Tweetable Mathematics

Question

Update: Please restrict your answers to "tweets" that give more than just the statement of the result, and give also the essence (or a useful hint) of the argument/novelty.

I am looking for examples that the essence of a notable mathematical development fits a tweet (140 characters in English, no fancy formulas).

Background and motivation: This question was motivated by some discussion with Lior Pachter and Dave Bacon (over Twitter). Going over my own papers and blog posts I found very few such examples. (I give one answer.) There were also few developments that I do not know how to tweet their essence but I feel that a good tweet can be made by a person with deep understanding of the matter.

I think that a good list of answers can serve as a good educational window for some developments and areas in mathematics and it won't be overly big.

At most 140 characters per answer, single link is permitted. Tweeting one's own result is encouraged.

Update: I learned quite a few things, and Noam's tweet that I accepted is mind-boggling.

I have discovered a truly remarkable proof of this theorem which this Tweet is too small to contain. — Glorfindel, Apr 26 '17 at 08:26
Misha, what is a soundbite? (Never mind I googled it, interesting idea. We did ask about pictures, and about formulas, but I am not aware of the top of my head of mathematical content via audio which is not via language.) — Gil Kalai, Apr 26 '17 at 08:28
I feel like most of the answers are mis-interpreting the question. This doesn't ask for a result whose statement is in 140 characters; that would be too broad: most paper titles fit in them. It asks for a result whose essence is tweetable: given the tweet alone, a mathematician with good knowledge of the field can fill in the details and complete a proof. So I am going to downvote almost all of them. — Federico Poloni, Apr 26 '17 at 10:16
@FedericoPoloni: what is it, a kind of joke? Could you please indicate to me where exactly in the question it is written that the tweet should be such that "a mathematician with good knowledge of the field can fill in the details and complete a proof"? Honestly, I do not think that your personal interpretation of the locution "essence of a notable mathematical development" should be taken as a rule here. — Francesco Polizzi, Apr 26 '17 at 10:21
@Ycor: yes, I agree. That said, I think that downvoting 14 answers on a personal interpretation basis is questionable, at best. — Francesco Polizzi, Apr 26 '17 at 10:33
@FrancescoPolizzi why do you tell me this? I haven't downvoted anything in this thread (so far) — YCor, Apr 26 '17 at 10:39
@Ycor: my comment on downvoting was intended to Federico Poloni, not to you. Sorry for the misunderstanding. — Francesco Polizzi, Apr 26 '17 at 10:41
This is something I strive for in my own tweets, and I think I have at least a few examples in my twitter feed: http://twitter.com/JDHamkins. — Joel David Hamkins, Apr 26 '17 at 12:01
While I don't quite agree with Federico Poloni's interpretation, I don't understand how this question is not far too broad? Lots of results can be distilled into aphorisms once you assume enough. Where does one make the cutoff? — Kimball, Apr 26 '17 at 12:50
Perhaps the next big MO question should be, "What is the essence of a mathematical result?" I myself lean toward Federico's interpretation- pithifying a theorem's statement does not necessarily clarify or illuminate the ideas at play. — Neal, Apr 26 '17 at 12:52
My initial intention was indeed that the "tweet" gives more than just the statement of the result but also the essence of the argument/novelty. To demand that a mathematician in the field can fill the details is too much to ask for. — Gil Kalai, Apr 26 '17 at 12:58
Answers have been posted as actual tweets in reply to: https://twitter.com/JDHamkins/status/857251368553373696. Please tweet your own answers there! — Joel David Hamkins, Apr 26 '17 at 18:21
There is a key difference between MO and Twitter: here, posts also have a lower bound of 30 characters, as I found by attempting to tweet Descartes. — Victor Protsak, Apr 26 '17 at 21:18
I think I'd enjoy the same question much more if it instead asked for theorems in the form of haiku. :-) — Todd Trimble, Apr 27 '17 at 01:23
Any particular reason why this question become protected, when in fact it specifically asks for "a good list" of answers? Not having posted to mo before, I am of course unable to post an answer as I don't have the requisite reputation. Well... maybe I'll just waste my time elsewhere, then! — Viktor Toth, Apr 27 '17 at 03:44
@ViktorToth it has been protected automatically by the system after having two answers by low-rep users (<10) deleted. A moderator or >15k user could unprotect it. — Glorfindel, Apr 27 '17 at 07:55
Gil, the tug-of-war over closing and re-opening this question would be my first candidate for a list of soundbites if you ever request this. — Mikhail Katz, Apr 27 '17 at 09:23
@Glorfindel I figured that much, I was just hoping that a moderator might notice my annoyed grunt :-) — Viktor Toth, Apr 27 '17 at 11:42
Lots of good examples in the history of the sadly defunct https://twitter.com/tinyproof — Plutor, Apr 27 '17 at 15:21
Inspired by Todd Trimble. "A single number, as a triplet of many, must be less than three." Fermat's last theorem. — , Apr 28 '17 at 04:18
@Viktor: Wasting your time elsewhere is a good idea. See you there. — Franz Lemmermeyer, Apr 30 '17 at 11:55
@FranzLemmermeyer Thank you for your kind words. You know the thing is, I foolishly composed an answer to this interesting question before I noticed that I am not welcome here on account of not having posted to this particular site before. Your welcoming words are testament to the quality and professionalism of this site and its moderators, I presume. Thank you for introducing yourself. — Viktor Toth, Apr 30 '17 at 13:55
I guess the unique goal of the edit is to get new reopening votes? — YCor, May 05 '17 at 14:42
Dear Yves, regarding reopening, in view of the very board interpretation of the question, I myself had mixed feelings about it, and, for this question, did not vote. However, now that the initial stream of answers anyway ended, I like the idea of useful real-time mathematical tweets in rare cases when (and if) a major tweetable advances will occur. — Gil Kalai, May 05 '17 at 15:40
Maybe there should be a parallel "commentable mathematics" -- results that can fit in an MO comment. And its arcane cousin "previewable mathematics" -- mathematical questions which are short enough that the full text of the question fits into the question preview that you see on the MO question page (you have to select a tag to see this -- otherwise you only see the question title). — Tim Campion, Sep 23 '19 at 18:17
Someone said that 140 characters of Tweet did not give enough space for a certain marvelous proof, his name started like Ferma... — Jérôme JEAN-CHARLES, Oct 04 '23 at 19:57

score 180 · Answer 1 · edited Apr 26 '17 at 11:26

180

Four color theorem: any planar graph can be colored with 4 colors. Only proof by computer. SAD.

edited Apr 26 '17 at 11:26

GNiklasch

2,391

answered Apr 26 '17 at 10:36

Anthony Quas

22,470

5

Expressing the essence of how it can be reduced to a huge finite problem for computer proof would be nice, though... – R.. GitHub STOP HELPING ICE Apr 26 '17 at 16:37
11

As years pass by (and more so if we go outside USA) the probability that someone gets the pun tends to zero. SAD. – leonbloy Apr 29 '17 at 13:14
2

