# cardinality

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• Apr 21st 2008, 10:00 PM
jzellt
cardinality
Let O be the set of odd integers and E be the set of even integers.

Prove that |Z| = |E|

Any advice?
• Apr 21st 2008, 10:09 PM
flyingsquirrel
Hi

What about finding a bijection from $Z$ into $E$ ?
• Apr 21st 2008, 10:11 PM
jzellt
I did figure that part out but I'm having trouble proving 1-1
• Apr 21st 2008, 10:21 PM
Jhevon
Quote:

Originally Posted by jzellt
I did figure that part out but I'm having trouble proving 1-1

what was the bijection you found?

also, did you learn that $|\mathbb{Z}| = |\mathbb{N}|$ and $|E| = |\mathbb{N}|$?
• Apr 21st 2008, 10:26 PM
jzellt
The way I was taught to solve a problem like this: |Z| = |E| was to find a bijection. Finding a bijection means proving that the function is 1-1 and onto. Right? I just don't know how to prove that it's 1-1.
• Apr 22nd 2008, 12:30 AM
Jhevon
Quote:

Originally Posted by jzellt
The way I was taught to solve a problem like this: |Z| = |E| was to find a bijection. Finding a bijection means proving that the function is 1-1 and onto. Right? I just don't know how to prove that it's 1-1.

you have to have the bijection first in order to prove it is 1-1. how can you show a function is 1-1 if you don't know what the function is? That is why I asked you to tell us your bijection, we can then help you to show it is 1-1
• Apr 22nd 2008, 08:33 PM
jzellt
I guess I'm going at the wrong way then. Can you show me how to find the bijection?
• Apr 22nd 2008, 08:51 PM
ThePerfectHacker
Quote:

Originally Posted by jzellt
I guess I'm going at the wrong way then. Can you show me how to find the bijection?

$f(x) = 2x$. Show that is a bijection.
• Apr 22nd 2008, 08:59 PM
jzellt
I did get that part then.

Is this right? To show it is onto I said that any x plugged into the function would come out even. Then the image of f is E and thus onto.

To show 1-1, I said for every x plugged into the function, there would be exactly one output, thus 1-1.

How does that sound?
• Apr 22nd 2008, 10:38 PM
Isomorphism
Quote:

Originally Posted by jzellt
I did get that part then.

Is this right? To show it is onto I said that any x plugged into the function would come out even. Then the image of f is E and thus onto.

Not quite. You have to show that for any even number e(that is for any element in E) , there exists an x such that e = 2x.
• Apr 23rd 2008, 04:24 PM
Jhevon
Quote:

Originally Posted by jzellt
I did get that part then.

Is this right? To show it is onto I said that any x plugged into the function would come out even. Then the image of f is E and thus onto.

To show 1-1, I said for every x plugged into the function, there would be exactly one output, thus 1-1.

How does that sound?

1-1 means, $f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1,x_2 \in \mathbb{Z}$.

So assume $f(x_1) = f(x_2) \implies 2x_1 = 2x_2 \implies x_1 = x_2$ by dividing through by 2. hence $f(x)$ is one-to-one

Follow Isomorphisms directions to show $f(x)$ is onto