Mathematics - 2017-07-12 (page 2 of 4)

Semiclassical

4:00 AM

But my brain ain't working right now :(

What makes me curious, actually, is that Pell's equation is $x^2-ny^2=1$ for $n$ a positive square-free integer

@typhon oh, see this bit here: en.wikipedia.org/wiki/Pell%27s_equation#Algebraic_number_theory

BAYMAX

yeah ok :)

BAYMAX

4:26 AM

If $f : \Bbb{R} \rightarrow \Bbb{R}$ is bijective then what can we say about $f^{-1}$

is that too bijective?

and if $f$ is differentiable then is $f^{-1}$ differentiable in thatccase?

arctic tern

if you're asking if the inverse of a bijection is also a bijection, you should probably make sure you know what "bijection" means. the answer to your second question is no (for example f(x)=x^3 around x=0).

BAYMAX

yes inverse of a bijection is bijection :)

Like I am interested in exploring the second part

Semiclassical

Well, what's the formula for $(f^{-1})'(x)$?

Balarka Sen

That formula holds iff $f$ is continuously differentiable.

BAYMAX

if I said it differentiable that implied continuous ?

Balarka Sen

4:41 AM

"continuously differentiable" $\neq$ "continuous and differentiable"

BAYMAX

oh

Daminark

Continuously differentiable is when the derivative is continuous

BAYMAX

nice

Semiclassical

I think $x|x|$ would be an example of that distinction.

(it's continuous and differentiable, but that derivative isn't continuous)

BAYMAX

to give inverse of function in desmos?

f-1

?

Semiclassical

4:44 AM

no idea.

Balarka Sen

@Daminark Did you give that lecture?

Secret

Mar 12 '15 at 2:16, by David Wheeler

The space $\Bbb R^{\Bbb R}$ is ridiculously big. There are functions out there NO ONE HAS EVER SEEN

Balarka Sen

That's true ^

Secret

The notebook entry of this investigation is back in February

Steamy and I did managed to try to calculate its cardinality, which I don rmb on top of my head before we fall back to integrable functions

Daminark

I'm giving it tomorrow

Balarka Sen

4:48 AM

Ah ok

Secret

I wonder, if the busy beaver lives there...

BAYMAX

Unit sphere in $\Bbb{R}^2$ is circle

Daminark

I do have some lecture notes written up, I can send them if you'd like

BAYMAX

= $S^1$

Daminark

It's very technical, like most of the work was understanding the notation

BAYMAX

4:49 AM

oh

demona

Daminark

But it's pretty dank

BAYMAX

your summer school

how is it going?

Balarka Sen

@Daminark I'd like the notes

It's funny, how you feel now was how I felt when I was first learning foliations

I suppose anything dynamical requires a lot of symbolical flexibility to rigorously state

Daminark

Sorry I uploaded it in 3 separate pictures, but it seems like this chat can't upload files

Balarka Sen

Got'em. Is that it?

Daminark

4:51 AM

@Baymax it's going really well, how's everything with you?

Yeah, I write rather tersely

BAYMAX

yeah it is fine thanks for asking :)

Balarka Sen

Thanks a lot.

Daminark

If there's something you'd want to clarify, let me know

The general flow of this is basically that the first few theorems are basically trying to establish this recurrence property of homeomorphisms on compact spaces

So then what you do is translate a finite partition of $\mathbb{Z}$ into a sequence in a two-sided sequence space on some finite alphabet

That space is $\{1,\ldots,m\}^{\mathbb{Z}}$, which is compact, and then the left shift on it becomes a homeomorphism

(Since it's two sided)

So then you show that a recurrence of this becomes an arithmetic progression

Balarka Sen

Ah, alright

That's an interesting proof-technique

Daminark

Second theorem on affine spaces required you to invoke IP-systems directly and rehash the results in a specific context, since merely recurrence can't handle this. That's the proof where what I wrote down on here did not reflect what I wrote on the board to understand it, so if it's tricky to read, do tell

