A mapping $f:N ightarrow N$, where $N$ is the set of natural numbers is defined as $f(n) = \left\{ \begin{array}{r} n^{2},\ for\ n\ odd \ 2n + 1,\ for\ n\ even \end{array} ight.\$ For $n \in N$. Then, $f$ is

Question

A mapping $f:N ightarrow N$, where $N$ is the set of natural numbers is defined as $f(n) = \left\{ \begin{array}{r} n^{2},\ for\ n\ odd \ 2n + 1,\ for\ n\ even \end{array} ight.\$ For $n \in N$. Then, $f$ is

Dr. Abubacker · Accepted Answer

(d)
Since, $f(n) = \left\{ \begin{array}{r}
n^{2},\ \ \ \ \ \ \ \ if\ n\ odd \
2n + 1,\ \ \ if\ n\ is\ even
\end{array} ight.\$
$f(1) = 1^{2} = 1\ \ \ \ \ \ \ \ f(2) = 2(2) + 1 = 5$
$f(3) = 3^{2} = 9\ \ \ \ \ \ \ \ f(4) = 2(4) + 1 = 9$
$	herefore\ \ \ \ f(3) = f(4)$
$	herefore f$ is not injective.
Also, $f$ is not surjective as every element of $N$ is not the image of any element of $N$

Q. A mapping $f:N \rightarrow N$, where $N$ is the set of natural numbers is defined as $f(n) = \left\{ \begin{array}{r} n^{2},\ for\ n\ odd \\ 2n + 1,\ for\ n\ even \end{array} \right.\$ For $n \in N$. Then, $f$ is

Explanation

Reviewed by Dr. Abubacker

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