Find all solutions to $$[2x]-[x+1]=2x$$ where $[x]=\lfloor x\rfloor$
$$$$ I divided this into 2 cases:
$$Case 1:x=[x]+\{x\}\text{ where } 0\le\{x\}<0.5$$ $$Case 1:x=[x]+\{x\}\text{ where } 0.5\le\{x\}<1$$ $$$$ $$Case1$$$$$$ $[2x]=2[x], [x+1]=[x]+1. 2x=2[x]+2\{x\}$ $$\Rightarrow [2x]-[x+1]=2[x]-[x]-1=2[x]+2\{x\}$$ $$-[x]-1=2\{x\}$$ Now $0\le\{x\}<1,$ thus $0\le 2\{x\}=-[x]-1<2$ $$\Rightarrow -1\ge[x]>-3$$
Thus either $[x]=-1$ or $[x]=-2$. $$$$If $[x]=-1\Rightarrow 2\{x\}=-(-1)-1=0\Rightarrow\{x\}=0$. Thus $x=[x]+\{x\}=-1$$$$$ If $[x]=-2\Rightarrow 2\{x\}=-(-2)-1=1\Rightarrow\{x\}=0.5$. Thus $x=[x]+\{x\}=-1.5$$$$$ However, it is clear that $x=-1.5$ does not satisfy the original equation. Similarly in case 2, I am getting one of the 2 possible values of $x$ as $0$, which also does not satisfy the original equation.$$$$ Could somebody please explain $where$ I've gone wrong? Many thanks in anticipation!
Since you only did Case 1 - it is clear that you needed $\{x\} <0.5$ and thus $=0.5$ will not satisfy Case 1. Thus, this is not a solution.
Similarly, in case 2 - "I am getting one of the 2 possible values of $x$ as 0" - you need $\{x\}\ge 0.5$, and thus $=0$ does not satisfy the case.
So, the solution should be intersected with the case.