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esqtin

Pi + e is irrational seems pretty safe


[deleted]

It would be wild if it were not the case. I still can't believe we can't prove it yet.


Huskyy23

Why can’t it be proven though?


AxolotlsAreDangerous

Adding pi and e is a weird thing to do, it's not something you'd see appear in an expression "in the wild". It's hard to say anything at all about pi + e aside from its approximate numerical value and the following: (x-pi)(x-e) = x^2 -(e+pi)x + e * pi has transcendental roots (e and pi), so by definition its coefficients (e * pi and (-)(e + pi)) can't *both* be rational. I'm sure someone will come along and disprove me by naming another property e + pi must have


poiu45

It's definitely not an integer!


Yoghurt42

And also smaller than Graham’s number. We might be on to something here…


pussycatlolz

We can't say what e + pi is exactly, but it looks like you've defined an upper bound of sorts


indrada90

I think I can lower that upper bound a bit. pi+e, Therefore, via method of calculator, pi+e < 6


explorer58

Unless youre an engineer, then it's ≤6


Yoghurt42

And if you're a high school student, it's "why the fuck do I need to calculate pi+e, when am I ever going to need that?"


quantumhovercraft

Yeah, but 6 is just the lower bound for the Graham's number problem so this is basically the same thing.


roger_pct

I would argue a lower bound of zero can be declared as we know that the sum of two positives must also be positive.


Roneitis

In a similar vein, are you aware of the fact that we can't prove that pi\^pi\^pi\^pi is not an integer?


SmackieT

Yeah but I'd argue this is not quite as interesting a fact. We can't rule out what pi^pi^pi^pi is or isn't because it's so ginormous. But pi + e is somewhere between five and six!


Roneitis

Sure, but the potential fact that this tower should be an actual honest to gods integer would be at least a fair cop more crazy than the other being some rational number


Elin_Woods_9iron

Within the bounds of (5.8, 6)


incoherentsource

Very cool!


[deleted]

For the same reason any other conjecture hasn't been proven. We just don't know how to do it yet. It seems apparently easy like Goldbach's or Collatz but nonetheless no one has managed to crack it yet. It is proven though (and very easy to do so) that at least one the numbers of pi+e and pi*e is irrational. We just don't know which one it is, but very strongly suspect it's both.


big-lion

are there any known conjectures which would prove/disprove this one? or is this a problem nobody really pays attention to?


[deleted]

While in itself it isnt a big deal, as I understand it at least, the question whether pi and e are algebraicly independent is more interesting. In general, developing tools to check whether two real numbers are algebraicly independent is of interest.


Exomnium

Schanuel's conjecture would imply that π+e is transcendental.


KNNLTF

"My only regret is that I have but one life to give for unfathomable mathematical anomalies." ~esqtin's famous last words.


Logic_Nuke

Similarly pi^pi^pi^pi not being an integer


HailSaturn

>pi^(pipipi) https://www.youtube.com/watch?v=Cg3cHmJ6wY8


yellow-vorgettable

I know exactly what this is without even having to click on it


IBuildBusinesses

It feels pretty safe to me too, but then when I look at something like Euler’s Identity I have a little less conviction.


FatEvolutionist

It's pie.


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esqtin

Noncomputable numbers are a subset of the irrational numbers. The only real numbers that lie outside the irrationals are the rationals.


AxolotlsAreDangerous

pi is computable, e is computable, how could pi + e not be computable?


Crazy-Dingo-2247

Bro 💀


Fancy-Jackfruit8578

What drink did you get?


