Let g : N [] N be defined as g(3n + 1) = 3n + 2, g(3n + 2) = 3n + 3, g(3n + 3) = 3n + 1, f…

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?

Accepted Answer

g : N $$\to$$ N

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

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

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

$$g(x) = \left[ {\begin{pmatrix} {x + 1} & {x = 3k + 1}  \
   {x + 1} & {x = 3k + 2}  \
   {x - 2} & {x = 3k + 3}  \

\end{pmatrix} } \right.$$

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

\end{pmatrix} } \right.$$

$$g\left( {g\left( {g\left( x \right)} \right)} \right) = \left[ {\begin{pmatrix} x & {x = 3k + 1}  \
   x & {x = 3k + 2}  \
   x & {x = 3k + 3}  \

\end{pmatrix} } \right.$$

If f : N $$\to$$ N, if is a one-one function such that f(g(x)) = f(x) $$\Rightarrow$$ g(x) = x, which is not the case

If f : N $$\to$$ N f is an onto function

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

one possibility is

$$f(x) = \left[ {\begin{pmatrix} x & {x = 3n + 1}  \
   x & {x = 3n + 2}  \
   x & {x = 3n + 3}  \

\end{pmatrix} } \right.$$ n$$\in$$N0

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

Solutions · 1