Mathematics - 2019-12-19 (page 1 of 3)

Thorgott

00:01

Oh, I hadn't looked at it yet. Yeah, that's a much nicer way of going about it.

Simply Beautiful Art

What's the probability that you pick the right answer to this question at random?
a) 25%
b) 50%
c) 66%
d) 25%

Balarka Sen

clever

you sneaky you

Thorgott

not this again

Ultradark

00:17

What's the probability that the sun will rise tomorrow

Semiclassical

Hello, problem of philosophical induction

Ultradark

What's the probability that A survey of a group's viewing habits over the last year revealed the following information:

28% watched gymnastics
29% watched baseball
19% watched soccer
14% watched gymnastics and baseball
12% watched baseball and soccer
10% watched gymnastics and soccer
8% watched all three sports.
Calculate the percentage of the group that watched none of the three sports during the last year.

24
36
41
52
60

Leaky Nun

@BalarkaSen I private messaged you on lichess

Semiclassical

Oh hey, PIE

Simply Beautiful Art

oh hey, I finally got that gold sequences and series badge

Balarka Sen

00:24

@LeakyNun lmao what a game

Semiclassical

@LeakyNun I feel like one needs to specify the conditional in more detail. For instance, one procedure for removing the card is “they look through the deck, remove the king of spades, and shuffle”

I guess that should be directed at @ted tho

Balarka Sen

@LeakyNun I can try playing retarded like that and you can try to mate me as fast as possible

Wanna try

Actually I should sleep lmao

Leaky Nun

lmao

Balarka Sen

Did you see Duda's double queen sac against Hikaru

Leaky Nun

no

come on let's go

Balarka Sen

00:27

Fuckin bullet are u serious

Fine

Shine On You Crazy Diamond

@Ultradark I proved twin primes using elementary theory

-1

Q: Yet another elementary take on the twin prime conjecture.

Let $F(S)$ be the free commutative monoid on countably many symbols $S$. Then it's obvious that $F(S) = \{1\} \uplus S \uplus S^2 \uplus \dots$ One can take $S = $ the prime numbers in $\Bbb{N}$ in which case $F(S) = \Bbb{N}$ itself. It's also obvious that $S^i S^j$ done elementwise is equal t...

It's nothing advanced. The twin primes have to be infinite because there is no natural polynomial map (such as $f(n) = (n + 2)n$) that can leave behind completely the $k$-semiprimes.

Ultradark

why does it have 2 downvotes?

and 3 stars

interesting

it's getting downvotes fast

Shine On You Crazy Diamond

I'ma leave it up see what happens

send $1 mill to this address...

Semiclassical

Dec 7 at 1:03, by Semiclassical

Yes, the 5 people who have viewed it since you posted it 10 minutes ago haven't found anything wrong with it.

Shine On You Crazy Diamond

@Semiclassical it's up for grabs, please do find something wrong. But when you're right you're right

so..

Ultradark

00:32

usually downvotes mean the proof is discredited just sayin

Shine On You Crazy Diamond

I can prove it here if you have questions

Ultradark

I've had downvotes before

Shine On You Crazy Diamond

Let $P = $ the primes

$P^i = $ all $i$-semiprimes.

$P^{i + j} = P^i P^j$ holds without proof because obvious

where ops are all elementwise

@Ultradark does that make sense so far?

Ultradark

yeah

Semiclassical

“I’ve persuaded myself” is not equivalent to “I can persuade others”

Rithaniel

00:34

Unless of course you aren't yourself right now

Shine On You Crazy Diamond

Well, then find me a polynomial map $f : \Bbb{N} \to \Bbb{N}$ that leaves behind all $P^2$ semi-primes eventually meaning $f(n) \notin P^2$ for all $n \geq$ some $N_f \in \Bbb{N}$.

You can't, because I proved you can't.

Ultradark

How did you do it with so few words?

Semiclassical

For clarification, what is an i-semiprime?

