Me and my study buddies are having a tough time with this problem and need help. It's probably a lot simpler than we're making it out to be but we need help.
Please prove that the sets E = {even integers}, $\mathbb{N}$, and $\mathbb{Z}$ are infinite sets and that they have the same cardinality.
On
To show that two sets have the same cardinality, you need two find a bijective map between them. In your case, there exist bijections between E and $\mathbb{N}$ and between $\mathbb{Z}$ and $\mathbb{N}$. Hence $E$ and $\mathbb{Z}$ have the same cardinality as $\mathbb{N}$. One usually says that a set that has the same cardinality as $\mathbb{N}$ is countable.
The bijection between $\mathbb{N}$ and $E$ is given by $n \mapsto 2n$ and the bijection between $\mathbb{N}$ and $\mathbb{Z}$ is given by $n \mapsto \frac{n}{2}$ if $n$ is an even number and $n \mapsto \frac{-(n+1)}{2}$ if $n$ is an odd number.
I'll let you check that theses maps are actually bijections ;)
Hints:
To show the sets are infinite, a contradiction will always be nice. Suppose they're finite, and try to derive a contradiction. For example, you could argue that $\Bbb N$ has only $n$ elements, for $n$ finite. Let those elements be $\{1,2,3,...,n\}$. But $n+1$ is also a natural number, showing a contradiction.
Another way to do this would be to establish bijections between each of the groups. I imagine it's given that the cardinality of either the naturals or integers is $\aleph_0$ at this point in your coursework - in that case, any other set bijective with either also has that cardinality. What remains is to establish and prove that such a function is indeed bijective.