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 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.

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: 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.

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.

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.

Step 1: Analyzing the One-One (Injective) Property

A function $f: X \to Y$ is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2 \in X$. To test if $f(x) = [x]$ is one-one, we investigate whether different inputs can yield the same output.

Let $x_1 = 1.2$ and $x_2 = 1.7$. Both $x_1, x_2 \in \mathbb{R}$.

Calculating the images:

• $f(1.2) = [1.2] = 1$ [By definition of the Greatest Integer Function]
• $f(1.7) = [1.7] = 1$ [By definition of the Greatest Integer Function]

Since $f(1.2) = f(1.7)$ but $1.2 \neq 1.7$, the function fails the condition for injectivity. Therefore, $f(x) = [x]$ is not one-one.

Step 2: Analyzing the Onto (Surjective) Property

A function $f: X \to Y$ is onto if for every $y \in Y$, there exists at least one $x \in X$ such that $f(x) = y$. Here, the codomain is $\mathbb{R}$.

Consider $y = 0.5$, where $0.5 \in \mathbb{R}$.

For $f(x) = [x] = 0.5$, we analyze the nature of the output of the Greatest Integer Function. By definition, $[x] \in \mathbb{Z}$ for any $x \in \mathbb{R}$ [Property of Floor Function].

Because the range of $f$ is the set of all integers $\mathbb{Z}$, and $\mathbb{Z} \subset \mathbb{R}$, there exists no $x \in \mathbb{R}$ such that $[x] = 0.5$. Since there are elements in the codomain $\mathbb{R}$ that have no pre-image in the domain $\mathbb{R}$, the function is not onto.

Visual Representation of the Greatest Integer Function

Step 3: Logical Summary

Final Solution: The Greatest Integer Function $f(x) = [x]$ is neither one-one nor onto, as it maps multiple real values to the same integer and its codomain $\mathbb{R}$ contains values (non-integers) that cannot be reached by the function.

Step 1: Formal Definition of the Function

The function is defined as $f: \mathbb{Z} \to \mathbb{Z}$, where $\mathbb{Z}$ represents the set of all integers ($\dots, -2, -1, 0, 1, 2, \dots$). The rule of correspondence is $f(x) = x^3$.

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

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

Let $x_1, x_2 \in \mathbb{Z}$ such that $f(x_1) = f(x_2)$.

Substituting the function rule: $x_1^3 = x_2^3$.

Taking the cube root on both sides: $x_1 = x_2$.

Since the cube root of an integer is unique in the domain of real numbers and preserves the integer property, $x_1 = x_2$ is the only possible solution. Therefore, $f$ is injective.

Step 3: Checking for Surjectivity (Onto)

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

Let $y \in \mathbb{Z}$ be an arbitrary element in the codomain. For the function to be surjective, $x^3 = y$ must have an integer solution $x \in \mathbb{Z}$ for every $y \in \mathbb{Z}$.

Consider $y = 2$. Since $y \in \mathbb{Z}$ (the codomain), we check if there exists an $x \in \mathbb{Z}$ such that $x^3 = 2$.

$x = \sqrt[3]{2}$.

However, $\sqrt[3]{2} \approx 1.2599$, which is not an integer. Since $\sqrt[3]{2} \notin \mathbb{Z}$, there is no pre-image for the element $2$ in the domain.

Step 4: Conclusion of Mapping Properties

The element $y=2$ (and many others like $y=3, 4, 5, \dots$) has no corresponding $x \in \mathbb{Z}$ such that $f(x)=y$. Thus, the range of $f$ is not equal to the codomain $\mathbb{Z}$.

Final Solution: The function $f(x) = x^3$ is injective but not surjective.

Step 1: Formal Definition of the Function

The function $f: \mathbb{N} \to \mathbb{N}$ is defined by the mapping $f(x) = x^3$, where $\mathbb{N}$ represents the set of all natural numbers $\{1, 2, 3, \dots \}$.

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

A function $f$ is injective if $f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2 \in \mathbb{N}$.

• Assume $f(x_1) = f(x_2)$.
• By the definition of the function, this implies $x_1^3 = x_2^3$.
• Taking the real cube root on both sides: $\sqrt[3]{x_1^3} = \sqrt[3]{x_2^3} \implies x_1 = x_2$ [By the uniqueness of roots for real numbers].
• Since $x_1 = x_2$, the function satisfies the condition for injectivity.