??????????????? – Lutz Mattner Mar 03 '18 at 23:45
4

@LutzMattner It's a political joke about Donald Trump, president of America (2016-202?). He likes to say "SAD!" and it's made into jokes. – MaudPieTheRocktorate May 28 '19 at 05:13
Many such cases! – C7X Jul 30 '23 at 20:37
Okay... what is the pun? Six years have passed by, I am inside the USA, and I don't get it. Is SAD an acronym/initialism for something? – Zach Teitler Sep 21 '23 at 17:51
Not really a pun. As @MaudPieTheRocktorate says, this is just a from of "speech" that Trump used on Twitter in his (first) electoral campaign: e.g. "The Democrats don't have a plan to plunder Mars... SAD" (SAD just being all caps sad). i.e. he is asserting that something is true, and then saying that there is something lacking about it. – Anthony Quas Sep 22 '23 at 00:50

score 122 · Answer 2 · edited Dec 29 '22 at 20:02

122

27⁵ + 84⁵ + 110⁵ + 133⁵ = 144⁵. Nice try, Euler. link

#Counterexamples

edited Dec 29 '22 at 20:02

Martin Sleziak

4,608

answered Apr 26 '17 at 14:21

Will Brian

17,444

8

You can shorten it and make it friendlier to read using Unicode superscript numeral characters. – R.. GitHub STOP HELPING ICE Apr 26 '17 at 16:34
2

How do we do it? (without using Tex, say on Twitter) – Gil Kalai Apr 26 '17 at 17:34
5

27⁵ + 84⁵ + 110⁵ + 133⁵ = 144⁵. – Ahmed Fasih Apr 26 '17 at 20:13
1

@GilKalai you can express x to the n as x^n with the use of the caret – Restioson Apr 29 '17 at 07:31
2

@Restioson: I originally typed it out that way, but R.. commented that it is in fact possible to tweet superscript numerals, so I changed it to make it more readable. I was able to type out the superscript 5's in my answer -- without MathJax -- by copy/pasting from this website: http://txtn.us/tiny-text. – Will Brian May 01 '17 at 13:46
@GilKalai are there still exponents? For me on the stack exchange app I just see the bases... – Restioson May 02 '17 at 10:05

score 117 · Answer 3 · answered Apr 26 '17 at 20:51

117

The sentence "The first positive integer that cannot be specified in a 140 character tweet" doesn't specify a well defined integer.

answered Apr 26 '17 at 20:51

Count Iblis

138
1
1
13

45

In other words, "There are non-tweetable positive integers, but there is no smallest non-tweetable positive integer. Contradiction." – Fedor Petrov Apr 27 '17 at 14:17
2

Forgive my ignorance, but assuming a base-10 representation, is 10^140 not the smallest non-tweetable positive integer? – ayane_m Apr 28 '17 at 05:42
3

We mean tweetability in plain languages and mathematical formulae, not just base-10 representations. – Fedor Petrov Apr 28 '17 at 06:23
15

You can tweet: "Ten to the power 140" – Gil Kalai Apr 28 '17 at 06:24
8

@彩音M https://en.wikipedia.org/wiki/Berry_paradox – Harrison Chen Apr 28 '17 at 06:26
An integer $x$ whose Kolmogorov complexity $K(x)=141$. – Jackson Walters Jul 17 '22 at 04:53

score 104 · Accepted Answer · edited Dec 29 '22 at 20:02

104

Every rational r is xyz(x+y+z) for some rational x,y,z.
Proof: Euler (1749) found x(r),y(r),z(r). Nobody knows how.

I have a guess. https://people.math.harvard.edu/~elkies/euler_14t.pdf

edited Dec 29 '22 at 20:02

Martin Sleziak

4,608

answered Apr 27 '17 at 00:10

Noam D. Elkies

77,218

10

And there are some simpler than Euler's x(r), y(r), and z(r). See https://mathoverflow.net/q/302933/24165 . – Anton Klyachko Jul 20 '18 at 04:05
10

Thank you for that reference. I already found x,y,z a bit simpler than Euler's, but not nearly that simple! Now I should revisit this to figure out if and how the Klyachko-Mazhuga-Ponfilenko fits into the elliptic-fibration picture . . . – Noam D. Elkies Jul 20 '18 at 05:01

score 91 · Answer 5 · edited Apr 27 '17 at 04:31

91

Not deep, but if [0,1]² is cut in N triangles of equal area, N is even. If not, extend 2-adic valuation on Q to R, tricolor plane and apply Sperner.

edited Apr 27 '17 at 04:31

JSCB

1,610

answered Apr 26 '17 at 14:00

paul Monsky

5,412
2
25
45

17

This is my favorite! Perhaps it would even benefit from removing the humble "Not deep". – Moritz Firsching Apr 26 '17 at 16:44
4

I like it too, and it suggests a nice general pattern for tweetable mathematics: (1) Statement, (2) Setup, (3) "Apply [known result]." – R.. GitHub STOP HELPING ICE Apr 27 '17 at 02:43
7

#MonskysTheorem – Igor Rivin May 06 '17 at 03:28

score 89 · Answer 6 · edited Apr 29 '17 at 23:08

89

Chebyshev said, and I say it again,
there is always a prime between n and 2n.
#BertrandsPostulate

edited Apr 29 '17 at 23:08

Peter LeFanu Lumsdaine

18,946

answered Apr 26 '17 at 12:56

Moritz Firsching

10,663

2

Where n = 0? If you have space, maybe include that n > 0 – Restioson Apr 29 '17 at 07:32
2

Well, its not really me who invented this tweet... (Also, then I should perhaps also include that $n$ is a natural number and not from some unique factorization domain or whatever) – Moritz Firsching Apr 29 '17 at 07:40
yeah, there just might not be enough chars for that... – Restioson May 01 '17 at 13:16
11

I wish Paul Erdos had a twitter account. – Santi Spadaro Apr 27 '19 at 23:56

score 85 · Answer 7 · edited Apr 27 '17 at 04:35

85

Integral of exp(-x²) dx over R = Γ(1/2) = √π. Proof: square the Gaussian integral and use polar coordinates!

edited Apr 27 '17 at 04:35

JSCB

1,610

answered Apr 26 '17 at 20:15

Noam D. Elkies

77,218

5

This is a great example! – André Henriques Apr 27 '17 at 21:29

score 65 · Answer 8 · edited Apr 27 '17 at 04:34

65

Erdos: You can color edges between √2ᵏ vertices red/blue so no monochrome size k subgraphs (Pick a random coloring. It probably works)

edited Apr 27 '17 at 04:34

JSCB

1,610

answered Apr 26 '17 at 20:56

Kevin P. Costello

5,751

3

Perfect example! I love this one. – Steven Stadnicki Apr 27 '17 at 00:04
8

It's even a counterexample to Gil Kalai's comment that "To demand that a mathematician in the field can fill the details is too much to ask for." – Noam D. Elkies Apr 29 '17 at 14:12

score 58 · Answer 9 · edited Apr 27 '17 at 04:32

58

The n-th binary digit of π is calculable without calculating all the previous digits.

edited Apr 27 '17 at 04:32

JSCB

1,610

answered Apr 26 '17 at 09:56

Martín-Blas Pérez Pinilla

631

1

https://math.stackexchange.com/a/2173610/272831 – Simply Beautiful Art Apr 26 '17 at 17:15
Amazing to know! – DS R May 07 '17 at 14:24

