Given Data & Visual Interpretation

Based on the standard representation of the provided function diagram (Figure 2.7), we observe the following mapping between set $P$ and set $Q$:

• Set $P = \{5, 6, 7\}$
• Set $Q = \{3, 4, 5\}$
• The arrows indicate mapping as follows: $5 \to 3$, $6 \to 4$, $7 \to 5$.

Step 1: Expressing the Relation in Roster Form

A relation in roster form is represented as a set of ordered pairs $(x, y)$ such that $x \in P$ and $y \in Q$, and $x$ is related to $y$ [Per Definition of a Relation in Set Theory].

From the visual mapping:

• $5$ is related to $3 \implies (5, 3)$
• $6$ is related to $4 \implies (6, 4)$
• $7$ is related to $5 \implies (7, 5)$

Thus, the Roster Form $R$ is:

$R = \{(5, 3), (6, 4), (7, 5)\}$

Step 2: Determining the Domain

The domain of a relation is the set of all first components of the ordered pairs in $R$ [Per Definition of Domain].

Extracting the first components from $R$:

$\text{Domain} = \{5, 6, 7\}$

Step 3: Determining the Range

The range of a relation is the set of all second components of the ordered pairs in $R$ [Per Definition of Range].

Extracting the second components from $R$:

$\text{Range} = \{3, 4, 5\}$

Final Solution: The relation in roster form is $\{(5, 3), (6, 4), (7, 5)\}$. The domain is $\{5, 6, 7\}$ and the range is $\{3, 4, 5\}$.

Step 1: Determining the Cardinality of Sets

Given the sets $A = \{x, y, z\}$ and $B = \{1, 2\}$, we define the cardinal number (the number of elements in a set) as follows:

• $n(A) = 3$ [Since there are 3 distinct elements: $x, y,$ and $z$]
• $n(B) = 2$ [Since there are 2 distinct elements: $1$ and $2$]

Step 2: Calculating the Cartesian Product $A \times B$

The Cartesian product $A \times B$ is the set of all possible ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$. The total number of elements in $A \times B$ is given by the product of the cardinalities of the individual sets:

$n(A \times B) = n(A) \times n(B)$

$n(A \times B) = 3 \times 2 = 6$

[Per the definition of the Cartesian product of two finite sets]

Step 3: Defining a Relation and its Power Set

A relation $R$ from set $A$ to set $B$ is defined as any subset of the Cartesian product $A \times B$. [By the definition of a relation: $R \subseteq A \times B$].

The total number of relations from $A$ to $B$ is equivalent to the number of possible subsets of the Cartesian product $A \times B$.

If a set has $m$ elements, the number of its subsets is given by $2^m$. Here, $m = n(A \times B) = 6$.

Step 4: Final Calculation

Total number of relations = $2^{n(A \times B)}$

Total number of relations = $2^6$

$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$

Visual Representation of Mapping

Final Solution: The total number of relations from set $A$ to set $B$ is $64$.

Step 1: Formal Definition of the Relation

Given set $A = \{1, 2, 3, 4, 6\}$. The relation $R$ is defined as $R = \{(a, b) : a, b \in A \text{ and } b \equiv 0 \pmod{a}\}$, where the notation $b \equiv 0 \pmod{a}$ denotes that $b$ is exactly divisible by $a$. This implies that the quotient $\frac{b}{a}$ must be an integer, which is equivalent to saying $a$ is a factor of $b$ [Definition of Divisibility in Number Theory].

Step 2: Systematic Verification of Ordered Pairs

We test each element $a \in A$ against every element $b \in A$ to determine if $\frac{b}{a} \in \mathbb{Z}$:

Step 3: Constructing the Roster Form

By collecting all ordered pairs identified in the systematic verification, we define the relation $R$ in roster form. We ensure that every pair satisfies the condition $b = k \cdot a$ for some $k \in \{1, 2, 3, \dots\}$ [Property of Divisibility].

The collection of all valid pairs $(a, b)$ constitutes the roster form of the relation $R$.

Final Solution: $R = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)\}$

Step 1: Analyzing the Definition of the Relation

The relation $R$ is defined on the set of integers $\mathbb{Z}$ as follows:

$R = \{(a, b) : a, b \in \mathbb{Z}, a - b \in \mathbb{Z}\}$

[By the property of closure under subtraction for integers: If $a, b \in \mathbb{Z}$, then $(a - b)$ is always an integer $\mathbb{Z}$.]

Step 2: Determining the Domain of R

The domain of a relation $R$ consists of all the first elements of the ordered pairs $(a, b)$ that belong to $R$.

