relations, and the transition from a general relation to a special relation like a function, an equivalence relation, or an order relation. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

define a relation as a subset of a Cartesian product

find domain, codomain, and range of a relation

test whether a relation is reflexive, symmetric, antisymmetric, and transitive

distinguish between a general relation and a function

identify equivalence relations and partial orders

---

Core Definition

📖 Relation Between Two Sets

Let $A$ and $B$ be sets.

A relation $R$ from $A$ to $B$ is any subset of the Cartesian product $A\times B$ .

So,

$\qquad R \subseteq A\times B$

If $(a,b)\in R$ , we write

$\qquad aRb$

which means that $a$ is related to $b$ under the relation $R$ .

📐 Cartesian Product

If $A$ and $B$ are sets, then

$\qquad A\times B = \{(a,b): a\in A,\ b\in B\}$

So every relation from $A$ to $B$ is built from ordered pairs taken from $A\times B$ .

---

Domain, Codomain, and Range

📐 Basic Components of a Relation

Suppose $R\subseteq A\times B$ .

The domain of $R$ is the set of all first components that appear in $R$ :

\qquad \operatorname{Dom}(R)=\{a\in A:\exists b\in B \text{ such that } (a,b)\in R\}

The codomain is the target set $B$

The range or image set is the set of all second components that actually appear:

\qquad \operatorname{Ran}(R)=\{b\in B:\exists a\in A \text{ such that } (a,b)\in R\}

❗ Very Important Distinction

codomain = full target set

range = only the values actually hit

These are equal only in special cases.

---

Relations on a Set

📖 Relation on a Set

If $R\subseteq A\times A$ , then $R$ is called a relation on the set $A$ .

This is the setting where we study:

reflexive

symmetric

antisymmetric

transitive

---

Four Main Properties

📐 Reflexive

A relation $R$ on $A$ is reflexive if

$\qquad (a,a)\in R \quad \text{for every } a\in A$

That is, every element is related to itself.

📐 Symmetric

A relation $R$ on $A$ is symmetric if

$\qquad (a,b)\in R \Rightarrow (b,a)\in R$

for all $a,b\in A$ .

📐 Antisymmetric

A relation $R$ on $A$ is antisymmetric if

$\qquad (a,b)\in R \text{ and } (b,a)\in R \Rightarrow a=b$

for all $a,b\in A$ .

This does not mean “not symmetric”.
It means two distinct elements cannot be related in both directions.

📐 Transitive

A relation $R$ on $A$ is transitive if

$\qquad (a,b)\in R \text{ and } (b,c)\in R \Rightarrow (a,c)\in R$

for all $a,b,c\in A$ .

---

How to Read These Quickly

💡 Fast Testing Method

When checking a relation on a finite set:

Reflexive: check whether all diagonal pairs $(a,a)$ are present

Symmetric: every arrow $a\to b$ must be matched by $b\to a$

Antisymmetric: if both $a\to b$ and $b\to a$ occur, then $a=b$

Transitive: whenever $a\to b$ and $b\to c$ are present, $a\to c$ must also be present

---

Equivalence Relations

📖 Equivalence Relation

A relation $R$ on a set $A$ is an equivalence relation if it is:

reflexive

symmetric

transitive

Examples:

equality on any set

congruence modulo $n$ on integers

“has the same remainder upon division by $n$ ”

📐 Equivalence Class

If $R$ is an equivalence relation on $A$ , the equivalence class of $a\in A$ is

$\qquad [a]=\{x\in A: xRa\}$

Equivalent elements belong to the same class.
The set $A$ gets partitioned into disjoint equivalence classes.

---

Partial Orders

📖 Partial Order

A relation $R$ on a set $A$ is a partial order if it is:

reflexive

antisymmetric

transitive

Examples:

$\le$ on real numbers

divisibility on positive integers

subset relation $\subseteq$ on a power set

❗ Difference Between Equivalence and Order

equivalence relation uses symmetric

partial order uses antisymmetric

These are very different ideas.

---

Inverse Relation

📐 Inverse of a Relation

If $R\subseteq A\times B$ , then the inverse relation $R^{-1}\subseteq B\times A$ is defined by

$\qquad R^{-1}=\{(b,a):(a,b)\in R\}$

So the ordered pairs are reversed.

---

Composition of Relations

📐 Composition

Suppose
$\qquad R\subseteq A\times B$
and
$\qquad S\subseteq B\times C$

Then the composition $S\circ R$ is the relation from $A$ to $C$ defined by

$\qquad (a,c)\in S\circ R \iff \exists b\in B \text{ such that } (a,b)\in R \text{ and } (b,c)\in S$

This idea is very useful in function language and multi-step mappings. ---

Relation vs Function

📖 When is a Relation a Function?

A relation $f\subseteq A\times B$ is a function from $A$ to $B$ if for every $a\in A$ , there exists exactly one $b\in B$ such that

$\qquad (a,b)\in f$

So a function is a very special kind of relation:

every input must be used

every input must have exactly one output

⚠️ Do Not Confuse These

A general relation may:

leave some elements of the domain unused

assign several outputs to one input

A function cannot do either of these.

---

Minimal Worked Examples

Example 1 Let

\qquad A=\{1,2,3\}

and

\qquad R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}

Check its properties.

reflexive: yes, because $(1,1),(2,2),(3,3)$ are all present

symmetric: yes, because $(1,2)$ and $(2,1)$ both occur

antisymmetric: no, because $(1,2)$ and $(2,1)$ occur with $1\ne 2$

transitive: yes

R

is an equivalence relation. --- Example 2 On positive integers, define

aRb

\qquad aRb \iff a \mid b

Then:

reflexive: yes, since $a\mid a$

symmetric: no

antisymmetric: yes

transitive: yes

So divisibility is a partial order. ---

Common Ready-to-Use Relations

📐 Standard Examples

Equality:

\qquad aRb \iff a=b

This is both an equivalence relation and a partial order.

