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: Formal Definition of the Relation

The set $P(X)$ is the power set of a non-empty set $X$. The relation $R$ is defined on $P(X)$ such that for any two elements $A, B \in P(X)$, $(A, B) \in R$ if and only if $A \subseteq B$. To determine if $R$ is an equivalence relation, it must satisfy three properties: Reflexivity, Symmetry, and Transitivity.

Step 2: Testing for Reflexivity

A relation $R$ is reflexive if for all $A \in P(X)$, $(A, A) \in R$.

• By the definition of subsets, every set is a subset of itself ($A \subseteq A$).
• Therefore, $(A, A) \in R$ holds for all $A \in P(X)$.
• Conclusion: The relation $R$ is reflexive.

Step 3: Testing for Symmetry

A relation $R$ is symmetric if $(A, B) \in R \implies (B, A) \in R$ for all $A, B \in P(X)$.

• Consider a set $X = \{1, 2\}$. Let $A = \{1\}$ and $B = \{1, 2\}$.
• Since $A \subset B$, the condition $ARB$ is satisfied.
• However, $B \not\subset A$ because the element $2 \in B$ is not present in $A$.
• Thus, $(B, A) \notin R$.
• Conclusion: The relation $R$ is not symmetric.

Step 4: 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 $A, B, C \in P(X)$.

• Assume $(A, B) \in R$, which implies $A \subseteq B$.
• Assume $(B, C) \in R$, which implies $B \subseteq C$.
• By the transitive property of set inclusion [Axiom of Set Theory], if $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$.
• Therefore, $(A, C) \in R$.
• Conclusion: The relation $R$ is transitive.

Step 5: Visual Representation of Non-Symmetry

Step 6: Final Logical Assessment

Since the relation $R$ fails the symmetry requirement, it does not fulfill the criteria for an equivalence relation [Definition: An equivalence relation must be reflexive, symmetric, and transitive].

Final Solution: No, $R$ is not an equivalence relation on $P(X)$ because it is not symmetric.

1. Analysis of the Function

Given the function $f: \mathbb{R} \to (-1, 1)$ defined by $f(x) = \frac{x}{1+|x|}$. To prove $f$ is a bijection, we must demonstrate both injectivity (one-one) and surjectivity (onto).

2. Proof of Injectivity (One-One)

Assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in \mathbb{R}$. We analyze three cases for the signs of $x_1$ and $x_2$:

• Case I: $x_1, x_2 \geq 0$
  $f(x) = \frac{x}{1+x}$. Since $f'(x) = \frac{(1+x)-x}{(1+x)^2} = \frac{1}{(1+x)^2} > 0$, the function is strictly increasing. Thus, $f(x_1) = f(x_2) \implies x_1 = x_2$.
• Case II: $x_1, x_2 < 0$
  $f(x) = \frac{x}{1-x}$. Since $f'(x) = \frac{(1-x) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2} > 0$, the function is strictly increasing. Thus, $f(x_1) = f(x_2) \implies x_1 = x_2$.
• Case III: $x_1 \geq 0$ and $x_2 < 0$
  If $x_1 \geq 0$, then $f(x_1) = \frac{x_1}{1+x_1} \geq 0$. If $x_2 < 0$, then $f(x_2) = \frac{x_2}{1-x_2} < 0$. Since the outputs have different signs, $f(x_1) \neq f(x_2)$.

[By the definition of an injective function, since $f(x_1) = f(x_2) \implies x_1 = x_2$ in all valid domains, $f$ is one-one].

3. Proof of Surjectivity (Onto)

Let $y \in (-1, 1)$. We seek $x \in \mathbb{R}$ such that $f(x) = y$.

For any $y \in (-1, 1)$, the calculated $x$ is a real number. Specifically:

• If $y \in [0, 1)$, $x = \frac{y}{1-y} \geq 0$.
• If $y \in (-1, 0)$, $x = \frac{y}{1+y} < 0$.

