# Prove that $\forall x\in\Bbb R:\lfloor x^2\rfloor-\lfloor rx\rfloor\ge-1\iff|r|\le2$.

Prove that the inequality $\lfloor x^2\rfloor-\lfloor rx\rfloor\ge-1$ holds true for all real numbers $x$ if and only if $|r|\le2$.

My (proposed) solution, which took me around 2 hours:

First, we reframe the inequality to $\lfloor x^2\rfloor-\lfloor rx\rfloor+1\ge0$.Suppose $r=a$ satisfies the inequality for all real numbers $x$. Consequently, it also does for all real numbers $-x$: $$\lfloor(-x)^2\rfloor-\lfloor a(-x)\rfloor+1\ge0$$ $$\lfloor x^2\rfloor-\lfloor (-a)x\rfloor+1\ge0$$ And we get that $r=-a$ satisfies the condition if and only if $r=a$ does too. Then, we only need to consider cases where $r\ge0$.

Furthermore, let us consider cases where $r\gt2$. The inequality holds for all $x$, so we test it for $0\lt x=\frac2r\lt1$. $$\lfloor x^2\rfloor-\lfloor2\rfloor+1\ge0$$ $$0-2+1\ge0$$, contradiction. We are left with $r\in[0,2]$.

Another observation. If $a\ge0$ and $r=a$ fulfills the condition, $$\lfloor x^2\rfloor-\lfloor rx\rfloor+1\ge0$$ For $x\ge0$, we find that $r\lt a$ also fulfills the condition. All three terms are positive when $x\lt0$, and therefore for $a\ge0$, all $r\lt a$ fulfill the condition if $r=a$ does. Therefore, all that's left to do is to prove the case for $r=2$, and we are done.

Notice that $\lfloor a+b\rfloor-1\le\lfloor a\rfloor+\lfloor b\rfloor$, and $(x-1)^2\ge0\implies\lfloor x^2-2x+1\rfloor=\lfloor x^2-2x\rfloor+1\ge0$. Combining the two, $$0\le\lfloor x^2-2x\rfloor-1+2\le\lfloor x^2\rfloor+\lfloor-2x\rfloor+2$$. And as long as $a\notin\Bbb Z$, $$\lfloor-a\rfloor=-\lfloor a\rfloor-1$$. Applying that, for $2x\notin\Bbb Z$, $$0\le\lfloor x^2\rfloor+\lfloor-2x\rfloor+2=\lfloor x^2\rfloor-\lfloor2x\rfloor+1\ \blacksquare$$ And if $n=2x\in\Bbb Z$, $$0\le\left\lfloor\frac14n^2\right\rfloor-n+2$$. Notice that $n\equiv1\pmod4$ or $n\equiv3\pmod4\iff n^2\equiv1\pmod4$. So, $$0\le\frac14(n^2-1)-n+2$$ $$0\le\frac14n^2-n+\frac74$$ $$0\le\left(\frac12n-1\right)^2+\frac34\ \blacksquare$$.

Is my proposed solution correct? And better yet, is there a faster (and shorter) way to find this proof?

Thank you in advance, and feel free to correct any errors I made!

## 1 answer

This is definitely more complicated than it needs to be.

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

### 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 $|r| \leq 2$. (If you want to formalize this more, you can take the limit with $|x| = 1-1/n$ as $n \to \infty$.) This gives the left to right implication.

Since floor is monotonic, we also have $x^2 + 1 \geq rx$ implies $\lfloor x^2 + 1 \rfloor \geq \lfloor rx \rfloor$. Thus we have the constraint $$x^2 - rx + 1 \geq 0$$ We know the minimum of this function is at $x = r/2$ and substituting that into the inequality gives $$r^2/4 - r^2/2 + 1 \geq 0$$ which can easily be solved for $r^2 \leq 4$ or $|r| \leq 2$ 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 $\iota \dashv \lfloor{-}\rfloor$ where $\iota : \mathbb Z \hookrightarrow \mathbb R$ is the inclusion which I'll suppress going forward. Explicitly, we have $m \leq \lfloor x\rfloor \iff m \leq x$ 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 $x \leq y \iff \forall z. z \leq x \to z \leq y$. Using these in our problem leads to the following proof.

### Proof 2

For the $\implies$ direction: $$\begin{align} \lfloor rx \rfloor \leq \lfloor x^2 + 1 \rfloor & \iff \forall m. m \leq \lfloor rx \rfloor \to m \leq \lfloor x^2 + 1 \rfloor \\ & \iff \forall m. m \leq rx \to m \leq x^2 + 1 \\ & \iff \forall m. m > x^2 + 1 \to m > rx \end{align}$$ 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 $\impliedby$ direction: If we can show that $rx \leq x^2 + 1$ 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 $2 \leq x + 1/x$, 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.

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