Show that $f: \mathbf{N} \rightarrow \mathbf{N}$, given by $$f(x)=\begin{aligned} & x+1, \text { if } x \text { is odd }, \\ & x-1, \text { if } x \text { is even } \end{aligned}$$ is both one-one and onto. The image of 1 and -1 under $f$ is 1 . Fig 1.4 Solution Suppose $f\left(x_{1}\right)=f\left(x_{2}\right)$. Note that if $x_{1}$ is odd and $x_{2}$ is even, then we will have $x_{1}+1=x_{2}-1$, i.e., $x_{2}-x_{1}=2$ which is impossible. Similarly, the possibility of $x_{1}$ being even and $x_{2}$ being odd can also be ruled out, using the similar argument. Therefore, both $x_{1}$ and $x_{2}$ must be either odd or even. Suppose both $x_{1}$ and $x_{2}$ are odd. Then $f\left(x_{1}\right)=f\left(x_{2}\right) \Rightarrow x_{1}+1=x_{2}+1 \Rightarrow x_{1}=x_{2}$. Similarly, if both $x_{1}$ and $x_{2}$ are even, then also $f\left(x_{1}\right)=f\left(x_{2}\right) \Rightarrow x_{1}-1=x_{2}-1 \Rightarrow x_{1}=x_{2}$. Thus, $f$ is one-one. Also, any odd number $2 r+1$ in the co-domain $\mathbf{N}$ is the image of $2 r+2$ in the domain $\mathbf{N}$ and any even number $2 r$ in the co-domain $\mathbf{N}$ is the image of $2 r-1$ in the domain $\mathbf{N}$. Thus, $f$ is onto. \\

Step 1: Prove f is one-one (Case 1: x1 odd, x2 even)

To prove that the function $$f$$ is one-one, we assume $$f(x_1) = f(x_2)$$ and show that $$x_1 = x_2$$ . We consider different cases for $$x_1$$ and $$x_2$$ . First, let's assume $$x_1$$ is odd and $$x_2$$ is even. In this case, $$f(x_1) = x_1+1$$ and $$f(x_2) = x_2-1$$ . Setting them equal gives $$x_1+1 = x_2-1$$ , which simplifies to $$x_2 - x_1 = 2$$ .

Step 2: Prove f is one-one (Case 2: x1 even, x2 odd)

Next, let's consider the case where $$x_1$$ is even and $$x_2$$ is odd. Here, $$f(x_1) = x_1-1$$ and $$f(x_2) = x_2+1$$ . Setting them equal gives $$x_1-1 = x_2+1$$ , which simplifies to $$x_1 - x_2 = 2$$ . This is also impossible because if $$x_1$$ is even and $$x_2$$ is odd, then $$x_1 - x_2$$ must be an odd number, not an even number like 2.

Step 3: Prove f is one-one (Case 3: x1, x2 both odd or both even)

Since the previous two cases are impossible, $$x_1$$ and $$x_2$$ must either both be odd or both be even. If both $$x_1$$ and $$x_2$$ are odd, then $$f(x_1) = x_1+1$$ and $$f(x_2) = x_2+1$$ . Setting $$f(x_1) = f(x_2)$$ implies $$x_1+1 = x_2+1$$ , which simplifies to $$x_1 = x_2$$ . Similarly, if both $$x_1$$ and $$x_2$$ are even, then $$f(x_1) = x_1-1$$ and $$f(x_2) = x_2-1$$ . Setting $$f(x_1) = f(x_2)$$ implies $$x_1-1 = x_2-1$$ , which simplifies to $$x_1 = x_2$$ . In all valid cases, $$x_1 = x_2$$ , so $$f$$ is one-one.

Step 4: Prove f is onto (Case 1: y is odd)

To prove that the function $$f$$ is onto, we need to show that for every element $$y$$ in the codomain $$\mathbf{N}$$ , there exists an element $$x$$ in the domain $$\mathbf{N}$$ such that $$f(x) = y$$ . Let $$y$$ be an odd number in the codomain, so $$y = 2r+1$$ for some integer $$r \ge 0$$ . If $$y=1$$ , then $$r=0$$ , so $$y=2(0)+1=1$$ . We need to find an $$x$$ such that $$f(x)=y$$ . Consider $$x = 2r+2$$ . Since $$r \ge 0$$ , $$x = 2r+2$$ is an even natural number (e.g., if $$r=0$$ , $$x=2$$ ; if $$r=1$$ , $$x=4$$ ). Since $$x$$ is even, $$f(x) = x-1 = (2r+2)-1 = 2r+1 = y$$ . Thus, every odd number in the codomain has a pre-image.

Step 5: Prove f is onto (Case 2: y is even)

Now, let $$y$$ be an even number in the codomain, so $$y = 2r$$ for some integer $$r \ge 1$$ (since $$y \in \mathbf{N}$$ , $$y$$ must be at least 2). We need to find an $$x$$ such that $$f(x)=y$$ . Consider $$x = 2r-1$$ . Since $$r \ge 1$$ , $$x = 2r-1$$ is an odd natural number (e.g., if $$r=1$$ , $$x=1$$ ; if $$r=2$$ , $$x=3$$ ). Since $$x$$ is odd, $$f(x) = x+1 = (2r-1)+1 = 2r = y$$ . Thus, every even number in the codomain also has a pre-image. Since every element in the codomain has a pre-image, $$f$$ is onto.

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