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 One-One (Injective) Mapping

A function $f: X \to Y$ is defined as one-one if and only if for all $x_1, x_2 \in X$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$ [Definition of Injective Mapping].

Step 2: Testing for One-One Property

Let us select two distinct values from the domain $R$, such as $x_1 = 2$ and $x_2 = 3$.

• According to the definition of the Signum Function, since $2 > 0$, $f(2) = 1$.
• Since $3 > 0$, $f(3) = 1$.

Here, $f(2) = f(3)$ but $2 \neq 3$. [By contradiction of the injective criterion].

Since multiple distinct elements in the domain map to the same image (1), the function is not one-one.

Step 3: Formal Definition of Onto (Surjective) Mapping

A function $f: X \to Y$ is onto if for every $y \in Y$, there exists at least one $x \in X$ such that $f(x) = y$. In this case, the codomain is the set of all real numbers, $R$ [Definition of Surjective Mapping].

Step 4: Testing for Onto Property

The range of the signum function is restricted to the set $\{-1, 0, 1\}$.

Consider any real number $y \in R$ that is not in the set $\{-1, 0, 1\}$, for example, $y = 2$.

There exists no $x \in R$ such that $f(x) = 2$, because the output of $f(x)$ is strictly limited to $\{1, 0, -1\}$ by definition. Therefore, the codomain $R$ is not equal to the range $\{-1, 0, 1\}$.

Since there are elements in the codomain with no pre-image in the domain, the function is not onto.

Visual Representation of the Signum Function

Final Solution: The Signum function $f(x)$ is neither one-one because $f(2) = f(3)$, nor onto because the range $\{-1, 0, 1\} \subset R$ does not cover the codomain $R$.

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 Function

The function is defined as $f: \mathbb{R} \to \mathbb{R}$ where $f(x) = x^2$. Here, the domain is the set of all real numbers $\mathbb{R}$, and the codomain is the set of all real numbers $\mathbb{R}$.

Step 2: Checking for Injectivity (One-to-One)

A function $f$ is injective (one-to-one) if for every $x_1, x_2$ in the domain, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

• Consider two distinct elements in the domain, e.g., $x_1 = 1$ and $x_2 = -1$.
• Calculate the images: $f(1) = (1)^2 = 1$ and $f(-1) = (-1)^2 = 1$.
• Since $f(1) = f(-1)$ but $1 \neq -1$, the definition of injectivity is violated [By the definition of an injective function].

Conclusion: The function $f(x) = x^2$ is not injective.

Step 3: Checking for Surjectivity (Onto)

A function $f: A \to B$ is surjective (onto) if for every $y \in B$ (codomain), there exists at least one $x \in A$ (domain) such that $f(x) = y$.

• The codomain is given as $\mathbb{R}$.
• For any real number $x$, $x^2 \geq 0$ [Property of real numbers: squares are non-negative].
• Consider a negative real number in the codomain, for instance, $y = -4$.
• If $f(x) = -4$, then $x^2 = -4$. This implies $x = \pm \sqrt{-4}$, which is not a real number ($x \notin \mathbb{R}$).
• Since there exists no $x$ in the domain $\mathbb{R}$ such that $f(x) = -4$, the range of the function is $[0, \infty)$, which is a proper subset of the codomain $\mathbb{R}$.

Conclusion: The function $f(x) = x^2$ is not surjective.

Step 4: Visual Verification

Step 5: Summary Table of Findings

Final Solution: The function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^2$ is neither injective nor surjective.

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 $\{1\}$.

Given Variables & Initial Setup

Consider the function $f: \mathbb{Z} \to \mathbb{Z}$ defined by $f(x) = x^2$, where $\mathbb{Z}$ represents the set of all integers ($\mathbb{Z} = \{..., -2, -1, 0, 1, 2, ...\}$).

Step 1: Testing for Injectivity (One-to-One)

A function $f: A \to B$ is injective (one-to-one) if $f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2 \in A$.

