Greatest Integer function

2) As you can see L.H.S is always an integer so RHS also has to be integer.

a) when x is itself an integer we can surely remove the [ ] from L.H.S

so the equation becomes 2x-x-1=2x

which gives x=-1 (the only integer solution)

b)the other and only possibility such that RHS is integer is x has to be of the form y+1/2
,where y is an integer
(say 1/2, 3/2, 5/2 etc)

so the equation becomes [2y+1] - [ y+1+1/2] = 2y+1

as y is an integer [y+1+1/2]= y+1

so we can write 2y+1 -y-1=2y+1 ,solving y=-1

so x=-1+1/2=-1/2

(2):\Rightarrow [2x]-[x+1]=2x$\\\\ Let $x=I+f$, where $\boxed{\0\leq f<1}$\\\\ $[2I+2f]-[I+f+1]=2I+2f\Leftrightarrow 2I+[2f]-I-[f]-1=2I+2f$\\\\ $[2f]-[f]=I+2f+1.......................(1)$\\\\ Now Here\\\\ (i)\underline{\underline{CASE}}:\Rightarrow $0\leq f<\frac{1}{2}\Leftrightarrow 0\leq 2f<1$\\\\ Then $[2f]-[f]=I+2f+1$\\\\ $0-0=I+2f+1\leftrightarrow 2f=-(I+1)$\\\\ but $0\leq f<\frac{1}{2}\Leftrightarrow 0\leq 2f <1$ OR $0\leq-(I+1)<1\Leftrightarrow -2<I\leq -1$\\\\ So $I=-1$\\\\ So If $I=-1$, Then $2f=-(I+1)=-(-1+1)=0\Leftrightarrow f=0$\\\\ So $\boxed{\boxed{x=I+f = -1+0=-1}}$\\\\
$Similarly \\\\ (ii)\underline{\underline{CASE}}:\Rightarrow $\frac{1}{2}\leq f <1\Leftrightarrow 1\leq 2f<2$\\\\ Put into eqn..(1), we get\\\\ $1+0=I+2f+1\Leftrightarrow 2f=-I\leftrightarrow f=-\frac{I}{2}$\\\\ Now Here $1\leq 2f<2$\\\\ $1\leq -I<2\Leftrightarrow -2<I\leq -1$\\\\ So $I=-1$\\\\ So $f=-\frac{I}{2}=-\frac{-1}{2}=\frac{1}{2}$\\\\ So $\boxed{\boxed{x = I+f = -1+\frac{1}{2}=-\frac{1}{2}}}$

We have [2x]-[x+1]=2x \Rightarrow [2x]-[x] = 2x+1 where x is of the form \frac{k}{2} for some integer k

Now, by Hermite's Identity, [2x]-[x] = \left[x+\frac{1}{2} \right]

Hence the equation is \left[x+\frac{1}{2} \right] = 2x+1

Thus 2x+1 \le x + \frac{1}{2} \Rightarrow x \le - \frac{1}{2}

Also, 2x+1> x + \frac{1}{2} -1 \Rightarrow x > - \frac{3}{2}

Thus -3 < 2x \le -1 and so 2x =-2 or -1.

Its easily verified that both x=-1, x=-\frac{1}{2} are solutions

@sir GREAT