• For any integer $a \in \mathbb{Z}$, we must determine if there exists an integer $b \in \mathbb{Z}$ such that $(a, b) \in R$.
• Per the definition, $(a, b) \in R$ if and only if $a - b = k$, where $k \in \mathbb{Z}$.
• Let $b = a - k$. Since $a \in \mathbb{Z}$ and $k \in \mathbb{Z}$, $b$ is necessarily an integer [$b \in \mathbb{Z}$].
• Therefore, every integer $a$ can be associated with at least one integer $b$ (in fact, infinitely many) such that $(a, b) \in R$.

Thus, the domain of $R$ is the set of all integers, $\mathbb{Z}$.

Step 3: Determining the Range of R

The range of a relation $R$ consists of all the second elements of the ordered pairs $(a, b)$ that belong to $R$.

• For any integer $b \in \mathbb{Z}$, we must determine if there exists an integer $a \in \mathbb{Z}$ such that $(a, b) \in R$.
• Per the definition, $(a, b) \in R$ if and only if $a - b = k$, where $k \in \mathbb{Z}$.
• Let $a = b + k$. Since $b \in \mathbb{Z}$ and $k \in \mathbb{Z}$, $a$ is necessarily an integer [$a \in \mathbb{Z}$].
• Therefore, every integer $b$ is the image of at least one integer $a$ under the relation $R$.

Thus, the range of $R$ is the set of all integers, $\mathbb{Z}$.

Step 4: Logical Mapping Illustration

Final Solution: The domain of $R$ is $\mathbb{Z}$ and the range of $R$ is $\mathbb{Z}$.

Step 1: Establishing the Domain and Codomain

Given the sets:

• $A = \{1, 2, 3, 5\}$
• $B = \{4, 6, 9\}$

The relation $R$ is defined as $R = \{(x, y) : |x - y| \in \{ \text{odd numbers} \}; x \in A, y \in B\}$. [Per the definition of relation in Cartesian product $A \times B$].

Step 2: Systematic Evaluation of Ordered Pairs $(x, y)$

To determine if the difference $|x - y|$ is odd, we utilize the parity property: The difference between two integers is odd if and only if one integer is even and the other is odd. [Parity Law: $Odd - Even = Odd$ and $Even - Odd = Odd$].

Step 3: Visual Representation of the Relation

Step 4: Writing the Roster Form

Collecting all identified pairs $(x, y)$ where the difference is odd:

$R = \{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\}$

Final Solution: $R = \{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\}$

Given Variables & Initial Mapping

Based on the provided relationship diagram (Fig 2.7), we identify the sets $P$ and $Q$ and their corresponding mapping pairs $(x, y)$ where $x \in P$ and $y \in Q$:

• Set $P = \{5, 6, 7\}$
• Set $Q = \{3, 4, 5\}$
• Mapping pairs: $(5, 3), (6, 4), (7, 5)$

Step 1: Identifying the Logical Relation

Observe the numerical correspondence between each input $x \in P$ and output $y \in Q$:

• For $x=5$, $y=3$ $\implies$ $5 - 2 = 3$
• For $x=6$, $y=4$ $\implies$ $6 - 2 = 4$
• For $x=7$, $y=5$ $\implies$ $7 - 2 = 5$

The universal rule governing this relation is $y = x - 2$, or equivalently, $x - y = 2$. [Per the Definition of a Relation as a subset of the Cartesian Product $P \times Q$].

Step 2: Defining Set-Builder Form

The set-builder form defines the relation $R$ by specifying the properties that the elements $(x, y)$ must satisfy.

$R = \{ (x, y) : y = x - 2, x \in P, y \in Q \}$

Step 3: Determining Domain and Range

The Domain is defined as the set of all first components of the ordered pairs in the relation, and the Range is the set of all second components [Per the Set-Theoretic definition of functions/relations].

Final Solution:

Set-Builder Form: $R = \{ (x, y) : y = x - 2, x \in \{5, 6, 7\}, y \in \{3, 4, 5\} \}$

Domain: $\{5, 6, 7\}$

Range: $\{3, 4, 5\}$

Step 1: Identifying the Set and Relation Condition

Given the set $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$. The relation $R$ is defined on $A$ as $R = \{(x, y) : 3x - y = 0, \text{ where } x, y \in A\}$.

The condition $3x - y = 0$ can be rewritten as $y = 3x$. This establishes that for every element $x \in A$, the corresponding image $y$ must be three times the value of $x$, provided $y$ is also an element of set $A$ [Per the definition of a relation from set $A$ to set $A$].

