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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!

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2 Answers

up vote 4 down vote accepted

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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I'm still having trouble figuring the rest out. I understand what you have written so far, but I'm stuck at not knowing what operations I can perform when dealing with the floor function. The only thing I can see is to subtract 2n from both sides to get $\lfloor{2a}\rfloor = n$ –  Sarathi J. Hansen Feb 20 '13 at 1:43
    
@Sarathi: And that’s exactly what you should do. Now remember that $0\le\alpha<1$ (why?), so $0\le 2\alpha<2$, and $\lfloor 2\alpha\rfloor$ must therefore be either $0$ or $1$; consider the two cases separately. –  Brian M. Scott Feb 20 '13 at 1:53
    
Ok, so because $\lfloor{2a}\rfloor = n = \lfloor{x}\rfloor$, then $0 \le x < 2$ and $\lfloor{x}\rfloor = $ 0 or 1. So do I have to find what values of x would make $\lfloor{2x}\rfloor$ equal to 0 or 3? –  Sarathi J. Hansen Feb 20 '13 at 2:16
    
@Sarathi: If $n=0$, then $x=\alpha$, and it must be true that $0\le\alpha<\frac12$ (why?); this case therefore gives you every $x\in\left[0,\frac12\right)$ as a solution. What can you say about $\alpha$ and $x$ when $n=1$? –  Brian M. Scott Feb 20 '13 at 2:21
    
If $n=1$, then $x=a+1$, so $1 \le{a+1} < \frac{3}{2}$. Therefore $x\in\left[0,\frac{1}{2}\right)\cup\left[1,\frac{3}{2}\right)$ (but wouldn't $\frac{3}{2}$ be included in the solution?) Is $0\le{a}<\frac{1}{2}$ true because $\lfloor{2a}\rfloor$ would step to the next integer and would no longer equal $n$ and $\lfloor{x}\rfloor$? (I'm not sure how to should that) –  Sarathi J. Hansen Feb 20 '13 at 2:44
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$

  1. If $x$ $\large\tt is$ an integer, we have $2x = 3x\quad\imp\quad \color{#0000ff}{\large x = 0}$.
  2. 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:
    1. $0 < \delta < 1/2$: $\pars{1}$ is reduced to: $$ 2n = 3n\quad\imp\quad n = 0\quad\imp\quad \color{#0000ff}{\large x\ \in\ \pars{0,{1 \over 2}}} $$
    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.

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