Yeah I like it, it's pretty slick

Balarka Sen

4:58 AM

Ok, I'm reading the first page now

what does "IP" stand for? :P

Secret

[To be investigated after all function spaces] $\Bbb{R}^{\Bbb{R}^{\Bbb{R}}}$

Daminark

So, I think IP sets are a thing to describe sets of finite sums of some infinite set of natural numbers

Semiclassical

$\mathbb{R}\uparrow\mathbb{R}$

(no, i don't know what that would actually mean)

Daminark

I'm p sure this has to do with that, that name is of disputed origin, either idempotent (since such sets have to do with idempotent ultrafilters) or infinite-dimensional parallelepiped

Balarka Sen

huh

Semiclassical

5:01 AM

though I guess the better starting point would be $\mathbb{R}\uparrow\mathbb{N}$

Balarka Sen

I guess I'll stop being a Irritating Pedant and move on to actual math :P

Semiclassical

(i'm talking nonsense)

Daminark

Kek @Balarka

Semiclassical

(should probably be two uparrows, lol

Secret

One up arrow is exponentiation, thus that is the same as $\Bbb{R}^{\Bbb{R}}$. You might mean $\Bbb{R}\uparrow^{\mathfrak{c}}\Bbb{R}$. I have no idea, I need to check whether cardinal hyperoperations make sense first. But likely we don need to go to such vast space

Semiclassical

5:04 AM

Well, $\mathbb{R}\uparrow\uparrow 3$ should be R^R^R

Secret

Indeed, 2 uparrows is tetration . The space of all set valued functions might live there

Balarka Sen

@Daminark I am not sure if I understand the order. So I can say $\{1, 2\} < \{3, 4\}$, but nothing about $\{2, 3, 4\}$ and $\{1, 7, 8\}$?

Daminark

Yeah, it's only a partial ordering

Balarka Sen

Got it.

Secret

and set valued function might be related to second order logic

NB I don't know whether uncountable order logic make sense, but countable order logic can be as the sup of nth order logic. Likely only n cat people will need to deal with that

Mar 12 '15 at 2:53, by David Wheeler

So: $$\int 2^x\ dx = \int e^{\ln(2)x} = \dfrac{1}{\ln(2)}\int \ln(2)e^{\ln(2)x}\ dx = \dfrac{1}{\ln(2)}\int (e^{\ln(2)x})'\ dx = $$?

I wanna put a lambert W in the 3rd step

Semiclassical

5:18 AM

you'd need another x.

Secret

Oops, yeah

Hmm... $\int \frac{x(\ln 2)e^{x\ln 2}}{x}dx=\int (e^{x\ln2})'dx=\int \frac{W(?)}{x}dx$

Semiclassical

nah. remember, it's $W(x)e^{W(x)}=x$.

So you'd want to substitute $x=W(u)/(\ln 2)$.

Typhon

@Semiclassical I was thinking more of the fact that if a and b are the results of norms of that integers then ab is also a norm

there's a corollary which gives factoring, but I have yet to prove it for all quadratic rings. i need to get around to doing that.

Secret

5:34 AM

$\int \frac{\ln 2}{(1+W(u))}du$ hmm, it gets a nontrivial look, interesting

Semiclassical

low blow, smbc, low blow

Balarka Sen

that smbc is actually a little scary

Secret

I knew the sugar coated lie of growing up since child (unlike most of my peers) but whether getting a career with a degree (insert suitable sentence) is a lie, not yet

Regardless, my dream job is research, so that's should be ok I guess...

Balarka Sen

5:52 AM

@Daminark Oh, by the way, did you mean the big union = X in lemma 1?

Typhon

@AkivaWeinberger @SimplyBeautifulArt @Semiclassical @Secret @BalarkaSen here is a riddle for you all to ponder over. I know you all liked the last number theory question I made. Might interest you all. math.stackexchange.com/questions/2355747/…

and im heading out

so goodnight ya'll

@AkivaWeinberger fyi the factoring thing won't work here as directly as not all of them are unique factoring sets.

you'd have to build the necessary proofs up first. Also, no special cheat sheet for this one. This was basically the tail end of the project I did in the class. I examined the integers of the rational sets in correlation with the euclidean domains.

so... there's no proofs for this one. Have to prove from scratch.