Step 2: Tabulating Ordered Pairs $(x, y)$

We test each value of $x \in A$ to determine if $y = 3x$ results in an element within $A$:

For all $x \geq 5$, $y \geq 15$, which exceeds the upper bound of set $A$. Thus, the relation in Roster form is: $R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$.

Step 3: Defining Domain, Codomain, and Range

• Domain: The set of all first components of the ordered pairs in $R$. [Per the definition of Domain in set theory].
  Domain = $\{1, 2, 3, 4\}$
• Codomain: The target set $A$ from which the images are drawn. [Per the definition of a relation $R: A \to A$].
  Codomain = $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$
• Range: The set of all second components of the ordered pairs in $R$. [Per the definition of Range in set theory].
  Range = $\{3, 6, 9, 12\}$

Step 4: Visualization of the Relation

Final Solution: Domain = {1, 2, 3, 4}; Codomain = {1, 2, 3, ..., 14}; Range = {3, 6, 9, 12}

Step 1: Identification of Domain Constraints

The relation $R$ is defined as $R = \{(x, x^3) : x \text{ is a prime number less than } 10\}$. To determine the elements of the domain, we must identify all prime numbers $p$ such that $p < 10$.

• A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. [By Definition of Prime Numbers]
• The set of prime numbers less than 10 is: $S = \{2, 3, 5, 7\}$.

Step 2: Mapping Elements to Codomain via the Function $f(x) = x^3$

For each element $x \in S$, we calculate the corresponding output $y = x^3$ to form the ordered pairs $(x, y)$.

Step 3: Visual Representation of the Relation Mapping

Step 4: Writing in Roster Form

The roster form is the representation of a set by listing all its elements between curly braces. [By Definition of Roster/Tabular Form of Sets]

Combining the ordered pairs derived in Step 2:

$R = \{(2, 8), (3, 27), (5, 125), (7, 343)\}$

Final Solution: $R = \{(2, 8), (3, 27), (5, 125), (7, 343)\}$

Step 1: Identifying the Set-Builder Form and Domain

The relation $R$ is defined as $R = \{(x, x + 5) : x \in \{0, 1, 2, 3, 4, 5\}\}$.

The domain of a relation consists of all the first elements (the inputs $x$) of the ordered pairs present in the relation. [By the definition of the domain of a relation in set theory, $\text{Dom}(R) = \{x : (x, y) \in R\}$].

Given the set of values for $x$ is $\{0, 1, 2, 3, 4, 5\}$, the domain is explicitly stated as the set of these input values:

$\text{Domain}(R) = \{0, 1, 2, 3, 4, 5\}$

Step 2: Deriving the Set of Ordered Pairs

To find the range, we must map each element of the domain to its corresponding image under the relation $y = f(x) = x + 5$.

The relation $R$ in roster form is: $R = \{(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)\}$.

Step 3: Determining the Range

The range of a relation consists of all the second elements (the outputs $y$) of the ordered pairs in $R$. [By the definition of the range of a relation in set theory, $\text{Ran}(R) = \{y : (x, y) \in R\}$].

Based on the calculations in Step 2, the set of all second coordinates is $\{5, 6, 7, 8, 9, 10\}$.

Visual Representation of the Mapping

Final Solution: The domain of the relation $R$ is $\{0, 1, 2, 3, 4, 5\}$ and the range is $\{5, 6, 7, 8, 9, 10\}$.

Step 1: Formal Definition of the Relation

The relation $R$ is defined on the set $A = \{1, 2, 3, 4, 6\}$ such that $R = \{(a, b) : a, b \in A \text{ and } b \text{ is exactly divisible by } a\}$. This implies that $a$ must be a factor of $b$ for the ordered pair $(a, b)$ to belong to the set $R$ [Per the definition of divisibility in integers, $b$ is divisible by $a$ if $b = ka$ for some $k \in \mathbb{Z}$].

Step 2: Systematic Identification of Ordered Pairs

We evaluate each element $a \in A$ to determine its multiples $b$ that are also contained within set $A$:

• For $a = 1$: $1$ divides $1, 2, 3, 4, 6$. Thus, pairs are $(1,1), (1,2), (1,3), (1,4), (1,6)$.
• For $a = 2$: $2$ divides $2, 4, 6$. Thus, pairs are $(2,2), (2,4), (2,6)$.
• For $a = 3$: $3$ divides $3, 6$. Thus, pairs are $(3,3), (3,6)$.
• For $a = 4$: $4$ divides $4$. Thus, the pair is $(4,4)$.
• For $a = 6$: $6$ divides $6$. Thus, the pair is $(6,6)$.

