Solving equations involving the floor function
HINT: Let $n=\lfloor x\rfloor$, so that $n\le x<n+1$. Let $\alpha=x-n$, the fractional part of $x$, so that $x=n+\alpha$. You’re looking for those $x$ such that $\lfloor 2x\rfloor=3\lfloor x\rfloor$, i.e., such that $\lfloor 2(n+\alpha)\rfloor=3n$.
Clearly $\lfloor 2(n+\alpha)\rfloor=\lfloor 2n+2\alpha\rfloor$, and because $2n$ is an integer, $\lfloor 2n+2\alpha\rfloor=2n+\lfloor 2\alpha\rfloor$. Can you finish it from here?
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- If $x$ $\large\tt is$ an integer, we have $2x = 3x\quad\imp\quad \color{#0000ff}{\large x = 0}$.
- if $x$ $\large \tt\mbox{is not}$ an integer: $x = n + \delta$ where $n$ is an integer and $0 < \delta < 1$. Then,
$$
\floor{2x} = 3\floor{x}
\quad\imp\quad
\floor{2n + 2\delta} = 3\floor{n + \delta} = 3n
\tag{1}
$$
We have two sub-cases:
- $0 < \delta < 1/2$: $\pars{1}$ is reduced to: $$ 2n = 3n\quad\imp\quad n = 0\quad\imp\quad \color{#0000ff}{\large x\ \in\ \pars{0,{1 \over 2}}} $$
- $1/2 \leq \delta < 1$: $\pars{1}$ is reduced to: $$ 2n + 1 = 3n\quad\imp\quad n = 1\quad\imp\quad \color{#0000ff}{\large x\ \in\ \pars{1,2}} $$
Then, the solution becomes $\ds{\color{#0000ff}{x \in \left[0,{1 \over 2}\right) \bigcup \pars{1,\vphantom{1 \over 2}2}}}$.
How about $x < 0$ ?. I left it to the OP.