Less than or equal:

\qquad aRb \iff a\le b

This is a partial order, not an equivalence relation.

Congruence modulo $n$ :

\qquad aRb \iff a\equiv b \pmod n

This is an equivalence relation.

Divisibility:

\qquad aRb \iff a\mid b

This is a partial order on positive integers.

---

Common Mistakes

⚠️ Avoid These Errors

❌ confusing range with codomain

❌ thinking antisymmetric means “not symmetric”

❌ forgetting that reflexive requires every $(a,a)$

❌ checking transitivity using only one example instead of all relevant chains

❌ assuming every relation is automatically a function

✅ The safest method is to test each definition directly.

---

CMI Strategy

💡 How to Solve Relation Questions

first identify whether the relation is from $A$ to $B$ or on $A$

if properties are asked, write the exact definition before checking

for finite sets, inspect ordered pairs systematically

distinguish carefully between symmetric and antisymmetric

if the question involves classes or partitions, suspect an equivalence relation

if it involves comparison, divisibility, or inclusion, suspect a partial order

---

Practice Questions

:::question type="MCQ" question="Let

A=\{1,2,3\}

and

R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}

. Which of the following is correct?" options=["

R

is reflexive and antisymmetric","

R

is symmetric but not reflexive","

R

is reflexive and symmetric but not antisymmetric","

R

is transitive but not symmetric"] answer="C" hint="Check diagonal pairs first, then compare

(1,2)

and

(2,1)

." solution="The relation contains

(1,1),(2,2),(3,3)

, so it is reflexive. It also contains both

(1,2)

and

(2,1)

, so it is symmetric. But antisymmetry fails because

(1,2)

and

(2,1)

are both in

R

while

1\ne 2

. Hence the correct statement is that

R

is reflexive and symmetric but not antisymmetric. Therefore the correct option is

\boxed{C}

." ::: :::question type="NAT" question="How many relations are there from a 2-element set to a 3-element set?" answer="64" hint="A relation is any subset of the Cartesian product." solution="If set

A

has 2 elements and set

B

has 3 elements, then

\qquad |A\times B| = 2\cdot 3 = 6

A relation from

A

B

is any subset of

A\times B

. A set with 6 elements has

\qquad 2^6 = 64

subsets. Hence the number of relations is

\boxed{64}

." ::: :::question type="MSQ" question="Which of the following relations on the set of integers are equivalence relations?" options=["

aRb \iff a-b

is divisible by

5

","

aRb \iff a\le b

","

aRb \iff a

and

b

have the same parity","

aRb \iff a\mid b

"] answer="A,C" hint="An equivalence relation must be reflexive, symmetric, and transitive." solution="1. True. 'Difference divisible by 5' is congruence modulo 5, which is reflexive, symmetric, and transitive.

False. The relation

\le

is reflexive and transitive, but not symmetric.

True. Having the same parity is an equivalence relation: each integer has the same parity as itself, parity is symmetric, and parity agreement is transitive.

False. Divisibility on integers is not symmetric, so it is not an equivalence relation.

Hence the correct answer is

\boxed{A,C}

." ::: :::question type="SUB" question="On the set of positive integers, define

aRb

a \mid b

. Prove that

R

is a partial order but not an equivalence relation." answer="It is reflexive, antisymmetric, and transitive, but not symmetric." hint="Check the three properties of a partial order directly, then test symmetry." solution="We must show that divisibility is reflexive, antisymmetric, and transitive. Reflexive: For every positive integer

a

, we have

\qquad a\mid a

because

a=a\cdot 1

. Antisymmetric: Suppose

\qquad a\mid b \quad \text{and} \quad b\mid a

Then there exist positive integers

m,n

such that

\qquad b=am \quad \text{and} \quad a=bn

Substituting,

\qquad a=(am)n=a(mn)

Since

a>0

, we get

\qquad mn=1

Hence

m=n=1

, so

\qquad a=b

Therefore the relation is antisymmetric. Transitive: Suppose

\qquad a\mid b \quad \text{and} \quad b\mid c

Then there exist positive integers

m,n

such that

\qquad b=am,\quad c=bn

\qquad c=(am)n=a(mn)

Hence

\qquad a\mid c

Thus the relation is transitive. So

R

is a partial order. It is not an equivalence relation because symmetry fails. For example,

\qquad 2\mid 4

but

\qquad 4\nmid 2

Hence

R

is a partial order but not an equivalence relation." ::: ---

Summary

❗ Key Takeaways for CMI

A relation from $A$ to $B$ is any subset of $A\times B$ .

Domain, codomain, and range are different concepts.

Reflexive, symmetric, antisymmetric, and transitive are the main structural properties.

Equivalence relation = reflexive + symmetric + transitive.

Partial order = reflexive + antisymmetric + transitive.

A function is a special relation with exactly one output for each input.

---

💡 Next Up

Proceeding to Domain.

---

Part 4: Domain

Domain

Overview

The domain of a function is the set of all inputs for which the function is defined. In basic problems this may look straightforward, but CMI-style questions often test deeper understanding: the difference between domain, codomain, and range; the role of quantifiers like “for each” and “there exists”; and hidden restrictions coming from denominators, square roots, logarithms, or inverse trigonometric expressions. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

Define the domain of a function precisely.

Distinguish clearly between domain, codomain, and range.

Find the natural domain of algebraic, radical, rational, and logarithmic expressions.

Interpret domain statements written using quantifiers.

Avoid common mistakes involving undefined inputs.

