Smallest non-commutative ring with unity 
Find the smallest non-commutative ring with unity. (By smallest it means it has the least cardinal.) 

I tried rings of size 4 and I found no such ring.

In an earlier version of this post I caused an accident by giving a wrong answer. 
Thanks to those who pointed out the error! Here is a little update:
The ring $M_2(\Bbb F_2)$ of $2 \times 2$-matrices with entries in $\Bbb F_2$ is a non-commutative ring with 16 elements, because $$\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} \neq \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$$
It has a subring of order $8$, namely the upper triangular matrices which are non-commutative by the example above.
We show that $8$ is minimal:
Let $R$ be a finite ring with unity with $n$ elements.
We need two preliminaries:
1.) If the additive group is cyclic, then $R$ is commutative.
Proof: If the additive group of $R$ is cyclic, we can choose $1$ as a generator: If we have $0 = 1+\dots+1 = m \cdot 1$ for some $m\in \Bbb N$, then $0=(m\cdot 1) \cdot g = m\cdot(1\cdot g)=m\cdot g$ so the additive order of $1$ is maximal. Thus, the multiplication table of $R$ is determined by $1 \cdot 1 = 1$, showing $R \cong \Bbb Z/n \Bbb Z$ and $R$ is commutative.
2.) All rings of order $4$ are commutative.
Proof: As a general result, all rings with order equal to a squared prime are commutative: Ring of order $p^2$ is commutative.
Thus any ring of order $1,2,3,5,6,7$ is ruled out by 1.) using the Sylow-theorems and $4$ is ruled out by 2.).
So $8$ is the minimal cardinality a non-commutative ring can have.
