Let g : N → N be defined as g(3n + 1) = 3n + 2, g(3n + 2) = 3n + 3, g(3n + 3) = 3n + 1, ​ for all n 0.

Question

Let g : N → 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 ? 

(1) There exists an onto function f : N → N such that fog = f 

(2) There exists a one–one function f: N → N such that fog = f 

(3) gogog = g (4) There exists a function f : N → N such that gof = f

Answer 1

Correct option (1) There exists an onto function f : N → N such that fog = f

g : N → N g(3n + 1) = 3n + 2

g(3n + 2) = 3n + 3 

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

g(x) = \(\begin{bmatrix} x + 1 & x = 3k + 1 \\[0.3em] x + 1 & x = 3k + 2 \\[0.3em] x - 2 & x = 3k + 3 \end{bmatrix}\)

g(g(x)) = \(\begin{bmatrix} x + 2 & x = 3k + 1 \\[0.3em] x - 1 & x = 3k + 2 \\[0.3em] x - 1 & x = 3k + 3 \end{bmatrix}\)

g(g(g(x))) = \(\begin{bmatrix} x & x = 3k + 1 \\[0.3em] x & x = 3k + 2 \\[0.3em] x & x = 3k + 3 \end{bmatrix}\)

If f : N → N, f is a one-one function such that 

ƒ(g(x)) = ƒ(x) \(\Rightarrow\) g(x) = x, which is not the case 

If f ƒ : N → N ƒ is an onto function such that ƒ(g(x)) = ƒ(x), 

one possibility is

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