---

Core Meaning

📖 Domain

If $f : X \to Y$ is a function, then:

$X$ is the domain

$Y$ is the codomain

the range or image is the set

\qquad f(X)=\{f(x):x\in X\}

So:

every input must come from the domain

every output lies in the codomain

the range is the part of the codomain that is actually attained

❗ Function Condition

A relation $f : X \to Y$ is a function if for each $x \in X$ , there exists a unique $y \in Y$ such that
$\qquad f(x)=y$

This is the most basic domain-based property of a function.

---

Domain, Codomain, and Range Are Different

📐 Do Not Mix These Up

For a function $f : X \to Y$ :

Domain = set of allowed inputs

Codomain = target set declared in advance

Range = actual outputs produced

Always:

\qquad f(X) \subseteq Y

But usually:

\qquad f(X) \ne Y

unless the function is onto.

Example If

f : \mathbb{R} \to \mathbb{R}

is given by

\qquad f(x)=x^2

then:

domain = $\mathbb{R}$

codomain = $\mathbb{R}$

range = $[0,\infty)$

So the codomain is not the same as the range. ---

Natural Domain of an Expression

📖 Natural Domain

If a formula is given without an explicit domain, the natural domain is the largest set of real numbers for which the expression makes sense.

---

Standard Restrictions

📐 Restriction from a Denominator

If a denominator appears, it cannot be zero.

Examples:

$\dfrac{1}{x-3}$ requires $x \ne 3$

$\dfrac{x+1}{x^2-4}$ requires $x \ne 2,-2$

$\dfrac{1}{(x-1)(x+5)}$ requires $x \ne 1,-5$

📐 Restriction from an Even Root

For real-valued functions, an even root requires the quantity inside to be non-negative.

Examples:

$\sqrt{x-2}$ requires $x \ge 2$

📐 Restriction from a Logarithm

For real logarithms, the argument must be strictly positive.

Examples:

$\log x$ requires $x>0$

$\log(x-2)$ requires $x>2$

$\log\left(\dfrac{x+1}{x-3}\right)$ requires

\qquad \dfrac{x+1}{x-3}>0

📐 Restriction from Combined Expressions

When several restrictions appear together, the domain is their intersection.

Example:
For
$\qquad f(x)=\dfrac{\sqrt{x-1}}{x^2-9}$

we need:

$x-1 \ge 0 \Rightarrow x \ge 1$

$x^2-9 \ne 0 \Rightarrow x \ne 3,-3$

Combining,

\qquad \text{Domain} = [1,\infty)\setminus\{3\}

---

Quantifier Language

❗ Reading Function Statements Correctly

Statements about domain often use quantifiers such as:

“for each $x$ in $X$ ”

“there exists $y$ in $Y$ ”

“there exists a unique $y$ in $Y$ ”

These have different meanings.

For a function

f:X\to Y

For each $x\in X$ , there exists $y\in Y$ such that $f(x)=y$

This is always true for a function.

For each $y\in Y$ , there exists $x\in X$ such that $f(x)=y$

This means

f

is onto.

If for each $x_1,x_2\in X$ , $f(x_1)=f(x_2)$ implies $x_1=x_2$

This means

f

is one-to-one.

💡 Very Important Distinction

“for each $x$ , there exists $y$ ” is about whether the function is defined at every input in the domain.

“for each $y$ , there exists $x$ ” is about whether every codomain value is attained.

These are not the same statement.

---

Minimal Worked Examples

Example 1 Find the domain of

\qquad f(x)=\dfrac{1}{x^2-5x+6}

Factor the denominator:

\qquad x^2-5x+6=(x-2)(x-3)

So we must exclude

\qquad x=2,3

Hence,

\qquad \text{Domain}=\mathbb{R}\setminus\{2,3\}

--- Example 2 Find the domain of

\qquad g(x)=\sqrt{4-x^2}

We need

\qquad 4-x^2 \ge 0

\qquad x^2 \le 4

Hence,

\qquad -2 \le x \le 2

Therefore,

\qquad \text{Domain}=[-2,2]

---

Fast Domain Rules

📐 Quick Reference

For real-valued functions:

Polynomial $\Rightarrow$ domain is all real numbers

Rational expression $\Rightarrow$ denominator $\ne 0$

Square root $\Rightarrow$ inside $\ge 0$

Fourth root, sixth root, etc. $\Rightarrow$ inside $\ge 0$

Logarithm $\Rightarrow$ argument $>0$

$\dfrac{1}{\sqrt{A(x)}}$ requires
$\qquad A(x)>0$
because denominator cannot be zero

⚠️ Watch This Carefully

There is a big difference between:

$\sqrt{A(x)}$ requires $\qquad A(x)\ge 0$

---

Set Notation for Domain

📐 Useful Forms

Domain answers are often written as:

$\mathbb{R}\setminus\{a,b\}$

$(-\infty,2)\cup(2,\infty)$

$[1,\infty)$

$(-\infty,-3]\cup[3,\infty)$

You should be comfortable moving between condition form and interval form.

---

Common Mistakes

⚠️ Avoid These Errors

❌ confusing range with codomain

❌ forgetting to exclude denominator zeros

❌ using $\ge 0$ instead of $>0$ inside logarithms

❌ using $\ge 0$ instead of $>0$ for $\dfrac{1}{\sqrt{A(x)}}$

---

CMI Strategy

💡 How to Attack Domain Questions

First identify the type of expression: polynomial, rational, radical, logarithmic, or mixed.

Write every restriction separately.

Intersect all restrictions.

Convert the final answer into clean interval or set notation.

In function-language questions, track the order of quantifiers carefully.

Keep domain, codomain, and range separate in your head.

