Matrix ring $\cong \Bbb F_2[x]/x^2 = $ dual numbers over $\Bbb Z/2$ For no real reason, I did a classification of (associative, with a 1) rings with less than 8 elements (they all happen to be commutative).  Most of the rings I got were of a type I knew - namely: cyclic, field, or Boolean - but one wasn't.
$R \hspace{.04 in} := \hspace{.04 in} \Bigg\langle \bigg\{\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\bigg\} \text{ , matrix operations over the field } \mathbb{Z}/2\mathbb{Z} \Bigg\rangle$
Does $R$ arise in any remotely natural situations?
 A: $R$ is a certain subring of the ring $B$ of upper triangular $2 \times 2$ matrices over $\mathbb{F}_2$.  $B$ arises as the endomorphism ring of a complete flag in $(\mathbb{F}_2)^2$, so in that sense it's a natural geometric object to look at.  The construction generalizes to a  vector space over any field.  
I'm not sure exactly what you mean by "natural situation," though, or why you should expect that every finite ring arises in such a situation.  For any finite field $\mathbb{F}_q$, every subring of $\mathcal{M}_n(\mathbb{F}_q)$ is a finite ring, for example.  As George Lowther says, for finite commutative rings we can also consider quotients $\mathbb{F}_q[x]/p(x)$ for arbitrary polynomials $p(x)$.  These break up into products of rings of the form $\mathbb{F}_q[x]/r(x)^n$ for irreducible $r$, which is a form of the Chinese remainder theorem and is related to the structure theorem for finitely generated modules over a principal ideal domain (in this case, $\mathbb{F}_q[x]$).  Is this "natural"?
A: It's the case $\rm\ R = \mathbb F_2$ of $\rm R[t]/t^2\:.\ $ This is known as the algebra of dual numbers over the ring $\rm R$.  Such rings and higher order analogs $\:\rm R[t]/t^n \;$  prove quite useful when studying (higher) derivations algebraically since such rings provide very convenient algebraic models of tangent / jet spaces. For example, they permit easy transfer of properties of homomorphisms to derivations -- see for example section 8.15 in Jacobson, Basic Algebra II. See this post for further discussion and links.
A: I'm not sure what would be natural in this context, but I would find it natural to denote $0 = \begin{pmatrix}0&0\\0&0\end{pmatrix}$,
$1 = \begin{pmatrix}1&0\\ 0&1\end{pmatrix}$, $\epsilon = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ , $1+\epsilon = \begin{pmatrix}1&1\\0&1\end{pmatrix}$
Where 0 is a additive unit, $1$ is a multiplicative unit and  $\epsilon^2 = 0$.
So this ring arises as $\mathbb Z[\epsilon]/(2,\epsilon^2)$.
A: Let $R$ be an associative ring with $1 \neq 0$, and let $M$ be an $(R,R)$-bimodule.  Then the following set of matrices forms a ring with respect to the usual operations for matrix addition and matrix multiplication.
  \begin{equation}
    T(R,M) = \left\{\begin{pmatrix}r & x \\ 0 & r\end{pmatrix} \mid r \in R,~ x \in M\right\}
  \end{equation}
Any such ring is known as a trivial extension.  The ring described in the question has the form $T(R,R)$ for $R = \mathbb{Z}/2\mathbb{Z}$.