Richard Stanley · Answer 10 · 2017-04-26T17:02:31.197

48

There are infinitely many primes p for which there is a prime between p+1 and p+246.

edited Apr 26 '17 at 17:02

answered Apr 26 '17 at 16:51

Richard Stanley

49,238

24

One can add "Still want to know about p+3". – Gregory Arone May 06 '17 at 15:00
1

The reference would fit too: https://arxiv.org/abs/1407.4897 – Apr 16 '19 at 13:36

score 42 · Answer 11 · answered Apr 27 '17 at 02:44

42

Euclid: there are infinitely many primes, if not, multiply all and add 1. Link

If only they had Twitter back then. They had so many tweetable proofs :)

answered Apr 27 '17 at 02:44

amakelov

987

60

There are infinitely many composites: Multiple the first n of them. Don't add 1 (attributed to Hendrik Lenstra) – David Apr 27 '17 at 20:43
9

@JavaMan: that proves that if there are at least two composites, then there are infinitely many… – Peter LeFanu Lumsdaine Apr 29 '17 at 23:09
3

Also, don't forget xkcd. – Ivo Terek Apr 30 '17 at 22:18
2

Better: "Euclid: there are arbitrarily many primes, by multiplying those given and adding 1." – May 01 '17 at 18:56
8

This phrasing can support a common misunderstanding of the proof though. I would end with: "...and add 1; then factor." – usul Nov 16 '17 at 17:45

score 41 · Answer 12 · edited Sep 06 '19 at 17:59

41

33 = 8866128975287528³ + (-8778405442862239)³ + (-2736111468807040)³. Andrew Booker! link.

Update 2019-09-05: $(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3 = 42$, https://people.maths.bris.ac.uk/~maarb/. This completes all the numbers less than $114$.

edited Sep 06 '19 at 17:59

José Hdz. Stgo.

8,732

answered Mar 09 '19 at 14:15

Gil Kalai

24,218

4

This is fantastic!!!!! (Read the link if you didn't already know why.) Honestly, this looks like a winning answer. – Todd Trimble Mar 09 '19 at 14:20
2

Tim Browning deserves a tradable digital trophy for this solution. This is a computational problem which is easy to verify, difficult to solve, and of mathematical significance. One should be able to construct a cryptocurrency smart contract that can reward someone for solving this kind of problem. https://mathoverflow.net/questions/322022/what-solutions-to-useful-computational-problems-could-be-rewarded-through-crypto – Joseph Van Name Mar 09 '19 at 14:30
4

@JosephVanName I do really like and like to think about your linked question - however, as you kind of hinted at, if Andrew Booker were to be awarded merely for his solution, he would be demotivated to actually publish his methods, lest others copy! – Mark S Mar 10 '19 at 23:45
3

https://mathoverflow.net/a/223121/88133 – Zach Teitler Sep 06 '19 at 01:21

score 39 · Answer 13 · answered Apr 26 '17 at 11:52

39

The most beautifully useless theorem of mathematics IMO:

"This statement is a theorem (and moreover, I can prove it)" — link

answered Apr 26 '17 at 11:52

Gro-Tsen

29,944
4
80
335

10

Can I get you interested in joining the tautology club? – Mindwin Remember Monica Apr 26 '17 at 15:36
11

@Mindwin Ah, but the amazing thing about these "Henkin sentences" is that they are not so trivial to prove. – Gro-Tsen Apr 26 '17 at 15:50
3

This made my world that much bigger. Thank you. – Akshat Mahajan May 02 '17 at 02:52

score 38 · Answer 14 · answered May 14 '17 at 23:35

38

$AB-BA=I$ has no solution in finite matrices. Pf: Trace it!

answered May 14 '17 at 23:35

T. Amdeberhan

41,802

20

Except void matrices ($n=0$ :) – Duchamp Gérard H. E. May 15 '17 at 00:04
3

$AB - BA= I$ not solvable in Banach algebras. Pf: Spectra of AB and BA. – David Handelman Apr 16 '19 at 13:04
11

Except when the dimension is equal to zero in the coefficient ring (e.g. $M_{2 \times 2}(\mathbb{F}_2)$). – WhatsUp Apr 16 '19 at 14:12

score 35 · Answer 15 · edited Sep 01 '20 at 22:02

35

$n!$ divides $(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1})$.

Proof: $\mathfrak{S}_n$ embeds in $\mathrm{GL}_n(\mathbf{F}_2)$. $\quad \square$

(I have read this elegant justification on MO someday, yet I do not find it now.)

edited Sep 01 '20 at 22:02

M. Winter

12,574

answered Jun 07 '18 at 13:53

Desiderius Severus

5,324

1

What does the fancy $\mathfrak{S}_n$ stand for, symmetric group on $n$ letters? – Stefan Perko Jun 07 '18 at 14:24
2

@StefanPerko Absolutely! – Desiderius Severus Jun 07 '18 at 14:49
4

Nice! But I think it should be $\operatorname{GL}_n(\mathbf{F}_2)$ rather than $\operatorname{GL}_2$. – Zach Teitler Jun 07 '18 at 18:16
1

@ZachTeitler Thanks for the correction – Desiderius Severus Jun 08 '18 at 07:58
1

For someone without a clue I do not know whether the third term in the product is supposed to be $2^n-3$ or $2^n-4$. I guess the latter makes more sense. – M. Winter Mar 22 '20 at 12:34
3

@M.Winter Good question. It is $2^n-4$ (then $2^n-8$, and so on). The reason is that $|\operatorname{GL}_n(\mathbf{F}_2)| = (2^n-1)(2^n-2)(2^n-4)(2^n-8)\dotsm(2^n-2^{n-1})$. And the reason for that is that if you build an invertible matrix column by column, after choosing the first $k$ columns, the next column to be chosen out of the $2^n$ possible vectors must not lie in the span of the first $k$ columns, which is a $k$-dimensional subspace so it has $2^k$ elements, leaving $2^n-2^k$ choices for the $(k+1)$st column. – Zach Teitler Sep 01 '20 at 21:55
The factorial can be expressed as a q-Pochhammer product or series as well is confirmed with this very statement. This factors a $2^n^n$. – Steffen Jaeschke Jun 10 '23 at 12:12

score 31 · Answer 16 · answered Apr 26 '17 at 09:42

31

Halting problem - there is no computer program which can determine if an arbitrary computer program halts on a specified input. Link

answered Apr 26 '17 at 09:42

Glorfindel

2,743

score 29 · Answer 17 · answered Apr 26 '17 at 19:48

29

You cannot comb a hairy ball, should you wish to.