Shine On You Crazy Diamond

I proved an even more general result that there is no polynomial map that does that for leaving behind any $P \cup P^2 \cup \dots \cup P^i$ other than $1(n) = 1$ the constant map

$i$-semiprime is a semiprime composed of $i$ primes.

Ultradark

a 1-semiprime is a semiprime right

Shine On You Crazy Diamond

00:36

$1$-semiprime = prime

Ultradark

ohh yeah

Shine On You Crazy Diamond

I guess party at my place since I'm a millionaire of the future

Ultradark

I'm looking forward to learning and seeing what feedback comes

Semiclassical

Do you mean “at most $i$ primes”?

Ultradark

but I don't understand much of it tbh

Shine On You Crazy Diamond

00:37

No I mean $i$ primes

I use the union $P \cup P^2 \cup \dots P^i$ to mean what you said at most

Like I said, all elementary. You can check that it works.

It's unlike Fermat's proof, where only one in a billion people can actually read it. This one is attainable within a day.

Semiclassical

Then P^1 P^1 would contain P^1 itself.

Shine On You Crazy Diamond

No

$P \not\ni 1$

Semiclassical

Oh, wait.

Mike Miller

@BalarkaSen Twin primes got proved up above

Shine On You Crazy Diamond

Please don't joke :)

Semiclassical

00:39

@MikeMiller kazoo

Shine On You Crazy Diamond

Yaya! we did it

lol

This is kinda funny, but my proof is super serious mon

Ultradark

okay what's the main idea behind the proof

Semiclassical

“ As an analogy, suppose your friend in Boston blindfolded you, drove you around for twenty minutes, then took the blindfold off and claimed you were now in Beijing. Yes, you do see Chinese signs and pagoda roofs, and no, you can’t immediately disprove him — but based on your knowledge of both cars and geography, isn’t it more likely you’re just in Chinatown?”

Rithaniel

What're you guys trying to prove?

Thorgott

the usual, twin primes

Rithaniel

00:41

Wait, actually? I thought Mike was just being silly

Balarka Sen

Amazing

Shine On You Crazy Diamond

The main idea is that you let $X_i = \{ $ polynomial natural maps $f : \Bbb{N} \to \Bbb{N} : $ that eventually leave behind all the $k$-semiprimes for $k \leq i \}$.

Ultradark

yeah we proved it

Shine On You Crazy Diamond

These sets form monoids and they're all equal to $\{ 1(n)\}$ the trivial monoid

Ultradark

but wait...

Shine On You Crazy Diamond

00:42

That's the proof in its entirety but I fleshed out the details for you in the post

Rithaniel

Well, you guys should work on a dissertation releasing it to the world.

Shine On You Crazy Diamond

I already released it to the world

Ultradark

oh I actually had no part in it lol

Shine On You Crazy Diamond

it's publicly avail on MSE

It's true that for any polynomial map, $f(n)$ eventually leaves behind the set (or value) $\{1\}$. I used that fact

So for all polynomial natural maps

Rithaniel

(Probably not the correct way to release a result like that)

Shine On You Crazy Diamond

00:43

$f(n) \neq 1$ eventually for all $n \geq N$ for some $N$.

This is in fact the way of the future

It's called the internet

and instant publishing

Balarka Sen

@LeakyNun opting out mane ur 2 good 5 me

Ultradark

it's kinda like the airxiv

Balarka Sen

also need to sleep and witness twin primes before

Ultradark

lol

Leaky Nun

lmao

Thorgott

00:44

there's an answer!

Ultradark

yea@!

imma check it out

Rithaniel

You should probably work with a respected professor who will recognize any potential mistakes, work it up into a proper, thorough paper, and then announce the result during a number theory conference

Then the people in the chat might take the result more seriously

Ultradark

it's hard to just work with a respected professor though tbh

Mike Miller

i encourage you not to take any amount of mathematics seriously

Balarka Sen

I am upvoting the question for pure lulz

I encourage everyone to as well

Thorgott

00:47

get rid of the concept of "seriousness" altogether

Ultradark

I voted today!

