Animated Solution for Mathematics - Functions: Let $g:\mathbf{N}\to \mathbf{N}$ 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 ?

The function $g:\mathbb{N}\to \mathbb{N}$ is defined by the following rules: $g(3n+1)=3n+2$ $g(3n+2)=3n+3$ $g(3n+3)=3n+1$

This structure partitions the set of natural numbers $\mathbb{N}$ into disjoint cycles of length 3. Specifically, for any $n\ge 0$, the set $\{3n+1,3n+2,3n+3\}$ forms a cycle under the mapping $g$.

Composition Analysis

We test the statement $g\circ g\circ g=g$. Let us evaluate this at $x=1$:

However, the definition of the function gives $g(1)=2$. Since $1 eq 2$, the composition $g\circ g\circ g$ is actually the identity function $I(x)=x$, not $g$. Thus, option C is false.

Fixed Points and Injective Mappings

Consider the condition $g(f(x))=f(x)$. This equation implies that $f(x)$ must be a fixed point of the function $g$.

By examining the cycle structure, we see that for any $x\in \mathbb{N}$, $g(x) eq x$. Because there are no fixed points in the domain, there exists no function $f$ such that $g(f(x))=f(x)$. This invalidates option D.

Furthermore, if $f$ were an injective (one-to-one) function, the condition $f(g(x))=f(x)$ would imply $g(x)=x$ by the definition of injectivity. As established, $g(x)=x$ has no solutions in $\mathbb{N}$, rendering option B false.

We now examine option A, which asks if there exists a function $f:\mathbb{N}\to \mathbb{N}$ such that $f(g(x))=f(x)$ and $f$ is onto.

We can construct such a function by assigning the same value to every element within a specific cycle. Define $f$ as follows:

For $n=0$, we have $f(1)=f(2)=f(3)=1$. For $n=1$, we have $f(4)=f(5)=f(6)=2$, and so on.

This function $f$ is onto because its range is the entire set $\mathbb{N}$. Additionally, it satisfies the condition $f(g(x))=f(x)$ because $f$ maps all elements of any given cycle $\{3n+1,3n+2,3n+3\}$ to the same constant value $n+1$. Therefore, option A is true.