Let $x_1 = 1$ and $x_2 = -1$. Both $1 \in \mathbb{Z}$ and $-1 \in \mathbb{Z}$.

Calculating the images:

• $f(1) = (1)^2 = 1$
• $f(-1) = (-1)^2 = 1$

[By the definition of an injective function: If $f(x_1) = f(x_2)$ but $x_1 \neq x_2$, the function is not injective.]

Since $f(1) = f(-1) = 1$ but $1 \neq -1$, the function $f(x) = x^2$ is not injective.

Step 2: Testing for Surjectivity (Onto)

A function $f: A \to B$ is surjective (onto) if for every $y \in B$, there exists at least one $x \in A$ such that $f(x) = y$.

Let the codomain be $B = \mathbb{Z}$. Choose an element $y = -2 \in \mathbb{Z}$.

We solve for $x$: $f(x) = x^2 = -2$.

$x = \pm\sqrt{-2}$.

[Per the definition of the set of integers, $\sqrt{-2} \notin \mathbb{Z}$.]

Because there is no integer $x$ such that $x^2 = -2$, there exists an element in the codomain that has no pre-image in the domain.

Therefore, the function $f(x) = x^2$ is not surjective.

Visual Representation

Summary of Findings

Final Solution: The function $f: \mathbb{Z} \to \mathbb{Z}$ given by $f(x) = x^2$ is neither injective nor surjective.

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).

Step 1: Analyzing Injectivity (One-One Nature)

To determine if $f(x)$ is one-one, we assume $f(x_1) = f(x_2)$ for any $x_1, x_2 \in A$ and verify if $x_1 = x_2$ [Definition of Injectivity].

Given: $f(x) = \frac{x-2}{x-3}$

$\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}$

Cross-multiplying: $(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$

Expanding both sides: $x_1x_2 - 3x_1 - 2x_2 + 6 = x_2x_1 - 3x_2 - 2x_1 + 6$

Subtracting $x_1x_2$ and $6$ from both sides: $-3x_1 - 2x_2 = -3x_2 - 2x_1$

Rearranging terms: $x_2 - x_1 = 0 \implies x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$, the function $f$ is one-one.

Step 2: Analyzing Surjectivity (Onto Nature)

To determine if $f$ is onto, we check if for every $y \in B$, there exists an $x \in A$ such that $f(x) = y$ [Definition of Surjectivity].

Let $y = \frac{x-2}{x-3}$. Solving for $x$ in terms of $y$:

$y(x-3) = x-2$

$yx - 3y = x - 2$

$yx - x = 3y - 2$

$x(y - 1) = 3y - 2$

$x = \frac{3y - 2}{y - 1}$

For $x$ to be defined, $y - 1 \neq 0$, which means $y \neq 1$. Since the codomain $B = R - \{1\}$, this condition is satisfied for all $y \in B$. Furthermore, we must ensure $x \neq 3$ so that $x \in A$.

If $x = 3$, then $3 = \frac{3y - 2}{y - 1} \implies 3y - 3 = 3y - 2 \implies -3 = -2$, which is a contradiction. Thus, for every $y \in B$, there exists a unique $x \in A$. The function $f$ is onto.

Step 3: Graphical Visualization

Step 4: Comparative Summary

Final Solution: The function $f$ is both one-one and onto (bijective).

Step 1: Formal Definition of Bijectivity

A function $f: X \to Y$ is bijective if and only if it satisfies two conditions:

• Injectivity (One-to-One): For any $x_1, x_2 \in X$, $f(x_1) = f(x_2) \implies x_1 = x_2$.
• Surjectivity (Onto): For every $y \in Y$, there exists at least one $x \in X$ such that $f(x) = y$.