ÍgjøgnumMeg

all of mathematics is a meme

Mike Miller

:this:

ÍgjøgnumMeg

also proof by proof-verification tag is not equivalent to proof

Balarka Sen

the meme cohomology

that should be name of my blog

Shine On You Crazy Diamond

00:49

So you would agree that any polynomial natural map eventually leaves behind the value of $1$, right?

Rithaniel

I'm probably the person in this chat with the silliest sense of humor, and yet I feel strange about this notion of not taking any amount of math seriously

Thorgott

except the constant $1$ map

Rithaniel

Like my brain is saying "don't joke about that, that's srs bsns"

ÍgjøgnumMeg

Alright I amend my statement to "all of number theory is a meme"

Semiclassical

Hello Chinatown

Shine On You Crazy Diamond

00:50

@Thorgott yep, that's addressed in the question. In fact that's the only map that can leave behind any of the $i$-semiprimes for $i \leq k$.

ÍgjøgnumMeg

also I just got 3 entire bowls of cereal out of one-bowl-of-cereal's amount of milk and I'm strangley proud of this

Rithaniel

Now that is a major accomplishment, Meg

Ultradark

I parameterized a curve yesterday

ÍgjøgnumMeg

trvth

Thorgott

why are you eating cereal at 2am

ÍgjøgnumMeg

00:52

Because

Balarka Sen

askin the right questions

ÍgjøgnumMeg

I

just returned from a party

that I'm about to return to

Mike Miller

nerd

ÍgjøgnumMeg

hahaha

but the controversial twin prime shitposting interested me

Thorgott

you returned from the party just to eat some cereal?

2

ÍgjøgnumMeg

00:53

No I returned from the party to pee and then I saw that i had cereal

3

Semiclassical

Let me pose a question: How would this result work for, say f(n)=n(n+3)

Balarka Sen

I am laughing audibly

Thorgott

same

ÍgjøgnumMeg

and then I ate the cereal and decided to check mathstack

and now I'm in this mess

Shine On You Crazy Diamond

app.ziteboard.com/?code=f1bc2375-c5f6-465a-a1f1-e16947e66fc5

Rithaniel

00:54

Wait, laughing . . . out loud?

lol

Shine On You Crazy Diamond

I've opend up a whiteboard if anyone wants more proof

Ultradark

I lvoe math

ÍgjøgnumMeg

anyway I should go back, the cereal hath sated my hunger

Semiclassical

If it works for all polynomial maps, then that example should work as well

Shine On You Crazy Diamond

feck just had the biggest belly laugh

2

woulnd't it be great if my proof weren't a fraud ...

:D

Semiclassical

00:55

So, what would your result state in that case?

Shine On You Crazy Diamond

J/k

I'm 100% serious

@Semiclassical

Thorgott

hf at the party, now with the additional satisfaction from knowing twin primes is no more

Shine On You Crazy Diamond

@Semiclassical not sure what you're asking

ÍgjøgnumMeg

tbf it'd be cool but I don't have the conciousness to read it atm as I am intoxicated and I'm just taking for granted that the proof is false because the proof might be false

I'll check back in a bit, i'll cross all parts of my body that occur in pairs

Mike Miller

That seems dumb

I'm taking for granted that the proof is correct

Semiclassical

00:57

Well, you claim that the twin prime conjecture would follow by applying your argument to the case of f(n)=n(n+2)

Shine On You Crazy Diamond

So you would agree that if $X_i = \{ $ polynomial natural maps that eventually leave behind $ P^1 \cup P^2 \cup \dots \cup P^i\}$ that $X_i$ is a monoid, right?