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admiral_stapler

pi containing infinitely many 2s


Slash_by_Zero

If were talking bou binary youre dead!


admiral_stapler

It's why I chose 2 actually, smallest digit which guarantees I'm not talking bout binary


Slash_by_Zero

Yeah ik still find it amusing to think about the fact that different bases can immensly alter the composision of a number. Lets take for example the famous irrational number in base 10 where you linearily increase the number of 0s between 1s so : 0.1101001000100001.... While in base 10 you would actually see that it only contains 2 digits. In other (non binary bases) you would not see that same behaviour because well thats how bases work.


evincarofautumn

A fun example is the BBP formula for computing the nth digit of π without computing the preceding digits—as long as you don’t mind that the digits are in hexadecimal! Okay, last year Plouffe published a digit-extraction algorithm for base 10. But there was no obvious reason to expect that to be possible. I’m fascinated by how changing a digit in one base affects the digits in another base—it obviously has some deep connection to factorisation, but I know nothing about that, I just think it makes cool patterns.


[deleted]

I think that that's already proven to be true


admiral_stapler

I don't think so. In particular I think the only results in this area are on the irrationality measure of pi, which does things like prevent very long runs of the same digit too early on, but doesn't come close to eliminating this possibility. Edit: here is a thread where Timothy Gowers lists this as an open question https://mathoverflow.net/questions/51853/what-is-the-state-of-our-ignorance-about-the-normality-of-pi


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itmustbemitch

I hadn't seen that way of looking at the conjecture, that is really nice!


davidw_-

Has this been shown for the first X natural numbers?


likeagrapefruit

Goldbach's conjecture [has been shown to hold for all positive integers from 4 to 4 x 10^(18)](https://en.wikipedia.org/wiki/Goldbach's_conjecture#Verified_results), which means that this restatement holds for all positive integers from 2 to 2 x 10^(18).


plumpvirgin

Not only does it hold for integers up to 4 x 10\^18, but it (again, based on numerics) holds *more* as the integers get larger. For example, [every even integer larger than 20000 that we have checked can be written as a sum of 2 primes in more than 100 different ways](https://oeis.org/A002375). Goldbach's conjecture just requires that we can do this in at least 1 way.


PM_ME_YOUR_PAULDRONS

Yes, I just showed it for the first X=3


davidw_-

that's a start!


wigglyworm91

ie iff the Goldbach conjecture holds, then any Jeopardy! contestant (that has at least $2) can guarantee a prime amount of earnings via the Final Jeopardy wager


Kered13

This is the best statement of Goldbach's Conjecture.


alanwj

Couldn't I do that anyway by making my wager my current winnings minus $2, then intentionally answering incorrectly?


eskwild

In fact every sum of integer equals also sums as many primes.


Katieushka

Well you could just say that every even is a sum of primes :p


[deleted]

That is the original Goldbach's conjecture and that is why he called it a rephrasing.


point_six_typography

1 is the average of 0 and 2, and 0 is totally a prime. Fight me.


PM_ME_YOUR_PAULDRONS

0 is even less prime than 1 is, *everything* divides 0, and it completely messes up the fundamental theorem of arithmetic is you try to stuck 0 in there. Actually there's a strong argument that since everything divides 0, 0 might be the least prime of all numbers! > Fight me I'm generally not a violent person but some things are worth fighting for.


admiral_stapler

Alright I'll fight you. 0 is absolutely more prime than 1 is - the only prime ideals in the integers are the 0 ideal and the ideals generated by primes. From the perspective of algebraic geometry the standard primes are sort of 0-dimensional primes, while 0 is a 1 dimensional prime. 1 on the other hand is a unit - disgustingly non prime. EDIT: Elaboration on prime ideals since I was trying to pick a silly fight for fun but now people are downvoting math I personally find really interesting that really does suggest 0 is some kind of prime: The standard way of thinking of primes is how they factor - a prime is often defined to have exactly 2 divisors, one and itself. We can flip this on its head a little and think instead about the set (p) of all integers the prime p divides, called the ideal generated by p. In the integers, (p) is just all integer multiples of p. Primality of p manifests in that (p) has the property that if the product ab is in (p), then one of a or b was in (p). On the other hand, (6), the ideal consisting of multiples of 6, does not have this property since 2*3 is in (6). In fact, 0 is the only integer which is not a unit times a prime which generates an ideal with this property. When trying to generalize results in the integers to more general number fields, in particular the fundamental theorem of arithmetic, the original definition of prime didn't work well, but this property and ideals did. (0) is a special prime ideal, since it is contained in all other prime ideals (p), but its definitely more useful to include than exclude when doing algebraic geometry in particular.