Step 2: Proving Injectivity

Let $(a_1, b_1)$ and $(a_2, b_2)$ be elements in the domain $A \times B$. Assume $f(a_1, b_1) = f(a_2, b_2)$.
By the definition of the function $f(a, b) = (b, a)$, we have: $(b_1, a_1) = (b_2, a_2)$
By the definition of equality of ordered pairs, the corresponding components must be equal [Axiom of Ordered Pairs]: $b_1 = b_2$ and $a_1 = a_2$
Therefore, $(a_1, b_1) = (a_2, b_2)$.
Since $f(a_1, b_1) = f(a_2, b_2) \implies (a_1, b_1) = (a_2, b_2)$, the function $f$ is injective.

Step 3: Proving Surjectivity

Let $(b, a)$ be an arbitrary element in the codomain $B \times A$.
By definition of the Cartesian product $B \times A$, $b \in B$ and $a \in A$.
Since $a \in A$ and $b \in B$, the ordered pair $(a, b)$ exists in the domain $A \times B$.
Now, evaluate the function at $(a, b)$: $f(a, b) = (b, a)$
Since for every element $(b, a) \in B \times A$, there exists an element $(a, b) \in A \times B$ such that $f(a, b) = (b, a)$, the function $f$ is surjective.

Conclusion

Having satisfied both the conditions of injectivity and surjectivity, the function $f: A \times B \to B \times A$ is confirmed to be a bijection.
Final Solution: The function $f(a, b) = (b, a)$ is bijective as it satisfies the requirements for both one-to-one and onto mappings.

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 $A$.

Step 1: Formal Definition of the Function

The function is defined as $f: \mathbb{N} \to \mathbb{N}$, where $\mathbb{N} = \{1, 2, 3, \dots\}$. The rule governing the mapping is given by $f(x) = x^2$.

Step 2: Analysis of Injectivity (One-to-One)

A function is injective if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2$ in the domain. [Per the definition of Injective Functions].

Consider $f(x_1) = f(x_2)$:

$x_1^2 = x_2^2$

Taking the square root of both sides: $\sqrt{x_1^2} = \sqrt{x_2^2} \implies |x_1| = |x_2|$.

Since the domain is $\mathbb{N}$ (natural numbers), both $x_1$ and $x_2$ must be positive. Therefore, $x_1 = x_2$.

Example check: Let $x_1 = 2, x_2 = 3$. $f(2) = 4, f(3) = 9$. Since $2 \neq 3 \implies 4 \neq 9$.

Conclusion: The function is injective.

Step 3: Analysis of Surjectivity (Onto)

A function is surjective if the Range of $f$ is equal to the Codomain of $f$. [Per the definition of Surjective Functions].

The codomain is $\mathbb{N} = \{1, 2, 3, 4, 5, \dots\}$.

The range of $f(x) = x^2$ for $x \in \mathbb{N}$ is $\{1^2, 2^2, 3^2, \dots\} = \{1, 4, 9, 16, \dots\}$.

We observe that there exist elements in the codomain that have no pre-image in the domain. For example, $2 \in \text{Codomain}$, but there is no $x \in \mathbb{N}$ such that $x^2 = 2$ (as $\sqrt{2} \notin \mathbb{N}$).

Since $\text{Range} \subset \text{Codomain}$ but $\text{Range} \neq \text{Codomain}$, the function is not surjective.

Step 4: Summary Table

Final Solution: The function $f: \mathbb{N} \to \mathbb{N}$ given by $f(x) = x^2$ is injective but not surjective.

Step 1: Analyzing Injectivity (One-to-One Property)

A function $f: A \to B$ is injective if $f(n_1) = f(n_2) \implies n_1 = n_2$. We evaluate this by testing values for $n \in \mathbb{N}$.

• Consider $n_1 = 1$ (odd): $f(1) = \frac{1+1}{2} = 1$.
• Consider $n_2 = 2$ (even): $f(2) = \frac{2}{2} = 1$.

