Let O be the set of odd integers and E be the set of even integers.

Prove that |Z| = |E|

Any advice?

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- Apr 21st 2008, 10:00 PMjzelltcardinality
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 PMflyingsquirrel
Hi

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

- Apr 21st 2008, 10:21 PMJhevon
- Apr 21st 2008, 10:26 PMjzellt
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 AMJhevon
- Apr 22nd 2008, 08:33 PMjzellt
I guess I'm going at the wrong way then. Can you show me how to find the bijection?

- Apr 22nd 2008, 08:51 PMThePerfectHacker
- Apr 22nd 2008, 08:59 PMjzellt
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 PMIsomorphism
- Apr 23rd 2008, 04:24 PMJhevon
1-1 means, $\displaystyle f(x_1) = f(x_2) \implies x_1 = x_2$ for all $\displaystyle x_1,x_2 \in \mathbb{Z}$.

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

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