Animated Solution for Mathematics - Functions: Let a function $f:\mathbb{N}\to \mathbb{N}$ be defined by $f(n)=\{\begin{array}{ll}2n,& n=2,4,6,8,\dots \\ n-1,& n=3,7,11,15,\dots \\ \frac{n+1}{2},& n=1,5,9,13,\dots \end{array}$ then, $f$ is

We are given a function $f:\mathbb{N}\to \mathbb{N}$ defined on the set of natural numbers $\mathbb{N}=\{1,2,3,\dots \}$. The mapping is defined piecewise based on the form of the input $n$:

1. If $n$ is even, $f(n)=2n$. 2. If $n=4k-1$, $f(n)=n-1$. 3. If $n=4k-3$, $f(n)=\frac{n+1}{2}$.

To determine if the function is onto (surjective), we examine the image of each subset of the domain:

For even numbers ($n\in \{2,4,6,\dots \}$), the range is $\{4,8,12,16,\dots \}$, which consists of all multiples of $4$. For numbers of the form $4k-1$ ($n\in \{3,7,11,\dots \}$), the range is $\{2,6,10,14,\dots \}$, which consists of all even numbers not divisible by $4$. * For numbers of the form $4k-3$ ($n\in \{1,5,9,\dots \}$), the range is $\{1,3,5,7,\dots \}$, which consists of all odd numbers.

By taking the union of these three sets, we obtain:

Since the union of these ranges covers the entire codomain $\mathbb{N}$, the function is onto.

To determine if the function is one-one (injective), we check if distinct inputs map to distinct outputs.

First, note that the three ranges identified above are mutually disjoint: The first range contains multiples of $4$. The second range contains even numbers not divisible by $4$. * The third range contains all odd numbers.

Because these sets have no elements in common, an output value uniquely identifies which rule was used to generate it. Furthermore, each individual rule is a strictly increasing function within its respective domain.

Since the function is both one-one and onto, we conclude that $f$ is a bijection. This demonstrates that the set of natural numbers can be partitioned and remapped while preserving its fundamental structure.