Mathematics - 2017-08-16 (page 1 of 4)

Karl Kronenfeld

00:08

I seem to remember it being necessary to represent sets of integers as well.

Jasper Loy

Holy shit, it's Karl again!

Karl Kronenfeld

Hey

Jasper Loy

Hello @KarlKronenfeld long time no see.

Karl Kronenfeld

@JasperLoy Yeah, I thought I would stop by. :)

Tanner Swett

My intention is to exclude statements which quantify over sets of integers, so maybe "arithmetic statement" isn't the correct term.

Karl Kronenfeld

00:14

I was trying to ensure that the incompleteness theorem based argument worked for the paradox

Tanner Swett

Now, there's a particular theory I'm thinking of.

Consider the theory Q, which is made by starting with Peano arithmetic, and then, whenever Q proves a statement S, adding the statement "PA + S is consistent" to Q.

Karl Kronenfeld

Is Q really different from PA?

Tanner Swett

Q seems to be arithmetically sound. All of its axioms are true, after all, right?

Yes, Q proves that PA is consistent, but PA does not.

Karl Kronenfeld

True.

I doubt I can decide whether it is permissible to take the limit like that.

Tanner Swett

We can define Q0 = PA, then Q1 = Q0 plus "PA is consistent with S" for all statements S proved by Q0, then Q2 = Q1 + "PA is consistent with S" for all statements S proved by Q1, and so on.

Then Qn is defined for all natural numbers n. Q itself is just the union of all of these.

Karl Kronenfeld

00:23

You unfortunately still do not gain the provability of the statement that Q is consistent.

Tanner Swett

If Qn is consistent for all n, then Q is consistent by compactness.

Since a proof of a contradiction in Q must also be a proof of a contradiction in some Qn, since a proof can only use finitely many axioms.

Karl Kronenfeld

Ah, I'm just showing how much I have forgotten in mathematical logic.

Q now sounds good to me, but who am I to say? :)

Oh, but it's not "good" in the original sense, of course.

Tanner Swett

So Q doesn't prove its own consistency—presumably—but Q does prove the consistency of PA + any finite subset of Q.

Now, I certainly believe that Q is consistent. Is the statement "Q is consistent" an arithmetic statement?

Karl Kronenfeld

That sounds like Q would have to have its own Godel numbering system.

Tanner Swett

Let's see. There's a computer program which enumerates all of the statements Q proves. So the consistency of Q is equivalent to the statement that this computer program never outputs "0 = 1".

Right? That makes "Q is consistent" an arithmetic statement.

Karl Kronenfeld

00:38

Right.

Tanner Swett

Well, I'm a little tired of thinking about this.

So lemme pivot a bit. Really, there are only two things I don't like about PA. The minor problem is that it doesn't prove Goodstein's theorem, but I don't care that much about Goodstein's theorem.

The major problem is that it doesn't allow you to talk about infinite sets.

I did a Google search to try to find out if there's a set theory which is a conservative extension of PA.

The good news is, apparently there is. "ALPO", defined by H Friedman, 1980.

The bad news is, I haven't yet found a freely accessible description of ALPO.

Faust7

Morning

Karl Kronenfeld

lol, some reason I don't have access to the original @TannerSwett

Tanner Swett

00:56

I found a document that describes ALPO, although not in words I can easily understand!

It's a set of lecture notes by Friedman, titled "Aspects of constructive set theory and beyond".

Karl Kronenfeld

A listing of axioms (by name, not formula) is found in sec. 7.2 of pdfs.semanticscholar.org/dbb1/…

Jasper Loy

@Faust7 It's actually morning here, but I guess for most people in this chat, it's night.

Tanner Swett

I see that the axioms are numbered the same way in that document as in the lecture notes I found.

Karl Kronenfeld

I see they're found in the lecture notes.

Yeah, I do not know what $\Delta_0$ means.

I am supposed to be writing, so I will get back to that. This all is interesting to me, thank you.

Tanner Swett

No problem, thanks for your help.

Faust7

01:03

@JasperLoy you feeling better?

Tanner Swett

@KarlKronenfeld Looks to me like a $\Delta_0$ formula is one that contains only bounded quantifiers.

Jasper Loy

02:26

@Faust7 Nope, it has been many years, but I hope it will end soon, thanks. =D

user147690

03:25

@BalarkaSen I love that Ulver album you mentioned. Thanks for that :D.

Balarka Sen

04:51

@AlexClark Good to hear, I like it a lot too.

Abcd

05:31

0

Q: Looking for a more efficient way to solve the Straight Line Equation problem.

