Suppose that $R$ is a ring and $a,b\in R$. Prove that if $a+b = ab$, then $ab=ba$.
I think for this problem because we are given that $a,b$ are elements of the ring we need to show that $a$ and $b$ are invertible. However, in this particular case I don't really know how to do it using $a+b=ab$.
The statement as given does not seem to be true. For a counterexample, let $\mathbb{Z}^{\omega}$ denote the Abelian group of sequences of integers, and let $R$ be the ring of endomorphisms of $\mathbb{Z}^{\omega}$ (with multiplication given by composition). Now define two elements $$a, (x_1, x_2, \ldots) \mapsto (x_1 + x_2, x_2 + x_3, x_3 + x_4, \ldots), \\ b, (x_1, x_2, \ldots) \mapsto (x_1, x_1 + x_2, x_2 + x_3, \ldots).$$ Then $ab = a + b$ (both map $(x_1, x_2, \ldots)$ to $(2x_1 + x_2, x_1 + 2x_2 + x_3, x_2 + 2x_3 + x_4, \ldots$), but $ab \ne ba$ (since $ba$ maps $(x_1, x_2, \ldots)$ to $(x_1 + x_2, x_1 + 2x_2 + x_3, x_2 + 2x_3 + x_4, \ldots)$).