ÍgjøgnumMeg

@MikeMiller that's probably a better attitude

Shine On You Crazy Diamond

Where $1(n) = 1$ is the multiplicative identity

it's a monoid under pointwise mul of maps

Thorgott

I'm taking for granted that the proof is either correct or fallacious

Semiclassical

So, I’m asking what result you’d obtain if you use another polynomial like f(n)=n(n+3)

Thorgott

00:58

actually delete the either*

Balarka Sen

I think I am dying

Shine On You Crazy Diamond

The result is that no such polynomial exists other than $1(n)$

Thorgott

I don't wanna take the consistency of our axiomatic system for granted

Semiclassical

So, what does that say about my f(n) as far as primes go?

Mike Miller

@Thorgott I don't take for granted that any axiom system A is correct

However, given any axiom system A, I do assume that A+Choice is just as consistent as A is

I'm not too worried about the consequences

Shine On You Crazy Diamond

01:00

Well my construction starts off by letting $S = $ the primes. This has certain properties including $S^{i+j} = S^i S^j$.

Balarka Sen

so its a graded monoid

very interesting

Thorgott

how brave

Shine On You Crazy Diamond

It's actually a topological monoid

with $S^i$ as basis

$S^i \cap S^j = \varnothing$

when $i \neq j$

Balarka Sen

nice topology

Shine On You Crazy Diamond

Yes, but multiplication is indeed continuous since if $a \in P^k$ and $f(x) = xa$ then $f^{-1}(P^{k+i}) = P^i$.

Continuity on a basis implies continuity on all open sets or unions thereof

Balarka Sen

01:05

proceed with your proof

its extremely intriguing

Semiclassical

My question can be put like this: Would you conclude, based on your result, that no matter how large $n$ is, I can always find $n$ such that $n(n+3)$ is a product of two primes?

Shine On You Crazy Diamond

I think I found a flaw

$f(n) = n^k$ leaves behind eventually $P^1 \cup \dots \cup P^{k-1}$

So maybe all monoids equal powers of $n$ or something.

I should delete

Semiclassical

what I was concerned with was the following: If $n>2$, then at least one of $n$ and $n+3$ is a multiple of 2 which is greater than 2

hmm. maybe I overlooked something there

Shine On You Crazy Diamond

I'm not going to delete but make an answer pointing out this flaw I found

Mike Miller

Yeah why not

Ultradark

01:12

I'm off to parametrize another curve

Balarka Sen

01:22

@Shine It seems joriki is really hoping the proof is correct

Shine On You Crazy Diamond

@BalarkaSen I know he's holding onto hope for me

I think though you could fix it

Since $f(n) = (n + 2)n$ is not a power map

Balarka Sen

I think we should look at how the dynamics differs if you add a linear factor to a power map

Maybe power maps are unstable, you know what I mean?

Mike Miller

@BalarkaSen There are things studied in algebra called quadratic-linear-constant algebras which behave differently when it's just quadratic, quadratic-linear, and quadratic-constant, separately

So which powers you have matters quite a lot

Balarka Sen

Yeah its when you have mixed gradings, I feel you

The unstability comes from being concentrated on a single grade

Shine On You Crazy Diamond

01:37

Does this seem correct?

Let $P$ be the natural primes and define $Q_i = \bigcup_{k=1}^i P^k$ where $P^k = \{ q_1 \cdots q_k : q_j \in P\}$. Let $M_i = \Bbb{Z} \setminus Q_i$. Each is a monoid since $1 \in M_i$ for all $i$ and if $x, y \in M_i$ then $xy \in M_{2i} \subset M_i$ since set complement is inclusion reversing.

Leaky Nun

does this mean trump is impeached? or are there sneaky terms and conditions?

ÍgjøgnumMeg

Twin primes true $\implies$ Trump impeached

6

Leaky Nun

8:24 p.m. ET, December 18, 2019
Majority of House votes to impeach President Trump
A majority of the US House of Representatives has voted to support the first article of impeachment against President Trump.

House Democrats have 216 votes, which is the number needed to impeach the President.
Voting is still happening on the House floor.

