Q1:

Show that the function $f : R_{*} \to R_{*}$ defined by $f(x) = \frac{1}{x}$ is one-one and onto, where $R_{*}$ is the set of all non-zero real numbers. Is the result true, if the domain $R_{*}$ is replaced by $N$ with co-domain being same as $R_{*}$?

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.