Step 1: Formal Definition of the Relation

Let $A$ be the set of all polygons in a plane. The relation $R$ is defined on $A$ as: $R = \{ (P_1, P_2) : \text{number of sides of } P_1 = \text{number of sides of } P_2 \}$. To prove $R$ is an equivalence relation, we must demonstrate that it is Reflexive, Symmetric, and Transitive [Per the definition of an Equivalence Relation].

Step 2: Proof of Equivalence Relation Properties

• Reflexivity: For any polygon $P \in A$, the number of sides of $P$ is equal to the number of sides of $P$. Thus, $(P, P) \in R$. [Reflexive Property holds].
• Symmetry: Let $(P_1, P_2) \in R$. This implies that $P_1$ and $P_2$ have the same number of sides. Consequently, $P_2$ and $P_1$ must also have the same number of sides. Thus, $(P_2, P_1) \in R$. [Symmetric Property holds].
• Transitivity: Let $(P_1, P_2) \in R$ and $(P_2, P_3) \in R$. This implies $P_1$ and $P_2$ have $n$ sides, and $P_2$ and $P_3$ have $n$ sides. Therefore, $P_1$ and $P_3$ must both have $n$ sides. Thus, $(P_1, P_3) \in R$. [Transitive Property holds].

Since the relation is reflexive, symmetric, and transitive, $R$ is an equivalence relation.

Step 3: Determining the Equivalence Class

We are given a right-angled triangle $T$ with sides $3, 4, 5$. A triangle, by definition, is a polygon with $3$ sides. The equivalence class $[T]$ consists of all elements $P \in A$ such that $(P, T) \in R$.

$[T] = \{ P \in A : P \text{ has the same number of sides as } T \}$.

Since $T$ has $3$ sides, any polygon related to $T$ must also have exactly $3$ sides. Therefore, the set of all elements in $A$ related to $T$ is the set of all triangles in the plane.

Step 4: Visualizing the Relation

Step 5: Tabular Summary of Relation Conditions

Final Solution: The relation is an equivalence relation because it satisfies reflexivity, symmetry, and transitivity. The set of all elements in $A$ related to the right-angled triangle $T$ is the set of all triangles in plane $A$.

Step 1: Formal Definition of the Relation

The set $A$ is defined as $A = \{1, 2, 3, 4, 5, 6\}$. The relation $R$ on set $A$ is defined as $R = \{(x, y) : y \text{ is divisible by } x\}$, where $x, y \in A$.

By enumerating the ordered pairs $(x, y)$ such that $x$ divides $y$ without remainder, we obtain the set $R$:

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

Step 2: Testing for Reflexivity

A relation $R$ is reflexive if for every $a \in A$, $(a, a) \in R$. [Definition of Reflexive Relation].

• For all $a \in \{1, 2, 3, 4, 5, 6\}$, $a$ is divisible by $a$ because $\frac{a}{a} = 1$ (an integer).
• Specifically, $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \in R$.

Conclusion: Since $\forall a \in A, (a, a) \in R$, the relation $R$ is reflexive.

Step 3: Testing for Symmetry

A relation $R$ is symmetric if $(x, y) \in R \implies (y, x) \in R$ for all $x, y \in A$. [Definition of Symmetric Relation].

• Consider the element $(1, 2) \in R$ because $2$ is divisible by $1$.
• Test the converse: Is $(2, 1) \in R$? For $(2, 1)$ to be in $R$, $1$ must be divisible by $2$. Since $\frac{1}{2} = 0.5 \notin \mathbb{Z}$, $1$ is not divisible by $2$.
• Thus, $(1, 2) \in R$ but $(2, 1) \notin R$.

Conclusion: Since there exists at least one pair $(x, y) \in R$ such that $(y, x) \notin R$, the relation $R$ is not symmetric.

Step 4: Testing for Transitivity

A relation $R$ is transitive if $(x, y) \in R$ and $(y, z) \in R \implies (x, z) \in R$ for all $x, y, z \in A$. [Definition of Transitive Relation].

• Let $(x, y) \in R \implies y = k_1 x$ for some integer $k_1$.
• Let $(y, z) \in R \implies z = k_2 y$ for some integer $k_2$.
• Substituting the first into the second: $z = k_2 (k_1 x) = (k_1 k_2) x$.
• Since $k_1, k_2$ are integers, their product $(k_1 k_2)$ is also an integer.
• Therefore, $z$ is divisible by $x$, implying $(x, z) \in R$.

Example verification: $(2, 4) \in R$ and $(4, 4) \in R \implies (2, 4) \in R$. Or $(1, 2) \in R$ and $(2, 6) \in R \implies (1, 6) \in R$.

Conclusion: Since the condition holds for all triplets, the relation $R$ is transitive.

Step 5: Visual Representation

Final Solution: The relation $R$ is reflexive and transitive, but not symmetric.

