# Let g : N → N be defined as

Question:

Let $g: \mathbf{N} \rightarrow \mathbf{N}$ be defined as

$g(3 n+1)=3 n+2$

$g(3 n+2)=3 n+3$

$\mathrm{g}(3 \mathrm{n}+3)=3 \mathrm{n}+1$, for all $\mathrm{n} \geq 0$

Then which of the following statements is true?

1. There exists an onto function $f: \mathbf{N} \rightarrow \mathbf{N}$ such that fog $=\mathrm{f}$

2. There exists a one-one function $\mathrm{f}: \mathbf{N} \rightarrow \mathbf{N}$ such that fog $=\mathrm{f}$

3. gogog $=\mathrm{g}$

4. There exists a function $\mathrm{f} / \mathrm{N} \rightarrow \mathrm{N}$ such that gof $=\mathrm{f}$

Correct Option: 1

Solution:

$\mathrm{g}: \mathrm{N} \rightarrow \mathrm{N} \quad \mathrm{g}(3 \mathrm{n}+1)=3 \mathrm{n}+2$

$g(3 n+2)=3 n+3$

$g(3 n+3)=3 n+1$

$g(x)=\left[\begin{array}{ll}x+1 & x=3 k+1 \\ x+1 & x=3 k+2 \\ x-2 & x=3 k+3\end{array}\right.$

$g(g(x))=\left[\begin{array}{ll}x+2 & x=3 k+1 \\ x-1 & x=3 k+2 \\ x-1 & x=3 k+3\end{array}\right.$

$g(g(g(x)))=\left[\begin{array}{ll}x & x=5 k+1 \\ x & x=3 k+2 \\ x & x=3 k+3\end{array}\right.$

If $f: \mathrm{N} \rightarrow \mathrm{N}, f$ is a one-one function such that

$f(\mathrm{~g}(\mathrm{x}))=f(\mathrm{x}) \Rightarrow \mathrm{g}(\mathrm{x})=\mathrm{x}$, which is not the case

If $\mathrm{f} f: \mathrm{N} \rightarrow \mathrm{N} f$ is an onto function

such that $f(g(x))=f(x)$

one possibility is

$f(\mathrm{x})=\left[\begin{array}{ll}\mathrm{n} & \mathrm{x}=3 \mathrm{n}+1 \\ \mathrm{n} & \mathrm{x}=3 \mathrm{n}+2 \\ \mathrm{n} & \mathrm{x}=3 \mathrm{n}+3\end{array} \quad \mathrm{n} \in \mathrm{N}_{0}\right.$

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

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