True / False

Let A = {0, 1} and N the set of all natural numbers. Then the mapping $$f : N \rightarrow A$$ defined by $$f(2n - 1) = 0, f (2n) = 1 \forall n \epsilon N$$ is many-one onto.

ATrue
Correct Answer
BFALSE

Solution

Let $$A={0,1}$$ and N the set of all natural numbers. Then the mapping $$f:N \longrightarrow A$$ defines by $$f(2n-1)=0,$$ $$f(2x)=1$$ $$\forall$$ $$n$$ $$\epsilon$$ $$N$$ is many-one onto. The above statement is absolutely true.

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