I'm reading Linear Equations in Non-Commutative Fields by Oystein Ore in the Annals of Mathematics. The papers is available here or here if one has access to jstor.
Ore is working in a noncommutative domain, such that any two nonzero elements have a common right multiple. Ore defines equality of fractions as follows:
Let $\frac{a}{b}$ and $\frac{a_1}{b_1}$ be two arbitrary fractions. The elements $\beta\neq 0$ and $\beta_1\neq 0$ can be determined so that $ b\beta_1=b_1\beta $ and we say $ \frac{a}{b}=\frac{a_1}{b_1} $ when $a\beta_1=a_1\beta$.
On page 468, (page 6 of the paper), Ore defines multiplication as $ \frac{a}{b}\cdot\frac{a_1}{b_1}=\frac{a\alpha_1}{b_1\beta} $ where $b\alpha_1=a_1\beta,\beta\neq 0$, and states that "simple calculations" show this is well defined. Can someone please help me show it is well defined? I let $ \frac{a}{b}=\frac{a'}{b'},\qquad\frac{a_1}{b_1}=\frac{a_1'}{b_1'}. $ Then $ \frac{a}{b}\cdot\frac{a_1}{b_1}=\frac{a\lambda}{b_1\mu},\qquad\text{ where }b\lambda=a_1\mu $ and $ \frac{a'}{b'}\cdot\frac{a_1'}{b_1'}=\frac{a'\lambda'}{b_1'\mu'},\qquad\text{ where }b'\lambda'=a_1'\mu'. $
I use the common multiple property to find $\sigma$ and $\rho$ such that $b_1\mu\sigma=b_1'\mu'\rho$. It then follows that the above two fractions are equal if $a\lambda\sigma=a'\lambda'\rho$, which would follow from $b\lambda\sigma=b'\lambda'\rho$, since $a/b=a'/b'$. However, I don't know how to get further. What is the simple calculation I'm overlooking? Thanks.