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

Concept: One-One and Onto Functions

A function is one-one (injective) if $f(a) = f(b) \implies a = b$. It is onto (surjective) if for every $y$ in the codomain, there exists some $x$ in the domain such that $f(x) = y$.

Step 1: One-one

Assume $f(a) = f(b)$ for $a, b \in \mathbf{R}_*$. Then $\frac{1}{a} = \frac{1}{b}$. Cross-multiplying gives $b = a$. Hence $f$ is one-one.

Step 2: Onto

Take any $y \in \mathbf{R}_*$. We need $x$ such that $\frac{1}{x} = y$. Choose $x = \frac{1}{y}$, which is also in $\mathbf{R}_*$ (since $y \neq 0$). Then $f(x) = y$. So $f$ is onto.

Thus $f$ is bijective on $\mathbf{R}_*$.

Step 3: Replace domain with $\mathbf{N}$

Now domain is $\mathbf{N} = \{1,2,3,\dots\}$, codomain still $\mathbf{R}_*$. For onto, we need every non-zero real to be of the form $\frac{1}{n}$ for some natural $n$. But numbers like $2$ or $\frac{1}{2.5}$ are not reciprocals of any natural number. So $f$ is not onto. It remains one-one (since $\frac{1}{m} = \frac{1}{n} \implies m=n$), but not surjective.

We need to check two things for the function $f: \mathbf{R}_* \to \mathbf{R}_*$ given by $f(x) = 1/x$: whether it is one-one (injective) and onto (surjective). The set $\mathbf{R}_*$ means all real numbers except zero.

The core idea is that $f$ is its own inverse. If you take a number, take its reciprocal, and then take the reciprocal again, you get back the original number. That property alone guarantees both injectivity and surjectivity — but let's verify step by step.

1. Checking one-one (injectivity)

A function is one-one if different inputs always give different outputs. Equivalently, if two outputs are equal, the inputs must be equal.

Assume $f(x_1) = f(x_2)$ for some $x_1, x_2 \in \mathbf{R}_*$.

That means:

Since $x_1$ and $x_2$ are non-zero, we can cross-multiply:

So $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Hence $f$ is one-one.

2. Checking onto (surjectivity)

A function is onto if every element in the co-domain is actually reached by some input from the domain. Here the co-domain is $\mathbf{R}_*$ — all non-zero real numbers.

Take any $y \in \mathbf{R}_*$. We need an $x \in \mathbf{R}_*$ such that $f(x) = y$.

Set $f(x) = y$:

Solving for $x$:

Since $y \neq 0$, $1/y$ is also a non-zero real number, so $x \in \mathbf{R}_*$. And indeed $f(1/y) = y$.

Thus every $y$ in the co-domain has a pre-image in the domain. So $f$ is onto.

3. The function is bijective

Since $f$ is both one-one and onto, it is a bijection. In fact, $f$ is its own inverse: $f(f(x)) = x$ for all $x \in \mathbf{R}_*$.

4. What happens if the domain is $\mathbf{N}$ instead?

Now consider $f: \mathbf{N} \to \mathbf{R}_*$ defined by $f(x) = 1/x$, where $\mathbf{N} = \{1, 2, 3, \dots\}$.

Is it one-one?

Yes. The same reasoning applies: if $1/x_1 = 1/x_2$, then $x_1 = x_2$. Different natural numbers give different reciprocals.

Is it onto?

No. The co-domain is still $\mathbf{R}_*$, which includes numbers like $1/2$, $2/3$, $\pi$, $-5$, etc. But the outputs of $f$ are only reciprocals of natural numbers: $\{1, 1/2, 1/3, 1/4, \dots\}$.

For example, $y = 2$ is in $\mathbf{R}_*$, but there is no natural number $x$ such that $1/x = 2$ (that would require $x = 1/2$, which is not a natural number). So the function is not onto.

Method: Testing one-one / onto, then re-testing after changing the domain

For a function given by a formula, prove one-one algebraically and onto by solving for a preimage — then, if the domain or codomain changes, re-run only the step the change affects.

Steps

Step 1: One-one — set $f(x_1)=f(x_2)$ and simplify to $x_1=x_2$. For $f(x)=\frac{1}{x}$, $\frac{1}{x_1}=\frac{1}{x_2}\Rightarrow x_1=x_2$.

Step 2: Onto — take an arbitrary target $y$ in the codomain, solve $f(x)=y$ for $x$, and check the solution lies in the domain. For $\frac{1}{x}=y$, $x=\frac{1}{y}$.

Step 3: Changing the domain — injectivity usually survives shrinking the domain (fewer inputs cannot create a clash). Surjectivity is what breaks: the candidate preimage $x=\frac{1}{y}$ may no longer lie in the smaller domain (e.g. $\frac{1}{y}\notin\mathbf{N}$ for most $y$), so re-check onto specifically.

Common Mistakes

Mistake 1: Assuming that because $f$ is onto on $\mathbf{R}_*$ it stays onto on $\mathbf{N}$.

Why it's wrong: shrinking the domain to $\mathbf{N}$ removes almost all preimages; the candidate $x=\frac{1}{y}$ is rarely a natural number (e.g. $y=2$ needs $x=\frac{1}{2}\notin\mathbf{N}$). Correct approach: re-test surjectivity after any domain change.

Mistake 2: Fearing that narrowing the domain could break one-one.

Why it's wrong: fewer inputs can only remove clashes, never create them, so $\frac{1}{x}$ stays injective on $\mathbf{N}$. Correct approach: note injectivity is preserved and focus the re-check on onto.