---

Practice Questions

:::question type="MCQ" question="The domain of

f(x)=\dfrac{1}{x^2-4}

is" options=["

\mathbb{R}

","

\mathbb{R}\setminus\{-2,2\}

","

\mathbb{R}\setminus\{2\}

","

[-2,2]

"] answer="B" hint="The denominator must be nonzero." solution="We need

\qquad x^2-4 \ne 0

\qquad (x-2)(x+2)\ne 0

Hence

\qquad x \ne 2,-2

Therefore the domain is

\boxed{\mathbb{R}\setminus\{-2,2\}}

. So the correct option is

\boxed{B}

." ::: :::question type="NAT" question="Find the number of integers in the domain of

f(x)=\sqrt{9-x^2}

." answer="7" hint="First find the interval on which the square root is defined." solution="For the square root to be defined, we need

\qquad 9-x^2 \ge 0

\qquad x^2 \le 9

Hence

\qquad -3 \le x \le 3

The integers in this interval are

\qquad -3,-2,-1,0,1,2,3

There are

7

such integers. Therefore, the answer is

\boxed{7}

." ::: :::question type="MSQ" question="Which of the following statements are always true for a function

f:X\to Y

?" options=["For each

x\in X

, there exists

y\in Y

such that

f(x)=y

.","For each

x\in X

, there exists a unique

y\in Y

such that

f(x)=y

.","For each

y\in Y

, there exists

x\in X

such that

f(x)=y

.","

f(X)\subseteq Y

."] answer="A,B,D" hint="Separate the basic function property from onto-ness." solution="1. True. A function must assign an output in

Y

to every input in

X

True. The output must also be unique for each input.

This is not always true; it is true exactly when the function is onto.

True. By definition, every output of the function lies in the codomain.

Hence the always true statements are

\boxed{A,B,D}

." ::: :::question type="SUB" question="Find the domain of

f(x)=\dfrac{\sqrt{x-1}}{x^2-5x+6}

." answer="

[1,\infty)\setminus\\{2,3\\}

" hint="Use the square-root restriction and denominator restriction together." solution="We need the square root to be defined:

\qquad x-1 \ge 0

\qquad x \ge 1

Also the denominator must be nonzero:

\qquad x^2-5x+6 \ne 0

Factor:

\qquad x^2-5x+6=(x-2)(x-3)

\qquad x \ne 2,3

Combining all restrictions, the domain is

\qquad [1,\infty)\setminus\{2,3\}

Therefore, the required domain is

\boxed{[1,\infty)\setminus\{2,3\}}

." ::: ---

Summary

❗ Key Takeaways for CMI

The domain is the set of all valid inputs of a function.

Domain, codomain, and range are three different objects.

Denominators must be nonzero, logarithm arguments must be positive, and even-root contents must be non-negative.

In mixed expressions, the domain is the intersection of all restrictions.

“For each $x$ there exists a unique $y$ ” is the basic function condition.

Careful reading of quantifiers is essential in abstract function-language questions.

---

💡 Next Up

Proceeding to Range.

---

Part 5: Range

Range

Overview

The range of a function is the set of all values actually taken by the function. In exam problems, finding the range is rarely about plotting alone. It usually depends on algebraic rewriting, inequalities, monotonicity, domain restrictions, and inverse-style reasoning. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

distinguish clearly between domain and range,

find the range of standard algebraic, rational, radical, exponential, logarithmic, and modulus functions,

use substitution and inequalities to determine the exact set of output values,

handle transformed functions such as $a f(bx+c)+d$ ,

avoid common mistakes involving hidden domain restrictions.

---

Core Idea

📖 Range of a Function

If a function is written as

$\qquad y = f(x)$

then the range of $f$ is the set of all real numbers $y$ for which there exists at least one allowed value of $x$ such that

$\qquad y = f(x)$

So:

domain asks: which inputs are allowed?

range asks: which outputs actually occur?

---

Domain vs Range

❗ Do Not Mix These Up

For a function $f$ :

Domain = all admissible values of $x$

Range = all values of $f(x)$ obtained from those admissible values

Example:

\qquad f(x)=\sqrt{x-2}

Then:

domain is $\qquad x \ge 2$

range is $\qquad y \ge 0$

---

Standard Range Facts

📐 Basic Functions and Their Ranges

Constant function:

\qquad f(x)=c

Range:

\qquad \{c\}

Linear function:

\qquad f(x)=ax+b,\ a\ne 0

Range: all real numbers

Square function:

\qquad f(x)=x^2

Range:

\qquad [0,\infty)

Cubic function:

\qquad f(x)=x^3

Range: all real numbers

Reciprocal function:

\qquad f(x)=\dfrac{1}{x}

Range: all real numbers except

0

Absolute value:

\qquad f(x)=|x|

Range:

\qquad [0,\infty)

Square root:

\qquad f(x)=\sqrt{x}

Range:

\qquad [0,\infty)

Exponential:

\qquad f(x)=a^x,\ a>0,\ a\ne 1

Range:

\qquad (0,\infty)

Logarithm:

\qquad f(x)=\log_a x,\ a>0,\ a\ne 1

Range: all real numbers

---

Main Methods for Finding Range

💡 CMI Strategy

To find the range of a function, try these in order:

Set $y=f(x)$ and solve for $x$

Then determine for which values of

y

the obtained

x

is valid.

Use known inequalities

Such as

x^2 \ge 0

|x| \ge 0

, AM-GM, or

\qquad x+\dfrac{1}{x} \ge 2

\le -2

Rewrite the expression

Complete the square, factor, or substitute.

Use monotonicity

If the function is increasing or decreasing on its domain, the range often comes from endpoint behavior.

Study transformations

Use the known range of a base function and then shift/stretch/reflect.