Initial Setup & Defining the Relation

To construct a relation $R$ on a set $A$ that is symmetric but neither reflexive nor transitive, we define $A = \{1, 2, 3\}$. Let us define the relation $R$ as follows:

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

Step 1: Testing for Reflexivity

A relation $R$ on set $A$ is reflexive if $(a, a) \in R$ for all $a \in A$. Here, $A = \{1, 2, 3\}$. For $R$ to be reflexive, we require $(1, 1), (2, 2),$ and $(3, 3) \in R$. [Per the definition of Reflexive Relation].

• $(1, 1) \notin R$
• $(2, 2) \notin R$
• $(3, 3) \notin R$

Since the required elements are absent, the relation $R$ is not reflexive.

Step 2: Testing for Symmetry

A relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$ for all $a, b \in A$. [Per the definition of Symmetric Relation].

• Given $(1, 2) \in R$, we check for $(2, 1)$. Since $(2, 1) \in R$, this condition is satisfied.
• Given $(2, 1) \in R$, we check for $(1, 2)$. Since $(1, 2) \in R$, this condition is satisfied.

Therefore, the relation $R$ is symmetric.

Step 3: Testing for Transitivity

A relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. [Per the definition of Transitive Relation].

In our set $R = \{(1, 2), (2, 1)\}$, consider $a=1, b=2, c=1$:

• $(1, 2) \in R$
• $(2, 1) \in R$
• However, $(1, 1) \notin R$

Since $(1, 2) \in R$ and $(2, 1) \in R$, but $(1, 1) \notin R$, the relation fails the transitivity criterion. Thus, the relation $R$ is not transitive.

Visual Representation

Final Solution: The relation $R = \{(1, 2), (2, 1)\}$ on the set $A = \{1, 2, 3\}$ is symmetric but neither reflexive nor transitive.

Step 1: Formal Definition of the Set and Relation

The set $A$ is defined as $\{x \in \mathbb{Z} : 0 \leq x \leq 12\}$, which yields $A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$. The relation $R$ is defined as the identity relation $R = \{(a, b) : a = b\}$ for all $a, b \in A$.

Step 2: Proving Equivalence Relation Properties

A relation is an equivalence relation if and only if it is reflexive, symmetric, and transitive.

• Reflexivity: For any $a \in A$, we observe that $a = a$ is always true [Reflexive Property of Equality]. Thus, $(a, a) \in R$ for all $a \in A$. Hence, $R$ is reflexive.
• Symmetry: Let $(a, b) \in R$. By definition, $a = b$. This implies $b = a$. Therefore, $(b, a) \in R$ [Symmetric Property of Equality]. Hence, $R$ is symmetric.
• Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$. This implies $a = b$ and $b = c$. By the transitive property of equality, $a = c$. Therefore, $(a, c) \in R$ [Transitive Property of Equality]. Hence, $R$ is transitive.

Since $R$ satisfies reflexivity, symmetry, and transitivity, $R$ is an equivalence relation on set $A$.

Step 3: Visualizing the Identity Relation on a Small Subset

To visualize the identity relation for a subset $A' = \{0, 1, 2, 3\}$, we plot the points where $a=b$.

Step 4: Finding the Equivalence Class of 1

The set of all elements related to $1$, denoted as $[1]$, is defined as the set of all $b \in A$ such that $(1, b) \in R$.

Per the definition $R = \{(a, b) : a = b\}$, we require $1 = b$.

Thus, the only element in $A$ satisfying this condition is $b = 1$.

Therefore, $[1] = \{1\}$.

Final Solution: The relation $R$ is an equivalence relation as it satisfies reflexivity, symmetry, and transitivity. The set of all elements related to 1 is $\{1\}$.

Step 1: Formal Definition of the Relation

The relation $R$ is defined on the set of all integers $\mathbb{Z}$. The relation is given by the set of ordered pairs:

$R = \{(x, y) : x - y \in \mathbb{Z}\}$, where $x, y \in \mathbb{Z}$.

We must evaluate $R$ against the three criteria: Reflexivity, Symmetry, and Transitivity.

Step 2: Checking for Reflexivity

A relation $R$ on a set $A$ is reflexive if for every $a \in A$, $(a, a) \in R$.

• Consider any integer $x \in \mathbb{Z}$.
• We evaluate the condition $x - x = 0$.
• Since $0$ is an integer, $0 \in \mathbb{Z}$ [Per the definition of the set of integers].
• Therefore, $(x, x) \in R$ for all $x \in \mathbb{Z}$.

Conclusion: The relation $R$ is reflexive.

Step 3: Checking for Symmetry

A relation $R$ is symmetric if $(x, y) \in R$ implies that $(y, x) \in R$ for all $x, y \in \mathbb{Z}$.

• Assume $(x, y) \in R$. This implies that $(x - y) = k$, where $k$ is some integer $k \in \mathbb{Z}$.
• To check if $(y, x) \in R$, we must verify if $y - x$ is an integer.
• $y - x = -(x - y)$.
• Since $(x - y) = k$, then $y - x = -k$.
• Because the set of integers is closed under additive inversion, if $k \in \mathbb{Z}$, then $-k \in \mathbb{Z}$.