Leaky Nun

6:10 AM

12 hours ago, by Akiva Weinberger

I think it's provable you can't define $\Bbb R$ in $\Bbb C$ with just the language of fields

Can we prove this?

Secret

How to write the condition of completeness in symbols, perhaps there's a symbol that is not used in the language of fields? Also reals need total order which the complex don't have (they only have a partial order)

Secret

i need to find out what modular forms are later. They still give me an impression that they are floating donuts in space

Leaky Nun

@Secret one idea I have is that we can prove that every formula that defines $\Bbb R$ must also define _

but it doesn't work

because I only know that every formula that defines a set of number must also define their conjugates

I could say $\Bbb R$ is the numbers unchanged under conjugate, but I can't define conjugate in that language

Daminark

6:38 AM

@Balarka whoops, yeah

@Leaky I don't know algebra and thus am not sure if this would work

Leaky Nun

@Daminark well, I don't seem to make it work

Daminark

But I think $\mathbb{C}$ is isomorphic to $\mathbb{R}[t]/\langle x^2 + 1\rangle$

Leaky Nun

@Secret and the results turned out to be fine, if you care enough

@Daminark hmm...

$x \mapsto i$ seems to be ok

I agree with your thought

and I can define $\{i,-i\}$ in the language by $x^2+1=0$

but I can never separate $i$ from $-i$

Daminark

I mean, is there a canonical identification of the two?

Secret

-i is additive inverse of I, thus the existence of negative elements should uniquely define -i?

Daminark

6:44 AM

Like, I thought it was just that you get two things, and you choose one of them to be $i$

Secret

-i also have the property of being the reciprocal of i

Kane

Can somebody give me a hint on this question:
Explain why for $18 \leq n \leq 400$, $n \neq 361$, that $n$ is prime iff $n$ does not have proper factors 2 or 5 and GCD(n, 51051) = 1

Secret

That is, i has a negative which coincides with its multiplicative inverse

Perturbative

Hey everyone

Kane

I think its saying n is prime so long as it is not divisible by 2, 3 or 5 between 18 - 400

Am I on the right track?

Alessandro Codenotti

6:48 AM

@Daminark what does canonical mean?

Perturbative

@AlessandroCodenotti I think he means natural

Leaky Nun

@Secret no it doesn't

any formula that $i$ satisfies, $-i$ also satisfies

@Kane 51051 = 3x7x11x13x17

Kane

@LeakyNun ahhh

Secret

What about their minimal polynomials, it is always unique for each algebraic number?

Kane

so n is prime if it doesnt have factors 2, 3, 5, 7, 11, 13, 17

Leaky Nun

6:53 AM

@Kane yes

Kane

and they are all the primes up to 20

Leaky Nun

you forgot 19

Kane

oh crap

Leaky Nun

which is why it says "except 361"

because 361 = 19 x 19

Kane

@LeakyNun thanks for your help

Leaky Nun

6:54 AM

@Secret of course it is unique; that is what minimal means

@Kane no problem

Secret

Wait, then we might be able to use the minimal polynomials to separate out i and -i, unless x is not in the language of fields

Tobias Kildetoft

@LeakyNun No, "minimal" need not imply unique

Leaky Nun

@Secret you can't

@TobiasKildetoft you're right

Alessandro Codenotti

Hmm, you can map $X$ to $i$ and $1$ to $1$ to get an isomorphism @Daminark

Leaky Nun

18 mins ago, by Leaky Nun

$x \mapsto i$ seems to be ok

Secret

7:00 AM

O great, so we cannot uniquely specify each algebraic number. I thought they are all computable...

Leaky Nun

@Secret a minimal polynomial of an algebraic number must also be satisfied by its conjugate(s)

if $2+\sqrt3$ satisfies a polynomial, so must $2-\sqrt3$

Alessandro Codenotti

Oh, I missed it

Leaky Nun

@AlessandroCodenotti do you have any ideas?

Alessandro Codenotti

What's the question exactly?

Leaky Nun

59 mins ago, by Leaky Nun

12 hours ago, by Akiva Weinberger

I think it's provable you can't define $\Bbb R$ in $\Bbb C$ with just the language of fields

Can we prove this?