answered Apr 26 '17 at 19:48

Massimo Ortolano

263

10

The wind always stands still somewhere. – user1504 May 06 '17 at 03:15
4

Hairy donuts on the other hand... – David Feb 24 '19 at 04:16

score 19 · Answer 18 · edited Dec 29 '22 at 20:00

19

Sphere packing-The E8 lattice provides the optimal packing in eight-dimensional space. Link

edited Dec 29 '22 at 20:00

Martin Sleziak

4,608

answered Apr 26 '17 at 08:33

Francesco Polizzi

65,122

6

Still within the twitter limit :) : Sphere packing-The E8 lattice provides the optimal packing in eight-dimensional space, Pf: LP+modular forms https://arxiv.org/abs/1603.04246 – Gil Kalai Apr 26 '17 at 08:58

score 19 · Answer 19 · answered Apr 26 '17 at 09:32

19

Mihăilescu's theorem (ex Catalan conjecture) - $8$ and $9$ are the only consecutive proper powers of natural numbers. Link

answered Apr 26 '17 at 09:32

GNiklasch

2,391

1

first proved in 2002? WOW – Forever Mozart May 06 '17 at 03:28

score 18 · Answer 20 · answered Apr 26 '17 at 08:20

18

Borsuk's conjecture - a counterexample for DIM>2000, the tensor product of the unit sphere with itself. Link

answered Apr 26 '17 at 08:20

Gil Kalai

24,218

Francesco Polizzi · Answer 21 · 2017-04-26T14:54:18.837

18

Wedderburn's theorem-Every finite skew field is a field. Pf: class equation on centralizers and cyclotomic irreducibility. Link

edited Apr 26 '17 at 14:54

answered Apr 26 '17 at 09:59

Francesco Polizzi

65,122

1

Edited to give some idea on the "essence" of the proof. – Francesco Polizzi Apr 26 '17 at 14:54
6

Alternative proof (not to be confused with "alt-proof"): apply Chevalley(-Warning) to the norm form. – Noam D. Elkies Apr 27 '17 at 01:00

score 17 · Answer 22 · answered Apr 26 '17 at 13:21

17

Turing discovers problems that even computers cannot solve. Link

answered Apr 26 '17 at 13:21

Benjamin Steinberg

37,275

38

You won't believe the second, and the fifth blew my mind! – Peter Samuelson Apr 26 '17 at 13:38
2

@PeterSamuelson I don't get it. Second what, fifth what? – Todd Trimble Apr 26 '17 at 16:07
13

@Todd, it is an internetism, to get people to click on a link to a list of, oh, money saving incredible mathematical tweets? Gerhard "Using This One Weird Tip" Paseman, 2017.04.26. – Gerhard Paseman Apr 26 '17 at 19:34
@GerhardPaseman Oh yes, I've seen that. Makes sense now. – Todd Trimble Apr 26 '17 at 19:38

Saal Hardali · Answer 23 · 2018-06-12T12:29:44.017

17

Breaking News: A mathematician blew up 6 points on a plane! You won't believe what happened next...

edited Jun 12 '18 at 12:29

answered May 01 '17 at 18:34

Saal Hardali

7,549

4

BREAKING NEWS: a plane got blown up by a mathematician. At least five cups of coffee are thought to be dead. – Michael Freimann Sep 21 '18 at 19:15

Asaf Karagila · Answer 24 · 2019-03-11T16:03:55.000

16

Cantor's theorem!

The cardinality of the reals is greater than that of the natural numbers.

Also, various incompleteness results along the lines of the Mathematical T-Rex:

The continuum hypothesis is neither provable nor disprovable from ZFC, unless ZFC is inconsistent to begin with!

(The above can be extended to include a myriad of choice-related statements and ZF, for example.)

edited Mar 11 '19 at 16:03

answered Apr 26 '17 at 09:54

Asaf Karagila

38,140

5

You still have room to mention "diagonalization" and "forcing", respectively, if we are to focus on the "essence of a notable mathematical development" aspect. If you split "CH isn't provable..." and "CH isn't disprovable…" into two tweets, you could accommodate the constructible universe, too. – GNiklasch Apr 26 '17 at 12:44

score 16 · Answer 25 · answered Apr 26 '17 at 10:24

16

One more:

CFSG - The largest sporadic simple group has order 808017424794512875886459904961710757005754368000000000.

answered Apr 26 '17 at 10:24

GNiklasch

2,391

Oops, that was two links, so it doesn't fit the conditions. Feel free to ignore either - the two Wikipedia pages have direct links to each other. – GNiklasch Apr 26 '17 at 10:47
5

Is there any intrinsic definition of sporadic group? Out of curiosity, I'd like to know if any other finite simple group was once considered as sporadic (= not abelian, alternating, or in any "list" of Lie-flavor type) before it was extrapolated as Lie type. – YCor Mar 09 '19 at 15:55

score 15 · Answer 26 · answered Apr 26 '17 at 12:29

15

Of course, every finite group of odd order is solvable.

answered Apr 26 '17 at 12:29

Francois Ziegler

29,500

26

I'd like to see the essence of this development on a tweet. :) – Gil Kalai Apr 26 '17 at 12:33
2

I have discovered a truly remarkable proof of this theorem which this ~~margin~~tweet is too small to contain. – chx Apr 30 '17 at 00:35
1

Savage one .... – DS R May 07 '17 at 14:28
2

Proof: Apply CFSG ;-) – Zach Teitler Sep 20 '19 at 20:04

score 15 · Answer 27 · answered May 06 '17 at 02:46

15

Invert 2×2 matrix A? Easy: switch diagonal entries, negate off-diags, & divide by Δ=det(A).
Pf: divide A²-tA+ΔI=0 by A, solve for 1/A !

MO link

answered May 06 '17 at 02:46

Noam D. Elkies

77,218

GNiklasch · Answer 28 · 2017-04-26T15:32:06.187

14

Lindemann's theorem - You need more than a compass and a ruler to square a circle. Link

Edit Or if we really want the essence of the argument, on which a whole lot of stronger results have been patterned since:

Exploiting the properties of the exponential function, if $\pi$ was algebraic there'd be rational integer strictly between 0 and 1.

edited Apr 26 '17 at 15:32

answered Apr 26 '17 at 09:37

GNiklasch

2,391

OK, I'll stop now… :) – GNiklasch Apr 26 '17 at 09:37
Well, Lindemann's theorem is much more than this. "Lindemann's theorem: pi is transcendental" would be better. – Francesco Polizzi Apr 26 '17 at 09:42
2

Sure. I intentionally toned it down so as to mean something to a broader audience, who might not want to expend the extra time to follow a link... – GNiklasch Apr 26 '17 at 09:51
Lindemann's theorem does not say $\pi$ can be approximated too well by rationals to be algebraic. As far as we know, $\pi$ is not particularly well approximated by rationals. – Robert Israel Apr 26 '17 at 15:23
Thank you @Robert Israel. Fixed the toned-up version (I hope). – GNiklasch Apr 26 '17 at 15:33

score 14 · Answer 29 · answered Apr 27 '19 at 23:57

14

One among $\pi+e$ and $\pi e$ is irrational. Proof: If not, then $(x+e)(x+\pi) \in \mathbb{Q}[x]$ and hence $e$ and $\pi$ would be algebraic.

NOTE: It is still open whether either of $\pi+e$ and $\pi e$ is irrational.

answered Apr 27 '19 at 23:57

Santi Spadaro

4,343

6

"One" meaning "at least one". – Gerry Myerson Apr 28 '19 at 06:34
6

Yes, but that would get me over the 140 character limit. ;-) – Santi Spadaro Apr 28 '19 at 08:09
2

It might be better to say that it is still open whether either of $\pi + e$ and $\pi e$ is rational. – Zach Teitler Mar 21 '20 at 18:46

score 14 · Answer 30 · answered Jul 03 '19 at 09:11

14

This is actually a real tweet by Ryan O'Donnell on Huang's proof of the sensitivity conjecture.