Conclusion: Since $y - x \in \mathbb{Z}$, the relation $R$ is symmetric.

Step 4: Checking for Transitivity

A relation $R$ is transitive if $(x, y) \in R$ and $(y, z) \in R$ implies $(x, z) \in R$ for all $x, y, z \in \mathbb{Z}$.

• Assume $(x, y) \in R$ and $(y, z) \in R$.
• By definition, $x - y = k_1$ and $y - z = k_2$, where $k_1, k_2 \in \mathbb{Z}$.
• We examine $x - z$:
  $x - z = (x - y) + (y - z)$ [By algebraic identity].
  $x - z = k_1 + k_2$.
• Since the sum of two integers is always an integer [Closure property of integers under addition], $k_1 + k_2 \in \mathbb{Z}$.

Conclusion: Since $x - z \in \mathbb{Z}$, $(x, z) \in R$. Therefore, the relation $R$ is transitive.

Visual Representation of Integer Mapping

Final Solution: The relation $R$ is reflexive, symmetric, and transitive (it is an equivalence relation).

Given Variables & Initial Setup

Let $A$ be the set of all human beings in a town at a particular time. The relation $R$ is defined on set $A$ as: $R = \{ (x, y) : x, y \in A \text{ and } x = y + 7 \}$, where heights are measured in centimeters.

Step 1: Testing for Reflexivity

A relation $R$ on set $A$ is reflexive if $(a, a) \in R$ for every $a \in A$. For $(a, a)$ to be in $R$, it must satisfy the condition $a = a + 7$. [Per the axiom of equality in arithmetic, $a = a + 7$ implies $0 = 7$, which is a contradiction.] Since no human being can be 7 cm taller than themselves, $(a, a) \notin R$ for any $a \in A$. Conclusion: The relation is not reflexive.

Step 2: Testing for Symmetry

A relation $R$ on set $A$ is symmetric if $(x, y) \in R \implies (y, x) \in R$ for all $x, y \in A$. Assume $(x, y) \in R$. This implies $x = y + 7$. If we test for $(y, x)$, the condition requires $y = x + 7$. Substituting $x = y + 7$ into the second equation: $y = (y + 7) + 7 \implies y = y + 14$, which implies $0 = 14$. [Since $0 \neq 14$, the assumption that $(y, x) \in R$ is false.] Conclusion: The relation is not symmetric.

Step 3: Testing for Transitivity

A relation $R$ on set $A$ is transitive if $(x, y) \in R$ and $(y, z) \in R \implies (x, z) \in R$. Let $(x, y) \in R \implies x = y + 7$. Let $(y, z) \in R \implies y = z + 7$. For the relation to be transitive, we must check if $(x, z) \in R$, which requires $x = z + 7$. Substituting $y$ from the second equation into the first: $x = (z + 7) + 7$ $x = z + 14$ [By the transitive property of equality, since $14 \neq 7$, $x$ is 14 cm taller than $z$, not 7 cm.] Therefore, $(x, z) \notin R$. Conclusion: The relation is not transitive.

Final Solution: The relation $R$ is neither reflexive, nor symmetric, nor transitive.

1. Definitions and Setup

Let $A$ be the set of all books in a library. The relation $R$ on $A$ is defined as:

$R = \{(x, y) : x, y \in A \text{ and } \text{pages}(x) = \text{pages}(y)\}$

To prove that $R$ is an equivalence relation, we must demonstrate that it satisfies the three properties of an equivalence relation: Reflexivity, Symmetry, and Transitivity.

2. Verification of Reflexivity

A relation $R$ is reflexive if for every $x \in A$, $(x, x) \in R$.

• Consider any book $x \in A$.
• Clearly, the number of pages in book $x$ is equal to the number of pages in book $x$ itself. [Reflexive Property of Equality]
• Thus, $(x, x) \in R$ for all $x \in A$.

Therefore, $R$ is reflexive.

3. Verification of Symmetry

A relation $R$ is symmetric if $(x, y) \in R \implies (y, x) \in R$ for all $x, y \in A$.

• Assume $(x, y) \in R$. This implies that $\text{pages}(x) = \text{pages}(y)$.
• By the symmetric property of equality, if $a = b$, then $b = a$. Therefore, $\text{pages}(y) = \text{pages}(x)$. [Symmetric Property of Equality]
• This implies that $(y, x) \in R$.

Therefore, $R$ is symmetric.

4. Verification of Transitivity

A relation $R$ is transitive if $(x, y) \in R$ and $(y, z) \in R \implies (x, z) \in R$ for all $x, y, z \in A$.

• Assume $(x, y) \in R$ and $(y, z) \in R$.
• From $(x, y) \in R$, we have $\text{pages}(x) = \text{pages}(y)$.
• From $(y, z) \in R$, we have $\text{pages}(y) = \text{pages}(z)$.
• By the transitive property of equality, if $a = b$ and $b = c$, then $a = c$. Thus, $\text{pages}(x) = \text{pages}(z)$. [Transitive Property of Equality]
• This implies that $(x, z) \in R$.