"the language of fields" include the first-order logic symbols and $0$, $1$, $+$, $\times$, and $=$.

Alessandro Codenotti

7:11 AM

@TobiasKildetoft isn't it unique if you work in a decent extension of a field and ask for minimal polynomials to be monic?

Leaky Nun

@AlessandroCodenotti yes but it isn't what "minimal" means

Tobias Kildetoft

@AlessandroCodenotti I objected to the claim that the term minimal itself implied unique

Alessandro Codenotti

@LeakyNun Not by trying to define it

Leaky Nun

@AlessandroCodenotti surely, but how can we prove that we can't actually define it?

Alessandro Codenotti

Minimal was defined as the monic generator of the kernel of a morphism in the abstract algebra course I took

Daminark

7:13 AM

Oh define $\mathbb{R}$ in $\mathbb{C}$

Seems like I can't read

Tobias Kildetoft

@AlessandroCodenotti Minimal means the same in all contexts, it just depends on the choice of a partial order (or occasionally preorder)

@LeakyNun Good question. The only thing a quick search turned up is the comments on math.stackexchange.com/questions/1402011/… stating that we can't

But it does not give any indication of why

Secret

@LeakyNun Using what you said, then the minimal and monic polynomial of $\pm i$ is both $x^2+1=0$?

Alessandro Codenotti

I guess my partial order is by grade first and then ordering the leading coefficents by divisibility amonh those with the same grade

Leaky Nun

> A subset $H$ of $S$ is minimal with respect to the property if $H$ has the property, and no subset $K \subset H$, $K \ne H$, has the property. If $H$ has the property and $H \subset K$ for every subset $K$ with the property, then $H$ is the smallest subset with the property. There may be many minimal subsets, but there can be only one smallest subset.

@AlessandroCodenotti this is what "minimal" means to me

P.53, A First Course In Abstract Algebra, John B. Fraleigh

Seventh Edition

2003, so quite outdated

Tobias Kildetoft

@LeakyNun Yeah, that is with the partial order given by inclusion

Leaky Nun

7:17 AM

the order would be the degree here, I think

@Secret yes

Secret

Great that means... what I said yesterday is all wrong...

Leaky Nun

@Secret what did you say yesterday?

Secret

16 hours ago, by Secret

I think one for each algebraic number is enough, since every algebraic number can be uniquely defined by its minimal polynomial

It also means user107952's conjecture is still unresolved

17 hours ago, by user107952

My conjecture is that every algebraic number except 0 needs a single identity besides commutativity, associativity, and distributivity of addition and multiplication, and 0 itself needs 2 identities, namely $(x*0)=0$ and $(x+0)=x$

because if you cannot separate an algebraic number from its conjugate, then you cannot have an identity (in the context of universal algebra) to define $r$

Leaky Nun

I don't actually understand what "identity" means there

Secret

It's pretty much what we called an axiom which takes the form of an equation

Leaky Nun

7:21 AM

Could you demonstrate using an example?

Always helps^

^

:-D

Rings have $+$,$*$,$0$,$1$,$-$,${}^{-1}$, and the ring axioms. Thus $\langle+,*,0,1,^{-1},->\rangle$ is the signature of rings. The ring axioms are called identities in the context of universal algebra since those are the sentences you need to establish your structure has the required signature

e.g. x+0=x is an identity of rings. 0 is called a constant (because it is not an arbitrary element taken from an underlying set

Leaky Nun

@Secret Can you demonstrate using an example of an algebraic number?

