Finding an example of a bijection from $\Bbb N$ to $E^+$.

Question

Give an example of a bijection $h$ from $\Bbb N$ to $E^+$ such that $h(1) = 16, h(2) =12, \text{ and } h(3) = 2. $ $\Bbb N = \text{ natural numbers }$ , $E^+= \text{ positive even integers. }$

So in order to solve this problem we must find a one to one correspondence. Where each element of 1 set is paired exactly 1 element of the other.

$f: \Bbb N \rightarrow E^+, \begin{cases} 16, & \text{n} = 1 \\ 12 & \text{n}=2 \\ 2 & \text{n}=3 \\ \color{red}4 & \color{red}{ n=8} \\ \color{red}{6} & \color{red}{\text{n}=6} \\ \color{red}{2n} &\color{red}{\text{n} \neq 1,2,3,6,8} \\ \end{cases}$

My question for this problem is how does someone derive everything after $2$ if $n =3$ ? My guess is that one looks at all the even positive elements between $2$ and $16$ and list the ones that was not displayed. If that is the case how does one get $n$? Also the definition of an even number is $2n$.

Answer 1

There are many correct answers; the one in the question is designed to make $f$ follow a simple formula as much as possible. One could also follow your suggestion of first taking care of the missing even numbers between $2$ and $16$, but this time just doing so in order:

$$\begin{array}{rcc} n:&1&2&3&4&5&6&7&8&9&10&\ldots&n&&\\ f(n):&\color{blue}{16}&\color{blue}{12}&\color{blue}2&4&6&8&10&14&\color{red}{18}&\color{red}{20}&\ldots&\color{red}{2n} \end{array}$$

Here the blue values are those that we were given, the black are the missing ones between $2$ and $16$, and the red are ones after that taking in their natural order. This function is

$$f(n)=\begin{cases} 16,&\text{if }n=1\\ 12,&\text{if }n=2\\ 2,&\text{if }n=3\\ 4,&\text{if }n=4\\ 6,&\text{if }n=5\\ 8,&\text{if }n=6\\ 10,&\text{if }n=7\\ 14,&\text{if }n=8\\ 2n,&\text{if }n\ge 9\;. \end{cases}$$

If you look closely, you’ll see that it could equally well be written

$$f(n)=\begin{cases} 16,&\text{if }n=1\\ 12,&\text{if }n=2\\ 2n-4,&\text{if }3\le n\le 7\\ 14,&\text{if }n=8\\ 2n,&\text{if }n\ge 9\;. \end{cases}$$

The version in your question was chosen to maximize the set of $n$ for which $f(n)=2n$.

Answer 2

The simplest bijection, as you say, is $n \in \Bbb N \to 2n \in E^+$, but that doesn't respect the required values. One easy way to fix that is just to permute the smallest values in some way. As the largest required value is $16$, we can just permute the lowest $8$ values and leave everything above the same. So $$f(n)=\begin {cases} 16&n=1\\12&n=2\\2&n=3\\4&n=4\\6&n=5\\8&n=6\\10&n=7\\14&n=8\\2n&n \gt 8 \end {cases}$$ You could use any other permutation for $4 \le n \le 8$ that you like. You could also extend the explicit list if you want, but it would be more work.