---

Method 1: Set $y=f(x)$ and Eliminate $x$

📐 Inverse-Style Range Method

Suppose

$\qquad y=f(x)$

Try to solve for $x$ in terms of $y$ .

Then ask:

for which $y$ is the resulting $x$ real?

for which $y$ does it satisfy the original domain?

This is one of the safest general methods.

Example 1 Find the range of

\qquad y=x^2+4x+7

Complete the square:

\qquad y=(x+2)^2+3

Since

\qquad (x+2)^2 \ge 0

, we get

\qquad y \ge 3

So the range is

\qquad \boxed{[3,\infty)}

--- Example 2 Find the range of

\qquad y=\dfrac{1}{x-2}

Since

x\ne 2

, the denominator is never zero, so

\qquad y\ne 0

Also every nonzero real value of

y

is possible by taking

\qquad x=2+\dfrac{1}{y}

So the range is

\qquad \boxed{\mathbb{R}\setminus\{0\}}

---

Method 2: Use Inequalities

📐 High-Value Range Inequalities

Some standard facts:

$x^2 \ge 0$

$(x-a)^2 \ge 0$

$|x| \ge 0$

if $x>0$ , then

\qquad x+\dfrac{1}{x} \ge 2

if $x<0$ , then

\qquad x+\dfrac{1}{x} \le -2

therefore, for real $x\ne 0$ ,

\qquad x+\dfrac{1}{x} \in (-\infty,-2] \cup [2,\infty)

Example 3 Find the range of

\qquad y=x+\dfrac{1}{x},\qquad x\ne 0

For

x>0

, by AM-GM,

\qquad x+\dfrac{1}{x} \ge 2

For

x<0

, write

x=-t

with

t>0

\qquad x+\dfrac{1}{x}=-(t+\dfrac{1}{t}) \le -2

Hence the range is

\qquad \boxed{(-\infty,-2]\cup[2,\infty)}

---

Method 3: Transform Known Ranges

📐 Effect of Transformations

Suppose the range of $f(x)$ is $R$ .

Then:

$f(x)+k$ shifts the range upward by $k$

$a f(x)$ scales the range by factor $a$

- if

a<0

, it also reflects vertically

$f(x-h)$ usually changes the domain, but not the set of output values if the new domain still covers the original behavior

$|f(x)|$ forces all outputs to be non-negative

Example 4 If the range of

x^2

[0,\infty)

, then the range of

\qquad -3x^2+5

is obtained by:

multiplying by $-3$ : gives $\qquad (-\infty,0]$

adding $5$ : gives $\qquad (-\infty,5]$

So the range is

\qquad \boxed{(-\infty,5]}

---

Rational Functions and Range

❗ Do Not Confuse Missing Value with True Range

For a rational function, the range is not found just from asymptotes. You must check whether the apparently missing value is actually attained.

Example:
$\qquad y=\dfrac{x-1}{x+1}$

Solve for $x$ :

$\qquad y(x+1)=x-1$

$\qquad yx+y=x-1$

$\qquad x(y-1)=-(1+y)$

$\qquad x=\dfrac{-(1+y)}{y-1}$

This is valid for every $y\ne 1$ .

So the range is
$\qquad \boxed{\mathbb{R}\setminus\{1\}}$

---

Radical Functions and Range

📐 Square Root Type

If
$\qquad y=\sqrt{g(x)}$

then automatically
$\qquad y\ge 0$

But not every non-negative value must always occur. After writing

$\qquad y=\sqrt{g(x)}$

square carefully:

$\qquad y^2=g(x)$

Then check which $y\ge 0$ are actually possible.

Example 5 Find the range of

\qquad y=\sqrt{4-x^2}

We need

\qquad 4-x^2 \ge 0 \Rightarrow -2 \le x \le 2

Also

\qquad y\ge 0

Maximum occurs at

x=0

\qquad y=2

Minimum occurs at

x=\pm 2

\qquad y=0

So the range is

\qquad \boxed{[0,2]}

---

Range via Monotonicity

📐 When Increasing/Decreasing Helps

If a function is continuous and strictly increasing on an interval, then its range on that interval is determined by its endpoint behavior.

Similarly for strictly decreasing functions.

This is especially useful for:

exponential functions,

logarithmic functions,

simple rational functions on one interval,

compositions of monotone functions.

---

Range of Composite Functions

❗ Composition Rule

For
$\qquad y=f(g(x))$

first find the range of
$\qquad u=g(x)$

Then apply $f$ only to those allowed values of $u$ .

This is often much easier than treating the whole expression directly.

Example 6 Find the range of

\qquad y=\sqrt{x^2+1}

Since

\qquad x^2 \ge 0

, we get

\qquad x^2+1 \ge 1

Hence

\qquad y=\sqrt{x^2+1} \ge 1

Also

y=1

occurs at

x=0

. So the range is

\qquad \boxed{[1,\infty)}

---

Modulus and Range

📐 Absolute Value Effects

If
$\qquad y=|f(x)|$

then:

always $\qquad y\ge 0$

all negative values in the range of $f$ are reflected upward

if range of $f$ is $(-\infty,a]$ , then range of $|f|$ must be analyzed carefully depending on whether $0$ lies inside that interval

Example 7 Find the range of

\qquad y=|x^2-4|

Since

\qquad x^2-4 \in [-4,\infty)

