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

(D) To show that $f$ is invertible,we must find a function $g: Y \rightarrow N$ such that $g \circ f = I_N$ and $f \circ g = I_Y$.
Consider an arbitrary element $y \in Y$. By the definition of $Y$,$y = 4x + 3$ for some $x \in N$.
Solving for $x$,we get $x = \frac{y-3}{4}$.
Define $g: Y \rightarrow N$ by $g(y) = \frac{y-3}{4}$.
Now,calculate the composition $g \circ f(x)$:
$g(f(x)) = g(4x+3) = \frac{(4x+3)-3}{4} = \frac{4x}{4} = x$.
Thus,$g \circ f = I_N$.
Next,calculate the composition $f \circ g(y)$:
$f(g(y)) = f\left(\frac{y-3}{4}\right) = 4\left(\frac{y-3}{4}\right) + 3 = (y-3) + 3 = y$.
Thus,$f \circ g = I_Y$.
Since $g \circ f = I_N$ and $f \circ g = I_Y$,the function $f$ is invertible and its inverse is $f^{-1}(y) = \frac{y-3}{4}$.