An algebra problem by Department 8

Consider cases to solve the problem . $case1$ $x\geq 0$ . Now try to put integer values and check where a disparity arises. Here disparity means that when we put certain integer values change of direction of > sign between LHS & RHS occurs . Like if we put x=1 we find $LHS>RHS$ and if we put x=2 $LHS<RHS$ . Hence disparity . Put $x$ =1+ $\theta$ where $\theta$ is fractional part of x. Now since RHS must be an integer $\theta$ can be either 0,1,1/3,2/3. But none satisfies. $case2$ $x\leq0$ . Here we find disparity between -1 and -2. Hence put $x$ =-2+ $\theta$ . Here we find $\theta$ = $2/3$ satisfies. Hence answer is $\boxed{-4/3}$ .