@Scmbradley #ProofInAToot!

Let \(\newcommand{\vrp}{\varepsilon}\vrp_{AJ},\vrp_{JK},\vrp_{AK}\) be pairwise timelike events in a flat region, and \(\vrp_{AP}\) another event timelike and straight wrt. \(\vrp_{AJ},\vrp_{AK}\) and lightlike wrt. \(\vrp_{JK}\). Consequently, these four events are plane wrt. each other:
\[\newcommand{\sqv}{s^2[\vrp} 0=\begin{vmatrix}0&1&1&1&1\\1&0&\sqv_{AJ},\vrp_{AP}]&\sqv_{AJ},\vrp_{AK}]&\sqv_{AJ},\vrp_{JK}]\\1&\sqv_{AJ},\vrp_{AP}]&0&\sqv_{AJ},\vrp_{AK}]+\sqv_{AJ},\vrp_{AP}]-2\sqrt{\sqv_{AJ},\vrp_{AK}]\,\sqv_{AJ},\vrp_{AP}]}&0\\1&\sqv_{AJ},\vrp_{AK}]&0&\sqv_{AJ},\vrp_{AK}]+\sqv_{AJ},\vrp_{AP}]-2\sqrt{\sqv_{AJ},\vrp_{AK}]\,\sqv_{AJ},\vrp_{AP}]}&\sqv_{JK},\vrp_{AK}]\\1&\sqv_{AJ},\vrp_{JK}]&0&\sqv_{JK},\vrp_{AK}]&0\end{vmatrix}.\]
Thus:\[\frac{\sqv_{AJ},\vrp_{JK}]}{\sqv_{AJ},\vrp_{AK}]}=1-\sqrt{\frac{\sqv_{AJ},\vrp_{AP}]}{\sqv_{AJ},\vrp_{AK}]}}+\frac{\sqv_{JK},\vrp_{AK}]}{\sqv_{AJ},\vrp_{AK}]}\left(1-\sqrt{\frac{\sqv_{AJ},\vrp_{AK}]}{\sqv_{AJ},\vrp_{AP}]}}\right).\]

Now adding the following inequality:
\[2\sqrt{\frac{\sqv_{JK},\vrp_{AK}]}{\sqv_{AJ},\vrp_{AK}]}}\le\frac{\sqv_{JK},\vrp_{AK}]}{\sqv_{AJ},\vrp_{AK}]}\sqrt{\frac{\sqv_{AJ},\vrp_{AK}]}{\sqv_{AJ},\vrp_{AJ}]}}+\sqrt{\frac{\sqv_{AJ},\vrp_{AP}]}{\sqv_{AJ},\vrp_{AK}]}}\]

and collecting yields:

\[\frac{\sqv_{AJ},\vrp_{JK}]}{\sqv_{AJ},\vrp_{AK}]}+2\sqrt{\frac{\sqv_{JK},\vrp_{AK}]}{\sqv_{AJ},\vrp_{AK}]}}\le1+\frac{\sqv_{JK},\vrp_{AK}]}{\sqv_{AJ},\vrp_{AK}]},\]

Therefore
\[
\sqrt{\frac{\sqv_{AJ},\vrp_{JK}]}{\sqv_{AJ},\vrp_{AK}]}}+\sqrt{\frac{\sqv_{JK},\vrp_{AK}]}{\sqv_{AJ},\vrp_{AK}]}}\le1,\]

a.k.a. reverse triangle inequality of (timelike) Lorentzian distances, https://en.wikipedia.org/wiki/Triangle_inequality#Reversal_in_Minkowski_space

#ProofInAToot #TwinFauxParadox #Relativity #Spacetime

@johncarlosbaez #ProofInAToot

@MoritzFirsching #ProofInAToot. No, better: #FormalProofInAToot!

@elboadi #ProofInAToot !

\(t^*:=(L/c)-(c/a)(\sqrt{1+((a/c)\, t^*)^2}-1),\)

\((a/c)\, T_{ABA}:=2\, \text{ArcSinh}[\, (a/c)\, t^*\, ].\)

\(t_m-t_i:=(c/b)(\sqrt{1+((b/c)\, t_m)^2}-1)+(L/c)-(c/a)(\sqrt{1+((a/c)\, t_i)^2}-1),\)

\(t_f-t_m:=(c/b)(\sqrt{1+((b/c)\, t_m)^2}-1)+(L/c)-(c/a)(\sqrt{1+((a/c)\, t_f)^2}-1).\)

\((a/c)\, t_f:=\text{Sinh}[\, \text{ArcSinh}[\, (a/c)\, t_i\, ]+(a/c)\, T_{ABA}\, ]\)

\(\implies (b/a)=\frac{1}{(a/c^2)\, L}=\text{Exp}[\, -(a/c)\, T_{ABA}\, ].\)

#ProofInAToot ? #ClaimInAToot #Acceleration

@pustam_egr Nice. There is a hash tag, #ProofInAToot, for such proofs that fit into a toot. You could edit your toot to contain that toot to make it more discoverable.

