In a noncommutative ring, is there always a pair $x,y$ such that $xy-yx=1$? Let $R$ be a non-commutative ring. Are there two element $x,y\in R$ such that $xy-yx=1_{R}$? 
I have proved it is true for $R$ being an algebra with finite dimension. 
Sorry, I made a mistake, should be not true for $R$ being a finite dimension  algebra over a field with char $0$. 
 A: For an algebra of positive finite dimension over a field of characteristic $0$, the answer is always no, since $1=xy-yx$ implies that the trace of multiplication by $1$, which is the dimension of your algebra, vanishes.
However, there is one very prominent example of an algebra which has the desired property, namely the Weyl algebra $k\langle x,\partial\rangle/([x,\partial]=1)$ which acts faithfully on the polynomial ring by multiplication and differentiation.
A: I have tried with this non-commutative ring and found the answer to be no. Please check.
Consider the ring $R$ of all elements of the form $a_0+a_1i+a_2j+a_3k$ where $a_i's \in \mathbb R$. Addition is defined componentwise. Also multiplication is defined as usual by following the relation $i^2=j^2=k^2=-1; ij=jk=ki=1; kj=ji=ik=-1$. But I can not find elements in $R$ satisfying given relation.
A: In unital Banach algebras, the unit is not a commutator.  This is mentioned without proof on the Wikipedia page for Banach algebras.  The proof I know relies on spectral theory: since $z\mapsto 1+z$ is entire, the spectral mapping theorem gives $\sigma(1+ba) = 1+\sigma(ba)$; but $\sigma(ab)\cup\{0\}=\sigma(ba)\cup\{0\}$, so if $ab=1+ba$ then it follows easily that $\sigma(ab)$ is unbounded, which is impossible.
