Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
6  
$\mathbb{N} = \text{odd}+\text{even}$ and $\mathbb{Z}=\text{negative} + \text{positive}$ –  Cortizol Feb 24 '13 at 13:24
6  
"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
add comment

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.

share|improve this answer
    
You kill the poor problem. + and + for Gigili –  B. S. Feb 24 '13 at 16:13
add comment

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 $f$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.