@bmreiniger #ProofInAToot challenge accepted. Take a level table with 4 legs in a square. Define h(𝑑) as the height of the wobbly leg off the ground when the other three legs are grounded. 𝑑 is the degrees of rotation from the table's start. Innitially, we have h(0)>0. Rotating the table 90°, keeping the same three legs grounded forces the wobbly leg to be below ground. Thus, h(90°)<0. When the ground is continuous, h(⋅) is too, so the Intermediate Value Theorem gives ∃𝑑*∈(0,90°) s.t. h(𝑑*)=0.

@samurai
#ProofInAToot ?

@divbyzero
by induction!
1 station ✅

n+1 stations: There is some station A with at least enough gas to reach its CCW neighbor B. Remove the section of track between them, merging into AB, and removing an amount of gas that covers that track section. The I.H. applies.

If we start the reduced track at AB, then starting at A works. Otherwise, starting at the same station for the full track as the reduced track works. :qed:

#ProofInAToot

@bgalehouse Put large rectangle with one corner at the origin. We will count pairs (subrectangle, lattice point corner), in two different ways. Firstly, any subrect has either 0, 2 or 4 lattice point corners, so the total is even. Secondly, each lattice point is a corner of 0, 2 or 4 subrects, unless it is a corner of the large rect, in which case it's the corner of only one. Certainly one such corner exists, but the total is even, so there must be another QED #ProofInAToot

Simplemente maravilloso: #proofinatoot

Assume \(\mathbb{R}\) is enumerable.

But \(\forall\) enumerations of \(\mathbb{R}\), \(\exists r \in \mathbb{R}\) s.t. \(\forall i, digit_i(r) == (digit_i(x_i) + 1 \mod 10)\).

Ergo \(\mathbb{R}\) is not enumerable.

#ProofInAToot

This proof has now driven 2 people insane.

@mattp $$ \begin{align} S_k & := \sum_{n=1}^k \frac{n^3}{2^n} \\ & = 26 -\frac{k^3 + 6k^2 + 18k+26}{2^k} \end{align} $$ which can be seen via induction from $$S_{k+1} - S_k = \frac{(k+1)^3}{2^{k+1}},$$ then $$\lim_{k\to\infty} S_k= 26.$$ #proofinatoot

Unfortunately, the #LaTeX here does not seem to support the (ab)use of emojis. #ProofInAToot

Given H,L compact, pick some h in H. Cover L with neighborhoods of each point that miss neighborhoods of h. Take finite subcover of these L point nbhds, and the corresponding finite intersection of h nbhds. We have h in U_h disjoint from V_h containing L.

Now take the finite subcover of U_h's for H. Then the union U of this subcover contains H, and the coresponding finite intersection V of V_h's contains L, and these are disjoint. #proofInAToot

To see something really cool (technologically and for online culture), you should check out the #proofinatoot tag and open it in your browser. The geniuses over at mathstodon.xyz have not only gotten the best instance name on here, they also managed to enable MathJax for toots on their instance!

https://mathstodon.xyz/tags/proofinatoot

Claim. A \(2\times 5\times 6\) box does not fit into a \(3\times 3\times 7\) box.

#ProofInAToot: If the 1st box fits into the 2nd, then any \(\epsilon\)-neighborhood of the first would fit into an \(\epsilon\)-neighborhood of the second. But an \(\epsilon\)-neighborhood of the 1st box has volume \[4/3\pi\epsilon^3+13\pi\epsilon^2+104\epsilon+60\] and of the 2nd has volume \[4/3\pi\epsilon^3+13\pi\epsilon^2+102\epsilon+63.\] For \(\epsilon>3/2\), the first volume exceeds the second. \(\square\)

#ProofInAToot

All natural numbers are interesting.

Proof by contradiction.

Assume not. Then the set of boring numbers should have a minimum.

That number is interesting and thus does not belong in the set of boring numbers.

#proofinatoot
Twitter sucks cuase it's owned by elon

Elon is bad