Solve for x: [2x] = [x] where [.] stands of greatest integer function.
I have solved this question graphically and the result seems to be [$\frac{-1}{2}, \frac{1}{2}$)
However I am unable to even think of a process to solve this without graph, can someone please help me with this.
$\endgroup$3 Answers
$\begingroup$Using Hermite's Identity
$$\lfloor {2x}\rfloor=\lfloor {x}\rfloor+\lfloor {x+\frac{1}{2}}\rfloor=\lfloor {x}\rfloor$$
$$\lfloor {x+\frac{1}{2}}\rfloor=0$$
$$0\leq {x+\frac{1}{2}}<1$$
$$-\frac{1}{2}\leq x <\frac{1}{2}$$
$\endgroup$ 4 $\begingroup$Let $k=\lfloor x\rfloor=\lfloor 2x\rfloor$. From this, we get $k\le x<k+1\quad,k \le 2x<k+1$. Hence $\max(k, \frac{k}{2})\le x <\min(k+1, \frac{k+1}{2})$. For $k\ge 1$, this gives $k \le x <\frac{k+1}{2} \implies k<\frac{k+1}{2} \implies k<1 $ which is false. For $k\le -2$, we get $\frac{k}{2} \le x <k+1 \implies \frac{k}{2}<k+1 \implies k>-2$ which is false again. So $k=0$ or $-1$, corresponding to which we get $x\in [0,\frac{1}{2})$ and $x \in [\frac{-1}{2}, 0)$. Hence our final answer is $$x \in [-\frac{1}{2}, \frac{1}{2})$$
$\endgroup$ $\begingroup$Write $x = n+ \alpha$ where $n=\lfloor x\rfloor$ and $\alpha\in[0,1)$.
The equation then is $n = \lfloor 2(n+\alpha)\rfloor= \lfloor2n+2\alpha\rfloor =2n+\lfloor2\alpha\rfloor$, so $n+\lfloor2\alpha\rfloor=0$.
If $\alpha \in [0,1/2)$ then $\lfloor 2\alpha\rfloor=0$ and $n=0$. So $x=\alpha\in[0,1/2)$.
If $\alpha \in [1/2,1)$ then $\lfloor 2\alpha\rfloor=1$ and $n=-1$. So $x=-1+\alpha\in[-1/2,0)$.
The union of the two cases is $x\in [-1/2,1/2)$
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