If the straight line through the point $P(3,4)$ makes an angle $\dfrac{\pi}{6}$ with the $x$ axis and meets the line $12x +5y + 10=0$ at Q, find the length of PQ. My method: Equation of the given line is; $y- \sqrt3x+3\sqrt3-4=0$ // using Point slope form. Now, solving the simultaneous ...

Hi @LeakyNun, needed an advice from you (if you don't mind...)

Leaky Nun

use vectors lol

Abcd

@LeakyNun No.... Not this question

Leaky Nun

why not?

Abcd

@LeakyNun Trigonometry was intuitive to me but Coordinate Geometry is not, what should I do? (This is my question to you.) Need your advice...

@LeakyNun I wasn't referring to the question that I posted/.

Leaky Nun

@Abcd do more questions :P

Abcd

05:37

@LeakyNun What if I am unable to solve them?

Leaky Nun

ask us

Abcd

Ohkay...

Secret

Thoughts?

arxiv.org/abs/1706.08908

Leaky Nun

06:00

@Secret Let's say I define an order in the infinite subsets of $\Bbb N$ as such: Let i=0. If i is in A but not in B, then A<B. If i is in B but not in A, then B<A. Otherwise, increment i by 1 and repeat.

Formally, $A < B \iff \operatorname{min}((A \setminus B) \cup (B \setminus A)) \in A$

Is this a total order? Is this a well-order?

Abcd

06:28

If the sum of the distances of a point from 2 perpendicular lines in a plane is 1, then it's locus is a square. How @LeakyNun?

Leaky Nun

@Abcd try drawing the two lines on an actual paper

Abcd

@LeakyNun How do I prove it?

Leaky Nun

@Abcd WLOG let the two lines be the x-axis and the y-axis. The equation of the locus becomes |x| + |y| = 1

Abcd

@LeakyNun How did you get that equation?

$|x|+|y|=1 $ How did you obtain this @LeakyNun?

Leaky Nun

@Abcd the distance between (x,y) and the x-axis is |y|, and the y-axis |x|

Daminark

07:01

Hey @Alessandro and @Tobias!

Tobias Kildetoft

@Daminark Hi

mercio

o..o

Alessandro Codenotti

Hi @Dami

Daminark

How's it going guys?

Tobias Kildetoft

Working on my grant proposal. It is really hard to figure out how to formulate things nicely

SteamyRoot

07:23

"Just give me the money, we both know you won't understand my project anyway" :^)

Tobias Kildetoft

@SteamyRoot Hmm, it does have a certain ring to it.

Alessandro Codenotti

"Please I have a wife and 17 children to feed"

Tobias Kildetoft

But there is the issue that one or more of the reviewers might actually be an expert in the subject

Daminark

Oh sorry I didn't see this earlier, and lolol

@Alessandro's suggestion is bulletproof

They might just give you tenure after that tbh

SteamyRoot

@TobiasKildetoft "If you do understand it, I'll cite all your papers."

Problem solved

@AlessandroCodenotti Bonus points for a prime number of children

Tobias Kildetoft

07:25

@SteamyRoot Sounds good. Then it is just a matter of finding out the identity of the reviewer'

Perturbative

07:38

Hi chat!

Secret

@LeakyNun for total order, still checking. For well order, guess not as one can pick infinite subsets forming a chain where the next element in the chain has more elements than the previous hence smaller and thus can form an indefinitely decreasing sequence. The same thing can be said in the other direction by building an increasing chain starting from any infinite proper subset that has no finite support

So one can make chains that go infinitely up or down starting at any proper subset with no finite support, hence there cannot be a well order

Concrete examples: Indefinite decreasing chain, start from $2\Bbb{N}$, and then for the next members, continue to union sets of multiples of prime $p\Bbb{N}$

Daminark

07:56

Hey @Perturbative!

Secret

Indefinite increasing chain. Start with $2\Bbb{N} \cup \{1,6,5,8,90,3,4,7,...\}$, and construct the next element by throwing away one element at a time

Perturbative

Heya! @Daminark

How you doing?

Daminark

Everything's alright, how about you?

Perturbative

Aight over here too

Balarka Sen

Hi chat

Daminark

08:03

Yo

@Balarka someone's diffgeo lecture notes for tomorrow:

Balarka Sen

instant quality hit

Perturbative

"Elegant"

Jolenealaska

Hey guys. I'm putting together an experiment with two artificial sweeteners (acesulfame potassium and sucralose). Ace-K is supposedly 200 times the sweetness of sugar, sucralose is supposedly 600 times the sweetness of sugar. I'm just about done putting the experiment together on paper, but I suspect I'm going to need a sanity check on the math. Is this a good place to get feedback about such an issue?