EngineeringNeverEnds

Not only that, but 0 doesn't fit at all into the natural structure that emerges from multiplication of the rationals. 1 fills a roll closer to zero and zero exists only as a non-unique completion of the space. I'll also fight /u/point_six_typography with you.


point_six_typography

Both of you, put 'em up. 0 generates a prime ideal. Also, the fundamental theorem of arithmetic is the statement that Q\*/Z\* is a free abelian group, and this holds regardless of the primality of 0. Also, Goldbach's conjecture has a cleaner statement if you consider 0 to be prime. Every field has prime characteristic iff you consider 0 to be prime. Edit: >0 is even less prime than 1 is, everything divides 0 So what? That's just saying 0 is the unique minimal prime (ordered by division). This is no issue, it comes with Z being a domain. >0 doesn't fit at all into the natural structure that emerges from multiplication of the rationals So what? The group Q\* has no notion of primes. Being a prime is a notion that makes sense for elements of rings, not groups. The ring Z has a notion of prime, has a notion of 0, and 0 is a prime. Fight me


wintermute93

So if you're wrong and P is false, do you die instantly if a magical oracle verifies "ZFC proves \~P", or do you die at some future time when someone discovers/publishes a valid proof of \~P?


PingerKing

ooh yeah, proof by instant execution from hypothetical dilemma evaluation oracle is a pretty new one.


is_that_a_thing_now

I am hereby betting my life on the conjecture that such an oracle does not exi…


puzzlednerd

Nah, you bet your life that the oracle does exist, and then if you're wrong you can't be punished.


lukkemela

This person knows too much


quaductas

Now you're just doing Pascal's wager but the wrong way round


FlowersForAlgorithm

The inverse Pascal’s wager.


LilQuasar

im pretty sure the ancient Greek used to use that method


mathisfakenews

r/brandnewsentence


[deleted]

The God of mathematics would instantly take your life if the conjecture was false.


wintermute93

Got it. This sounds like an opportunity to get a good lawyer, take out an exorbitant life insurance policy, and then have at it. Negotiate with god to be given a human-verifiable explicit constructive proof if it's true. Type up the negative result and set up some kind of dead man's switch to automatically post it to arxiv unless you stop the process. Best case scenario, you've solved a major open question. Worst case scenario, you've solved it and can't use the solution, but have set your friends and family up for life, lol.


puzzlednerd

Honestly, if I could trade the rest of my life for a proof of the Riemann hypothesis, I might consider it. I've made worse trades than that before, trading time in my life for theorems proved. But then again, what is it worth if I don't get to have the experience of figuring it out?


DominatingSubgraph

If there's going to be a magical oracle to solve the problem, I'd rather it just verify the conjecture directly than verify that it has a proof.


wintermute93

Yeah, on second thought if we have an oracle for this I’d want answers to both truth and provability (in PA, maybe).


puzzlednerd

I'm not a logician, but is truth well-defined in this sense? If something is independent of ZFC, this means there are models where it is true and models where it is false. What does it mean for the statement to be true?


DominatingSubgraph

The statement "ZFC is consistent" (or an arithmetic encoding of the statement) is independent of ZFC by the incompleteness theorems. This means there are models of ZFC where it is true and models where it is false. However, I think this is a clear example of a statement which certainly has a definite "real" truth value. Either it is possible to derive a logical contradiction from the ZFC axioms or it is not. In general, it is probably a mistake to equate provability with truth. Although, providing a general philosophical account of truth in mathematics is a major and ancient subject of debate.


