# Bijection from $\mathbb Z$ to $\{\pm2+\frac1{n+1}\}$

Let $A=\{ 2+\frac{1}{n+1} ,\quad -2+\frac{1}{n+1} \mid\quad n=0,1,\ldots \}$.

Find a bijection from $\mathbb{Z}$ to $A$.

I try to set $f : \mathbb{Z} \to A$ where $f(k)=2+\frac{1}{k+1}$ if $k\geq 0$ and $f(k)=-2+\frac{1}{k+1}$ if $k <0$. But this does not work.

Can you give some ideas?

Thank you!

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What do you mean by isomorphism? – Mariano Suárez-Alvarez Jan 18 '12 at 20:59
The word "isomorphism" is usually reserved for situations where the sets involved are closed under some operation. Is there an operation under which $A$ is closed? Or are you just asking for a one-one onto map between the two sets? – Gerry Myerson Jan 18 '12 at 20:59
do you mean $f(k)=2+\frac{1}{k+1}$? – Holdsworth88 Jan 18 '12 at 21:00
Do you mean bijection? – ncmathsadist Jan 18 '12 at 21:00
Yes i mean a one-to-one and bijection map from Z to A. – passenger Jan 18 '12 at 21:05

Hints:

1. Can you find a bijection between $\{m\in\mathbb{Z}\mid m\geq 0\}$ and $\{2 + \frac{1}{n+1}\mid n=0,1,2,\ldots\}$?

2. Can you find a bijection between $\{m\in\mathbb{Z}\mid m\geq 0\}$ and $\{-2+\frac{1}{n+1}\mid 0,1,2,\ldots\}$?

3. Can you find a bijection between $\{k\in\mathbb{Z}\mid k\lt 0\}$ and $\{m\in\mathbb{Z}\mid m\geq 0\}$?

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:I can find the above bijections but I still can't solve the problem. – passenger Jan 18 '12 at 21:18
@passenger: Put them together! Use the bijections in 3 and in 2 to get a bijection from the negative integers to $\{-2+\frac{1}{n+1}\mid n=0,1,2,\ldots\}$; use the bijection on 1 to get a bijection from the nonnegative integers to $\{2+\frac{1}{n+1}\mid n=0,1,2,\ldots\}$. Now define your bijection as being the first function on the negatives, and the second function on the nonnegatives. What is the problem? – Arturo Magidin Jan 18 '12 at 21:19
:No problem! Silly question! I do it. Thank you! – passenger Jan 18 '12 at 21:21

Your idea almost works. Make no change when $k \ge 0$. When $k<0$, send $k$ to $-2+\frac{1}{|k|}$.

Comment: Before we get to formulas, let's think about what the set $A$ looks like. It consists of

$2+\frac{1}{1}$, $2+\frac{1}{2}$, $2+\frac{1}{3}$, and so on, together with

$-2+\frac{1}{1}$, $-2+\frac{1}{2}$, $-2+\frac{1}{3}$, and so on.

It seems reasonable to send $0$ to $2+\frac{1}{1}$, to send $1$ to $2+\frac{1}{2}$, to send $2$ to $2+\frac{1}{3}$, and so on. The definition of $f(k)$ for $k \ge 0$ is then clear.

Then it seems natural to send $-1$ to $-2+\frac{1}{1}$, to send $-2$ to $-2+\frac{1}{2}$, to send $-3$ to $-2+\frac{1}{3}$, and so on. The definition of $f(k)$ for $k < 0$ is then clear. (Instead of writing $2-\frac{1}{|k|}$ we could write $f(k)=-2-\frac{1}{k}$. But I feel more comfortable with positive numbers.)

The fact that $f$ is bijective is intuitively absolutely obvious, and (almost) does not require proof. However, if proof is asked for, it is easy to write down.

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@@Andre Nicolas:I think you mean $-2+\frac{1}{|k|+1}$ when $k<0$. I try this too, but I could not prove that is one-to-one, when $k_1 \geq 0$ and $k_2<0$ why $f(k_1)=f(k_2)$? – passenger Jan 18 '12 at 21:33
@passenger: you mean $f(k_1)\neq f(k_2)$: because $f(k_2) = -2+\frac{1}{|k_2|+1}\lt 0$, but $f(k_1) = 2+\frac{1}{k_1+1} \gt 0$. – Arturo Magidin Jan 18 '12 at 21:37
@ArturoMagidin:Yes my mistake sorry! It is obvious! – passenger Jan 18 '12 at 21:42
@passenger: It is $1/|k|$, because we are looking at negative $k$. The first number we want to "hit" is $-2+\frac{1}{1}$, and it is nice to let this be $f(-1)$. We want also $f(-2)=-2+\frac{1}{2}$, and so on. The situation is not quite symmetric, because of $0$. If you want to show that if $k_1\ge 0$ and $k_2<0$, then $f(k_1)\ne f(k_2)$ [note typo in your comment above], this is obvious. It is clear that $f(k_1)$ is positive, indeed $f(k_1)>2$. And it is clear that $f(k_2)$ is negative, for $f(k_2)=-2+1/|k_2| \le -1$. So they can't be equal. – André Nicolas Jan 18 '12 at 21:45
@AndreNicolas: It is clear!Thank you for your answer! – passenger Jan 18 '12 at 21:54