[Since every $y$ in the codomain has a pre-image $x$ in the domain, $f$ is onto].

Conclusion

Because the function satisfies the criteria for both injectivity and surjectivity, it is a bijection.

Final Solution: The function $f$ is both one-one and onto, hence it is a bijection.

Step 1: Theoretical Foundation of Equivalence Relations

An equivalence relation $R$ on a set $A$ must satisfy three reflexive, symmetric, and transitive properties [Per the definition of Equivalence Relations]:

• Reflexivity: $(a, a) \in R$ for all $a \in A$. Since $A = \{1, 2, 3\}$, $R$ must contain $\{(1, 1), (2, 2), (3, 3)\}$.
• Symmetry: If $(a, b) \in R$, then $(b, a) \in R$.
• Transitivity: If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Step 2: Identifying Mandatory Elements

Given $A = \{1, 2, 3\}$ and the condition that $(1, 2) \in R$, we must satisfy the properties above:

1. By Reflexivity: $\{(1, 1), (2, 2), (3, 3)\} \subseteq R$.
2. By Symmetry: Since $(1, 2) \in R$, it follows that $(2, 1) \in R$.
3. Current subset $R_0 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$.

We check for Transitivity in $R_0$: All pairs $(a, b)$ and $(b, c)$ lead to $(a, c)$ already within the set. Thus, $R_0$ is an equivalence relation.

Step 3: Determining Additional Possible Equivalence Relations

An equivalence relation partitions a set into disjoint equivalence classes. The number of equivalence relations corresponds to the number of partitions of the set $\{1, 2, 3\}$ where $1$ and $2$ are in the same subset.

Any other partition would require $1$ and $2$ to be in different blocks, which contradicts the condition that $(1, 2) \in R$. For instance, in partition $\{\{1\}, \{2\}, \{3\}\}$, $(1, 2) \notin R$. In partitions like $\{\{1, 3\}, \{2\}\}$, $(1, 2) \notin R$.

Step 4: Visualizing the Partitions

The first case corresponds to the relation $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$. The second case corresponds to the universal relation $R_2 = A \times A = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}$.

Final Solution: 2

Step 1: Analyzing the Set and Required Conditions

Given set $A = \{1, 2, 3\}$. The total number of elements in the Cartesian product $A \times A$ is $|A|^2 = 3^2 = 9$. A relation $R$ is a subset of $A \times A$. We are given three conditions for $R$:

• Condition 1 (Reflexive): For all $a \in A$, $(a, a) \in R$. Thus, $\{(1, 1), (2, 2), (3, 3)\} \subseteq R$.
• Condition 2 (Symmetric): If $(a, b) \in R$, then $(b, a) \in R$.
• Condition 3 (Mandatory Elements): $(1, 2) \in R$ and $(1, 3) \in R$.

Step 2: Deriving the Minimal Symmetric Relation

By the symmetry condition, if $(1, 2) \in R$, then $(2, 1)$ must be in $R$. Similarly, if $(1, 3) \in R$, then $(3, 1)$ must be in $R$. Adding the reflexivity requirements, the minimal relation $R_{min}$ containing these elements is:

$R_{min} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\}$

Note that $R_{min}$ is already symmetric and reflexive. Let us check for transitivity in $R_{min}$: Since $(2, 1) \in R$ and $(1, 3) \in R$, transitivity requires $(2, 3) \in R$. If $(2, 3) \in R$, then by symmetry $(3, 2) \in R$.

Step 3: Constructing Potential Relations

The remaining elements available to be added to $R_{min}$ are $(2, 3)$ and $(3, 2)$. Let $S = \{(2, 3), (3, 2)\}$. Any relation $R$ satisfying the conditions must be of the form $R_{min} \cup X$, where $X \subseteq S$.

Step 4: Logical Evaluation

Per the transitivity definition: $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.

