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  • Let A = {0,1} and N be the set of natural numbers. Then the mapping f: N→A defined by f(2n - 1) = 0, f(2n) = 1, n ∈ N, is onto.

Let A = \{ 0,1 \} and N be the set of natural numbers. Then the mapping f: N \rightarrow A defined by f(2n - 1) = 0, f(2n) = 1, n \in N, \\is onto.

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True

Here, A = \{ 0,1 \} , and\: \: f(2n - 1) = 0, f(2n) = 1, \forall A n \in N\\

Therefore, the range of f is  \{ 0, 1 \} \\

Hence, f: N \rightarrow A  mapping is onto.

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