Q.Let $f: \mathbb{N} \to Y$ be a function defined as $f(x) = 4x + 3$, where $Y = \{y \in \mathbb{N}: y = 4x + 3 \text{ for some } x \in \mathbb{N}\}$. Show that $f$ is invertible. Find the inverse.

Concept: Bijection Proof — To show a function is invertible, we prove it is both one-one (injective) and onto (surjective).

Step 1 (One-one): Assume $f(a) = f(b)$. Then $4a + 3 = 4b + 3 \implies 4a = 4b \implies a = b$. Hence $f$ is injective.

Step 2 (Onto): By definition, $Y$ is exactly the set of all outputs $4x+3$ for $x \in \mathbb{N}$. For any $y \in Y$, there exists $x \in \mathbb{N}$ such that $y = 4x+3$, so $f(x) = y$. Hence $f$ is surjective onto $Y$. …

We need to show that $f$ is invertible — that is, it has an inverse function. For a function to be invertible, it must be bijective: both one-to-one (injective) and onto (surjective). The domain is $\mathbb{N}$ (the natural numbers, usually $\{1,2,3,\dots\}$) and the codomain is $Y$, defined as the set of all numbers of the form $4x+3$ where $x\in\mathbb{N}$. So $Y$ is exactly the range of $f$ by construction — that already tells us $f$ is onto $Y$. The real work is checking injectivity and then finding the inverse.

Let’s walk through it.

1. Onto (surjectivity) is immediate.

   The definition of $Y$ is: $Y = \{ y \in \mathbb{N} : y = 4x+3 \text{ for some } x \in \mathbb{N} \}$. That is precisely the set of all outputs of $f$. So for every $y \in Y$, there exists some $x \in \mathbb{N}$ such that $f(x)=y$. Hence $f$ is onto $Y$ by definition. No further work needed here.

2. One-to-one (injectivity) — the key check.

   Suppose $f(a) = f(b)$ for $a,b \in \mathbb{N}$. Then

Subtract 3 from both sides: $4a = 4b$. Divide by 4: $a = b$.

So $f(a)=f(b)$ implies $a=b$, which means $f$ is injective.

3. Since $f$ is both injective and surjective onto $Y$, it is bijective, hence invertible. …

Method: Proving invertibility and finding the inverse

Use this whenever you must show $f$ is invertible and produce $f^{-1}$.

Steps

Step 1: Prove one-one

Assume $f(x_1)=f(x_2)$ and simplify to $x_1=x_2$. For $f(x)=ax+b\ (a\ne0)$ this is immediate.

Step 2: Prove onto

Show every element of the codomain is an output. When the codomain is defined as the range of $f$, surjectivity is automatic — say so explicitly.

Step 3: Solve $y=f(x)$ for $x$ to get the inverse …

Common Mistakes

Mistake 1: Thinking $f(x)=4x+3$ is not onto because it is onto $Y$, not $\mathbb{N}$

Why it's wrong: the codomain is defined as $Y=\{4x+3:x\in\mathbb{N}\}$, i.e. exactly the range, so surjectivity is built in; using $\mathbb{N}$ as codomain would indeed make it not onto. Correct approach: state that $Y$ is the range, so $f$ is onto $Y$ automatically.

Mistake 2: Writing the inverse without solving for $x$ correctly …