Secret

7:26 AM

Well, say we have $i,-i$. It's minimal polynomial is $x^2+1$, thus to define $\pm i$ we use the identity $x^2+1=0$

or for a simpler example, $\sqrt{2}$, we define it by using the polynomial $x^2-2=0$

These are identities in order to define the above three algebraic numbers in your ring

In particular, $\sqrt{2}$ is uniquely defined by the identity $x^2-2=0$

uh wait a sec, $\pm \sqrt{2}$, forgot about unique, but yeah, $\pm \sqrt{2}$ can be distinguished from $\pm i$ by those identities

Leaky Nun

@Secret $x=\sqrt2 \equiv (x\times x = 1 + 1 \land \exists r: r \times r = x)$

if the model is $\Bbb R$

Secret

uh, that's more like a definition for $r={}^4\sqrt{2}$, but the point is, can we always do that for all real algebraic numbers, if yes, then user107952's conjecture may still hold (We already have shown above that it does not work for at least one pair of complex number, $\pm i$

Leaky Nun

@Secret no, that's a definition for $x = \sqrt2$

Secret

but what's the point of the line $r\times r = x$? why not $r\times r\times r = x$ or other possible things?

Tobias Kildetoft

@Secret because that is the way to distinguish the positive elements

Secret

7:37 AM

ah ok, right, since $r\times r \geq 0, \forall r \in \Bbb{R}$

so that means, we can pick $x=-\sqrt{2}$ by using $-(r\times r)=x$

Leaky Nun

@Secret you don't have $-$ in your language

Secret

that's easy to fix: $x+r\times r = 0$

Leaky Nun

yes

Secret

ok so user107952's conjecture should be fine

Now I think I see what's going on, the reason why the uniqueness of specifying each algerbraic number by its monic minimal polynomial in complex fails is we lost the identity(?) $r\times r \geq 0$

since $i^2=-1<0$, which screws up total ordering

Leaky Nun

yes

Secret

7:44 AM

hmm, now that makes me curious, if we cannot define $\Bbb{R}$ in $\Bbb{C}$ using the language of fields because we cannot distinguish between complex conjugates, then what extra letter(s) do we need to add into the language in order to be able to do so...?

Tobias Kildetoft

@Secret adding a constant for $i$ works

or one for complex conjugate

Leaky Nun

@TobiasKildetoft I don't think it does

@TobiasKildetoft this would work though

Tobias Kildetoft

@LeakyNun Either of those will give the other

Leaky Nun

@TobiasKildetoft how will $i$ work?

Tobias Kildetoft

@LeakyNun You mean for specifying the reals inside the complex numbers?

Leaky Nun

7:49 AM

@TobiasKildetoft yes

Secret

I thought i is already inside the complex numbers, uh..., maybe it is not present in the language of fields. I think I don't really understand what is meant to be a constant for i

Leaky Nun

I don't think defining $i$ would work @TobiasKildetoft

@Secret adding $i$ to our language

Secret

ah ok

6 mins ago, by Tobias Kildetoft

or one for complex conjugate

The identity $1=\bar{1}$?

Tobias Kildetoft

@LeakyNun Somehow I thought we could get complex conjugation from i as a constant, but I think you are right that it will not suffice

Leaky Nun

@TobiasKildetoft because I can actually get $i$: $x^2+1=0$

(I know $-i$ would also satisfy, but it wouldn't matter)

@Secret no, the "conjugate" into our language

Tobias Kildetoft

7:53 AM

@LeakyNun Well, the point of addint $i$ is to distinguish it from $-i$

Leaky Nun

if conjugate is in our language, then $x \in \Bbb R \equiv x = \bar{x}$ @Secret

@TobiasKildetoft no, $i$ and $-i$ won't matter if you can get conjugate from $i$.

if you can get conjugate from $i$, then so can you from $-i$, so it wouldn't matter

Secret

ah ok, I think I have a terrible misconception of thinking the language of complex numbers are $\{\text{1st order logic },+,-,*,{}^{-1},\bar{},0,1,i,\Bbb{C}\}$ and forget that the original question said I only have the language of fields

because the language of fields is only $\{\text{1st order logic },+,-,*,{}^{-1},\Bbb{F},0,1\}$

[Extended investigation, unrelated, to be done later] I wonder if the following is true in general: Given any commutative ring $R$ with units, is the following true:

(typo)

Leaky Nun

@Secret I would say it is false

Secret

> If R has a total ordering, then $\not\exists k$ unit such that $-k=k^{-1}$.