Shine On You Crazy Diamond

What year is it?

loch

i think it goes to the senate

Leaky Nun

01:39

so he’s still not impeached yet

Shine On You Crazy Diamond

It's just a show for entertainment

loch

@LeakyNun washingtonpost.com/politics/2019/12/18/…

Balarka Sen

the real donald :(

Ultradark

it goes to the senate

and the senate will not impeach him

Ultradark

02:00

@shi yeah

Max

@Thorgott This is exactly how I got $F(v) = 2P([1-v^2 /2 ]Z_1^2 - Z_2^2) \leq 0)$, which is where I am stuck.

@BalarkaSen The Rayleigh distribution hasn't been introduced yet in this book, so there must be another solution.

I think I'm going to post a normal question on MSE. Thank you both for your help.

Akiva Weinberger

05:07

If you close one eye the world becomes 2D

How many D is it if you close both your eyes

Shootforthemoon

06:28

@AkivaWeinberger Lol

0, cuz u don't see the world anymore

Rithaniel

06:41

Disclaimer: The dimension of a space is not tied to the perceptions of an entity in that space

Akiva Weinberger

Something so bright, if you look at it once you go blind in one eye, if you look at it twice you go blind in the second eye, what if you look at it three times

(I need to sleep)

@Rithaniel Although, the "perception is reality" idea has been the basis of many trippy video games

Rithaniel

Superliminal comes to mind

Akiva Weinberger

Oh is that the perspective one

where whenever you drop an object it's placed as far away from you as possible (at the corresponding size to make the apparent size make sense)

Rithaniel

Yeah, so you can pick up some tiny chess piece and make it larger than the Eiffel Tower by just placing it somewhere

Akiva Weinberger

Wait

That was released last month?!

Rithaniel

06:47

You can also make yourself larger or smaller, too

Akiva Weinberger

I saw a bunch of early Works-In-Progress I didn't realize it was done

Rithaniel

Oh yeah, I saw some videos popping up on YT. Didn't watch any, but that put it back on my radar

Akiva Weinberger

Just watched a trailer, much more polished than I previously had seen

Gonna check it out when I get the chance

Rithaniel

Was thinking about getting it myself

Akiva Weinberger

I still need to finish Antichamber as well

Rithaniel

06:49

Now I have a game to look up

I've been playing a lot of Into the Breach

Which is like chess, but where you only have three pieces, but they have guns

Akiva Weinberger

Is that not how you're supposed to play chess?

You know what's a mindfuck that I need to get? Baba Is You

A block-pushing puzzle game where you get to mess with the rules of the game

because the rules are made of blocks that you can push around

Rithaniel

Oh yeah, I still need to beat Baba Is You

I've managed to get one "major-ish" secret

Akiva Weinberger

I've had the ending spoiled unfortunately

(unless there's a bigger secret ending I'm unaware of)

Rithaniel

But I stopped shortly after that, to take up less-headache-inducing activities

Akiva Weinberger

An idea for a game would be one where messing with maps messes with the territory

Rithaniel

06:55

I don't know how deep the well is with Baba Is You

So, if you've been spoilered, you might know more about the game than me

I'd like to make games that are straight-forward, but use strange things not found in games

Akiva Weinberger

Puzzle games, or other?

Leaky Nun

@AkivaWeinberger yeah I bought it and played with it for a while but it was a long time ago

Rithaniel

Like, a game which works like TIS-100, but instead of being in Euclidean space, the computer is on a hyperbolic plane

Akiva Weinberger

Dunno what TIS-100 is

but did I show you HyperRogue

Roguelike game on a hyperbolic plane

Rithaniel

Yeah, I saw HyperRogue

Akiva Weinberger

06:59

OK I do legit need to sleep now

Rithaniel

Though, TIS-100 is this: store.steampowered.com/app/370360/TIS100

A game about assembly programming

Also, sleep well

Leaky Nun

heh I can't get Liminal to work

it requires too much memory

Rithaniel

Hmmm, should I go after Antichamber or Superliminal?

