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Step-by-Step Solution
Step 1: Understand the definition of the function
The given function is defined by
$f: \mathbb{N} \to Y$, where $f(x) = 4x + 3$.
The set $Y$ is defined as $Y = \{\,y \in \mathbb{N} \mid y = 4x + 3 \text{ for some } x \in \mathbb{N}\}$.
In simpler terms, every element of $Y$ can be expressed in the form $4x + 3$ for some natural number $x$.
Step 2: Show that $f$ is one-to-one (injective)
To prove $f$ is injective, assume $f(a) = f(b)$ for some $a,b \in \mathbb{N}$. That is,
$$
4a + 3 = 4b + 3.
$$
Subtracting 3 from both sides, we get
$$
4a = 4b.
$$
Dividing both sides by 4, it follows that
$$
a = b.
$$
Hence, no two distinct inputs map to the same output, which shows $f$ is one-to-one.
Step 3: Show that $f$ is onto (surjective)
We need to show that every element in $Y$ has a pre-image in $\mathbb{N}$. By definition, for each $y \in Y$,
there exists an $x \in \mathbb{N}$ such that
$$
y = 4x + 3.
$$
This means any $y$ in $Y$ is generated by plugging some $x \in \mathbb{N}$ into $4x + 3$. Therefore, for every $y$ in $Y$, there is an $x$ in $\mathbb{N}$ with $f(x) = y$. Hence, $f$ is onto $Y$.
Step 4: Conclude that $f$ is invertible
Because $f$ is both one-to-one and onto ($f$ is bijective), it must be invertible. An invertible function means there exists another function $g: Y \to \mathbb{N}$ such that $g(f(x)) = x$ for all $x \in \mathbb{N}$, and $f(g(y)) = y$ for all $y \in Y$.
Step 5: Find the inverse function $g$
Starting from
$$
y = f(x) = 4x + 3,
$$
we solve for $x$:
$$
4x = y - 3 \quad \Longrightarrow \quad x = \frac{y - 3}{4}.
$$
Thus, the inverse function is given by
$$
g(y) = \frac{y - 3}{4}.
$$
This $g(y)$ satisfies $g(f(x)) = x$ and $f(g(y)) = y$ for all $x \in \mathbb{N}$ and $y \in Y$, confirming it is the correct inverse.