Therefore, $R$ is transitive.

5. Conclusion

Since the relation $R$ is reflexive, symmetric, and transitive, it satisfies all the conditions of an equivalence relation.

Final Solution: The relation $R$ is an equivalence relation as it satisfies the properties of reflexivity, symmetry, and transitivity on the set $A$.

Given Information

Set $A = \{1, 2, 3\}$

Relation $R$ defined on $A$: $R = \{(1, 2), (2, 1)\}$

Step 1: Testing for Reflexivity

A relation $R$ on set $A$ is reflexive if and only if for every element $a \in A$, the ordered pair $(a, a) \in R$. [By definition of a Reflexive Relation]

• For $a = 1$: $(1, 1) \notin R$
• For $a = 2$: $(2, 2) \notin R$
• For $a = 3$: $(3, 3) \notin R$

Since $(1, 1), (2, 2), (3, 3) \notin R$, the relation $R$ is not reflexive.

Step 2: Testing for Symmetry

A relation $R$ on set $A$ is symmetric if for every $(a, b) \in R$, it implies $(b, a) \in R$. [By definition of a Symmetric Relation]

• Given $(1, 2) \in R$, we check for $(2, 1) \in R$. Since $(2, 1) \in R$, the condition holds for this pair.
• Given $(2, 1) \in R$, we check for $(1, 2) \in R$. Since $(1, 2) \in R$, the condition holds for this pair.

Since for all $(a, b) \in R$, $(b, a) \in R$, the relation $R$ is symmetric.

Step 3: Testing for Transitivity

A relation $R$ on set $A$ is transitive if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$. [By definition of a Transitive Relation]

Consider the pairs $(1, 2) \in R$ and $(2, 1) \in R$ where $a=1, b=2, c=1$.

If $R$ were transitive, $(a, c)$ must be in $R$, meaning $(1, 1)$ must be in $R$.

However, $(1, 1) \notin R$. Since the condition $aRb$ and $bRc \implies aRc$ fails for $a=1, b=2, c=1$, the relation $R$ is not transitive.

Visual Representation of the Relation

Conclusion Summary

Final Solution: The relation $R$ is symmetric but neither reflexive nor transitive.

Step 1: Formal Definition of the Relation

Let $A$ be the set of all human beings in a town at a particular time. The relation $R$ is defined on set $A$ as:

$R = \{ (x, y) \in A \times A : x \text{ is the wife of } y \}$

For a relation $R$ to be defined as such, it implies that for any ordered pair $(x, y) \in R$, $x$ must be a female and $y$ must be a male. [Per the biological and linguistic definition of 'wife'].

Step 2: Testing for Reflexivity

A relation $R$ is reflexive if for every $x \in A$, $(x, x) \in R$.

According to the definition, $(x, x) \in R$ would mean "$x$ is the wife of $x$".

Analysis: No human being can be their own wife because it is biologically impossible for an individual to stand in the marital relationship of 'wife' to themselves. [By the Principle of Non-Contradiction].

Conclusion: $R$ is not reflexive.

Step 3: Testing for Symmetry

A relation $R$ is symmetric if $(x, y) \in R \implies (y, x) \in R$.

Suppose $(x, y) \in R$. This implies that $x$ is the wife of $y$. This necessitates that $x$ is female and $y$ is male.

Now, consider $(y, x)$. For $(y, x) \in R$, $y$ would have to be the wife of $x$. However, since $y$ is male, he cannot be the wife of $x$. [By the definition of the relation $R$].

Conclusion: $R$ is not symmetric.

Step 4: Testing for Transitivity

A relation $R$ is transitive if $(x, y) \in R$ and $(y, z) \in R \implies (x, z) \in R$.

Suppose $(x, y) \in R$ and $(y, z) \in R$.

• $(x, y) \in R \implies x$ is the wife of $y$ ($x$ is female, $y$ is male).
• $(y, z) \in R \implies y$ is the wife of $z$.

Analysis: For $(y, z) \in R$, $y$ must be a female. However, from the first condition, we established that $y$ is male. It is impossible for $y$ to be both male and female simultaneously. Thus, the condition $(y, z) \in R$ can never be satisfied if $(x, y) \in R$ is true. Since the antecedent $(x, y) \in R \land (y, z) \in R$ is always false, the relation is vacuously transitive in a logical sense, but generally categorized as not transitive because no such chain exists. [By the Law of Empty Antecedent in Predicate Logic].

Step 5: Final Evaluation

Final Solution: The relation $R$ is neither reflexive, nor symmetric, nor transitive.

Step 1: Theoretical Definition of the Required Relation

To construct a relation $R$ on a set $A$ that satisfies the given conditions, we must adhere to the following formal definitions:

• Reflexive: For every $a \in A$, $(a, a) \in R$. [Per the reflexive property definition].
• Symmetric: If $(a, b) \in R$, then $(b, a) \in R$ for all $a, b \in A$. [Per the symmetric property definition].
• Not Transitive: There must exist elements $a, b, c \in A$ such that $(a, b) \in R$ and $(b, c) \in R$, but $(a, c) \notin R$. [Per the negation of transitivity].

