Let $ f: N \rightarrow N $ defined by $ f(n)=\left\{\begin{array}{cl}\frac{n+1}{2} ; & \text { if } n \text { is odd } \ \frac{n}{2} ; & \text { if } n \text { is even }\end{array}\right. $ then $ \mathrm{f} $ is

Question

Let $ f: N \rightarrow N $ defined by $ f(n)=\left\{\begin{array}{cl}\frac{n+1}{2} ; & \text { if } n \text { is odd } \ \frac{n}{2} ; & \text { if } n \text { is even }\end{array}\right. $ then $ \mathrm{f} $ is, one-one and onto, one-one but not onto, onto but not one-one, neither one-one nor onto

MARKS App · Accepted Answer

Given that, $ f: N \rightarrow N $ defined by $ f(n)=\left\{\begin{array}{cc}\frac{n+1}{2} & n \text { is odd } \ \frac{n}{2} & n \text { is even }\end{array}\right. $ For $ n=1 $, we have $ f(1)=\frac{1+1}{2}=1 $ and, if $ n=2 $, we have $ f(2)=\frac{2}{2}=1 $ So, $ f(1)=f(2) $ but $ 1 \neq 2 $. Therefore, $ f(x) $ is not one-one. Now, $ f(x)=\frac{n+1}{2} $ if $ \mathrm{n} $ is odd if $ y=\frac{n+1}{2} $, then $ n=2 y-1, \forall y $ Also, $ f(x)=\frac{n}{2} $ if $ n $ is even. That is, $ y=\frac{n}{2} $ or $ n=2 y, \forall y $ Therefore, $ f(x) $ is onto.

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