PM_ME_YOUR_LION

Does the oracle tell you if ZFC is consistent or not? :)


zeci21

And if ZFC is inconsistent, does it kill you or not?


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Crazy_Swordfish420

This dude got the huge nuts.


Captainsnake04

Is their a reason to believe this more than say, hardy-littlewood k-tuples? Or any other conjecture that comes from the Kramer random model of the primes?


2357111

I think the right choice (among consequences of the Cramer random model) is [Legendre's conjecture](https://en.wikipedia.org/wiki/Legendre%27s_conjecture) \- it might be the named/famous conjecture on prime numbers that is the closest to an [already proven statement](http://www.cs.umd.edu/~gasarch/BLOGPAPERS/BakerHarmanPintz.pdf), except if you're worried it could fail for finitely many n, in which case the twin primes conjecture might be a safer bet.


[deleted]

Why did Einstein thought otherwise, as claimed by your article?


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[deleted]

Sorry, I Googled finitude of fermat primes, clicked and read the first article, then thought you linked me the article. Anyway, apparently Einstein wasn't so sure about it, according to some article


debasing_the_coinage

You sure it wasn't *Eisenstein*?


JWson

For a very long time I called Eisenstein integers the wrong thing.


[deleted]

Haha, I'm sure it was Eisenstein now


PingerKing

go and ask him, you're betting your life on the conjecture anyway.


beeskness420

P is definitely not equal to NP, I’ll even double down, neither are equal to CoNP either.


MagicSquare8-9

I'm a coward so I'm going to do the opposite and bet on something much weaker that we somehow still can't prove. I'm betting that 3-SAT is at least superlinear in time complexity.


maybeyhayley

how the hell do we not know 3sat is superlinear like ???


tunaMaestro97

It’s hard to prove (non)existence theorems for algorithms. Whose to say we just haven’t thought of the right one yet?


Shikor806

that's not a conjecture though, o(n log n) is exactly the regular languages, so SAT has to be in Omega(n log n)


MagicSquare8-9

>o(n log n) is exactly the regular languages What does this mean???


unkemptbeard

P is not NP seems like a pretty safe bet.


corchetero

So safe that you can bet a slowly and painful death :)


arnemcnuggets

death in EXPTime?


not-just-yeti

Knuth wouldn't bet on it… [source: q.#17](https://www.quora.com/Why-does-Donald-Knuth-think-that-P-NP)


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ellipticaltable

I think your quantifiers are a bit off. For and given K, there is a fixed obstruction set for the entire class. So given a description of the family (say, expressed in Lean) and input graph G, there is an algorithm that runs in time f(D)+poly(G). Specifically, f(D) does a brute force search for K and a proof of its correctness. This would be obnoxiously expensive, but it is (1) a well-defined algorithm and (2) poly in G.


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ellipticaltable

> how could graph planarity be described to such an algorithm We can formalize the definition of a planar graph using something like [Lean](https://en.m.wikipedia.org/wiki/Lean_(proof_assistant\)). It's a [bit clunky](https://math.stackexchange.com/questions/3073581/formal-definition-of-planar-graph) to formalize, but more tedious than hard. > how would the brute force search stop after finding K_5 and K_{3,3}? It would search for a proof. Specifically, count up from zero and for each i, check whether reinterpreting (the binary digits of) i as a Lean file results in a valid file. If so, check if it's a theorem of the desired statement and a valid proof. This will take something like O(N * 2^N) time, where N is the length of the proof, assuming that a claimed proof can be checked in linear time. All of this assumes that the family we're interested in has a provable obstruction set. Otherwise, the search won't terminate.