Hao Huang@Emory:

Ex.1: ∃edge-signing of n-cube with 2^{n-1} eigs each of +/-sqrt(n)

Interlacing=>Any induced subgraph with >2^{n-1} vtcs has max eig >= sqrt(n)

Ex.2: In subgraph, max eig <= max valency, even with signs

Hence [GL92] the Sensitivity Conj, s(f) >= sqrt(deg(f))

answered Jul 03 '19 at 09:11

Gil Kalai

24,218

See also: https://gilkalai.wordpress.com/2019/07/02/amazing-hao-huang-proved-the-sensitivity-conjecture/ – Gil Kalai Jul 03 '19 at 11:19

score 14 · Answer 31 · answered Jul 21 '19 at 22:04

14

det(exp(X)) = exp(tr(X)). Remarkably, the RHS does not involve off-diagonal elements of X.

answered Jul 21 '19 at 22:04

Mark L. Stone

1,462

score 13 · Answer 32 · answered Apr 26 '17 at 16:49

13

The prime numbers contain arbitrarily long arithmetic progressions.

answered Apr 26 '17 at 16:49

Richard Stanley

49,238

score 13 · Answer 33 · answered Apr 26 '17 at 20:08

13

1_000_000 $ problem solved: every simply connected, closed 3-manifold is homeomorphic to the 3-sphere. Hint: use the Ricci flow.

answered Apr 26 '17 at 20:08

coudy

18,537
5
74
134

score 13 · Answer 34 · answered Apr 27 '17 at 08:29

13

Lagrange: every integer is sum of 4 squares. Pf: 4-squares identity + every real prime splits in the ring of Hurwitz quaternions. Link

answered Apr 27 '17 at 08:29

Francesco Polizzi

65,122

Glorfindel · Answer 35 · 2017-04-26T10:05:27.040

12

Banach–Tarski paradox: a solid sphere can be divided in a finite # of parts which can be joined to form 2 spheres identical to the original.

edited Apr 26 '17 at 10:05

answered Apr 26 '17 at 09:52

Glorfindel

2,743

1

Too many characters. :-) – Gro-Tsen Apr 26 '17 at 10:02
@Gro-Tsen true. Let's cheat a bit on the grammar, and use # for something else than a #hashtag. – Glorfindel Apr 26 '17 at 10:06
10

"finite # of parts" -> "5 parts" save some more bites. Feels like code-golfing… – Dirk Apr 26 '17 at 12:07
12

Can the idea in the proof be distilled into 140 characters? "Banach-Tarski: AC + free group actions create unmeasurable ghost sets in the sphere; these become two same-volume spheres, from one!" – Neal Apr 26 '17 at 12:57
1

A long haired cat or a samoyed can be used very effectively for demonstration... – chx May 01 '17 at 19:46

Joe Silverman · Answer 36 · 2017-04-26T15:24:11.490

12

Here are three.

Wiles theorem: Every elliptic curve over Q is parameterized by modular functions.

Faltings Theorem/Mordell Conjecture: A curve of genus at least 2 has only finitely many rational points.

Faltings Theorem/Shafarevich Conj: There are only finitely many abelian varieties with good reduction outside a given finite set of primes.

edited Apr 26 '17 at 15:24

answered Apr 26 '17 at 11:12

Joe Silverman

45,660

3

"Wiles theorem" actually has several other authors; I think all five can be accommodated within the 140-character limit. – Noam D. Elkies Apr 29 '17 at 14:14

score 12 · Answer 37 · answered Apr 26 '17 at 22:19

12

Borsuk–Ulam theorem:

If you crumple a ball and run a steam roller over it there are at least two antipodal points touching.

answered Apr 26 '17 at 22:19

kyticka

101
3

Brendan W. Sullivan · Answer 38 · 2017-04-26T19:14:41.180

9

No surjection f from S to its powerset. If so, let T = {x in S | x notin f(x)}. Must be some y such that f(y)=T. Then y in T iff y notin T!#

(140 characters, nothing fancy)

edited Apr 26 '17 at 19:14

answered Apr 26 '17 at 17:14

Brendan W. Sullivan

321

score 9 · Answer 39 · answered Apr 29 '17 at 14:24

This question is actually much older than what it appears to be. The famous fundamental anagram of calculus (Newton, 1676)

6accdae13eff7i3l9n4o4qrr4s8t12ux

is an anagram of the Latin

Data aequatione quotcunque fluentes quantitates involvente, fluxiones invenire; et vice versa

(less than 140 symbols). In modern English it means

Given an equation involving any number of fluent quantities to find the fluxions, and vice versa

or, in Arnold's interpretation

It is useful to solve differential equations

Gil Kalai · Answer 40 · 2020-01-16T15:00:41.620

9

This is based on a real tweet of Ryan O'Donnell on the proof of MIP*=RE which disproves Connes' Embedding Conjecture from 1976.

MIP* = RE, by Zhengfeng Ji, Anand Natarajan, Thomas Vidick, John Wright, Henry Yuen: https://arxiv.org/abs/2001.04383 . There is a multiple-entagled-quantum-provers proof system for the Halting Problem, and Connes' Embedding Conjecture is false.

edited Jan 16 '20 at 15:00

answered Jan 16 '20 at 06:05

Gil Kalai

24,218

score 8 · Answer 41 · answered Apr 26 '17 at 10:00

8

Gödel's first incompleteness theorem - any consistent formal system capable of basic arithmetic is incomplete. Link

answered Apr 26 '17 at 10:00

Glorfindel

2,743

score 8 · Answer 42 · answered Apr 26 '17 at 12:45

8

If Ax = y underdetermined w unique sparse soln x0, minimizing L1(x) recovers x0 under mild conds. Shannon-Nyquist bad (or sick) thm! Link

answered Apr 26 '17 at 12:45

Steve Huntsman

15,258

score 8 · Answer 43 · answered Apr 26 '17 at 16:19

8

Few things multiply nicely (division rings usually noncommutative) but few things multiply nicely (finite ones are) #Wedderburn

answered Apr 26 '17 at 16:19

Noah Schweber

18,932

score 8 · Answer 44 · answered Apr 27 '17 at 02:58

8

A rectangle tiled by rectangles with an integral side has an integral side because it has mass zero under a certain periodic signed measure.

(Tweeted here)

answered Apr 27 '17 at 02:58

Evan Jenkins

7,107

score 8 · Answer 45 · answered Jul 08 '20 at 18:35

8

Every set of integers with divergent reciprocals contains 3-term arithmetic progressions. Erdos and Turan conjectured; Bloom and Sisask proved; Bateman and Katz's method for cap set used. link. #TWTMTH

answered Jul 08 '20 at 18:35

Gil Kalai

24,218

post: https://gilkalai.wordpress.com/2020/07/08/to-cheer-you-up-in-difficult-times-7-bloom-and-sisask-just-broke-the-logarithm-barrier-for-roths-theorem/ – Gil Kalai Jul 08 '20 at 18:35

score 7 · Answer 46 · answered Apr 26 '17 at 20:49

7

How about a proof that, given a set, there always exists a bigger set?