, taking modulus gives all values from

0

upward. So the range is

\qquad \boxed{[0,\infty)}

---

Common Mistakes

⚠️ Avoid These Errors

❌ Confusing domain with range

❌ Forgetting that square roots are non-negative

❌ Assuming every rational function misses its horizontal asymptote value

❌ Ignoring domain restrictions while solving for $x$ in terms of $y$

❌ Forgetting to check whether endpoint values are attained

❌ Using graph intuition without algebraic verification

---

Quick Recognition Sheet

📐 Fast Recall

$x^2+a \Rightarrow$ range starts at $a$

$-x^2+a \Rightarrow$ range ends at $a$

$\dfrac{1}{x-a}+b \Rightarrow$ range is all real numbers except $b$

$\sqrt{g(x)} \Rightarrow$ range is a subset of $[0,\infty)$

---

Practice Questions

:::question type="MCQ" question="The range of

f(x)=x^2+6x+11

is" options=["

[2,\infty)

","

[11,\infty)

","

(-\infty,2]

","

(-\infty,11]

"] answer="A" hint="Complete the square." solution="We write

\qquad x^2+6x+11=(x+3)^2+2

Since

\qquad (x+3)^2 \ge 0

, we get

\qquad f(x)\ge 2

Also

f(-3)=2

, so the minimum value is attained. Hence the range is

\boxed{[2,\infty)}

, so the correct option is

\boxed{A}

." ::: :::question type="NAT" question="Find the least value of

x+\dfrac{4}{x}

for

x>0

." answer="4" hint="Use AM-GM or set

t=\dfrac{x}{2}

." solution="For

x>0

, apply AM-GM to

x

and

\dfrac{4}{x}

\qquad x+\dfrac{4}{x} \ge 2\sqrt{x\cdot \dfrac{4}{x}}=2\sqrt{4}=4

Equality holds when

\qquad x=\dfrac{4}{x}

\qquad x^2=4 \Rightarrow x=2

Thus the least value is

\boxed{4}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["The range of

\sqrt{x+3}

[0,\infty)

","The range of

\dfrac{1}{x-1}

\mathbb{R}\setminus\{0\}

","The range of

x^3

is all real numbers","The range of

|x|

is all real numbers"] answer="A,B,C" hint="Think of what outputs are possible." solution="1. True. Since square root outputs are non-negative, and every non-negative value occurs.

True. Reciprocal form can never be

0

, and every other real value is achieved.

True. Cubic functions take every real value.

False.

|x|\ge 0

, so its range is

[0,\infty)

Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="Find the range of

f(x)=\dfrac{x-2}{x+3}

." answer="

\mathbb{R}\setminus\{1\}

" hint="Set

y=\dfrac{x-2}{x+3}

and solve for

x

." solution="Let

\qquad y=\dfrac{x-2}{x+3}

Then

\qquad y(x+3)=x-2

\qquad yx+3y=x-2

\qquad x(y-1)=-(2+3y)

\qquad x=\dfrac{-(2+3y)}{y-1}

This is possible for every real

y

except

\qquad y=1

Now check whether

y=1

can occur in the original equation:

\qquad \dfrac{x-2}{x+3}=1 \Rightarrow x-2=x+3

which gives

\qquad -2=3

, impossible. Hence

y=1

is not in the range, and every other real value is. Therefore the range is

\qquad \boxed{\mathbb{R}\setminus\{1\}}

." ::: ---

Summary

❗ Key Takeaways for CMI

Range means the set of all output values, not input values.

The most reliable method is to set $y=f(x)$ and solve for $x$ .

Inequalities and completing the square are central tools.

Transformations shift, scale, and reflect known ranges.

Rational and radical functions need careful algebra, not only graph intuition.

Domain restrictions always influence the range.

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💡 Next Up

Proceeding to Codomain.

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Part 6: Codomain

Codomain

Overview

The codomain of a function is one of the most basic but most misunderstood parts of function language. In many problems, students confuse codomain with range or image, but they are not the same. In CMI-style questions, this distinction becomes important in injectivity, surjectivity, invertibility, composition, and function interpretation. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

define codomain correctly

distinguish codomain from domain and range

decide whether a function is onto by comparing its image with its codomain

understand how codomain affects inverse functions and composition

avoid common language errors in basic function theory

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Basic Function Language

📖 Function as a Rule

A function is written as

$\qquad f: A \to B$

This means:

$A$ is the domain

$B$ is the codomain

for each element of $A$ , the function assigns exactly one element of $B$

📖 Domain, Codomain, Range

For a function

$\qquad f: A \to B$

domain = the set $A$ from which inputs are taken

codomain = the set $B$ into which outputs are declared to lie

range or image set = the actual set of values attained by $f(x)$ as $x$ runs over $A$

So,

\qquad \text{Range}(f) = \{f(x) : x \in A\}

and always

\qquad \text{Range}(f) \subseteq \text{Codomain}

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Core Idea

❗ Most Important Fact

The codomain is part of the definition of a function, while the range is determined after the function acts on the domain.

So two functions with the same rule and same domain can still be considered different if their codomains are different.

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Example: Codomain vs Range

📐 Classic Example

Consider

$\qquad f:\mathbb{R}\to\mathbb{R},\quad f(x)=x^2$

Then:

domain = $\mathbb{R}$

codomain = $\mathbb{R}$

range = $[0,\infty)$

So here the range is not equal to the codomain.

📐 Same Rule, Different Codomain

Now consider

$\qquad g:\mathbb{R}\to[0,\infty),\quad g(x)=x^2$

Then:

domain = $\mathbb{R}$

codomain = $[0,\infty)$

range = $[0,\infty)$

Now the range equals the codomain.

⚠️ Do Not Confuse These

For $f(x)=x^2$ on $\mathbb{R}$ :

codomain is not automatically $[0,\infty)$

codomain is whatever is declared in the function notation

range is what the function actually produces

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Why Codomain Matters

❗ Codomain Controls Surjectivity

A function $f:A\to B$ is onto or surjective if every element of the codomain is hit by some element of the domain.

So $f$ is surjective if

$\qquad \text{Range}(f)=B$

Thus surjectivity depends directly on the codomain.

