if you are asking about what the computer does it is like this: you have the variable

which means that out off all the integers that beneath take the largest.

now the computer doesn't has the function or the group or any of those stuff so he do it differently, the computer save memory with 's and 's, bits, integer he saves with 32-bits(usually)

for understanding with 8-bits it looks like this:

$1111~1111$bits

$1000~0000$bits

$0111~1111$bits

now for float he has a different method, 32-bit format looks like this:

$\underbrace{0}_{0=positive\\1=negative}\underbrace{00000000}_{the~exponent }~~\underbrace{00000000000000000000000}_{the~fraction~part}$

now how exactly this format works is not important now, but you can see from this format that if you have the float, for example, $0~~10000000~11000000000000000000000(=3.5)$ the computer can just ignore the last 22 bits and take only $0~~10000000~1$, the computer can extract all he needs from the first 10 bits if you do interested in how the float itself works:

the computer look at the first bit and put it in var name AXL(for this example) and do now he takes the last part and do $DX=1+\text{[the bit]}^\text{[the bit position]}+\text{[the bit]}^\text{[the bit position]}+...$

and the end result is:

$AX\times (DX\times 2^{\text{[the middle part value]}})$

now because that every part after the 10th bit is quarter or less you don't need them when you use floor

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$

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.