Prove that $\mathbb{Z}$ is an integral domain How would we prove that $\mathbb{Z}$ is an integral domain? That is, if $ca = cb$, where $a, b, c \in Z$ and $c \neq 0$, how would we prove that $a = b$? We cannot multiply by $c^{-1}$ as $c^{-1}$ may not exist.
 A: The answer depends on what you are assuming. You may have to go back to the definitions of sum and product on the natural numbers starting from the Peano axioms, and then appeal to the construction of the integers as equivalence classes of pairs of natural numbers. I may try and spell out the details, if you wish. 
If you know already that the integers are a subring of the rationals, though, you may multiply by the inverse of $c$ in the rationals. 

$\newcommand{\N}{\mathbb{N}}$$\newcommand{\Z}{\mathbb{Z}}$Here are the details. I assume some familiarity with the Peano axioms. I denote by $a+$ the successor of $a \in \N$.
The product of $a, b \in \N$ is defined recursively as
$$
a b =
\begin{cases}
0 & \text{if $b = 0$}\\
a c + a & \text{if $b = c+ \ne 0$.}
\end{cases}
$$
Assume $a, b \ne 0$. I will show that $a b \ne 0$. We have $ab = ac + a$ where $b = c+$. Now the sum of $x, y \in \N$ is defined recursively  as
$$
x + y =
\begin{cases}
x & \text{if $y = 0$}\\
(x + z)+ & \text{if $y = z+ \ne 0$.}
\end{cases}
$$
As $a \ne 0$, we have $a = d+$ for some $d$, so that
$$
a b = a c + a = (a c + d)+ \ne 0.
$$
Now the integers can be defined as the classes $[a, b]$ of pairs $(a, b) \in \N \times \N$ with respect to the equivalence relation $\sim$ given by $(a, b) \sim (c, d)$ iff $a + d = b + c$. One sees immediately that every class can be written either as $[n, 0]$ or $[0, n]$ for some $n \in \N$. ($[n, 0]$ will be identified with $n$, and $[0, n]$ with $-n$.) Moreover $[n,0] = [0,0] = 0$ iff $n = 0$ and similarly for $[0,n]$.
Now the product on $\Z$ is defined as
$$
[a,b] \cdot [c, d] = [ac + bd, ad + bc].
$$
Therefore if $0 \ne m, n \in \N$ we have that all possible products
$$
[m, 0] \cdot [n, 0] = [m n , 0] = [0, m] \cdot [0, n],
\qquad
[m, 0] \cdot [0, n] = [0, m n] = [0, m] \cdot [n, 0]
$$
are nonzero, by the corresponding result for $\N$.
A: There are certainly rings which are not integral domains, but the key property of the integers that you need to know is there are no non-zero zero-divisors, that is: If $ab=0$ then $a=0$ or $b=0$.
Then you have as Vincent points out, $ca=cb$ implies $c(a-b)=0$ since $c\neq 0$, then $a-b=0$ hence $a=b$ as desired.
I think the easiest way to see why there are no non-zero zero-divisors in the integers is to think about what multiplication means.  $ab$ represents the area of a rectangle of side length $a$ and width $b$.  If the area is $0$, one of the sides has to be $0$.
A: An approach: consider $\mathbb{Z}=\mathbb{N}\times \mathbb{N}/R$ where $R$ is the equivalence relation $$(x,y)R(x',y')\Leftrightarrow x+y'=x+x'.$$ We have to prove that $$\forall a, b, c, d \in \mathbb{N}: \left[{a, b}\right] \cdot \left[{c, d}\right] = \left[{0, 0}\right] \Rightarrow \left[{a, b}\right] = \left[{0, 0}\right] \vee \left[{c, d}\right] = \left[{0, 0}\right].$$ Using the well known definition of the sum and product on the quotient set $\mathbb{N}\times \mathbb{N}/R$ 
$$\left[{a, b}\right] \cdot \left[{c, d}\right]=[0,0]\Rightarrow [ac + bd, ad + bc]=[0,0]$$ $$\Rightarrow ac + bd + 0=ad + bc + 0$$ $$\Rightarrow ac + bd =ad + bc.$$ Suppose without loss of generality that $[c,d]\ne [0,0]$. Also suppose $c>d$ that is $c=d+k$ ($0\ne k\in\mathbb{N}$), then
$$ac+bd =ad+bc\Rightarrow a(d+k)+bd=ad+b(d+k)$$
$$\Rightarrow a(d+k)+bd =ad+b(d+k)$$
$$\Rightarrow ad+ak+bd =ad+bd+bk $$
$$\Rightarrow ak=bk$$
$$\underbrace{\Rightarrow}_{a,\;b,\;k\;\in\mathbb{N},\;k\ne 0}a=b$$
$$\Rightarrow [a,b]=[0,0].$$ Same arguments for $d>c$.
