Condition that satisfies inequality involving fractions and the floor function I want to solve the following inequality for $a$ :
$$\left \lfloor \frac{a}{b} \right \rfloor > \frac{a}{b+1}$$
Where both $a$ and $b$ are strictly positive integers.
My end goal with this is, for a given $b$, to be able to find the smallest $a$ of the form $6 \times 10^{7+p}$ ($p \in \mathbb{Z}$)
If you would like to known more about the context :

 I stumbled upon this problem while coding a musical file format converter.
  In one file format, the BPM of the song is stored as some integer version of $$\frac{6 \times 10^7}{\textrm{BPM}}$$ which I call the tempo value. 
 My code internally handles decimal BPMs, so it needs to convert back and forth between the two "representations". I wanted this conversion to be both "idempotent" and "minimal" in some way, "idempotent" as in : if I convert a tempo value to a BPM and back, I get the same value, and "minimal" meaning that the intermediate BPM is stored with the minimum number of decimal places needed to recover the exact tempo value if I just floor the "raw" conversion on the way back.
 if $v$ is the tempo value, the "raw" BPM can be computed as $$\textrm{BPM}_\textrm{raw} = \frac{6 \times 10^7}{v}$$ I then need to choose some number of decimal places $p$ to truncate it to, which (thanks wikipedia) can be expressed this way : $$\textrm{BPM}_\textrm{trunc} = \textrm{floor(}\textrm{BPM}_\textrm{raw}, p\textrm{)} = \frac{\left \lfloor 10^p \cdot \textrm{BPM}_\textrm{raw} \right \rfloor}{10^p}$$ And then to convert it back to an integer tempo value, I reuse the formula then floor the result again : $$v_\textrm{recovered} = \left \lfloor \frac{6 \times 10^7}{\textrm{BPM}_\textrm{trunc}} \right \rfloor$$ Plugging every formula together, I want to find a condition on p that satisfies : $$ \begin{align} v &= v_\textrm{recovered} \\ v &= \left \lfloor \frac{6 \times 10^7}{\textrm{BPM}_\textrm{trunc}} \right \rfloor \\ v &= \left \lfloor \frac{6 \times 10^7}{\frac{\left \lfloor 10^p \cdot \textrm{BPM}_\textrm{raw} \right \rfloor}{10^p}} \right \rfloor \\ v &= \left \lfloor \frac{6 \times 10^7}{\frac{\left \lfloor 10^p \cdot \frac{6 \times 10^7}{v} \right \rfloor}{10^p}} \right \rfloor \\ v &= \left \lfloor \frac{6 \times 10^{7+p}}{\left \lfloor \frac{6 \times 10^{7+p}}{v} \right \rfloor} \right \rfloor \\ \end{align} $$
 If we rename $v$ to $b$ and $6 \times 10^{7+p}$ to $a$ we need to find a condition on $a$ so that $$b = \left \lfloor \frac{a}{\left \lfloor \frac{a}{b}\right \rfloor}\right \rfloor$$
$b$ is equal to the floor of that fraction if and only if the value of that fraction sits between $b$ and $b+1$ exclusive, which translates to $$ b \leq \frac{a}{\left \lfloor \frac{a}{b}\right \rfloor} < b+1$$ The left part is true, this comes from the fact that $\left \lfloor x \right \rfloor \leq x$ for any $x$, plugging in $\frac{a}{b}$ gives $$\left \lfloor \frac{a}{b} \right \rfloor \leq \frac{a}{b} \Leftrightarrow \frac{a}{\frac{a}{b}} \leq \frac{a}{\left \lfloor \frac{a}{b} \right \rfloor} \Leftrightarrow b \leq \frac{a}{\left \lfloor \frac{a}{b} \right \rfloor}$$ So I just need to find a condition for the right side, which can be rewritten a bit : $$\frac{a}{\left \lfloor \frac{a}{b}\right \rfloor} < b+1 \Leftrightarrow \frac{a}{b+1} < \left \lfloor \frac{a}{b}\right \rfloor$$ And here we are ...


Let $a=bk+r$, where $0\leq r\leq {b-1}.$
Then we have, $\lfloor {\frac ab} \rfloor=k.$
So, $k>\frac {a}{b+1}.$
So that, $bk+k>bk+r.$
Which means, $k>r$. So, the range will be $a>b(r+1).$ So the range of $a$ for which your inequality is satisfied depends upon what $a$ mod $b$ is. But $a>b^2$ will invariably give safe values.