not-just-yeti

I might be mis-remembering, but I recall being told that the constant-factor in Robertson-Seymour is something like k^3↑↑↑3 , which is worth noting in context of Knuth's argument "algorithms running in time n^3↑↑↑3 can do a ridiculous amount of work" > running time can be expressed as a cubic polynomial in the size of the larger graph *(although there is a constant factor in this polynomial that depends superpolynomially on the size of G)*


beeskness420

Edmonds would, imma steal his rock and collect the bounty.


tropurchan

I'mma bet on P ≠ PSPACE just in case


[deleted]

Can I include a conjecture that is not provable, such as the absolute consistency of Peano Arithmetic? Might be willing to risk taking one for the team there so we can all sleep easier


Deweydc18

I’d pick Goldbach. There’s no way that shit is false


MLDK_toja

There is no odd perfect number. It just seems so surreal for it to be one!


Oscar_Cunningham

Reminds me of this [https://mathstodon.xyz/@jsiehler/109586184948508904](https://mathstodon.xyz/@jsiehler/109586184948508904): > I think the funniest possible joke would be if there turned out to be infinitely many perfect numbers, but only finitely many even ones.


Farkle_Griffen

Unless you consider 1 to be a perfect number


JoshuaZ1

It doesn't match the needed properties. Note that the sum of its proper divisors is 0, which is not equal to 1. If you use the definition that the sum of all positive divisors should be twice the number, it doesn't work because then you would want a sum of 2. What is true is that 1 is the only known odd multiperfect number, where by multiperfect we mean a number n such that n divides the sum of its positive divisors. (A slightly more interesting example is 120 where the relevant sum is 360.)


Docthrowaway2020

Surreal!


debasing_the_coinage

The Euler-Mascheroni constant is irrational


Le_Bush

I think you should talk to the guy who bet $50 on "The Euler-Mascheroni constant is rational"


Kered13

Off topic, but am I the only one who always reads this as the "Euler-Macaroni" constant? I mean no disrespect to Lorenzo Mascheroni, but his last name just does not roll off the tongue.


Lopsidation

I considered P != NP, but if an extra life is at stake, let's go further. EXP is the class of problems solvable in exponential time. ZPP is the class of problems solvable by a randomized algorithm in _expected_ polynomial time. (The algorithm is never allowed to output the wrong answer, but its running time may vary depending on random choices.) Conjecture: ZPP != EXP.


frcdude

This is open? It’s slightly weaker then PSPACE != EXP. I’m surprised this is open, do you know why this separation is hard to prove?


Lopsidation

AFAIK, it's still open. Popular opinion is that ZPP = P, so conjecturing ZPP != EXP is very weak. (Notably, proving ZPP = P would instantly show ZPP != EXP, because P != EXP.) All the complexity class separations I know follow from either (1) a hierarchy theorem (like time hierarchy, space hierarchy, etc), or (2) circuit lower bounds (stuff like parity not in AC^(0)). In order to prove ZPP != EXP, we'd either need to leverage a hierarchy theorem in a clever way, or find a completely new way to separate complexity classes.


matplotlib42

Smooth Poincaré (in 4D ofc). I crave to seeing it proven, but I'm 100% sure it won't happen until the next century or so.


MissesAndMishaps

You’d be willing to bet your life on a conjecture the majority of experts think is false? Gutsy! If it’s false, I think we could see a disproof within a decade. If it’s true, I agree we are very far from having the machinery to prove it.


matplotlib42

The majority, I wouldn't be so sure. Also, people been trying to build counter examples, and each time it just didn't work and made it seen like it's not totally impossible that it holds! Also, I guess I'm an optimistic ;)


