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I want to show that $|\mathbb{Z}|=|\mathbb{N}|$. FWIW, I think again that I must define a injective and surjective function from $\mathbb{Z}$ to $\mathbb{N}$. But how? Is there any proof as to how could one define such functions and based on what information?

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$\mathbb{N} = \text{odd}+\text{even}$ and $\mathbb{Z}=\text{negative} + \text{positive}$ –  Cortizol Feb 24 '13 at 13:24
"Is there any proof as to how could one define such functions and based on what information?" You made my brain hurt with this sentence... –  TonyK Feb 24 '13 at 13:25
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2 Answers

up vote 3 down vote accepted

Since you are to show that $\mathbb{N}$ and $\mathbb{Z}$ have the same cardinality, you're correct: you need to find a bijection (hence both injective and surjective) between $\mathbb Z $ and $\mathbb N$

One bijection between $\mathbb{Z}$ and $\mathbb{N}$ is the function $f: \mathbb{Z} \to \mathbb{N}$, defined by:

$$f(k) = \begin{cases} \\ \\ 2k & \quad k \in \mathbb Z,\; k>0 \\ \\ -2k + 1 & \quad k\in \mathbb Z, \; k \leq 0 \\ \\ \end{cases} $$

In words, you are simply mapping positive integers to positive even integers, and non-positive integers to positive odd integers.

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You kill the poor problem. + and + for Gigili –  B. S. Feb 24 '13 at 16:13
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You can do something like this, $f\colon\mathbb{Z} \to \mathbb{N}$. By $f(0) = 0,\; f(1) = 1, \;f(-1) = 2,\; f(2) = 3,\; f(-2) = 4,\;$ etc. This gives a bijection from $\mathbb{Z}$ to $\mathbb{N}$.

I leave it as an exercise for the reader to give an explicit formula for the function $f$.

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Oliver, I simply put your answer (the $f(x)'s$) enclosed in $$ signs. Feel free to roll back to your original post if you are offended by my edit. –  amWhy Feb 24 '13 at 15:09
Haha no problem! it looks way better now :) –  Oliver E. Anderson Feb 25 '13 at 8:29
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