Step 2: Defining the Set and Relation

Let set $A = \{1, 2, 3\}$. We define the relation $R$ as follows:

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

Step 3: Verification of Properties

Step 4: Visual Representation

Final Solution: The relation $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)\}$ on the set $A = \{1, 2, 3\}$ is reflexive and symmetric, but not transitive.

Given Variables & Initial Setup

Let $A$ be the set of all human beings in a town at a particular time. The relation $R$ is defined on set $A$ as:

$R = \{(x, y) : x, y \in A \text{ and } x \text{ is the father of } y\}$

Step 1: Testing for Reflexivity

A relation $R$ is reflexive if for every $x \in A$, $(x, x) \in R$.

• By definition, $(x, x) \in R$ if and only if $x$ is the father of $x$.
• It is logically and biologically impossible for a person to be their own father [Per the Law of Non-Contradiction].
• Since $x$ cannot be the father of $x$ for any $x \in A$, $(x, x) \notin R$.

Conclusion: The relation $R$ is not reflexive.

Step 2: Testing for Symmetry

A relation $R$ is symmetric if $(x, y) \in R \implies (y, x) \in R$ for all $x, y \in A$.

• Assume $(x, y) \in R$, which implies $x$ is the father of $y$.
• If $x$ is the father of $y$, then $y$ must be the child of $x$. It does not imply that $y$ is the father of $x$.
• Example: If $A$ contains a person 'John' and his daughter 'Mary', then $(\text{John, Mary}) \in R$, but $(\text{Mary, John}) \notin R$.

Conclusion: The relation $R$ is not symmetric.

Step 3: Testing for Transitivity

A relation $R$ is transitive if $(x, y) \in R$ and $(y, z) \in R \implies (x, z) \in R$ for all $x, y, z \in A$.

• Assume $(x, y) \in R$ and $(y, z) \in R$.
• This implies $x$ is the father of $y$, and $y$ is the father of $z$.
• If $x$ is the father of $y$ and $y$ is the father of $z$, then $x$ is the paternal grandfather of $z$, not the father of $z$.
• Therefore, $(x, z) \notin R$.

Conclusion: The relation $R$ is not transitive.

Visual Representation of Logical Failure

Final Solution: The relation $R = \{(x, y) : x \text{ is father of } y\}$ is neither reflexive, nor symmetric, nor transitive.

Step 1: Defining the Set and the Relation

Given set $A = \{1, 2, 3, ..., 14\}$. The relation $R$ is defined on $A$ as:

$R = \{(x, y) : 3x - y = 0\}$, which can be rewritten as $y = 3x$.

To determine the ordered pairs in $R$, we substitute values of $x \in A$ such that $y \in A$:

• If $x = 1, y = 3(1) = 3$
• If $x = 2, y = 3(2) = 6$
• If $x = 3, y = 3(3) = 9$
• If $x = 4, y = 3(4) = 12$
• If $x = 5, y = 3(5) = 15$ (Since $15 \notin A$, this pair is excluded.)

Therefore, the relation is represented by the set: $R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$.

Step 2: Testing for Reflexivity

A relation $R$ is reflexive if $(a, a) \in R$ for every $a \in A$ [Definition of Reflexive Relation].

Consider $1 \in A$. However, $(1, 1) \notin R$ because $3(1) \neq 1$ ($3 \neq 1$).

Since $(1, 1) \notin R$, the relation $R$ is not reflexive.

Step 3: Testing for Symmetry

A relation $R$ is symmetric if $(x, y) \in R \implies (y, x) \in R$ for all $x, y \in A$ [Definition of Symmetric Relation].

We observe that $(1, 3) \in R$ because $3(1) - 3 = 0$.

However, for the relation to be symmetric, $(3, 1)$ must also be in $R$. Substituting into the definition: $3(3) - 1 = 9 - 1 = 8 \neq 0$. Thus, $(3, 1) \notin R$.

Since $(1, 3) \in R$ but $(3, 1) \notin R$, the relation $R$ is not symmetric.

Step 4: Testing for Transitivity

A relation $R$ is transitive if $(x, y) \in R$ and $(y, z) \in R \implies (x, z) \in R$ for all $x, y, z \in A$ [Definition of Transitive Relation].

Consider the pairs $(1, 3) \in R$ and $(3, 9) \in R$.

For $R$ to be transitive, we must have $(1, 9) \in R$.

Checking the condition $y = 3x$ for the pair $(1, 9)$: $9 \neq 3(1)$. Thus, $(1, 9) \notin R$.

Since $(1, 3) \in R$ and $(3, 9) \in R$ but $(1, 9) \notin R$, the relation $R$ is not transitive.