For every set A, there is a set that doesn't inject into A. Take the set B of ordinals that inject into A. B is an ordinal and is not in B.

I think one can also fit a proof of Sylvester's theorem in 137 characters.

Take n points not on a line. Let L be a line containing >1 points minimizing the distance to a point off L. It contains exactly 2 points.

answered Apr 26 '17 at 20:49

Burak

4,135

Can you ensure that B is not a proper class within the tweet limit? – Hagen von Eitzen Apr 29 '17 at 12:53
@Hagen von Eitzen: No. The tweet assumes the familiarity of the reader with the axiom of replacement. – Burak Apr 29 '17 at 13:44

score 7 · Answer 47 · answered May 06 '17 at 08:45

7

Littlewood's example of a 2 line dissertation fits here, I believe.

An integral function never 0 or 1 is a constant:

$\exp(i\Omega(f(z)))$ is a bounded integral function

answered May 06 '17 at 08:45

მამუკა ჯიბლაძე

17,468

1

Just seen this. Shouldn't it be "entire function"? – Yemon Choi Mar 09 '19 at 19:48
@YemonChoi It is the way Littlewood writes, but you are right, current terminology is "entire". Shall I change it nevertheless? Can't decide... – მამუკა ჯიბლაძე Mar 10 '19 at 05:29

score 7 · Answer 48 · answered Mar 12 '19 at 01:11

7

Cole 1903: 2^67-1=147,573,952,589,676,412,927=193,707,721 × 761,838,257,287 #MoreThanThreeYearsOfSundays

answered Mar 12 '19 at 01:11

Mark S

2,143

score 6 · Answer 49 · answered Apr 26 '17 at 10:01

6

There is an algorithm to compute étale cohomology in finite time (but we don't know how long it takes) — link

answered Apr 26 '17 at 10:01

Gro-Tsen

29,944
4
80
335

score 6 · Answer 50 · answered Apr 26 '17 at 13:03

6

Weyl's law: pack a domain with tiny squares. In the limit, hear the domain's volume.

and

Polya's theorem: pack a square with tiny domains. In the limit, the domain is always higher-pitched than Weyl's law thinks it should be.

answered Apr 26 '17 at 13:03

Neal

826

Pardon my ignorance, but I can't make sense of this. How do I hear something by/after packing? – Dirk Apr 27 '17 at 05:39
Which Pólya's theorem does this refer to? Searching, I find the Pólya enumeration theorem. – cfh Apr 27 '17 at 07:06
@cfh "On the Eigenvalues of Vibrating Membranes," 1961, Proc LMS (http://onlinelibrary.wiley.com/doi/10.1112/plms/s3-11.1.419/abstract). A domain that tiles Euclidean space has every Dirichlet eigenvalue greater than the value predicted by Weyl's law. – Neal Apr 27 '17 at 12:59
2

@Dirk Curious ignorance needs no pardon :) c.f. xkcd 1053. This refers to the eigenvalues of the Laplacian on a domain with (Dirichlet) boundary condition. One can interpret them as squared frequencies of standing waves, so associate them with harmonic tones of a domain. Weyl's law says that the asymptotics of this sequence encodes the volume of the domain. As the Laplacian is elliptic, there's a variational principle that ends up allowing one to compare eigenvalues across domains; Weyl's law can be proven by tiling with squares (which have computable eigenvalues) and making that comparison. – Neal Apr 27 '17 at 13:12
That's pretty cool, thanks for the explanation! (Now try to explain that in a tweet...) – Dirk Apr 27 '17 at 14:10

score 6 · Answer 51 · answered May 01 '17 at 18:05

6

Cap set problem solved: polynomial method, punchline going back to (a+b)²=a²+b²+2ab. link

answered May 01 '17 at 18:05

Gil Kalai

24,218

2

I actually tweeted it and a few more under the hashtag #TWTMTH – Gil Kalai May 01 '17 at 18:06

score 6 · Answer 52 · answered May 06 '17 at 08:28

6

A finite(dimensional) domain has to be a field: all nonzero "multiply-by-an-element" maps are self-embeddings.

answered May 06 '17 at 08:28

მამუკა ჯიბლაძე

17,468

score 5 · Answer 53 · answered Apr 30 '17 at 11:07

5

There is a way of pretending that any reduced ring is Noetherian and a field. Grothendieck's generic freeness lemma is then quite easy. link (Section 11.5)

answered Apr 30 '17 at 11:07

Ingo Blechschmidt

3,152

score 5 · Answer 54 · answered May 07 '17 at 17:57

5

**ΕΥΡΗΚΑ num=Δ+Δ+Δ (Every positive integer can be written as the sum of three triangular numbers, Gauss, July 10, 1796)

answered May 07 '17 at 17:57

Mark S

2,143

N. de Rancourt · Answer 55 · 2019-04-16T19:26:47.230

5

$S$ unit sphere of an $\infty$-dim Banach isn't compact. Pf: for $H$ closed hyperplane, $\bigcap(S\cap H)=\varnothing$ (Hahn-Banach), but no finite subintersection is empty.

Note: $\infty$, $\bigcap$, $\cap$, $\neq$, and $\varnothing$ are unicode characters, so this is actually tweetable!

edited Apr 16 '19 at 19:26

answered Apr 16 '19 at 12:56

N. de Rancourt

1,005

2

Shouldn't $\ne$ be $=$? – Andrés E. Caicedo Apr 16 '19 at 13:13
2

Thanks @AndrésE.Caicedo, I correct! – N. de Rancourt Apr 16 '19 at 19:26

score 4 · Answer 56 · answered Apr 26 '17 at 09:28

4

Stark-Heegner theorem: There are nine imaginary quadratic fields with class number one. Link

answered Apr 26 '17 at 09:28

GNiklasch

2,391

score 4 · Answer 57 · answered Apr 26 '17 at 10:12

4

I can't resist. Birkhoff's theorem:

For an erg measure-pres trans, orbit-wise avgs of a fn agree with the spatial avgs.

answered Apr 26 '17 at 10:12

Anthony Quas

22,470

score 4 · Answer 58 · answered Apr 26 '17 at 12:34

4

Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos & generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.

Not in English, 268 characters so two tweets, @Glorfindel 's comment, but I couldn't not post it.

answered Apr 26 '17 at 12:34

Ethan Bolker

109

Dear Ethan, warm regards--Gil – Gil Kalai Apr 26 '17 at 16:04
The canonical answer – David Roberts Nov 05 '22 at 09:15

score 4 · Answer 59 · answered Apr 26 '17 at 12:59

4

Trichotomy: the rational points on an algebraic curve are parametrizable; form a finitely-generated abelian group; or form a finite set.

answered Apr 26 '17 at 12:59

R.P.