Tobias Kildetoft

@Jolenealaska Sure

Jolenealaska

Great. I need to call it a night, but I'll come back with what I think should work.

Faust7

08:14

ina finite group what is the order of the identity?

i always considered it to be 2

but is it 1 somehow or 0?

Tobias Kildetoft

@Faust7 It is $1$

Faust7

why

Tobias Kildetoft

the smallest positive integral power that gives the identity

Faust7

fail

thx

Perturbative

Quick sanity check, if $X$ and $Y$ are two smooth manifolds, with $\dim(X) = m < \dim(Y) = n$ the inclusion map isn't an immersion right? Because $i : X \to Y$ defined by $i(x) =x$, has derivative at a point $a \in X$ of $di_a(x) = 1$, and hence $T_a(Y)$ is the span of the identity matrix which is just $\mathbb{R}^n$. But because $di_a(x)$ maps all $x \in \mathbb{R}^m$ to the identity in $\mathbb{R}^n$, $i$ is not an immersion, correct?

Faust7

08:18

Sorry im fresh out of sanity

Daminark

Get yir pipin hot sanity right here

So unfortunately you're insane

Not actually insane but this isn't right

So $f$ being an immersion isn't saying that the map $x\mapsto df_x$ is injective

Balarka Sen

Uh, the inclusion map of a submanifold is an immersion.

Any embedding is automatically an immersion.

Daminark

It's that the maps $df_x$ are injective linear maps

But then even in your case, you need to keep in mind that $df_x$ is a map from $T_x(X)$, which may differ for each case even if though they're isomorphic

The train is instructions for how to read

Balarka Sen

I am addicted to that tune.

oHHoHoHoHOhHhHhHHHHSMACK

Daminark

Honestly I love it

Perturbative

08:27

Okay thanks @Dami and @BalarkaSen!

Balarka Sen

Glad to be of no help.

Daminark

@Balarka I guess I never got around to mentioning that the inclusion is an immersion

So you were of some help

Abcd

If$\lvert\dfrac{z-2}{z-3}\rvert= 2$ and $z=x +iy$, show that $3(x^2+y^2)-20x+32=0$

I have found $x^2+y^2 = 16$. How do I find the value of $x$?

Balarka Sen

@Daminark Have you seen this?

Very classic.

Daminark

Yeah

Abcd

08:30

Anyone?

Daminark

Have you ever played the Big Smoke stuff? I never did but I watched a friend try to play the train mission

@Abcd okay so

Gimme a sec to chalk it out

Balarka Sen

@Daminark Yeah

The train mission is super annoying

Faust7

Im going to take a nap now but it was good to see you @Daminark and @BalarkaSen maybe c u in the next morning

or in a hour if nur still up

Daminark

Okay so @Abcd what's the proof that $x^2 + y^2 = 16$?

See you @Faust7!

Abcd

@Daminark Just a minute. Uploading a pic

@Daminark ^ (Tell me if you can't understand my messy handwriting)

Kirill

08:40

Any chance to render math on mobile?

Abcd

is it correct @Daminark?

Daminark

Yeah that's absolutely false

$|z-2| \ne |z| - |2|$

Abcd

Oh

Daminark

Sorry :/

Abcd

@Daminark How do I solve it then?

Kirill

08:44

What's your question, @Abcd ?

Perturbative

Okay, so in the example I gave above, the immersion $i$ is trivially injective since $di_a$ is the identity matrix $\left(\frac{I_m}{0}\right)$ and maps $T_a(X)$ into $T_a(Y)$

Abcd

17 mins ago, by Abcd

If$\lvert\dfrac{z-2}{z-3}\rvert= 2$ and $z=x +iy$, show that $3(x^2+y^2)-20x+32=0$

@Kirill ^

Kirill

Do not know what \dfrac is..

Abcd

@Kirill ^

Kirill

Thanks

Abcd

08:48

There should be automatic rendering of Math Jax! I face the same problem while using mobile!

Kirill

Yes, I've asked about that, nb answered.

Abcd

@Kirill that's unfair. Why not post this as a meta suggestion?

Kirill

That's to epic for me. I'll solve it somehow.

Abcd

@Kirill What do you mean by "that's too epic for me"?

Kirill

You want an official post referring that here on the forum? @Abcd

Abcd

08:57

@Kirill yes

Leaky Nun

09:53

@abcd $|\dfrac{(x-2)+yi}{(x-3)+yi}| = 2$

@Secret nice

Alessandro Codenotti

@LeakyNun I'm not 100% sure but iirc it's consistent with $\sf ZF$ that the only well-orderable subsets of $\mathcal P(\Bbb N)$ are countable since it's consistent that $\omega_1$ doesn't inject into $\mathcal P(\omega)$

Jasper Loy

Hey does anyone here think that the blue colour used to highlight your own messages in this chat is too light?

