Question

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

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

Answer 1

Answer: (A)

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

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

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