If $f: \mathbf{N} \times \mathbf{N} \rightarrow \mathbf{N}$ is defined by $f((m, n))=2^{m-1}(2 n-1),…
If $f: \mathbf{N} \times \mathbf{N} \rightarrow \mathbf{N}$ is defined by $f((m, n))=2^{m-1}(2 n-1), \forall(m, n) \in \mathbf{N} \times \mathbf{N}$, then $f$ is
One-one but not onto
Onto but not one-one
Neither one-one nor onto
Both one-one and onto
Solution
The given function $f: \mathbf{N} \times \mathbf{N} \longrightarrow N$ is defined by $f(m, n)=2^{m-1}(2 n-1), \forall(m, n) \in \mathbf{N} \times \mathbf{N}$.
Now, let $f((a, b))=f((c, d))$, where $a, b, c, d \in \mathbf{N}$
$
\begin{aligned}
& \Rightarrow 2^{a-1}(2 b-1)=2^{c-1}(2 d-1) \\
& \Rightarrow 2^a \cdot b-2^{a-1}=2^c \cdot d-2^{c-1} \Rightarrow(a, b)=(c, d)
\end{aligned}
$
$\therefore f$ is a one-one function.
Now as $2^{m-1}$ is a even number for $\forall m \in \mathbf{N}-\{1\}$ and $(2 n-1)$ is a odd number for $\forall \mathbf{N} \in \mathbf{N}$, so the every natural number can be obtain by the for $2^{m-1} \cdot(2 n-1)$ for some combination of $(m, n)$, so $f$ is an onto function.
Hence, option (4) is correct