In $R_1$: We have $(2, 1) \in R_1$ and $(1, 3) \in R_1$. Transitivity requires $(2, 3) \in R_1$. Since $(2, 3) \notin R_1$, $R_1$ is not transitive. This satisfies our requirement.

In $R_2$: Since $(2, 3)$ and $(3, 2)$ are added, we must verify all combinations. It is known that the union of the identity relation with all pairs $(i, j)$ and $(j, i)$ for a partition of the set forms an equivalence relation (which is transitive). $R_2$ is the universal relation $A \times A$, which is reflexive, symmetric, and transitive.

Final Solution: The only relation that is reflexive, symmetric, and not transitive under these constraints is $R_1$ itself. Thus, there is exactly 1 such relation.

Step 1: Formal Definition of the Function

Let $A = \{1, 2, 3, \dots, n\}$. We are considering a function $f: A \to A$. For $f$ to be an onto function (surjective), every element in the codomain $A$ must have at least one pre-image in the domain $A$. [Per the Definition of Surjective Mapping].

Step 2: Analysis of Cardinality and Injectivity

Since the domain and codomain are the same finite set $A$ with $|A| = n$, a function $f: A \to A$ is onto if and only if it is also injective (one-to-one). [Per the Pigeonhole Principle for finite sets of equal cardinality]. A function that is both injective and surjective is defined as a bijective function or a permutation of the set $A$.

Step 3: Calculating the Permutations

To determine the number of onto functions, we must count the number of ways to map $n$ distinct elements of the domain to $n$ distinct elements of the codomain such that each element is used exactly once.

• The first element $1$ can be mapped to any of the $n$ elements in the codomain.
• The second element $2$ can be mapped to any of the remaining $(n-1)$ elements.
• The third element $3$ can be mapped to any of the remaining $(n-2)$ elements.
• Continuing this process, the $n$-th element has only $1$ remaining choice.

Step 4: Mathematical Formulation

The total number of onto functions is the product of the number of choices for each element, which is the definition of the factorial of $n$:

$N = n \times (n-1) \times (n-2) \times \dots \times 1 = n!$

Step 5: Theoretical Justification

The set of all bijections from a set of $n$ elements to itself is the Symmetric Group $S_n$. The order of this group, representing the number of permutations, is given by $n!$. Since every bijection is an onto function in this context, the total number of onto functions is exactly $n!$.

Final Solution: The number of onto functions is $n!$.

Step 1: Establishing Definitions of Equality for Functions

Two functions $f: A \to B$ and $g: A \to B$ are defined as equal if and only if:

• Their domains are identical (given here as $A = \{-1, 0, 1, 2\}$).
• Their codomains are identical (given here as $B = \{-4, -2, 0, 2\}$).
• For every element $x \in A$, the image satisfies $f(x) = g(x)$. [Per the Axiom of Extensionality in Function Theory].

Step 2: Evaluating the Function $f(x) = x^2 - x$ for all $x \in A$

Step 3: Evaluating the Function $g(x) = 2|\frac{x}{2} - 1| - 1$ for all $x \in A$

Step 4: Comparative Analysis and Verification

We compare the results generated in Step 2 and Step 3:

• For $x = -1$: $f(-1) = 2$ and $g(-1) = 2$. ($f = g$)
• For $x = 0$: $f(0) = 0$ and $g(0) = 1$. ($f \neq g$)
• For $x = 1$: $f(1) = 0$ and $g(1) = 0$. ($f = g$)
• For $x = 2$: $f(2) = 2$ and $g(2) = -1$. ($f \neq g$)

[By the condition of function equality, $f(x)$ must equal $g(x)$ for all $x \in A$. Since $f(0) \neq g(0)$ and $f(2) \neq g(2)$, the functions fail the equality criterion].

Step 5: Visual Representation

Final Solution: The functions $f$ and $g$ are not equal because $f(x) \neq g(x)$ for at least one $x \in A$ (specifically for $x=0$ and $x=2$).