Prove the following identity: $$\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +\lfloor x+\frac{2}{n}\rfloor +\lfloor x+\frac{3}{n}\rfloor+...+\lfloor x+\frac{n-1}{n}\rfloor =\lfloor nx\rfloor$$ where n is a Natural Number.
$$$$At first I thought of splitting it into 2 cases: when x is an integer, and when x isnn't an integer. The case of $x$ being an integer is quite simple: $\lfloor x+\frac{k}{n}\rfloor=x$ for $0\le k\le n-1$. Thus the LHS becomes $n\lfloor x\rfloor$ which is equal to the RHS. $$$$However I do not know how to go about the case of $x$ not being an integer.
Lastly, I would actually prefer a proof where it is not necessary to make cases based on the values of $x$, but to have one general proof which satisfies all $x$.
Could somebody please show me how to complete the proof in both ways (ie first, the case where $x$ isn't an integer, and secondly the general proof which doesn't involve breaking into cases)? Many thanks in advance!
Choose integers $a,b$ such that $0<b\le n$ and $$a-\frac bn\le x<a-\frac {b-1}n\ .$$ (In other words, round $x$ downwards to the nearest multiple of $1/n$.) Then $$\Bigl\lfloor x+\frac kn\Bigr\rfloor =\cases{a-1&if $k=0,1,\ldots,b-1$\cr a&if $k=b,b+1,\ldots,n-1$.}$$ So $$LHS=b(a-1)+(n-b)a=na-b\ ;$$ on the other hand, $$na-b\le nx<na-b+1$$ so $$RHS=na-b=LHS\ .$$