Visual Representation of the Relation

Final Solution: The relation $R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$ is neither reflexive, nor symmetric, nor transitive.

1. Definitions and Setup

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

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

To determine the nature of the relation, we analyze it against the criteria for reflexive, symmetric, and transitive relations.

2. Reflexivity Analysis

A relation $R$ is reflexive if for every $a \in A$, $(a, a) \in R$ [Reflexive Law].

• $(1, 1) \in R$
• $(2, 2) \in R$
• $(3, 3) \in R$
• $(4, 4) \in R$

Since $\forall a \in \{1, 2, 3, 4\}$, $(a, a) \in R$, the relation is reflexive.

3. Symmetry Analysis

A relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$ [Symmetric Law].

Consider the element $(1, 2) \in R$. For the relation to be symmetric, $(2, 1)$ must also be in $R$.

However, $(2, 1) \notin R$.

Since $(1, 2) \in R$ but $(2, 1) \notin R$, the relation is not symmetric.

4. Transitivity Analysis

A relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$ [Transitive Law].

Let us test the pairs:

• $(1, 3) \in R$ and $(3, 2) \in R$. If transitive, $(1, 2)$ must be in $R$. Since $(1, 2) \in R$, this condition is satisfied.
• $(3, 2) \in R$ and $(2, 2) \in R$. If transitive, $(3, 2)$ must be in $R$. Since $(3, 2) \in R$, this condition is satisfied.
• $(1, 2) \in R$ and $(2, 2) \in R$. If transitive, $(1, 2)$ must be in $R$. This is satisfied.

Upon verifying all combinations $(a, b)$ and $(b, c)$ where $a, b, c \in A$, we find that $(a, c) \in R$ is always satisfied.

Therefore, the relation is transitive.

5. Comparative Summary

Final Solution: The relation R is reflexive and transitive but not symmetric.

Definition of Relation $R$

The relation $R$ is defined on the set of real numbers $\mathbb{R}$ such that $R = \{(a, b) : a \leq b\}$, where $a, b \in \mathbb{R}$.

Step 1: Testing for Reflexivity

A relation $R$ is reflexive if for every $a \in \mathbb{R}$, $(a, a) \in R$.

• For any real number $a$, it is a fundamental property of the real number system that $a = a$.
• Since the condition $a \leq b$ includes the equality $a = b$, it follows that $a \leq a$ is always true for any $a \in \mathbb{R}$. [By the reflexive property of inequality].
• Therefore, $(a, a) \in R$ for all $a \in \mathbb{R}$.
• Conclusion: The relation $R$ is reflexive.

Step 2: Testing for Symmetry

A relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$ for all $a, b \in \mathbb{R}$.

• Consider a counter-example: Let $a = 1$ and $b = 2$.
• Since $1 \leq 2$, $(1, 2) \in R$.
• However, checking the symmetric pair $(2, 1)$, we see that $2 \leq 1$ is false.
• Since $(1, 2) \in R$ but $(2, 1) \notin R$, the condition for symmetry is violated.
• Conclusion: The relation $R$ is not symmetric.

Step 3: Testing for Transitivity

A relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$ for all $a, b, c \in \mathbb{R}$.

• Suppose $(a, b) \in R$ and $(b, c) \in R$.
• By definition, this implies $a \leq b$ and $b \leq c$.
• By the transitive property of inequality in real numbers, if $a \leq b$ and $b \leq c$, then $a \leq c$. [By the law of trichotomy and transitivity of order].
• Since $a \leq c$, it follows that $(a, c) \in R$.
• Conclusion: The relation $R$ is transitive.

Visualization of Order Relation

Final Solution: The relation $R = \{(a, b) : a \leq b\}$ is reflexive because $a \leq a$, transitive because $a \leq b$ and $b \leq c$ implies $a \leq c$, but not symmetric because $a \leq b$ does not necessarily imply $b \leq a$.

Given Relation and Domain

The relation $R$ is defined on the set of all real numbers $\mathbb{R}$ by the condition:

$R = \{(a, b) : a \leq b^3, \text{ where } a, b \in \mathbb{R}\}$

Step 1: Testing for Reflexivity

A relation is reflexive if $\forall a \in \mathbb{R}, (a, a) \in R$. This requires $a \leq a^3$ to hold for all $a \in \mathbb{R}$.

Consider the element $a = \frac{1}{2}$:

• The condition requires $\frac{1}{2} \leq (\frac{1}{2})^3$
• $\frac{1}{2} \leq \frac{1}{8}$
• This statement is False, as $0.5 > 0.125$.

[Definition of Reflexive Relation: $(a, a) \in R$ for all $a \in A$]. Since there exists at least one $a \in \mathbb{R}$ for which $a \not\leq a^3$, the relation is not reflexive.

Step 2: Testing for Symmetry

A relation is symmetric if $(a, b) \in R \implies (b, a) \in R$.

Consider the pair $(1, 2)$:

• Check if $(1, 2) \in R$: $1 \leq 2^3 \implies 1 \leq 8$. This is True.
• Check if $(2, 1) \in R$: $2 \leq 1^3 \implies 2 \leq 1$. This is False.

