When can the equality be satisfied? Take the equality given by $\alpha n = \beta m$ where $\alpha, \beta$ are irrational and $n, m$ are rational.  For any irrational $\alpha$ and $\beta$, must there exist a rational $n$ and $m$ that satisfy the equality?
If not, assume $\alpha n = \beta m$ holds.  Then, upon perturbing $\alpha$ by some $\epsilon$, how large must $\epsilon$ be such that there exists a rational $l, k$ that satisfy the equality $(\alpha +\epsilon) l = \beta k$?
 A: It is somewhat misleading to express this using four numbers. If $\beta \neq 0$ it is really about the single real number $\rho = \frac {\alpha}{\beta}$ and how closely $\rho$ can be approximated by a rational number $r$ ie the size of $|\rho-r|$.
This is answered by the fact that the rational numbers are dense in the reals, so you can approximate as closely as you desire. The rationals of denominator $N$ partition the reals into intervals of length $\frac 1N$ and $\rho$ will lie in one of those intervals and will be within $\frac 1{2N}$ of one of the endpoints.
The existence of efficient rational approximations has been explored - ones with small denominator relative to the error. Look up Diophantine Approximation if you are interested.
A: For any real $\alpha, \beta$ with $\beta \ne 0:$
(1). Given arbitrarily small $f> 0,$ let $x=f/|\beta|.$ Since $\Bbb Q$ is dense in $\Bbb R$ there exists $l\in \Bbb Z$ and  $k\in \Bbb Z$ such that $l\ne 0$ and $$|k/l-\alpha/\beta|<x.$$ Let $\epsilon=\beta k/l-\alpha.$ So $(\alpha +\epsilon)l=\beta k.$ We have $$|\epsilon|=|\beta k/l-\alpha|=|\beta|\cdot |k/l-\alpha / \beta|<|\beta|\cdot x=f.$$
(2).If  there exist $non$-$zero$ $m,n\in \Bbb Q$ such that $\alpha n=\beta m,$ then  $\alpha/\beta=m/n \in \Bbb Q.$  Such $n,m$ will not exist if $\alpha /\beta \not \in \Bbb Q.$ For example, if $\alpha =\sqrt 6$ and $\beta=\sqrt 2.$
