Mathematics

#6: Post edited by $user avatar$ Derek Elkins‭ · 2024-04-07T10:57:20Z (9 months ago)
Had my adjunction backward which made the proof nonsense

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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. I find this often happens when I'm more deliberate about taking a 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 left adjoint via $\lfloor{-} floor \dashv \iota$ where $\iota : \mathbb Z \hookrightarrow \mathbb R$ is the inclusion which I'll suppress going forward. Explicitly, we have $\lfloor x floor \leq m \iff x \leq m$ for any $x \in \mathbb R$ and $m \in \mathbb Z$. Finally, we have the full and faithfulness of the (contravariant) Yoneda embedding, which is to say $x \leq y \iff \forall z. y \leq z \to x \leq z$. Using these in our problem leads to the following proof.
### Proof 2
For the $\implies$ direction:
$$\begin{align}
\lfloor rx \rfloor \leq \lfloor x^2 + 1 \rfloor
~~& \iff \forall m. \lfloor x^2 + 1 floor \leq m \to \lfloor rx floor \leq m \\\\~~
& \iff \forall m. x^2 + 1 \leq m \to rx \leq m
\end{align}$$
~~We now simply choose $m = 2$ and $x = \pm 1$ with sign opposite of $r$ and we're done.~~
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.

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 $\iota \dashv \lfloor{-} floor$ where $\iota : \mathbb Z \hookrightarrow \mathbb R$ is the inclusion which I'll suppress going forward. Explicitly, we have $m \leq \lfloor x floor \iff m \leq x$ 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:
$$\begin{align}
\lfloor rx \rfloor \leq \lfloor x^2 + 1 \rfloor
& \iff \forall m. m \leq \lfloor rx floor \to m \leq \lfloor x^2 + 1 floor \\\\
& \iff \forall m. m \leq rx \to m \leq x^2 + 1 \\\\
& \iff \forall m. m > x^2 + 1 \to m > rx
\end{align}$$
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.

#5: Post edited by $user avatar$ Derek Elkins‭ · 2024-04-06T07:58:53Z (9 months ago)
Nicer proof by being more categorically inspired

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This is definitely more complicated than it needs to be.
First, we can rewrite the inequality as $$\lfloor x^2 + 1\rfloor \geq \lfloor rx \rfloor$$
$\lfloor x^2 + 1 \rfloor = 1$ 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 $\lfloor x^2 + 1 floor \geq \lfloor rx floor$. 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 $-4$.)

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. I find this often happens when I'm more deliberate about taking a 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 $\lfloor x^2 + 1 floor \geq \lfloor rx floor$. 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 left adjoint via $\lfloor{-}\rfloor \dashv \iota$ where $\iota : \mathbb Z \hookrightarrow \mathbb R$ is the inclusion which I'll suppress going forward. Explicitly, we have $\lfloor x\rfloor \leq m \iff x \leq m$ for any $x \in \mathbb R$ and $m \in \mathbb Z$. Finally, we have the full and faithfulness of the (contravariant) Yoneda embedding, which is to say $x \leq y \iff \forall z. y \leq z \to x \leq z$. Using these in our problem leads to the following proof.
### Proof 2
For the $\implies$ direction:
$$\begin{align}
\lfloor rx \rfloor \leq \lfloor x^2 + 1 \rfloor
& \iff \forall m. \lfloor x^2 + 1 \rfloor \leq m \to \lfloor rx \rfloor \leq m \\\\
& \iff \forall m. x^2 + 1 \leq m \to rx \leq m
\end{align}$$
We now simply choose $m = 2$ and $x = \pm 1$ with sign opposite of $r$ and we're done.
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.

#4: Post undeleted by $user avatar$ Derek Elkins‭ · 2024-04-06T01:43:52Z (9 months ago)

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#3: Post edited by $user avatar$ Derek Elkins‭ · 2024-04-06T01:42:49Z (9 months ago)
Correct mistake

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This is definitely more complicated than it needs to be.
First, we can rewrite the inequality as $$\lfloor x^2 + 1\rfloor \geq \lfloor rx \rfloor$$
The first interesting value of $x^2 + 1$ to consider is $m = 2$ corresponding to $x = \pm 1$. Choosing $x$ to be the opposite sign of $r$ leads to $2 \geq \lfloor |r| \rfloor$ and thus $|r| \leq 2$. 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 $-4$.)

This is definitely more complicated than it needs to be.
First, we can rewrite the inequality as $$\lfloor x^2 + 1\rfloor \geq \lfloor rx \rfloor$$
$\lfloor x^2 + 1 \rfloor = 1$ 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 $-4$.)

#2: Post deleted by $user avatar$ Derek Elkins‭ · 2024-04-06T01:18:52Z (9 months ago)

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#1: Initial revision by $user avatar$ Derek Elkins‭ · 2024-04-06T01:14:56Z (9 months ago)

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This is definitely more complicated than it needs to be.

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

The first interesting value of $x^2 + 1$ to consider is $m = 2$ corresponding to $x = \pm 1$. Choosing $x$ to be the opposite sign of $r$ leads to $2 \geq \lfloor |r| \rfloor$ and thus $|r| \leq 2$. 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 $-4$.)

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