Leaky Nun

go play chess lol

ÍgjøgnumMeg

Go do mathematics*

Rithaniel

07:03

It's been a while since I've played legit chess

Leaky Nun

@ÍgjøgnumMeg do you play chess?

@Rithaniel wanna play with me?

ÍgjøgnumMeg

Nah I suck at it

Rithaniel

I'm probably really out of practice, but sure

Leaky Nun

@Rithaniel how many minutes?

Rithaniel

Is there "no time limit?" Again, super out of practice

Shootforthemoon

07:05

Wanna play with me too?

Leaky Nun

lichess.org/ZtQRbPS2 @Rithaniel

Rithaniel

Though, if not, then 5 minutes?

Shootforthemoon

Not that good though

Leaky Nun

@Shootforthemoon unfortunately there's no 3-player chess

Shootforthemoon

Ahahahhaha

I mean in another match

Leaky Nun

07:10

@Shootforthemoon oh hey I can play two people together

Rithaniel

07:59

Well, those were two unfortunate games

According to the computer, in that second game I was at a disadvantage the instant I moved my queen forward

Leaky Nun

08:17

@Rithaniel lemme see...

@Rithaniel well it fluctuated before you stepped into the queen trap

Rithaniel

Yeah, but the instant that I did that, I never rose above -5 again, and I was only that high because of a mistake you made, not because of anything I did

Leaky Nun

08:33

@AlessandroCodenotti what bizarre games

Alessandro Codenotti

Too fast for me

Leaky Nun

@AlessandroCodenotti then why is your rating 18XX

provisional rating

maybe you should get a real rating

just play 12 games

Alessandro Codenotti

@LeakyNun I played a couple of 2+0 and won

One of them by flagging iirc lol

Leaky Nun

go get a real rating

it won't take you a lot of time

Alessandro Codenotti

I don't like those fast formats

My rating is probably like 1400 if I have to guess based on how much lower rated I am in 5+3 compared to 10+0

Adam

12:33

does anyone else have a cat that think they are people? like he has just for no reason started refusing to use the cat door and demanding the back door be opened for him and before he goes thru it he stops turns his head and glares at me for a few seconds

towc

12:54

anyone knows what "algorithmic heave-position" might refer to?

the context seems to be related to ASTs

user8469759

13:31

1

Q: Equivalence between two topologies generated by seminorms - Follow up

This is a follow up question to this one, I didn't feel like to have extended comments so I prefer to ask a separate question. There's this comment specifically: Suppose we have two families of seminorms $p_\alpha$ and $q_\beta$. If for every $\alpha$ there is $\beta$ and a constant $C_{\alp...

general-topology topological-vector-spaces

anakhro

15:18

woof

towc

aren't you?

Balarka Sen

@anakhro Looked at E-M?

Thorgott

@Max The answer that was given on main is what I had in mind with my suggestion. Also, you don't need to refer to the Rayleigh distribution explicitly to make Balarkas suggestion work. Just writing it as the square root of an Exp(1/2)-distributed variable suffices as it cancels out.

Balarka Sen

15:34

You don't even need any information about the radial random variable, just that the angle is uniformly distributed

Which is witness of the fact that standard two-variable normal is radially symmetric

This technique is used to simulate normal random variable, it's called the Box-Muller simulation

Thorgott

Coincidentally, I will have to explain Box-Muller to some students tomorrow

Balarka Sen

Hah nice

adesh mishra

15:51

I have a question on ellipse. Can you help me?

Show that the perpendiculars from the center upon all chords, which join the ends of perpendicular diameters, are of constant length.

Ajay Mishra

@adeshmishra Have you figured it out?

adesh mishra

Yes , let me give you my solution

Ajay Mishra

No, I don't need that, I was just asking in case you didn't did that yet.

I could've helped.

Transcript for

$Mathematics$ Mathematics