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

Let $f:N \rightarrow Y$ be a function defined as

$f(x)=4x+3,\: \: where\: \: Y=\left \{ y\equiv N:y=4x+3,for\: \: some\: \: x\equiv N \right \}.$Arbitrary Element Y

$y=4x+3$

$x=\frac{y-3}{4}$

Define $g:Y\rightarrow N\: \: by\: \: g(y)=\frac{y-3}{4}$

Now $gof(x)=f(f(x))=g(4x+3)=\frac{4x+3-3}{4}=x$

and $fog(y)=f(g(y))=f\left ( \frac{y-3}{4} \right )=\frac{4(y-3)}{4}+3=y-3+3=y$

This shows that $gof=I_x$ and $fog=Iy$ which implies that f is invertible and g is the inverse of f.

$\therefore Inverse\: \: of\: \: f'=g(y)=\frac{y-3}{4}$

Hence Proved.

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