Daminark

@AlessandroCodenotti Wai u drop choice?

Like aside from my GCH memeing, it seems like losing choice actually makes set theory really sketchy

Jasper Loy

0

Q: Blue colour used to highlight your own messages in math chat is too light

I frequent the SE chat rooms quite a bit and noticed that in room 36, the main room for Math SE, the blue colour used to highlight your own messages is a bit too light, to the point where it is almost invisible, at least for myself. In contrast, the orange used to highlight pings to you is very b...

discussion feature-request

Hey guys, I just typed a meta post talking about what I just did, please give your feedback there, thanks.

Alessandro Codenotti

10:10

Sometimes it's interesting to see what happens without, just for the sake of it

user91500

test

Daminark

Grade: F+

@AlessandroCodenotti I mean I guess but like... I dunno it just seems like so many things which are absolute nonsense become legal when you drop choice

Alessandro Codenotti

Sure, but it's interesting to see what can be proved in $\sf ZF$ alone

Daminark

@Jasper I'd say I'm okay with the color as is, and the issue is that if it were made darker, you might have an issue with text contrast

Jasper Loy

@Daminark Oh no, there are so many kinds of blue! That won't be a problem.

Daminark

10:19

@Alessandro I guess my point is, why not kill off one of the other axioms instead?

Alessandro Codenotti

$\sf AC$ is interesting because there are a lot of weird consequences, but I find equally interesting to know that, say, $\sf ZF$-$\sf P$ can't prove that there is an uncountable set

(Where $\sf P$ is the powerset axiom)

Daminark

P is pairing or power set?

Alessandro Codenotti

the latter

Daminark

Lol sniped

Alessandro Codenotti

as usual :P

Daminark

10:21

But yeah anyway, I dunno, there's a professor here who does proof complexity, and I've been told to take a class by him to try it out

Alessandro Codenotti

I find reverse mathematics quite interesting

Jasper Loy

I find reverse engineering quite interesting

Daminark

Until then I'm just gonna religiously go with AC because a world without it sounds horrible. I'd sooner drop infinite sets, and I refuse to drop infinite sets

Alessandro Codenotti

(I know nothing about it apart from the fact that they try to determine what's the weakest set of axioms a result can be proved from)

Jasper Loy

@Daminark AC is Alex Clark, lol.

Daminark

10:23

He has been officially replaced with the axiom of choice

Hacked, I think

@AlexClark let me know how it feels to have transformed into the axiom of choice

:P

Alessandro Codenotti

@Daminark I think ZF-I and ZFC-I are the same (I is the axiom of infinity)

user147690

Feels bad man :(.

Daminark

Yeah I think choice is provable in finite sets

Jasper Loy

@AlexClark Oh hey, I just sent you an email.

Alessandro Codenotti

ok, no, that's obviously wrong, there are models of the first which are not models of the second

I think I want ZF-I+not I and likewise for ZFC

Daminark

10:27

Wait no I think you're right here

In finite sets you can prove choice by induction probably

Oh I mean... okay sure

Alessandro Codenotti

sure, but say you have a model of $\sf ZFC$, that's also a model of $\sf ZFC$-$\sf I$ despite being full of infinite sets

Daminark

Lol get sniped m8

Alessandro Codenotti

fair revenge

Daminark

I mean I tend to be extremely loose in conversation, so I took ZF-I to mean, we're banning infinite sets

Technically not the same thing but it was implied

Alessandro Codenotti

yeah, but one needs to be very precise with axioms since there are a lot of subletlies everywhere

Daminark

10:31

We're not really doing any formal stuff though, so I think it's fair that certain things go without necessarily being said explicitly

Alessandro Codenotti

fair enough

anyway it's lunchtime, I'll be back later! Bye

Daminark

Aight, see you!

quid

This is basically just a test.

user147690

@quid You mean life?

user147690

Yeah...

quid

10:35

@Alex well, originally I meant me posting this particular comment. But I guess this is a trivial corollary of your theorem. :-)

user147690

@quid ;D

Jasper Loy

Now @quid can see the colours I talked about in my meta post. =D

quid

Right, that was the point. It appears the math theme is the issue. My take would be that the gray of the other messages should be more white-ish as it's on various other sites, but that's a detail.

Jasper Loy

I think Christian Blatter is a first rate mathematician. I have seen his posts and he has a brilliant mind.

quid

Yes. He wrote well known books too.

