If f: N times N rightarrow N is defined by f( | vedclass

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(D) The function $f: N 	imes N ightarrow N$ is defined by $f(m, n) = 2^{m-1}(2n-1)$.To check for one-one: Let $f(m_1, n_1) = f(m_2, n_2)$.Then $2^{

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Both one-one and onto

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(D) The function $f: N 	imes N ightarrow N$ is defined by $f(m, n) = 2^{m-1}(2n-1)$.To check for one-one: Let $f(m_1, n_1) = f(m_2, n_2)$.Then $2^{m_1-1}(2n_1-1) = 2^{m_2-1}(2n_2-1)$.Since any natural number $k$ can be uniquely written as $2^{m-1}(2n-1)$ where $2n-1$ is the odd part of $k$ and $2^{m-1}$ is the highest power of $2$ dividing $k$,we must have $m_1-1 = m_2-1$ and $2n_1-1 = 2n_2-1$.This implies $m_1 = m_2$ and $n_1 = n_2$. Thus,$f$ is one-one.To check for onto: For any $k \in N$,if $k$ is odd,$k = 2^{1-1}(2n-1)$ with $m=1$. If $k$ is even,$k = 2^p \cdot q$ where $q$ is odd,so $m-1 = p$ and $2n-1 = q$.Since every $k \in N$ has a unique representation of this form,$f$ is onto.Therefore,$f$ is both one-one and onto.

If $f: N \times N \rightarrow N$ is defined by $f(m, n) = 2^{m-1}(2n-1)$ for all $(m, n) \in N \times N$,then $f$ is

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