📐 Same Rule, Different Surjectivity

Consider $f(x)=x^2$ .

\qquad f:\mathbb{R}\to\mathbb{R}

then

f

is not onto, because negative numbers are not attained.

\qquad f:\mathbb{R}\to[0,\infty)

then

f

is onto, because every non-negative real number is attained.

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Codomain and Inverse Functions

❗ Inverse Needs the Right Codomain

A function has an inverse as a function only when it is both:

one-one

onto its codomain

So codomain matters in invertibility too.

📐 Example with Inverse

Take

$\qquad f:[0,\infty)\to[0,\infty),\quad f(x)=x^2$

This function is one-one and onto, so it has inverse

$\qquad f^{-1}(x)=\sqrt{x}$

But if we write

$\qquad f:[0,\infty)\to\mathbb{R},\quad f(x)=x^2$

then it is not onto its codomain, so it is not invertible as a map onto $\mathbb{R}$ .

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Codomain and Composition

📐 Composition Condition

If

$\qquad f:A\to B,\quad g:C\to D$

then $g\circ f$ is defined only when the outputs of $f$ fit into the domain of $g$ .

A standard safe condition is:

$\qquad \text{Codomain of }f \subseteq \text{Domain of }g$

or at least

$\qquad \text{Range of }f \subseteq \text{Domain of }g$

💡 Why Codomain Helps in Composition

When we write function chains carefully, codomain tells us the intended set into which the first function lands. This is why codomain is not decorative; it helps control whether composition is well-formed.

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Common Set-Theoretic View

📖 Image of a Set

If $f:A\to B$ and $S\subseteq A$ , then the image of $S$ under $f$ is

$\qquad f(S)=\{f(x):x\in S\}$

In particular,

$\qquad f(A)=\text{Range}(f)$

So

$\qquad f(A)\subseteq B$

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Key Relationships

📐 Must-Know Relationships

For a function $f:A\to B$ :

domain = $A$

codomain = $B$

range = $f(A)$

$f(A)\subseteq B$

$f$ is onto $\iff f(A)=B$

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Minimal Worked Examples

Example 1 Consider

\qquad f:\mathbb{R}\to\mathbb{R},\quad f(x)=x^2

Then the codomain is

\mathbb{R}

because that is declared in the function notation. But the actual outputs satisfy

x^2\ge 0

, so the range is

\qquad [0,\infty)

Hence range and codomain are different. --- Example 2 Consider

\qquad f:\mathbb{R}\to[0,\infty),\quad f(x)=x^2

Now the range is exactly

[0,\infty)

, so the function is onto. ---

Common Traps

⚠️ Avoid These Errors

❌ saying codomain means the actual outputs

❌ confusing codomain with range

❌ deciding surjectivity without looking at the codomain

❌ thinking a formula alone determines the codomain

❌ forgetting that the same algebraic rule can define different functions if codomains differ

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Recognition Guide

💡 Fast Questions to Ask

When a function is given, ask:

what is the domain?

what is the codomain?

what values are actually attained?

is the range equal to the codomain?

if yes, is the function onto?

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CMI Strategy

💡 How to Handle Function-Language Questions

read the notation $f:A\to B$ very carefully

do not infer codomain from the formula unless it is explicitly stated

compute the actual image set separately

compare image set with codomain to test onto-ness

when inverse or composition appears, check codomain and domain compatibility before doing algebra

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Practice Questions

:::question type="MCQ" question="For a function

f:A \to B

, which of the following is always true?" options=["

\text{Codomain}(f)=\text{Range}(f)

","

\text{Range}(f)\subseteq \text{Codomain}(f)

","

\text{Domain}(f)\subseteq \text{Codomain}(f)

","Every function is onto its codomain"] answer="B" hint="Think about actual outputs versus declared target set." solution="For a function

f:A\to B

, every actual output

f(x)

must lie in the codomain

B

. Therefore,

\qquad \text{Range}(f)\subseteq \text{Codomain}(f)

The range need not equal the codomain, and not every function is onto. Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="Let

f:\mathbb{R}\to\mathbb{R}

be defined by

f(x)=x^2+1

. Find the least value in the range of

f

." answer="1" hint="Use the fact that

x^2 \ge 0

for all real

x

." solution="We have

\qquad f(x)=x^2+1

Since

x^2\ge 0

for all real

x

\qquad x^2+1\ge 1

Equality occurs when

x=0

. So the least value in the range is

\boxed{1}

." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["Two functions with the same rule and same domain but different codomains may be treated as different functions.","A function is onto exactly when its range equals its codomain.","For

f:\mathbb{R}\to\mathbb{R}

defined by

f(x)=x^2

, the codomain is

[0,\infty)

.","For any function

f:A\to B

, we always have

f(A)\subseteq B

."] answer="A,B,D" hint="Separate declared target set from actual image." solution="1. True. Codomain is part of the function description, so changing codomain can change the function description.

True. A function is onto exactly when every element of the codomain is attained, that is,

\qquad \text{Range}(f)=\text{Codomain}(f)

False. If the function is written as

f:\mathbb{R}\to\mathbb{R}

, then the codomain is

\mathbb{R}

, not

[0,\infty)

True. By definition, all outputs of

f

lie in the codomain.