[By Definition of Injectivity]: Since $f(1) = f(2) = 1$ but $1 \neq 2$, the function $f$ fails the condition for injectivity. Therefore, $f$ is not one-to-one.

Step 2: Analyzing Surjectivity (Onto Property)

A function $f: A \to B$ is surjective if for every $y \in \text{Codomain}$, there exists an $x \in \text{Domain}$ such that $f(x) = y$. Here, the codomain is $\mathbb{N}$.

Case 1: If $y$ is odd, let $n = 2y - 1$. Since $y \geq 1$, $n$ is an odd natural number.
$f(2y-1) = \frac{(2y-1)+1}{2} = \frac{2y}{2} = y$.

Case 2: If $y$ is even, let $n = 2y$. Since $y \geq 1$, $n$ is an even natural number.
$f(2y) = \frac{2y}{2} = y$.

[By Definition of Surjectivity]: Since every element in the codomain $\mathbb{N}$ has a pre-image in the domain $\mathbb{N}$, the function $f$ is onto.

Visual Representation of Mapping

Step 3: Conclusion on Bijectivity

A function is defined as bijective if and only if it is both injective and surjective.
[Per the Definition of Bijective Functions]: Since $f$ is surjective but not injective, it cannot be bijective.

Final Solution: The function $f$ is not bijective because it is not injective (many-to-one), even though it is surjective.

Part 1: Proving $f(x) = \frac{1}{x}$ is One-One (Injective)

To determine if the function $f: R_{*} \to R_{*}$ is one-one, we assume two arbitrary elements $x_1, x_2 \in R_{*}$ such that $f(x_1) = f(x_2)$.

By definition of the function:
$\frac{1}{x_1} = \frac{1}{x_2}$ [Given $f(x) = \frac{1}{x}$]

Taking the reciprocal of both sides:
$x_1 = x_2$ [By the Algebraic Property of Equality for non-zero real numbers]

Since $f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2 \in R_{*}$, the function $f$ is one-one.

Part 2: Proving $f(x) = \frac{1}{x}$ is Onto (Surjective)

A function is onto if for every $y \in R_{*}$ (co-domain), there exists an $x \in R_{*}$ (domain) such that $f(x) = y$.

Let $y$ be an arbitrary element in the co-domain $R_{*}$.
$y = \frac{1}{x} \implies x = \frac{1}{y}$

Since $y \in R_{*}$ (where $R_{*}$ is the set of all non-zero real numbers), $y \neq 0$. Consequently, $x = \frac{1}{y}$ is also a non-zero real number, meaning $x \in R_{*}$.

Since for every $y \in R_{*}$, there exists a pre-image $x = \frac{1}{y} \in R_{*}$ such that $f(\frac{1}{y}) = \frac{1}{1/y} = y$, the function $f$ is onto.

Part 3: Analysis of Domain Replacement ($N \to R_{*}$)

Now consider the function $f: N \to R_{*}$ defined by $f(x) = \frac{1}{x}$.

• One-One Check: If $f(x_1) = f(x_2)$, then $\frac{1}{x_1} = \frac{1}{x_2}$, which implies $x_1 = x_2$. The function remains one-one because the domain $N$ is a subset of $R_{*}$.
• Onto Check: For a function to be onto, every element in the co-domain $R_{*}$ must have a pre-image in the domain $N$. Let $y = 2 \in R_{*}$. Then $f(x) = 2 \implies \frac{1}{x} = 2 \implies x = \frac{1}{2}$. However, $\frac{1}{2} \notin N$ (since $N = \{1, 2, 3, ...\}$).

Because there exists at least one element $y \in R_{*}$ (e.g., $y=0.5$ or $y=2$) that does not have a corresponding natural number as a pre-image, the function is not onto.

Visual Representation of Mapping

Final Solution: The function $f: R_{*} \to R_{*}$ is both one-one and onto. If the domain is replaced by $N$, the function remains one-one but is not onto.