Thus, $f(x) = x^3$ is injective.

Step 3: Testing for Surjectivity (Onto)

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

• Consider an element $y = 2$ in the codomain $\mathbb{N}$.
• For surjectivity, there must exist $x \in \mathbb{N}$ such that $x^3 = 2$.
• Solving for $x$: $x = \sqrt[3]{2}$.
• We observe that $\sqrt[3]{2} \approx 1.2599$, which is not an element of the set of natural numbers ($\sqrt[3]{2} \notin \mathbb{N}$).

Since there exists at least one element in the codomain that does not have a pre-image in the domain, the function is not surjective.

Step 4: Visualizing the Mapping

Final Solution: The function $f(x) = x^3$ defined from $\mathbb{N} \to \mathbb{N}$ is injective but not surjective.

Step 1: Analyzing the Function Definition

The function $f: \mathbb{R} \to \mathbb{R}$ is defined by the mapping rule $f(x) = 3x$. The domain and codomain are both the set of real numbers ($\mathbb{R}$).

Step 2: Testing for Injectivity (One-to-One Function)

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

• Let $x_1, x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2)$.
• Substituting the given rule: $3x_1 = 3x_2$.
• Dividing both sides by $3$ (a non-zero scalar): $x_1 = x_2$.

Since the condition $f(x_1) = f(x_2) \implies x_1 = x_2$ is satisfied, the function $f$ is one-to-one.

Step 3: Testing for Surjectivity (Onto Function)

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

• Let $y$ be any arbitrary real number in the codomain $\mathbb{R}$.
• Set $f(x) = y$, which implies $3x = y$.
• Solving for $x$: $x = \frac{y}{3}$.
• Since $y \in \mathbb{R}$, then $\frac{y}{3}$ must also be a real number. Thus, for every $y \in \mathbb{R}$, there exists an $x = \frac{y}{3} \in \mathbb{R}$ such that $f(\frac{y}{3}) = 3(\frac{y}{3}) = y$.

Since the codomain equals the range, the function $f$ is onto.

Step 4: Visualizing the Mapping

Step 5: Synthesizing the Properties

Final Solution: The function $f(x) = 3x$ is both one-to-one and onto (bijective).

Step 1: Formal Definition of the Function

The function is defined as $f: \mathbb{R} \to \mathbb{R}$ where $f(x) = 1 + x^2$. 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: Analysis of Injectivity (One-One)

A function is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2 \in \mathbb{R}$.

Let $x_1 = 1$ and $x_2 = -1$.

Calculate the functional values:

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

[By Definition of Injectivity: If $f(x_1) = f(x_2)$ but $x_1 \neq x_2$, the function is not one-one.]

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

Step 3: Analysis of Surjectivity (Onto)

A function is onto if for every $y \in \text{Codomain}$, there exists at least one $x \in \text{Domain}$ such that $f(x) = y$.

Given $f(x) = 1 + x^2$, we know that for all $x \in \mathbb{R}$, $x^2 \geq 0$.

[Per the property of squares in Real Numbers: The square of any real number is non-negative.]

Therefore, $1 + x^2 \geq 1 + 0 \implies f(x) \geq 1$.

The range of the function is $[1, \infty)$. The codomain is $\mathbb{R} = (-\infty, \infty)$.

Since the range $\subseteq$ codomain but Range $\neq$ Codomain (e.g., consider $y = 0 \in \mathbb{R}$, there is no $x$ such that $1 + x^2 = 0$ because $x^2 = -1$ has no real solution), the function is not onto.

Step 4: Graphical Representation

Step 5: Conclusion on Bijectivity

A function is bijective if and only if it is both one-one and onto.

• Not one-one: Fails the horizontal line test (two $x$ values yield the same $y$ value).
• Not onto: The negative real numbers in the codomain are not mapped by any $x$ in the domain.

Final Solution: The function $f(x) = 1 + x^2$ is neither one-one nor onto, and therefore it is not bijective.

Given Function & Domain Definition

Function $f: \mathbb{R} \to \mathbb{R}$ is defined by the algebraic expression:

$f(x) = 3 - 4x$

Here, the domain is the set of all real numbers $\mathbb{R}$ and the codomain is also the set of all real numbers $\mathbb{R}$.

Step 1: Testing for Injectivity (One-One)

A function $f$ is injective (one-one) if and only if $f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2$ in the domain.

