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

Step 1: Prove Injectivity (One-to-One)

To prove that a function is invertible, we first need to show it is injective (one-to-one). We assume that $f(x_1) = f(x_2)$ for two elements $x_1, x_2$ in the domain $\mathbf{N}$. By substituting the function definition, we get $4x_1 + 3 = 4x_2 + 3$. Simplifying this equation leads to $x_1 = x_2$, which confirms that $f$ is injective.

Step 2: Prove Surjectivity (Onto)

Next, we need to prove that the function is surjective (onto). This means that for every element $y$ in the codomain Y, there exists an element $x$ in the domain $\mathbf{N}$ such that $f(x) = y$. We set $y = f(x)$ and solve for $x$. The codomain Y is defined as all natural numbers $y$ such that $y = 4x+3$ for some $x \in \mathbf{N}$. This implies that for any $y \in Y$, $y-3$ must be a multiple of 4, and $y-3 \ge 0$. Since $x \in \mathbf{N}$, $x = \frac{y-3}{4}$ will always be a natural number for any $y \in Y$. Thus, $f$ is surjective.

Step 3: Conclusion of Invertibility

Since the function $f$ has been proven to be both injective (one-to-one) and surjective (onto), it satisfies the conditions for invertibility. Therefore, $f$ is an invertible function.

Step 4: Find the Inverse Function

To find the inverse function, we use the expression for $x$ that we derived when proving surjectivity. If $y = f(x)$, then $x = f^{-1}(y)$. So, the inverse function $f^{-1}$ maps an element $y$ from the codomain Y back to its corresponding element $x$ in the domain $\mathbf{N}$.

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