[Definition of Symmetric Relation: If $(a, b) \in R$, then $(b, a) \in R$]. Since $(1, 2) \in R$ but $(2, 1) \notin R$, the relation is not symmetric.

Step 3: Testing for Transitivity

A relation is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.

Consider the elements $a = 10, b = 3, c = 2$:

• Check $(10, 3)$: $10 \leq 3^3 \implies 10 \leq 27$ (True).
• Check $(3, 2)$: $3 \leq 2^3 \implies 3 \leq 8$ (True).
• Check $(10, 2)$: $10 \leq 2^3 \implies 10 \leq 8$. This is False.

[Definition of Transitive Relation: If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$]. Since $(10, 3) \in R$ and $(3, 2) \in R$, but $(10, 2) \notin R$, the relation is not transitive.

Visual Representation of the Boundary $a = b^3$

Final Solution: The relation $R$ defined by $a \leq b^3$ is neither reflexive, nor symmetric, nor transitive.

Step 1: Formal Definition of the Relation

The relation $R$ is defined on the set of natural numbers $\mathbb{N}$ as follows:

$R = \{ (a, b) : a = b - 2, b > 6 \}$

[Definition: A relation $R$ from a set $A$ to set $B$ is a subset of the Cartesian product $A \times B$. Here $A = B = \mathbb{N}$].

Step 2: Analysis of Constraints

The condition $b > 6$ implies that the possible values for the second element ($b$) in the ordered pair $(a, b)$ must be chosen from the set $\{7, 8, 9, 10, ... \}$.

Using the transformation equation $a = b - 2$, we map these values to $a$:

• If $b = 7$, then $a = 7 - 2 = 5$. Pair: $(5, 7)$
• If $b = 8$, then $a = 8 - 2 = 6$. Pair: $(6, 8)$
• If $b = 9$, then $a = 9 - 2 = 7$. Pair: $(7, 9)$
• If $b = 10$, then $a = 10 - 2 = 8$. Pair: $(8, 10)$

Step 3: Verification of Options

The standard multiple-choice options typically associated with this problem are:

Step 4: Geometric Representation of the Discrete Relation

Step 5: Theoretical Conclusion

By checking the element $(6, 8)$ against the set-builder form: 1. $6 \in \mathbb{N}$ and $8 \in \mathbb{N}$. 2. The condition $b > 6$ is satisfied as $8 > 6$. 3. The equality $a = b - 2$ is satisfied as $6 = 8 - 2$. [By definition of Set-Builder notation evaluation].

Final Solution: The correct answer is (6, 8).

Definition of the Required Relation

To construct a relation $R$ on a set $A$ that is reflexive and transitive but not symmetric, we define a set $A = \{1, 2, 3\}$. We propose the relation $R$ as follows:

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

Step 1: Verification of Reflexivity

A relation $R$ on set $A$ is reflexive if for every $a \in A$, $(a, a) \in R$ [By Definition of Reflexive Relation].

• For $a = 1$: $(1, 1) \in R$
• For $a = 2$: $(2, 2) \in R$
• For $a = 3$: $(3, 3) \in R$

Since $(a, a) \in R$ for all $a \in \{1, 2, 3\}$, the relation is reflexive.

Step 2: Verification of Transitivity

A relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$ [By Definition of Transitive Relation].

Exhaustive checking of all transitive triplets confirms that the condition holds. Thus, the relation is transitive.

Step 3: Testing for Symmetry

A relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$ [By Definition of Symmetric Relation].

Consider the element $(1, 2) \in R$. For the relation to be symmetric, $(2, 1)$ must also be an element of $R$. However, $(2, 1) \notin R$. Therefore, the relation is not symmetric.

Visual Representation of the Relation

Final Solution: The relation $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$ on set $A = \{1, 2, 3\}$ is reflexive and transitive but not symmetric.

Definition and Mathematical Setup

Let $A$ be the set of all human beings in a town at a particular time. The relation $R$ is defined on set $A$ as:

$R = \{(x, y) : x \text{ and } y \text{ live in the same locality}\}$

Step 1: Testing for Reflexivity

A relation $R$ is reflexive if for every element $a \in A$, $(a, a) \in R$.

• For any human being $x \in A$, it is a fundamental truth that $x$ lives in the same locality as themselves. [Identity Property]
• Therefore, $(x, x) \in R$ for all $x \in A$.
• Conclusion: The relation $R$ is reflexive.

Step 2: Testing for Symmetry

A relation $R$ is symmetric if $(x, y) \in R \implies (y, x) \in R$ for all $x, y \in A$.

• Assume $(x, y) \in R$. This implies $x$ and $y$ live in the same locality. [Definition of $R$]
• If $x$ lives in the same locality as $y$, it necessarily follows that $y$ lives in the same locality as $x$. [Commutative property of the 'same locality' relation]
• Thus, $(y, x) \in R$.
• Conclusion: The relation $R$ is symmetric.

