Solving equations involving the floor function I am trying to solve the following problem:
For what real numbers x is: $\lfloor{2x}\rfloor = 3\lfloor{x}\rfloor$?
I'm not sure how to deal with the floor functions, so I have no idea where to start. If someone could walk me through the process that would great!
 A: $\newcommand{\+}{^{\dagger}}%
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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.
A: 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?
A: *

*equation ( ⌊2x⌋ = 3⌊x⌋ )

*can be said ( 3⌊x⌋ ≤ 2x )

*then divide by 3 ( ⌊x⌋ ≤ 2x/3 )

*then divide by x ( ⌊x⌋/x ≤ 2/3 )

*so clearly the floor of x divided by x must be less then or equal to 2/3

*or x divided by the floor of x is greater then or equal to 3/2

*Of course there is another constraint that I have left out (3⌊x⌋ ≤ 2x < 3⌊x⌋+1) but I am sure it is simpler this way