Step 3: Constructing the Relation Set $R$

Combining all identified pairs, we define $R$ as:

$R = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)\}$

Step 4: Extraction of the Range

The range of a relation $R$ is defined as the set of all second elements in the ordered pairs of $R$ [By the definition of the range of a binary relation, $\text{Range}(R) = \{b : (a, b) \in R\}$].

Listing the second components of every ordered pair in $R$:

• From $(1,1)$: $1$
• From $(1,2), (2,2)$: $2$
• From $(1,3), (3,3)$: $3$
• From $(1,4), (2,4), (4,4)$: $4$
• From $(1,6), (2,6), (3,6), (6,6)$: $6$

Step 5: Visual Representation

Final Solution: The range of $R$ is $\{1, 2, 3, 4, 6\}$.

Step 1: Establishing the Set of Inputs ($x$)

The relation $R$ is defined on the set of natural numbers $\mathbb{N} = \{1, 2, 3, \dots\}$. The condition states that $x$ must be a natural number less than $4$. Therefore, the possible values for $x$ are:

$x \in \{1, 2, 3\}$ [Definition of Natural Numbers and inequality constraint].

Step 2: Calculating Corresponding Outputs ($y$)

Using the governing equation $y = x + 5$, we determine the image of each $x$ under the relation $R$:

• For $x = 1$: $y = 1 + 5 = 6$
• For $x = 2$: $y = 2 + 5 = 7$
• For $x = 3$: $y = 3 + 5 = 8$

Since $\{6, 7, 8\} \subset \mathbb{N}$, all calculated pairs satisfy the condition $x, y \in \mathbb{N}$ [Closure property of natural numbers under addition].

Step 3: Representing the Relation in Roster Form

The roster form of a relation is the set of all ordered pairs $(x, y)$ that satisfy the defined rule.

$R = \{(1, 6), (2, 7), (3, 8)\}$

Step 4: Determining Domain and Range

The Domain is the set of all first elements of the ordered pairs in $R$, and the Range is the set of all second elements of the ordered pairs in $R$.

• Domain: $\{1, 2, 3\}$
• Range: $\{6, 7, 8\}$

Step 5: Visual Representation of the Relation

Final Solution: The relation in roster form is $R = \{(1, 6), (2, 7), (3, 8)\}$. The Domain is $\{1, 2, 3\}$ and the Range is $\{6, 7, 8\}$.

Step 1: Formal Definition of the Relation

Given the set $A = \{1, 2, 3, 4, 6\}$, the relation $R$ is defined as:

$R = \{(a, b) : a, b \in A, b \text{ is exactly divisible by } a\}$

The condition "$b$ is exactly divisible by $a$" implies that $\frac{b}{a} \in \mathbb{Z}$, or equivalently, $a$ is a divisor of $b$. [By the definition of divisibility in number theory].

Step 2: Systematic Identification of Ordered Pairs

We test each element $a \in A$ to determine which elements $b \in A$ satisfy the condition $a|b$:

• For $a = 1$: $1$ divides $1, 2, 3, 4, 6$. Pairs: $(1, 1), (1, 2), (1, 3), (1, 4), (1, 6)$.
• For $a = 2$: $2$ divides $2, 4, 6$. Pairs: $(2, 2), (2, 4), (2, 6)$.
• For $a = 3$: $3$ divides $3, 6$. Pairs: $(3, 3), (3, 6)$.
• For $a = 4$: $4$ divides $4$. Pair: $(4, 4)$.
• For $a = 6$: $6$ divides $6$. Pair: $(6, 6)$.

The complete set $R$ is:

$R = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)\}$

Step 3: Determination of the Domain

The domain of a relation $R$ is defined as the set of all first components of the ordered pairs in $R$. [Per the definition of the Domain of a Relation: $Dom(R) = \{a \in A : \exists b \in A \text{ such that } (a, b) \in R\}$].

From the set $R$ derived in Step 2, the unique first components are $\{1, 2, 3, 4, 6\}$.

Step 4: Verification via Mapping Visualization

Since every element $a \in A$ appears as a first component in at least one ordered pair (specifically, $(a, a) \in R$ for all $a \in A$ due to the reflexive property of the "divides" relation), the domain is the entire set $A$.

Final Solution: The domain of $R$ is $\{1, 2, 3, 4, 6\}$.