Prove that the set of integers are infinite How can you prove that the set of integers are infinite? 
And can the proof be generalized to prove the set of natural numbers, rational numbers, and complex numbers are infinite?
 A: As noted in the comments, this depends a lot of exactly which definitions and axioms one is using,
The most mainstream set-theoretic approach, I think, is (slightly popularized) to define "infinite" as "not of the same size as $\{x\in\mathbb N_0\mid x<n\}$ for any $n\in\mathbb N_0$". Where "same size" is again defined by the existence of a bijective function between two set.
So $\mathbb Z$ is infinite iff for every natural number $n$ it holds that there is no bijection from $\mathbb Z$ to the set of natural numbers less than $n$. This is so intuitively obvious that it almost hurts to have to think there is anything to prove, but if we want to be formal about it we're not supposed to just shout "obvious!" and leave it at that. So we forge onwards.
The way to prove something about every natural number is to use mathematical induction (which itself is often an axiom, but in some systems can be proved from yet more primitive axioms. That shall not concern us here). Thus:
The base case. We must prove that there is no bijection $f$ from $\mathbb Z$ to the set of natural numbers less than 0. That is, in fact, obvious by contradiction: If such an $f$ existed, $f(42)$ would be a natural number less than 0, but such a number doesn't exist, so $f$ itself cannot exist.
The induction step. We must prove that there is no bijection $f$ from $\mathbb Z$ to the set of natural numbers less than $n+1$, given that we know that there can be no bijection from $\mathbb Z$ to the naturl numbers less than $n$. Again we proceed by contradiction, so assume that $f$ is a bijection $\mathbb Z\to\{0,1,...,n\}$. Because $f$ is a bijection, there must be a $k\in z$ such that $f(k)=n$ and $f(j)\ne n$ for every $j\ne k$. Now define
$$g(i) = \cases{f(i)&\text{if }i<k\\f(i+1)&\text{if }i\ge k}$$
A bit of thought convinces us that $g$ must be a bijection from $\mathbb Z$ to $\{0,1,...,n-1\}$. But the induction hypothesis was that no such $g$ can exist, so we have reached a contradiction.
Thus, $\mathbb Z$ is infinite.
A: If you are defining the naturals from the Peano axioms, the first two suffice.  Given $\forall x S(x)\neq 0$ and $\forall x,y S(x)=S(y)\implies x=y$ the universe must be infinite.  For if it were finite, you would have $S^m(0)=S^n(0)$ (where the superscript represents iteration) for $n \neq m$.  Then $|n-m|$ operations of the second axiom contradicts the first.
A: Suppose there were only finitely many integers.  Take the largest such integer, and add one to it.
To be more precise, this follows from the Peano Axioms regarding the successor function.  Namely that it is injective, that it is a natural number, that no natural number has $0$ as a successor, and lastly that $0$ is a natural number.
A: Because you tagged this with number theory, I'll give a number theoretic view. One can see that there are infinitely many natural numbers by seeing that there are infinitely many primes, for example. Once we have that, all we need to realize is that every prime is a natural number. So there are infinitely many natural numbers.
But every natural number is an integer, so there are infinitely may integers. Again, every integer is a rational, so there are infinitely many rationals. This is an old trick, but every rational is a complex number, and so there are infinitely many complex numbers.
So I just gave a big inclusion/injection chain. It's more interesting to show that each set has the 'same' cardinality, except for the complex numbers (which has more). One could do this with an injective chain in the other direction - it's a good exercise.