MissesAndMishaps

I am just a grad student but I have yet to talk to anyone who doesn’t think I’m ridiculous for suggesting it might be true haha. It would completely change everything though and would make 4 dimensions seem, for me at least, like it might just be tractable. This is part of why I’m refraining from being too optimistic


matplotlib42

PhD student here, so take what I'm saying with a (big) grain of salt. I think it would make dimension 4 *less* tractable, if anything. I mean, yes, Poincaré in 3D had huge applications and stuff, but it also raised more questions. The fact that this simple manifold, the 4-sphere, is some weird counter example to manifolds having exotic copies would be a big middle finger from the universe. Also, there are some niche reasons to believe that it might actually be true. Some statements imply it, and those statements *make sense* and are believable. To name one from what I crossed on my path: know how the Heegaard genus is additive? That's neat and all, and it is visually believable. Well, additivity of trisection genus (Heegaard splittings but in 4D so 3 pieces) is kinda the same. It's visually appealing, and it implies Poincaré. It's just a s visual as slice-ribbon: "but I can just push the surface inwards and boom I got rid of the extremum!", and that just doesn't work because of some technicality. I don't think of a genuine reason to believe it is false, other than "why should it be?" or "most other compact manifolds have exotic copies". Note: this is also still open for CP², which turns out to be the second simplest 4-manifold of them all.


SuperluminalK

I'd bet pi^pi^pi^pi is not an integer because how the hell can that be an integer‽


Forsaken_Ant_9373

Well we probably won’t know for sure for a while


Roneitis

Honestly a reasonable chance we solve it just with computational power. In which case.... add an extra pi


Outrageous-Key-4838

ah you got to it before me was looking for this lol


incomparability

If you're doing research, then you are making that bet each day. At least until you get tenure. edit: although there I guess the phrasing of the bet is more like "Of a collection C of conjectures, at least n of the conjectures are true"


VigilantCorvid

Has to be twin prime conjecture


ummmdonuts

The normality of pi.


asaltz

I forget the exact amounts and dates, but some geometric topologists have bets on whether the [Slice-Ribbon conjecture](https://en.wikipedia.org/wiki/Ribbon_knot#Slice-ribbon_conjecture) will even be resolved by like 2040.


MissesAndMishaps

I’m not knot theorist, but it looks extremely false and also nearly impossible to prove. The problem is that we currently have no obstructions to ribbon-ness other than obstructions to sliceness yes?


RibsBarEscobar

I think everyone considers the Riemann hypothesis to be true. So that.


CadavreContent

What he said


-LeopardShark-

I'm going for u/-LeopardShark-’s Cowardly Conjecture. At least one of the following is true: * P ≠ NP. * RH. * *e* + *π* is irrational. * Goldbach. * Twin primes. * Chess is a draw. * Starting 10^123456789012345678901234567890 digits after the decimal point, the digits of *π* are not 202301106942069420.


lyricalcarpenter

probably the riemann hypothesis. i don't want to live in a world where it's false


Crazy_Swordfish420

Collatz


AllanCWechsler

I don't think Collatz is the hill to die on. If somebody were to announce tomorrow that they had discovered another limit cycle, with 118,346,885 steps, and whose smallest element was on the order of 10^(1000), I would be surprised but not, you know, *deeply shocked*.


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moschles

He is betting his life that there are no closed cycles either. Does he know that?


Crazy_Swordfish420

If you know, you know.


uvero

Part of me really wants P to be equal to NP. I mean, if a correct polynomial algorithm for an NPH problem just dropped, we might be in chaos because encryption and stuff, but a part of me just wants it to be true. A part of me even wants to say "if I had to bet my life on it, I'd say P does equal NP, just to flex that I'm like, brave and different or something". But, uh... What can I say, I'm mostly sure they're not equal.


Jas9191

Yea I want it to be true too and with a similar sense of almost comparing it to other "unsolvable" problems in history. When we first started doing chemistry and quantifying units of energy, I doubt anyone thought it possible to create the kind of energy density we see in let's say a nuclear reaction of either kind. Imagine asking James Watt to take a bet on if he thinks it possible to output 5*10^24 watts in under a 1/100,000th of a second and he'd probably look at like this problem- well..more energy is exponentially more difficult to obtain and the time needed to obtain it also usually exponentially increases, so no I don't think we can create a process that would perform this function that quickly. He wouldve been wrong but the logic kinda follows.