(because you can have a total ordering but no such elements exist) I think I swap the if and then again... grr

BAYMAX

The integral equation $\phi(x) - \frac{2}{\pi}\int_{0}^{\pi}\cos(x+t)\phi(t)dt = cos(3x)$ has infinitely many solutions?

Leaky Nun

8:05 AM

@Secret replace it with its contrapositive

and I would suspect that it is true

Secret

Conjecture:

> If $\exists k$ unit such that $-k=k^{-1}$, then R does not have a total ordering.

Leaky Nun

@Secret I say it's true, just prove it

Secret

(Now to revise how a total ordering is defined. Pretty sure $k^2\geq 0$ does not seemed to suffice...)

$k\geq 0$ or $0 \geq k$

$k^2 \geq 0$ or $0 \geq k^2$

$-k < 0$

$(-k)k < 0$

$1 < 0$

(Now how to argue that 1 should be > 0... (sure, it is true for all inductive systems like peano arithmetic, but we are talking about very general commutative rings here, thus we might need to be careful...)

Leaky Nun

@Secret prove that $1 < 0 \implies 1 > 0$

Secret

8:28 AM

$1 < 0$

$-k=k^{-1}<0$

$-k<0 \implies k > 0$

$k(-k) > 0$

$1 > 0$ or $k^2(-1)>0$

$1 > 0$ or $k^2 < 0$

hmm... let me start it all over again...

Leaky Nun

@Secret you don't need $k$ here

Secret

uh, but without k, the conjecture only holds if $-1=1$?

Leaky Nun

@Secret I mean, you don't need $k$ for this step

this step being $1 < 0 \implies 1 > 0$

Secret

I don't know how to do that, cause in an arbitrary commutative ring, 1 is not necessary a successor of 0 as it is just the multiplicative and additive identity respectively and they are both subjected to a total ordering

Leaky Nun

@Secret $1 < 0$
$-1 > 0$
$(-1)(-1) > 0$
$1 > 0$

Secret

8:40 AM

Ah I see, flipping signs for the 2nd step

then contradiction, QED

Leaky Nun

@Secret you're essentially subtracting $1$ from both sides, instead of flipping signs

Secret

right, that's the step I miss, too focused on multiplying

I wonder what this proof look like in pictures... cause the fact that existence of units that are negatives can screw up total ordering should be pretty visible if we plot all the elements in the ring on some diagram...

If the intuition on complex number still holds, then I suspect it will look like some abstract rotation or cycles among some elements

Secret

9:03 AM

Prove that $(-1)^2=1$

Proof:
$(-1)+1=0$

$(-1)^2+(-1)=0$

$(-1)^2=-(-1)$

Prove that $-(-1)=1$
Proof:

$-(-1)+(-1)=0$
Uniqueness of additvie inverse

$-(-1)=1$

Hence

$(-1)^2=1$

Tobias Kildetoft

@Secret Not sure what you mean by the intuition on complex numbers as those are not ordered.

Leaky Nun

@Secret my proof: proofwiki.org/wiki/User:Kc_kennylau/sandbox

Secret

well the observation that $i=1,i^2=-1,i^3,-i,i^4=1$, i.e. these 4 elements form a 4 element cycle around 0 just by multiplying i alone, which on the argand plane, look like a rotation

I always have the impression that multiplication rotates complex numbers around is why total ordering is screwed up in complex numbers

Liad

Why does a vertical line $x \times I$ is open in the ordered square?

it is not an intersection of a rectangle and $I_0 \ ^ 2 $

Alessandro Codenotti

9:19 AM

what's the ordered square again? $[0,1]^2$ with the order topology induced by the lexicographic order?

Liad

i thought it is $[0,1] \ ^ 2$ as a subspace of $\Bbb R \ ^ 2 $

Alessandro Codenotti

in that case I agree that a vertical line won't be closed, but I don't think that's the ordered square

SteamyRoot

Ordered square is indeed the one with lexicographic order, afaik.

Liad

Thanks, my mistake

@AlessandroCodenotti you meant open ? because it is closed if we think of it as a subspace of $\Bbb R \ ^ 2$

Alessandro Codenotti

indeed, thanks for catching that

BAYMAX

9:43 AM

$A_{1},A_{2},...,A_{m}$ are distinct $n \times n$ matrices such that $A_{i}A_{j} = 0$ for $i \neq j$ then relation between $m$ and $n$ is ?

