Question

Let $g: N \rightarrow N$ be defined as
$g(3 n+1)=3 n+2$
$g(3 n+2)=3 n+3$
$g(3 n+3)=3 n+1$, for all $n \geq 0$
Then which of the following statements is true?

• A. There exists an onto function $f: N \rightarrow N$ such that fog $=f$
• B. There exists a one-one function $f: N \rightarrow N$ such that fog $=f$
• C. gogog=g
• D. There exists a function $f: N \rightarrow N$ such that gof $=f$

Answer 1

Answer: (A)

$g: N \rightarrow N g(3 n+1)=3 n+2$
$g(3 n+2)=3 n+3$
$g(3 n+3)=3 n+1$

If $f: N \rightarrow N, f$ is a one-one function such that $f(g(x))=f(x) r g(x)=x$, which is not the
case l $f f: N \rightarrow N f$ is an onto function such that $f(g(x))=f(x)$, one possibility is

Here $f(x)$ is onto, also $f(g(x))=f(x) \forall x \in N$