In my textbook's proof of Beatty's Theorem, it starts by proving that no integer can occur in both sequences ($\left \lfloor{\alpha}\right \rfloor,\left \lfloor{2\alpha}\right \rfloor,\left \lfloor{3\alpha}\right \rfloor,\dots$ and $\left \lfloor{\beta}\right \rfloor, \left \lfloor{2\beta}\right \rfloor,\left \lfloor{3\beta}\right \rfloor, \dots$) by contradiction:
Suppose that $\lfloor i\alpha\rfloor=\lfloor j\beta\rfloor=n$. Since $\alpha, \beta$ are irrational, then it cannont be the case that $i\alpha=n$ or $j\beta=n$, so we have:$$n<i\alpha <n+1 \phantom{hello }\text{and} \phantom{world }n<j\beta<n+1$$Rearranging these inequalities yields $$\frac{n}{\alpha}<i<\frac{n+1}{\alpha}\phantom{latex }\text{and}\phantom{mathjx} \frac{n}{\beta}<j<\frac{n+1}{\beta}$$which add to give $$n=\frac{n}{\alpha}+\frac{n}{\beta}<i+j<\frac{n+1}{\alpha}+\frac{n+1}{\beta}=n+1$$ $$\vdots$$
Why does $\frac{n}{\alpha}+\frac{n}{\beta}=n$ in this proof? I think it might have to do with us assuming that $\lfloor i\alpha\rfloor=\lfloor j\beta\rfloor=n$, or maybe it has something to do with the property of irrationals? I originally thought that it was unique to this section of the proof (that no integer occurs in both sequences) but this sum is also used later when proving that every integer occurs in one fo the sequences.
Any help would be appreciated - I won't need an explanation of the rest of the proof, just need some clarification on why $\frac{n}{\alpha}+\frac{n}{\beta}=n$. I assume the same explanation will apply to why $\frac{n+1}{\alpha}+\frac{n+1}{\beta}=n+1$, but if it doesn't, reasoning behind that would also be appreciated.
See in the link you provided, the first statement is