Let $A=\{x \in \mathbb{R} | 0<x<1\}=(0,1)$. Define the relation $L$ on $A$ by $xLy$ if $\left\lfloor {logx} \right\rfloor=\left\lfloor {logy} \right\rfloor$.
I know that $A/L$ is the set of all equivalence classes under the relation $L$.
$[\frac{1}{2}]_L=\{x \in A |\left\lfloor {logx} \right\rfloor\ =\left\lfloor {log\frac{1}2} \right\rfloor\}=\{x \in A | \left\lfloor {logx} \right\rfloor=-1\}$
$[\frac{1}{3}]_L=\{x \in A |\left\lfloor {logx} \right\rfloor=\left\lfloor {log\frac{1}3} \right\rfloor\}=\{x \in A | \left\lfloor {logx} \right\rfloor=-1\}$
$[\frac{1}{10}]_L=\{x\in A|\left\lfloor {logx} \right\rfloor=\left\lfloor {log\frac{1}{10}} \right\rfloor\}=\{x \in A | \left\lfloor {logx} \right\rfloor=-1\}$
$[\frac{1}{100}]_L=\{x\in A|\left\lfloor {logx} \right\rfloor=\left\lfloor {log\frac{1}{100}} \right\rfloor\}=\{x \in A | \left\lfloor {logx} \right\rfloor=-2\}$
We're only allowed to consider real numbers between $0$ and $1$, and $logx$ is the base $10$ logarithm, so that $y=logx$ $\Rightarrow$ $x=10^y$.
I can see there are infinitely many $x's$ for which $x \hspace{1mm} L \hspace{1mm} \left\lfloor {log\frac{1}{2}} \right\rfloor$ and $x \hspace{1mm} L \hspace{1mm} \left\lfloor {log\frac{1}{3}} \right\rfloor$.
I'm thinking the set of all equivalence classes under the relation L is the set containing all the rational numbers between $0$ and $1$? How would you consider the quotient given that every equivalence class has infinitely many elements in it?