4,745

score 4 · Answer 60 · answered Apr 27 '17 at 17:56

4

The square of the hypotenuse is equal to the sum of the squares of the other two sides.

answered Apr 27 '17 at 17:56

Giulio

2,324

score 4 · Answer 61 · answered Nov 16 '17 at 15:45

4

Every non-constant complex polynomial has a complex root : If not the inverse is bounded analytic. Use Liouville. #FundamentalTheoremOfAlgebra

answered Nov 16 '17 at 15:45

Vamsi

3,323

score 4 · Answer 62 · answered Nov 16 '17 at 17:11

4

Not a theorem but a good result on Mersenne integers using Lucas-Lehmer test (via @toorandom)

2^74207281-1 is #prime, they checked: Consider s_1=4, s_2=14,...,s_j=(s_{j-1}^2)-2... m=(2^n)-1 prime <=> s_{n-1} is multiple of m #mathchat

answered Nov 16 '17 at 17:11

Eduardo R. Duarte

785

Gil Kalai · Answer 63 · 2018-06-12T05:35:48.537

4

Aubrey de Grey's strategies for finding the chromatic number of the plane: A) prolonging life to 1000 years and waiting for a solution B) Constructing a unit-distance planar graph that requires 5 colors!!! Based on SAT-solvers and a lot of Moser Spindles. link

edited Jun 12 '18 at 05:35

answered Jun 07 '18 at 13:17

Gil Kalai

24,218

1

The above far exceeds 140 characters. A more straightforward version works: “Aubrey de Grey constructed a unit-distance planar graph that requires 5 colors!!! Based on SAT-solvers and a lot of Moser Spindles.” – Dec 30 '22 at 03:00

score 4 · Answer 64 · answered Mar 30 '19 at 19:50

4

"The Magic Words Are Squeamish Ossifrage" - to factor a 129-digit semiprime took way less than 40 quadrillion years when early-90's era computers work together using early-90's era factoring algorithms over the early-90's era Internet!

answered Mar 30 '19 at 19:50

Mark S

2,143

score 4 · Answer 65 · answered Sep 06 '19 at 01:55

4

One cannot hear the shape of a drum. link

Proof via Sunada's Theorem.

answered Sep 06 '19 at 01:55

score 4 · Answer 66 · edited Jan 11 '21 at 13:30

4

Lucas-Lehmer for Wagstaff. Let $p$ be an odd prime, $s_0=4,s_{n+1}=s_n^2-2$, then $N_p=(2^p+1)/3$ is prime implies $N_p$ divides $s_{p-1}-5-9\left( \frac{p}{3} \right) $.

edited Jan 11 '21 at 13:30

CHUAKS

964

answered Sep 02 '20 at 14:25

Chua KS

487

score 3 · Answer 67 · answered Apr 26 '17 at 09:51

3

ZFC + "exists a Reinhardt cardinal" is inconsistent.

V = L implies that measurable cardinals do not exist.

answered Apr 26 '17 at 09:51

Martín-Blas Pérez Pinilla

631

score 3 · Answer 68 · edited Apr 29 '17 at 13:55

3

f_N = (-d/dx)^N cos(x^½)
= A(y)f_0 + B(y)f_1 for y = 1/(4x); A,B∈Z[y], deg O(N)
~ N!/(2N)! as N → ∞ = o(ε^N)
f_1/f_0∈Q ⇒ y∉Q
x = π^2 ⇒ π^2∉Q

edited Apr 29 '17 at 13:55

EKons

101

answered Apr 27 '17 at 06:05

Sridhar Ramesh

5,642

score 3 · Answer 69 · answered Apr 27 '17 at 19:26

3

Elliptic curves produce a key exchange that may be safe against quantum computers. link

answered Apr 27 '17 at 19:26

Joseph Van Name

27,791

They do? That is awesome. – Asaf Karagila Apr 27 '17 at 19:35
5

"May be". That is, nobody knows of a sub-exponential quantum algorithm in the supersingular case, because the known sub-exponential algorithms for ordinary elliptic curves require the endomorphism ring to be commutative. But I see no reason for confidence that a sub-exponential quantum algorithm for this case does not exist. – Robert Israel Apr 27 '17 at 20:40

score 3 · Answer 70 · answered Apr 28 '17 at 08:30

3

The smallest positive integer not definable in under sixty letters.

answered Apr 28 '17 at 08:30

Arseniy Akopyan

1,431

score 3 · Answer 71 · edited Sep 02 '20 at 10:50

3

The fundamental theorem of algebra: every polynomial splits in the field of complex numbers.

edited Sep 02 '20 at 10:50

A. Bailleul

1,174

answered Nov 16 '17 at 16:18

Rémy Martin

81

score 3 · Answer 72 · answered Nov 16 '17 at 21:53

3

e^x is sum of x^k/k!, k=0,1,...: Binomial theorem on (1+x/n)^n; coefficient of x^k is binom(n,k)/n^k=(1-1/n)...(1-(k-1)/n)/k!→1/k! as n→∞

(137 characters, uses fancy Unicode characters)

answered Nov 16 '17 at 21:53

Itai Bar-Natan

1,075

Samantha Y · Answer 73 · 2017-11-16T22:16:41.370

3

First, my original tweet, on the essential content of IUT, i.e. how it intends to prove ABC, in 140 characters:

mochizuki: invented big deformation machine-tracks how deformed schemethry needs before HA-thry applies-amount of deformation IS Spziro ineq

Now, Twitter recently raised the character limit for all users to 280 characters, so with my whopping 140 extra characters, I will write a new style tweet of the same flavor:

Mochizuki: Invented deformation machine-IUT-which elim. obstruct'ns from applying fund.thm.of HAtheory to schemethry by deforming schemethry. By measuring distort. needed b4 FTHAT applies, ineq. appears-this is content of Spziro conj. hence ABC, modulo rigor check:IUT black box

edited Nov 16 '17 at 22:16

answered Nov 16 '17 at 21:54

Samantha Y

356

oh and I should say, HA theory is Hodge-Arakelov theory, Mochizuki's initial framework for solving ABC – Samantha Y Nov 17 '17 at 20:21
This looks a good piece of ciphetext! – vidyarthi Dec 10 '20 at 12:45

score 3 · Answer 74 · answered Mar 03 '18 at 18:59

Lattices with exponential kissing number discovered by Serge Vlăduţ. Another home run for algebraic-geometry codes. Link.

Actually, with the new 260 characters policy we can add:

Lattices with exponential kissing number discovered by Serge Vlăduţ. Another home run for algebraic-geometry codes. Will exponential improvement for Minkowski's 1905 lower bound for sphere packing be the next grand slam?

Martín-Blas Pérez Pinilla · Answer 75 · 2017-04-26T10:19:01.477

2

Every consistent first-order theory has a model of countable size.

A set of sentences is consistent if and only if every finite subset is consistent.

A set of sentences has a model if and only if every finite subset of it has a model.

