I am looking for an example of an associative noncommutative ring $R$ with the following property: for $r,s \in R$, $$ rs \neq 0, \text{ but } sr = 0. $$ Moreover, do rings for which this cannot happen have a name?
Example of a ring for which $rs \neq 0$ but $sr = 0$.
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Here are some general remarks.
First, some model theory. The claim that $(R, r, s)$ is a tuple consisting of a ring $R$ and two elements $r, s \in R$ such that $rs \neq 0$ but $sr = 0$ can be formalized in first-order logic. The completeness theorem then guarantees that such a statement has a model (namely, a ring exists with these properties) if and only if it is consistent (namely, one cannot give a first-order proof of a contradiction from it).
Next, some category theory. There is a universal tuple $(R, r, s)$ of a ring $R$ and two elements $r, s \in R$ such that $sr = 0$: it is the quotient of the free ring $\mathbb{Z} \langle r, s \rangle$ on two generators $r, s$ by the relation $sr = 0$. Any example must be a quotient of this ring, and hence examples exist iff $rs \neq 0$ in this particular ring. It's possible to show that the underlying abelian group of this ring is free abelian on elements of the form $r^n s^m$, which proves this result and various other stronger results too.
In the ring of $2\times 2$ matrices, say with real entries, look at $$\begin{pmatrix}1&0\\0&0\end{pmatrix}$$ and $$\begin{pmatrix}0&0\\1&0\end{pmatrix}$$