Muhammad Nouman

10:38

Hi

Bye :-()

Jasper Loy

Oh I didn't know you are a moderator on Math Ed.

Yeah, maybe they should change the whole colour theme in the room.

Balarka Sen

I need to learn chemistry.

Darnit

Jasper Loy

Chemistry is the science that ties all sciences together, I read.

But I don't like it either.

Alessandro Codenotti

@BalarkaSen don't drink anything in the lab and keep hydrogen away from fire and you should be fine

Daminark

I mean you'll have to learn the subject of chemistry as wel though

Jasper Loy

10:44

@AlessandroCodenotti Also, keep yourself away from fire.

Balarka Sen

@Alessandro Put hydrogen in the fire and drink everything in the lab, gottem.

Daminark

@Balarka become the hydrogen

Balarka Sen

BALEETED

That was an immature meme, pls delete that

Daminark

10:59

[DATA SPONGE]

Balarka Sen

so i can still get back data by squeezing the sponge?

doesn't sound very efficient

Daminark

shrug

Kirill

11:35

@Abcd then just do it. I am not a big fan of this.

SteamyRoot

@AlessandroCodenotti Store the water in a container made from Alkali metals :^)

Kirill

@SteamyRoot

if you have a characteristic polynomail $(x-4)^4$ of a $4 \times 4$ matrix, and your Jordan blocks of a diagonalised form are of the length 1 and 3, does it true, that your minimal polynomial is $(x-3)^3$?

Daminark

Okay I don't know anything about Jordan stuff

SteamyRoot

@Kirill Of course not

Jasper Loy

Michael Jordan is a basketball player.

Daminark

11:41

But the set of roots of the characteristic polynomial and minimal polynomials should be the same

SteamyRoot

If the minimal polynomial has a factor $(x-3)$, then the matrix has eigenvalue $3$

Kirill

@JasperLoy ... and he has a good form

SteamyRoot

The minimal polynomial has the same factors as the characteristic polynomial, just of lower degree.

Jasper Loy

@Kirill LOL, you are becoming as good as Balarka.

Kirill

sorry, it was nonsense, @SteamyRoot. It was meant $(x-4)^3$ for a polynomial.

Balarka Sen

11:44

I don't think I am particularly good at pun

3

I am good at trash meme jokes

Jasper Loy

Hi @MatsGranvik I missed your old picture.

Kirill

@JasperLoy what do you mean?

Jasper Loy

@Kirill Balarka makes good jokes.

Kirill

@JasperLoy oh, that's good!

SteamyRoot

@Kirill In that case, yes. The degree of a factor $(x-\lambda)$ is the size of the largest Jordan block with eigenvalue $\lambda$

Kirill

11:47

@SteamyRoot glad to hear that, thanks. It's all about the exam I had.

Daminark

@Kirill oh the exam already happened? How'd it go?

Kirill

@Daminark I think it was great, but I am aware of too positive feedback.

@Daminark I've still mulled the transformation matrix in the Jordan basis, but ok

the only thing I am really interested in: "is that true, that for $A,B \in \mathbb{C}^{n \times n}: (A-\lambda I_n)^n=0 \Rightarrow \lambda=1$ is the only existing eigenvalue"?

Perturbative

1

Q: Suppose $X$ and $Y$ are two smooth manifolds, with $\dim(X) = m < \dim(Y) = n$ is the inclusion map an immersion?

Suppose $X \subseteq \mathbb{R}^k$ and $Y \subseteq \mathbb{R}^l$ are two smooth manifolds, with $\dim(X) = m < \dim(Y) = n$ is the inclusion map an immersion (where $X$ is a submanifold of $Y$)? My Attempted Proof: The inclusion map $i : X \to Y$ defined by $i(x) = x$, has derivative at a poin...

differential-geometry proof-verification differential-topology

The only problem I have with my proof above is the comment I made here in the first answer

Tobias Kildetoft

@Kirill No, of course that does not imply that $\lambda = 1$ is the only eigenvalue. What if the matrix is diagonal?

Also, what does $B$ have to do there?

Perturbative

How can I show the derivative has the form I claimed rigorously? The answer hinted at codeminsion, but couldn't it be done in a much simpler way?

Also I don't want to use the theorem that every embedding is an immersion as Balarka mentioned earlier, because I sort of wanted to prove this 'by hand' (if that makes sense)

Kirill

11:59

@TobiasKildetoft I couldn't find any counter-example...

Tobias Kildetoft

@Kirill Take a scalar matrix with $2$'s on the diagonal

(and $\lambda = 2$)

Transcript for

$Mathematics$ Mathematics