Prove that $\mathrm{\forall }x\in \mathbb{R}:\lfloor x^2\rfloor -\lfloor rx\rfloor \ge -1⟺|r|\le 2$.

This is definitely more complicated than it needs to be.

First, we can rewrite the inequality as $$\lfloor x^2+1\rfloor \ge \lfloor rx\rfloor$$

Proof 1

(This is the first proof I wrote, but the second one is nicer and takes a more categorical perspective.)

$\lfloor x^2+1\rfloor =1$ for $|x|<1$ so choosing $x$ to be the opposite sign of $r$ leads to $|x||r|<2$ on the same domain. Since $|x|$ gets arbitrarily close to $1$, this implies that $|r|\le 2$. (If you want to formalize this more, you can take the limit with $|x|=1-1/n$ as $n\to \mathrm{\infty }$.) This gives the left to right implication.

Since floor is monotonic, we also have $x^2+1\ge rx$ implies $\lfloor x^2+1\rfloor \ge \lfloor rx\rfloor$. Thus we have the constraint $$x^2-rx+1\ge 0$$ We know the minimum of this function is at $x=r/2$ and substituting that into the inequality gives $$r^2/4-r^2/2+1\ge 0$$ which can easily be solved for $r^2\le 4$ or $|r|\le 2$ giving the right to left implication. (The change of order for the inequality is because we end up multiplying both sides by $-4$.)

I wanted to make a second proof that leveraged a(n even) more categorical perspective. To that end, in addition to noting that floor is functorial, i.e. monotonic, we also know that it is a right adjoint via $\iota ⊣\lfloor -\rfloor$ where $\iota :\mathbb{Z}\hookrightarrow \mathbb{R}$ is the inclusion which I'll suppress going forward. Explicitly, we have $m\le \lfloor x\rfloor ⟺m\le x$ for any $x\in \mathbb{R}$ and $m\in \mathbb{Z}$. Finally, we have the full and faithfulness of the Yoneda embedding, which is to say $x\le y⟺\mathrm{\forall }z.z\le x\to z\le y$. Using these in our problem leads to the following proof.

Proof 2

For the $⟹$ direction: $$\begin{array}{rl}\lfloor rx\rfloor \le \lfloor x^2+1\rfloor & ⟺\mathrm{\forall }m.m\le \lfloor rx\rfloor \to m\le \lfloor x^2+1\rfloor \\ & ⟺\mathrm{\forall }m.m\le rx\to m\le x^2+1\\ & ⟺\mathrm{\forall }m.m>x^2+1\to m>rx\end{array}$$ We now choose $m=2$ and $|x|\in (0,1)$ with the sign of $x$ opposite of $r$. We then have $2/|x|>|r|$ and thus $|r|$ is bounded by the greatest lower bound of $\{2/|x|\mid x\in (0,1)\}$ which is $2$.

For the $⟸$ direction: If we can show that $rx\le x^2+1$ then monotonicity would finish the proof. If $r$ and $x$ have opposite signs, this easily holds, so assume they have the same sign and, without loss of generality, they're both positive. We then have $2\le x+1/x$, but the right hand sides' minimum (with positive $x$) which you can find using the usual calculus approach occurs when $x=1$, finishing the proof.

posted 12 months ago

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12mo ago

Derek Elkins‭

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