Step 3: Testing for Transitivity

A relation $R$ is transitive if $(x, y) \in R$ and $(y, z) \in R \implies (x, z) \in R$ for all $x, y, z \in A$.

• Suppose $(x, y) \in R$ and $(y, z) \in R$.
• $(x, y) \in R \implies x$ and $y$ live in the same locality.
• $(y, z) \in R \implies y$ and $z$ live in the same locality.
• If $x$ is in the same locality as $y$, and $y$ is in the same locality as $z$, by the transitive property of equality/equivalence, $x$ and $z$ must live in the same locality. [Axiom of Transitivity]
• Thus, $(x, z) \in R$.
• Conclusion: The relation $R$ is transitive.

Summary Analysis Table

Final Solution: The relation $R$ is reflexive, symmetric, and transitive. Consequently, $R$ is an equivalence relation.

Step 1: Establishing the Relation Definition

Given set $A = \{1, 2, 3, 4, 5\}$. The relation $R$ is defined as:

$R = \{(a, b) : a, b \in A, |a - b| \text{ is even}\}$

Step 2: Proving the Equivalence Relation

To prove $R$ is an equivalence relation, we must demonstrate that it is reflexive, symmetric, and transitive.

(i) Reflexivity: For any $a \in A$, we observe $|a - a| = 0$. Since $0$ is an even number, $(a, a) \in R$ for all $a \in A$. [Per the definition of Reflexive Relations].

(ii) Symmetry: Let $(a, b) \in R$. This implies $|a - b|$ is even. Since $|a - b| = |-(b - a)| = |b - a|$, then $|b - a|$ is also even. Thus, $(b, a) \in R$. [Per the definition of Symmetric Relations].

(iii) Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$. Then $|a - b| = 2k_1$ and $|b - c| = 2k_2$ for some integers $k_1, k_2$. We express $a - c = (a - b) + (b - c)$. Since the sum of two even numbers is even, $(a - c)$ is even, hence $|a - c|$ is even. Thus, $(a, c) \in R$. [Per the definition of Transitive Relations].

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Step 3: Analyzing Subsets and Equivalence Classes

We analyze the two subsets $S_1 = \{1, 3, 5\}$ and $S_2 = \{2, 4\}$.

To show that no element of $S_1$ is related to $S_2$, let $a \in \{1, 3, 5\}$ (odd) and $b \in \{2, 4\}$ (even). The difference $a - b$ is always odd (e.g., $1-2 = -1, 3-4 = -1, 5-2 = 3$). Since an odd number is not even, $|a - b|$ is not even. Therefore, $(a, b) \notin R$.

Step 4: Visualizing the Equivalence Classes

Final Solution: The relation is proven to be an equivalence relation as it satisfies reflexivity, symmetry, and transitivity. The set $A$ is partitioned into two disjoint equivalence classes: $\{1, 3, 5\}$ and $\{2, 4\}$, where cross-set relations are impossible due to the parity properties of integers.

Step 1: Theoretical Definition of the Required Relation

To construct a relation $R$ on a set $A$ that is Symmetric and Transitive but not Reflexive, we must satisfy the following logical conditions:

• Symmetric: If $(a, b) \in R$, then $(b, a) \in R$ for all $a, b \in A$. [By definition of symmetry]
• Transitive: If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ for all $a, b, c \in A$. [By definition of transitivity]
• Not Reflexive: There exists at least one element $x \in A$ such that $(x, x) \notin R$. [By negation of reflexivity]

Step 2: Defining the Set and the Relation

Let the set $A = \{1, 2, 3\}$. We define the relation $R$ on $A$ as:

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

Step 3: Verification of Properties

A. Symmetry Analysis

• $(1, 1) \in R \implies (1, 1) \in R$ (True)
• $(1, 2) \in R \implies (2, 1) \in R$ (True)
• $(2, 1) \in R \implies (1, 2) \in R$ (True)
• $(2, 2) \in R \implies (2, 2) \in R$ (True)

Since all ordered pairs satisfy the condition, $R$ is Symmetric.

B. Transitivity Analysis

The condition for transitivity requires that if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

• For $(1, 2) \in R$ and $(2, 1) \in R$, we must have $(1, 1) \in R$. Since $(1, 1) \in R$, it holds.
• For $(2, 1) \in R$ and $(1, 2) \in R$, we must have $(2, 2) \in R$. Since $(2, 2) \in R$, it holds.

Thus, $R$ is Transitive.

C. Reflexivity Analysis

A relation $R$ on $A$ is reflexive if $(a, a) \in R$ for all $a \in A$.

Here, $3 \in A$, but $(3, 3) \notin R$.

Since $(3, 3) \notin R$, $R$ is not Reflexive.

Visual Representation

Final Solution: The relation $R = \{(1, 1), (1, 2), (2, 1), (2, 2)\}$ on set $A = \{1, 2, 3\}$ is symmetric and transitive but not reflexive.