• Assume $f(x_1) = f(x_2)$
• $3 - 4x_1 = 3 - 4x_2$ [By substitution into function definition]
• Subtracting $3$ from both sides: $-4x_1 = -4x_2$
• Dividing by $-4$: $x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is one-one.

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 $y \in \mathbb{R}$ (codomain). We set $y = f(x)$.
• $y = 3 - 4x$
• Solve for $x$: $4x = 3 - y \implies x = \frac{3 - y}{4}$
• Since $y$ is a real number, $3 - y$ is a real number, and $\frac{3 - y}{4}$ is also a real number.
• Thus, for every $y \in \mathbb{R}$, there exists an $x = \frac{3 - y}{4} \in \mathbb{R}$ such that $f(x) = 3 - 4(\frac{3 - y}{4}) = 3 - (3 - y) = y$.

Since the range of $f$ is equal to the codomain $\mathbb{R}$, the function $f$ is onto.

Step 3: Determining Bijectivity

A function is bijective if it is both injective and surjective. Since $f$ is one-one and onto, it satisfies the criteria for a bijection.

Analysis Summary

Final Solution: The function $f(x) = 3 - 4x$ is one-one, onto, and therefore a bijection.

Step 1: Formal Definition of a One-One (Injective) Function

A function $f: A \to B$ is defined as one-one (or injective) if distinct elements in the domain $A$ are mapped to distinct elements in the codomain $B$. Formally, $f$ is one-one if and only if:
For all $x_1, x_2 \in A$, $f(x_1) = f(x_2) \implies x_1 = x_2$.
[Per the definition of Injective Mapping in Set Theory]

Step 2: Analysis of the Given Function Mapping

Given the set $A = \{1, 2, 3\}$ and the set $B = \{4, 5, 6, 7\}$, the function $f$ is defined as the set of ordered pairs:
$f = \{(1, 4), (2, 5), (3, 6)\}$

We extract the image of each element in the domain $A$:

• $f(1) = 4$
• $f(2) = 5$
• $f(3) = 6$

Step 3: Verification of the Injective Property

To prove $f$ is one-one, we examine the images of all elements in $A$:
Since $4 \neq 5$, $5 \neq 6$, and $4 \neq 6$, it is evident that $f(1) \neq f(2)$, $f(2) \neq f(3)$, and $f(1) \neq f(3)$.

Comparing these results against the contrapositive of the injective definition ($x_1 \neq x_2 \implies f(x_1) \neq f(x_2)$):
Because every unique input $x \in \{1, 2, 3\}$ yields a unique output $y \in \{4, 5, 6\}$, no two distinct elements of $A$ share the same image in $B$. [By the definition of mapping consistency].

Visual Representation

Conclusion

Since the distinct elements $1, 2, 3$ in set $A$ correspond to distinct images $4, 5, 6$ in set $B$, the condition $f(x_1) = f(x_2) \implies x_1 = x_2$ is satisfied for all $x \in A$.

Final Solution: The function $f$ is confirmed to be one-one as every element in the domain has a unique image in the codomain.

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

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

Given $f(x) = x^4$, let us test two distinct values from the domain $R$, such as $x_1 = 1$ and $x_2 = -1$.

• $f(1) = (1)^4 = 1$
• $f(-1) = (-1)^4 = 1$

Since $f(1) = f(-1)$ but $1 \neq -1$, the function fails the horizontal line test. [By the failure of the Injectivity Criterion].

Conclusion: The function $f(x) = x^4$ is not one-to-one.

Step 2: Analyzing the Surjectivity (Onto Property)

A function $f: R \to R$ is surjective (onto) if and only if for every $y \in R$ (codomain), there exists an $x \in R$ (domain) such that $f(x) = y$. [Per the Definition of Surjective Functions].

For $f(x) = x^4$, since $x$ is raised to an even power, the output $f(x)$ will always be non-negative for any real input $x$. That is, $f(x) \geq 0$ for all $x \in R$.

Consider a value $y = -2$ in the codomain $R$. There is no real number $x$ such that $x^4 = -2$, because the fourth power of any real number cannot be negative. [By the property of Real Numbers and Exponents].

Conclusion: The function $f(x) = x^4$ is not onto.

Step 3: Visual Representation of the Function

Step 4: Comparative Summary

Final Solution: The function $f(x) = x^4$ is neither one-to-one nor onto.