Let g: N rightarrow N be defined asg(3n+1)=3n | vedclass

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(B) Given $g(3n+1)=3n+2$,$g(3n+2)=3n+3$,and $g(3n+3)=3n+1$ for $n \geq 0$.First,we calculate $g \circ g \circ g(x)$:For $x = 3n+1$,$g(g(g(3n+1))) = g(

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There exists an onto function $f: N \rightarrow N$ such that $f \circ g = f$

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(B) Given $g(3n+1)=3n+2$,$g(3n+2)=3n+3$,and $g(3n+3)=3n+1$ for $n \geq 0$.First,we calculate $g \circ g \circ g(x)$:For $x = 3n+1$,$g(g(g(3n+1))) = g(g(3n+2)) = g(3n+3) = 3n+1$.Similarly,$g(g(g(3n+2))) = 3n+2$ and $g(g(g(3n+3))) = 3n+3$.Thus,$g \circ g \circ g(x) = x$ for all $x \in N$.Now,consider the condition $f(g(x)) = f(x)$.If $f$ is a constant function,say $f(x) = c$ for all $x \in N$,then $f(g(x)) = c = f(x)$. However,a constant function is not onto on $N$.If we define $f$ such that it maps all elements in the same cycle of $g$ to the same value,$f$ will satisfy $f(g(x)) = f(x)$.For example,define $f(3n+1) = f(3n+2) = f(3n+3) = n+1$. This function $f$ is onto because for any $y \in N$,we can choose $n = y-1$ such that $f(3(y-1)+1) = y$.Thus,there exists an onto function $f$ such that $f \circ g = f$.

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

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