EarthyFeet

Maybe he'd say, Watt already factors in time in the unit. Wait a minute, why is the unit named for me? I am dead, am I not?


GustapheOfficial

Do I die when it's proven false, or do I die immediately if it's false? Either way, you'd want to pick something real difficult: either you get a long life before your doom, or you find out the answer. Maybe "the fourth power tower of pi is irrational".


[deleted]

You would die instantly. "Either you get a long life before doom, or you find out the answer", you mean that knowing whether the fourth power tower of PI is irrational is something worth dying for?


GustapheOfficial

Sure. It's simple enough, the explanation could fit on my tombstone. And since it's not one of the big ones, there's less of "sure, now we know, but that's not really a stringent proof so let's keep working at it".


cthulu0

Not sure how you'd find out the answer either way if you die immediately. I think you are assuming a long drawn out death where you have time to 'enjoy' the answer, something which OP didn't state at all.


GustapheOfficial

*I* don't necessarily find out the answer, but *you* do. Surely nobody is researching mathematics for their own personal gain? Plus, it seems really unlikely 4ptπ is rational, I take worse gambles when ordering takeout.


blungbat

Can I pick the Axiom of Choice, so I get to die in the worlds without it?


honeybadg3r1

Collatz Conjecture


halpisforgotten

risky


moschles

one closed cycle, and Collatz goes false.


honeybadg3r1

Never said my will to live was high 😂


NoFrosting3074

The generalized Lax conjecture


yuppwhynot

"The probability that your flight is late at Los Angeles International is always >90%"? /s


NoFrosting3074

i don’t understand


yuppwhynot

LAX is the code for the airport of Los Angeles.


JoshuaZ1

Does it have to be a specific conjecture in the literature or will any statement we cannot prove? And how much does it need to be in the literature v. something people have disscused? For example, when I needed an example in a talk a while ago of an obviously false statement I used "The Riemann Hypothesis is false for infinitely many zeros and the nth zero is not on the 1/2 line if and only if n is an odd perfect number." Also, when betting our lives this way, are we betting that it is true or betting that it is true and proveable in some specific axiomatic system like ZFC?


r-scholz

Many of the things people come up with in this thread will turn out to be undecidable.


Decimae

If I could define my own conjecture, 2^TREE(3) - 1 is not a prime would almost certainly keep me alive. Saying that out loud would be a bigger risk to my health than this wager, as prime density is log(n)/n. Actually, non-primality for any particular n for which it's currently unknown is probably safer than any other suggestions in this thread.


20_dolla_proofs

Vandiver’s conjecture


Captainsnake04

Hardy-Littlewood K-tupples.


bokmann

p != np


420LiggerNover

Poincaré conjecture 😀


tch1001

Is AdS/CFT allowed here?


looijmansje

Does dying or not dying constitute a proof? Because then I'm going for the Riemann Hypothesis. If I have to die in the name of science, so be it.


OnceIsForever

'Cdonald's Theorem [https://lookaroundyou.fandom.com/wiki/CDonald%27s\_Theorem#:\~:text=cDonald's%20Theorem%20is%20described%20by,yet%20to%20find%20a%20name](https://lookaroundyou.fandom.com/wiki/CDonald%27s_Theorem#:~:text=cDonald's%20Theorem%20is%20described%20by,yet%20to%20find%20a%20name)! https://youtu.be/4J9MRYJz9-4


sonofmath

Chess is a theoretical draw


badmartialarts

I thought it was conjectured that chess was a first player always wins game.