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="Consider the function

f:\mathbb{R}\to\mathbb{R}

defined by

f(x)=x^2

. Find its domain, codomain, and range. Is the function onto?" answer="Domain

=\mathbb{R}

, Codomain

=\mathbb{R}

, Range

=[0,\infty)

, not onto." hint="Read the notation carefully and then determine the actual values attained." solution="The function is given by

\qquad f:\mathbb{R}\to\mathbb{R},\quad f(x)=x^2

So:

the domain is $\mathbb{R}$

the codomain is $\mathbb{R}$

Now determine the range. Since

\qquad x^2\ge 0

for all real

x

, the set of actual outputs is

\qquad [0,\infty)

Thus:

Domain $=\mathbb{R}$

Codomain $=\mathbb{R}$

Range $=[0,\infty)$

To be onto, the range must equal the codomain. But

\qquad [0,\infty)\ne \mathbb{R}

Therefore the function is not onto. So the final answer is:

\qquad \boxed{\text{Domain }=\mathbb{R},\ \text{Codomain }=\mathbb{R},\ \text{Range }=[0,\infty),\ \text{not onto}}

." ::: ---

Summary

❗ Key Takeaways for CMI

codomain is the declared target set in $f:A\to B$

range is the actual set of values attained by the function

always $\text{Range}(f)\subseteq \text{Codomain}(f)$

surjectivity means range equals codomain

codomain affects onto-ness, invertibility, and composition

the same formula can behave differently depending on the codomain chosen

Chapter Summary

❗ Basic function language — Key Points

Sets and Subsets: The fundamental building blocks, defining collections of distinct objects. Understanding notation like $\in$ , $\subseteq$ , $\cup$ , $\cap$ , and $\emptyset$ is crucial for all subsequent topics.
Ordered Pairs and Cartesian Product: An ordered pair $(a, b)$ is distinct from $(b, a)$ unless $a=b$ . The Cartesian product $A \times B = \{(a, b) \mid a \in A, b \in B\}$ forms the universe for defining relations between sets.
Relations: A relation $R$ from set $A$ to set $B$ is any subset of the Cartesian product $A \times B$ . It formally describes a connection or correspondence between elements of $A$ and $B$ .
Domain of a Relation: The set of all first components of the ordered pairs in a relation $R$ . Formally, $\operatorname{Dom}(R) = \{a \mid (a, b) \in R \text{ for some } b\}$ .
Range of a Relation: The set of all second components of the ordered pairs in a relation $R$ . Formally, $\operatorname{Ran}(R) = \{b \mid (a, b) \in R \text{ for some } a\}$ .
Codomain: For a relation from $A$ to $B$ , $B$ is the codomain. It represents the entire set of possible second components, of which the range is a subset ( $\operatorname{Ran}(R) \subseteq B$ ). Understanding this distinction is vital for defining functions.

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Chapter Review Questions

:::question type="MCQ" question="Let $A = \{1, 2\}$ and $B = \{x, y\}$ . Which of the following is NOT a relation from $A$ to $B$ ?" options=[" $\{(1, x), (2, y)\}$ ", " $\{(1, y), (2, x), (1, x)\}$ ", " $\{(x, 1), (y, 2)\}$ ", " $\emptyset$ "] answer=" $\{(x, 1), (y, 2)\}$ " hint="Recall that a relation from $A$ to $B$ must be a subset of $A \times B$ ." solution="The Cartesian product $A \times B = \{(1, x), (1, y), (2, x), (2, y)\}$ . A relation from $A$ to $B$ is any subset of $A \times B$ .
Option A, B, and D are all subsets of $A \times B$ .
Option C, however, contains ordered pairs where the first element is from $B$ and the second from $A$ . This is a relation from $B$ to $A$ , not from $A$ to $B$ .
Therefore, $\emptyset$ is a valid relation from $A$ to $B$ (the empty relation).
The correct answer is $\{(x, 1), (y, 2)\}$ .
"
:::

:::question type="NAT" question="Consider the relation $R = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} \mid y = x^2 \text{ and } -1 \le x \le 2\}$ . What is the sum of all distinct elements in the range of $R$ ?" answer="6" hint="First, list all the ordered pairs in the relation $R$ . Then, identify the set of all second components to find the range. Finally, sum the distinct elements." solution="Given the relation $R = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} \mid y = x^2 \text{ and } -1 \le x \le 2\}$ .
We find the ordered pairs by substituting integer values for $x$ from $-1$ to $2$ :
If $x = -1$ , $y = (-1)^2 = 1$ . So, $(-1, 1) \in R$ .
If $x = 0$ , $y = (0)^2 = 0$ . So, $(0, 0) \in R$ .
If $x = 1$ , $y = (1)^2 = 1$ . So, $(1, 1) \in R$ .
If $x = 2$ , $y = (2)^2 = 4$ . So, $(2, 4) \in R$ .
Thus, $R = \{(-1, 1), (0, 0), (1, 1), (2, 4)\}$ .
The range of $R$ is the set of all second components: $\operatorname{Ran}(R) = \{1, 0, 4\}$ .
The sum of all distinct elements in the range is $1 + 0 + 4 = 5$ .
The correct answer is 5.
Wait, re-reading the question, the sum of all distinct elements in the range. My calculation was $1+0+4=5$ . Let me double check my thought process.
$x=-1 \implies y=1$
$x=0 \implies y=0$
$x=1 \implies y=1$
$x=2 \implies y=4$
Range = $\{0, 1, 4\}$ . Sum = $0+1+4=5$ .

Let me re-evaluate the hint and solution, as the provided answer is 6.
If the answer is 6, it means the range elements summed to 6.
Possible range elements that sum to 6: $\{0, 1, 5\}$ or $\{0, 2, 4\}$ or $\{1, 2, 3\}$ .
My range is $\{0, 1, 4\}$ .

Could it be a typo in my initial derivation or the expected answer?
The question states: "What is the sum of all distinct elements in the range of $R$ ?"
My calculation for the range is correct: $\{0, 1, 4\}$ .
My calculation for the sum is correct: $0+1+4=5$ .
The provided `answer="6"` seems incorrect based on the question. I will correct the answer to 5.