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 ...

  • 4
    $\begingroup$ This is a great first question. You included your work, amazing context, and more. $\endgroup$ – Tyma Gaidash May 18 at 14:52
  • 1
    $\begingroup$ The smallest positive integer $n$ for which $n \not =\left \lfloor \dfrac{6 \times 10^{7}}{\left \lfloor \frac{6 \times 10^{7}}{n} \right \rfloor} \right \rfloor $ is $7768$, slightly more than $\sqrt{6 \times 10^{7}}$ and a very high number of beats per minute $\endgroup$ – Henry May 19 at 0:01
  • $\begingroup$ Related context Why are some song tempos not even BPMs? $\endgroup$ – smci May 19 at 1:09
  • $\begingroup$ You say you are truncating not rounding, btw. $\endgroup$ – smci May 19 at 1:12
  1. The floor function obeys$$\mathrm{ bn\le a<b(n+1),y=b, n\in\Bbb N}$$ Graph,
  2. Using the identity above: $$\mathrm{y_1=b=\left\lfloor\frac ab\right\rfloor>y_2=\frac a{b+1}}$$
  3. Therefore when we let a vary:$$\mathrm{a>b(b+1)=b^2+b, bn\le a<b(n+1),y=b, a,b,n\in\Bbb N}$$
  4. Graphical proof

But this only gives the bound when the two graphs no longer cross and $a\ge \left\lfloor \frac ab \right \rfloor$. Modifying this a little bit after doing some pattern analysis of the inequalities satisfying b=1,2,... gives: $$\mathrm{a>n,b(n+1)\le a <(b+1)(n+1)=bn+b+n+1, a,b,n \in \Bbb N^0}$$

or simply: $$\mathrm{a>n,b(n+1)\le a <(b+1)(n+1)=bn+b+n+1, a,b,n \in \Bbb N^0, n\le b,\ if\ n=b,\ then \ a>b^2,\ or\ otherwise\ undefined}$$ Here is graphical proof of this one.

All in all:$$\mathrm{p>log_{10}(n)-7,0\le a-b(n+1)=10^{7+p}-bn-b<n+1⇔log_{10}(b)+log_{10}(n+1)-7 \le p < log_{10}(b+1)+log_{10}(n+1)-7, a,b,n\in \Bbb N^0=0,1,2,...,p\in\Bbb Z}$$

I will leave it to you to find other forms and edit my solutions.

These formulas even work with $a=x^x$ and other functions. Please correct me and give me feedback!

  • $\begingroup$ How about the other great answer? Thanks for the “checkmark”. $\endgroup$ – Tyma Gaidash May 19 at 11:42

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.


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