The first order logic is the only logic with a finitary syntax to possess the Löwenheim-Skolem property and be complete.

edited Apr 26 '17 at 10:19

answered Apr 26 '17 at 10:03

Martín-Blas Pérez Pinilla

631

score 2 · Answer 76 · answered Apr 26 '17 at 12:13

2

Only nonbinary nontrivial perfect code is ternary w parity matrix rows 11122010000 11210201000 12101200100 12012100010 10221100001. Link

answered Apr 26 '17 at 12:13

Steve Huntsman

15,258

score 2 · Answer 77 · edited Dec 29 '22 at 19:51

2

Sz(q) has two orbits more than PSL(2,q) under the action of its automorphism group - see https://doi.org/10.1081/AGB-120004501, Thm. 3.4.

edited Dec 29 '22 at 19:51

Martin Sleziak

4,608

answered Apr 26 '17 at 14:16

Stefan Kohl

19,498
21
73
136

Just to follow the encouragement to post a result of one's own! – Stefan Kohl Apr 26 '17 at 14:17

score 2 · Answer 78 · answered Apr 27 '17 at 05:20

2

You can always find a transversal line meeting a family of parallel line segments on the plane such that any 3 can be transversed.

answered Apr 27 '17 at 05:20

Shake Baby

1,638

yberman · Answer 79 · 2017-08-27T22:04:40.800

2

At a party with $4^n$ people and there are either $n$ people who all know each other, or $n$ who are all mutual strangers.

(I used crappy bounds so it could be easily stated. I know the upper bound for $R(n,n)$ is another answer.)

edited Aug 27 '17 at 22:04

answered Apr 30 '17 at 00:09

yberman

771

1

The upper bound, rather. – Andrés E. Caicedo Apr 30 '17 at 00:41
2

See also https://gilkalai.wordpress.com/2017/03/29/r55-%e2%89%a4-48/ straight from OP's blog for the recent result that R(5,5) ≤ 48. The tweet could be "You can't have 48 people on a party such that no 5 are all acquainted and no 5 are all strangers." – Zsbán Ambrus Aug 27 '17 at 19:15

score 2 · Answer 80 · answered Nov 16 '17 at 15:24

2

Erdös-Faber-Lovasz conjecture: If $n$ copies of $K_n$ have pairwise intersection of $\leq 1$, you can color all points with $n$ colors.

answered Nov 16 '17 at 15:24

Dominic van der Zypen

45,374

I dont think it is already proved. Do you have a complete proof – vidyarthi Dec 10 '20 at 12:43
That's right, this is an open problem. – Dominic van der Zypen Dec 11 '20 at 13:19

score 2 · Answer 81 · answered Nov 16 '17 at 19:54

2

$\bar{\rho}$ irreducible Galois rep has finitely many lifts $\rho$ unramified outside of $S$. Proof: $(\rho_2^{-1}\rho_1\rho_2-\rho_1)/\mathcal{l}^r$ is a cocycle.

answered Nov 16 '17 at 19:54

Watson Ladd

2,419
13
19

Sebastien Palcoux · Answer 82 · 2019-07-22T12:29:42.387

2

A group G≠A_n,S_n with a core-free maximal subgroup of index n∈{266,506,759,1045,1288,1463,3795} is sporadic. Proof by GAP. Any other index?

edited Jul 22 '19 at 12:29

answered Jul 21 '19 at 21:41

Sebastien Palcoux

25,922

score 2 · Answer 83 · answered Mar 21 '20 at 17:22

2

If $S^n$ x $S^n$ minus diag & antidiag self deforms and each (x,y) → (y,x) then n = 1, 3, 7, 15, 31, 63 or 127 (Kervaire invariant).

(Stolen from the epigraph of I. M. James, The topology of Stiefel manifolds (1976).)

answered Mar 21 '20 at 17:22

Francois Ziegler

29,500

JP McCarthy · Answer 84 · 2021-04-09T16:24:39.780

2

When playing poker with a quantum decks of cards, you can only look at one card at a time.

On first sight you might find three aces of hearts, and two of spades, but when you reveal your hand to claim the pot you suddenly have nothing but a pair of twos.

EDIT: More here: https://arxiv.org/abs/2104.02817

edited Apr 09 '21 at 16:24

answered Sep 02 '20 at 12:42

JP McCarthy

956

This does not fit within 140 characters – Dec 30 '22 at 02:32
@MattF. question out of date. https://twitter.com/JP_Mc_C/status/1608749294022201346?s=20&t=C7WtmBtGi0qnp7qhZr0qCQ – JP McCarthy Dec 30 '22 at 08:58

score 2 · Answer 85 · edited Dec 30 '22 at 02:46

2

Brooks' theorem and list-coloring variants can be proved using the combinatorial Nullstellensatz and a related theorem of Alon and Tarsi.

edited Dec 30 '22 at 02:46

answered Dec 10 '20 at 12:42

vidyarthi

2,027

score 1 · Answer 86 · answered Apr 26 '17 at 15:21

1

Graham's number is so big, that its digits contain more information than can be contained within the volume of a human brain

answered Apr 26 '17 at 15:21

Dunno

101

6

Understatement of the century... – Nate Eldredge Apr 26 '17 at 22:19
1

And of the last century. – Todd Trimble Apr 27 '17 at 01:15
2

But does it have a chance for the understatement of the millennium? – Asaf Karagila Apr 27 '17 at 10:45
1

It's less of an understatement when you realize the word "volume" plays an important role: "It's a number so insanely, absurdly huge that storing all the digits of Graham's number in the brain could create a black hole, said John Baez, a mathematical physicist at the University of California, Irvine, who is researching big numbers. (Only so much information can be stored in a given amount of space, and trying to squish more matter into that space creates a black hole, he said.)" http://www.livescience.com/26870-ginormous-numbers-boggle-the-mind.html – Dustin G. Mixon Apr 29 '17 at 23:24
this is perhaps three years too late but I wonder if John Baez is aware that he's "researching big numbers" – AlexArvanitakis Dec 10 '20 at 23:02

score 1 · Answer 87 · answered Oct 17 '17 at 05:38

1

There is no smooth surjection from $S^5$ to $S^6$. #Sard

answered Oct 17 '17 at 05:38

Vamsi

3,323

Sebastien Palcoux · Answer 88 · 2020-12-10T11:55:48.673

1

Let $u_0 \ge 2$ be a rational, and $u_{n+1}=⌊u_n⌋(u_n - ⌊u_n⌋ + 1)$.
Does the sequence $(u_n)$ reach an integer? Link.
#AlternativeToContinuedFraction

edited Dec 10 '20 at 11:55

answered Dec 10 '20 at 10:54

Sebastien Palcoux

25,922

Cool Tweetable Puzzle – Sebastien Palcoux Dec 10 '20 at 11:56

score -7 · Answer 89 · answered Apr 26 '17 at 21:10

-7

Cogito ergo sum #ReneDescartes

answered Apr 26 '17 at 21:10

Victor Protsak

14,347

I find it hilarious that my "tweet" was too short for MO. Andere Städtchen, andere Mädchen, I suppose? – Victor Protsak Apr 26 '17 at 21:14
8

Isn't that more in the area of Tweetable Philosophy? – Frieder Ladisch Apr 26 '17 at 21:54
2

Perhaps. Or metamathematics, as I would interpret it. Arguably, that sentence is also at the root of modern scientific method. To those downvoting this answer, I have another one for you: $$\ $$ You are not fun anymore! #MonthyPython – Victor Protsak Apr 27 '17 at 01:48
5

@Victor Protsak Wait, so "another roof, another proof" [attrib. Erdős] is a play on "andere Städtchen, andere Mädchen"? – Noam D. Elkies May 01 '17 at 18:49

Tweetable Mathematics

89 Answers89

Linked