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(C) We are given $f(n) = \left(r+1, \frac{q+1}{2}ight)$ where $n = q 	imes 2^r$, $q$ is an odd integer, and $r \geq 0$.For one-one mapping:Suppose

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$f$ is a bijection

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(C) We are given $f(n) = \left(r+1, \frac{q+1}{2}ight)$ where $n = q 	imes 2^r$, $q$ is an odd integer, and $r \geq 0$.For one-one mapping:Suppose $f(n_1) = f(n_2)$.Then $\left(r_1+1, \frac{q_1+1}{2}ight) = \left(r_2+1, \frac{q_2+1}{2}ight)$.This implies $r_1+1 = r_2+1 \Rightarrow r_1 = r_2$ and $\frac{q_1+1}{2} = \frac{q_2+1}{2} \Rightarrow q_1 = q_2$.Since $n_1 = q_1 	imes 2^{r_1}$ and $n_2 = q_2 	imes 2^{r_2}$, it follows that $n_1 = n_2$. Thus, $f$ is one-one.For onto mapping:Let $(a, b) \in N 	imes N$. We need to find $n \in N$ such that $f(n) = (a, b)$.$r+1 = a \Rightarrow r = a-1$. Since $a \in N$, $a \geq 1$, so $r \geq 0$.$\frac{q+1}{2} = b \Rightarrow q = 2b-1$. Since $b \in N$, $b \geq 1$, so $q \geq 1$ and $q$ is odd.Thus, for any $(a, b) \in N 	imes N$, there exists $n = (2b-1) 	imes 2^{a-1} \in N$ such that $f(n) = (a, b)$.Therefore, $f$ is onto.Since $f$ is both one-one and onto, $f$ is a bijection.

Given that for any $n \in N$, there exist an odd integer $q$ and a non-negative integer $r$ such that $n$ can be written uniquely as $n = q \times 2^r$. Let $f: N \rightarrow N \times N$ be a function defined by $f(n) = \left(r+1, \frac{q+1}{2}\right)$. Then,

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