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: Definition of Injectivity

A function $f: A \to B$ is said to be injective (or one-to-one) if and only if for every $x_1, x_2 \in A$, the condition $f(x_1) = f(x_2)$ implies that $x_1 = x_2$. [Definition of Injective Mapping].

Step 2: Formal Algebraic Setup

Given the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$, we assume two arbitrary elements $x_1, x_2$ in the domain $\mathbb{R}$ such that:

$f(x_1) = f(x_2)$

Substituting the function definition:

$x_1^3 = x_2^3$ [Definition of function $f$]

Step 3: Logical Derivation and Solving for Variables

To solve $x_1^3 - x_2^3 = 0$, we apply the algebraic identity for the difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

$(x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) = 0$ [Factoring via Algebraic Identity]

This equation holds if:

• $(x_1 - x_2) = 0 \implies x_1 = x_2$
• $(x_1^2 + x_1x_2 + x_2^2) = 0$

Step 4: Analyzing the Quadratic Term

The term $(x_1^2 + x_1x_2 + x_2^2)$ can be rewritten by completing the square or treating it as a quadratic in $x_1$:

$x_1^2 + x_1x_2 + x_2^2 = \left(x_1 + \frac{x_2}{2}\right)^2 + \frac{3x_2^2}{4}$

For any real numbers $x_1, x_2$, the sum of squares $\left(x_1 + \frac{x_2}{2}\right)^2 + \frac{3x_2^2}{4} = 0$ if and only if $x_1 = 0$ and $x_2 = 0$. In all other cases, this expression is strictly positive. Since the function must hold for all $x \in \mathbb{R}$, the only universal solution that satisfies the equality is $x_1 = x_2$.

Visual Representation of the Function

Step 5: Conclusion

Since $f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2 \in \mathbb{R}$, the function satisfies the criteria for an injective mapping.

Final Solution: The function $f(x) = x^3$ is injective as it passes the horizontal line test and satisfies the algebraic condition $f(x_1) = f(x_2) \implies x_1 = x_2$.

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.

Step 1: Formal Definition of the Modulus Function

The function $f: \mathbb{R} \to \mathbb{R}$ is defined as $f(x) = |x|$, where the domain is the set of real numbers $\mathbb{R}$ and the codomain is the set of real numbers $\mathbb{R}$. The rule of assignment is defined piecewise as follows:

$f(x) = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}$

Step 2: Analysis of Injectivity (One-One)

A function $f: A \to B$ is injective (one-one) if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$. [Definition of Injective Mapping]

Consider two distinct real numbers $x_1 = 1$ and $x_2 = -1$.

• $f(1) = |1| = 1$
• $f(-1) = |-1| = 1$

Since $f(1) = f(-1) = 1$, but $1 \neq -1$, there exist distinct elements in the domain that map to the same image in the codomain. [Violation of the definition of one-one function]

Conclusion: The function $f$ is not one-one.

Step 3: Analysis of Surjectivity (Onto)

A function $f: A \to B$ is surjective (onto) if for every $y \in B$, there exists at least one $x \in A$ such that $f(x) = y$. [Definition of Surjective Mapping]

The codomain of the function is $\mathbb{R}$. Consider a negative value in the codomain, for instance, $y = -2$.

By the definition of the modulus function, $f(x) = |x| \geq 0$ for all $x \in \mathbb{R}$. There is no real number $x$ such that $|x| = -2$, because the absolute value of any real number is non-negative by definition. [Property of Absolute Value]

Conclusion: Since there exists at least one element (all negative real numbers) in the codomain that has no pre-image in the domain, the function $f$ is not onto.

Step 4: Graphical Representation

Step 5: Theoretical Synthesis

Final Solution: The function $f(x) = |x|$ is neither one-one nor onto, as it fails the horizontal line test (multiple domain elements map to one codomain element) and the range does not cover the negative subset of the codomain.