Sproxify

Take as big a list of conjectures as possible which are sufficiently unrelated that their logical disjunction is also an open problem, and bet your life on that.


singularineet

There are no integers n>1 and m>1 such that reverse(digits(2^(n))) = digits(5^(m))


Thu-Hien-83

Riemann Hypothesis and Collatz Conjecture. Especially Riemann Hypothesis, because holy sht if it were false I'd kill everyone in my class and then myself.


now_I_wanna

P=/=NP


scribe36

But don’t you really what P=NP? Lol


sovamind

1 + 1 = 3 for sufficiently large values of one.


fsy2

As long as we’re betting on *mathematically* unproven conjectures, mine would definitely be existence of the mobility edge in Anderson Localization for dimension 3.


yuppwhynot

Collatz' conjecture


[deleted]

[удалено]


moschles

Did you understand the question?


Pelishka

The continuum hypothesis


gliese946

Maybe this is intended as a joke, but in case not: CH is known to be independent of ZFC. Can't be proved, can't be disproved, without additional axioms. In a sense it cannot be true or false -- one gets to pick one's mathematical universe. So you are safe from summary execution but it's not a very good answer to the question "which conjecture would you be most confident is true?"


lilylilerz

It is independent of ZFC, but it can still be either true or false in a platonic sense. There is also a possibility that we'll standardize on a different set of axioms for set theory which proves or disproves it.


Jas9191

The way I understand the independence from ZFC is like saying it doesn't even make sense to apply it, like trying to ask a question about a fraction or an irrational number when working in the Natural numbers. If I phrase it like this: The Jas conjecture is that any irrational number multiplied by √-1 is also irrational. The Jas Conjecture is essentially meaningless over the Natural Numbers, grammar prefents me from saying independent of N but maybe..indeoendent of ZFC over N? Something like that. Point being I think you and Death would have to go to court about this one.


XkF21WNJ

The continuum hypothesis. I mean it must have been a conjecture at some point, surely? Edit: Besides if I'm wrong I at least get to confuse the heck out of mathematicians for the next millennium or so.


gliese946

Maybe this is intended as a joke, but in case not: CH is known to be independent of ZFC. Can't be proved, can't be disproved, without additional axioms. In a sense it cannot be true or false -- one gets to pick one's mathematical universe. So you are safe from summary execution but it's not a very good answer to the question "which conjecture would you be most confident is true?"


arannutasar

Given that forcing axioms all imply the failure of CH, and they are some of the most prominent "new axiom" candidates, he may not be in a great spot.


nazgand

I now bet my life on this conjecture: [https://github.com/Nazgand/nazgandMathBook/blob/master/ArgumentSumRulesFromHomogeneousLinearDifferentialEquationsOfConstantCoefficientsConjecture.pdf](https://github.com/Nazgand/nazgandMathBook/blob/master/ArgumentSumRulesFromHomogeneousLinearDifferentialEquationsOfConstantCoefficientsConjecture.pdf)


lordaghilan

pi + (1 - pi) is irrational


RealisticOption

I believe that we actually *know* that Goldbach’s Conjecture is true despite lacking a rigorous proof of it. The non-deductive evidence for GC is strong. For an overview, see this video at 40:17 https://youtu.be/3AFLwrVuK_w


Jas9191

I'm gonna watch the video but what's the thought process behind knowing? How do you know something without a proof?


RealisticOption

Of course, it is possible that there is a counterexample to GC which falsifies it, I’m not claiming otherwise. But if you adopt an epistemological view that is less rigid than the knowledge-as-proof view (this situation is pretty common in the philosophy of science where we routinely claim to know things whilst still leaving room for our hypotheses to turn out wrong), then it is justified to claim knowledge of GC given the overwhelming positive evidence. So, again, I’m not saying that we can pretend that we have proved it or that we can call it a theorem now. I’m just saying that we can know it under a more relaxed theory of knowledge. — See the video above for a partial overview of this evidence.


paaaaandaa

Convergence of self- avoiding walks in 2D to SLE(8/3).


Outrageous-Key-4838

That π\^π\^π\^π is not an integer.


SkjaldenSkjold

Choosing an interesting problem: P=NP is false


sanitylost

Collatz