Astyx

$m\le n^2$

For a start

I think $m\le n(n-1)$

Tobias Kildetoft

There is no upper bound actually @Astyx

Astyx

Oh yeah, nonzero matrices

BAYMAX

why would you think $m \leq n^2$

Tobias Kildetoft

Just take multiples of any matrix that squares to zero

So there is no relation without further restrictions

Astyx

9:46 AM

Cause $Tr(MN)$ is a scalar product in the space of matrices ?

Alessandro Codenotti

you want nonzero and linear independent @Astyx (as vectors in $\Bbb R^{n^2}$)

Astyx

You get linear independant from non zero and orthogonal don't you ?

Tobias Kildetoft

@Astyx No, as there are matrices that square to zero

@Astyx Ohh, I missed orthogonal there

Which was not a requirement of @BAYMAX

Astyx

Oh I see why I'm wrong

It's $Tr(M^TN)$, not $Tr(MN)$

Forget what I said

When $n=1$ then $m\le 2$

Secret

Ok, it's not a rotation, but an extra sign flip introduced:

Some elaboration: Let $-a=a^{-1}$. Then $a(-a)=1>0$ but $\forall c < a, c(-a)<0$

user84215

9:55 AM

In these chat rooms, can a post be removed by others?

Secret

thus there's a discontinuity in the behavior when a $ c< 0$ element is multiplied to a $ c > 0$ element that is an inverse of the target element in question, and the one that does not behave as expected screws up the total ordering

Astyx

@aminliverpool I doubt so, perhaps by mods at worst

If it is possible, it's not common practice anyway AFAIK

user84215

10:13 AM

It seems that my last post in a room was removed. Can it be done by the owner of the room?

Secret

Unlikely, the owners of this room are so inactive that we have not seen them in the past few weeks

user84215

10:28 AM

I did not mean this room. It happened in other room.

Martin Sleziak

@aminliverpool AFAIK moderators can delete messages. Room owners can move messages to another room. This is most frequently done to move messages to Trash.

Is this the message you are talking about: chat.stackexchange.com/transcript/message/38704973#38704973 ?

user84215

yes

Martin Sleziak

You can even see in the room transcript: 1 message moved to trash

So, mystery solved. I'll have to go off-line, have a nice day!

Liad

I want to prove that every order topology is regular. I saw a proof online using the equivalent definition of regularity ($X$ is regular iff for each $x \in U $ there is $x \in V \subset \overline V \subset U$ ) and i proved it in another way so i want to know if im correct :

Let $x \in X $ and $A \subset X $ be closed set that does not contains $x$.
$\{x\}$ is close because it's complement is open.
now , $ x \in X - A$ , and $X - A$ is open so there is an interval $(a,b) $ s.t $x \in (a,b) \subset X - A$.

Secret

11:20 AM

Regarding the observation of $-k=k^{-1}$ I have a few ideas for relaxed structures:

1. We can generalise that discontinuity further by having an interval $I=[a,b]$ (or countably more elements if the underlying set is discrete) such that the typical rule $(+)(-)=(-)$ violates besides for the element $k$. This will likely not change much since the resulting structure is already not totally ordered. This choice has the advantage of being able to retain the commutative ring structure

2. We can also relax the ring into a semiring by demanding that 1 has no additive inverse. (In the absence of other pathways), this will nuke the proof that $(-1)^2=1$ completely and thus preventing the contradiction $1 < 0 \implies 0 < 1$ to be proven. One consequence is that the resulting structure will have a bizarre property where the multiplicative identity is less than the additive identity, unlike the typical cases, though it is likely it is not really as bizarre as it seems

Secret

11:40 AM

[Is an exercise until I typed it up] However, a bypass that invalidates proposal 1 exists if there are idemponents. Guess what they are

Akiva Weinberger

@LeakyNun There's only countable many sets we could possibly define total, right? Because there are countable many formulas

So like if you define "$x$ is almost definable" to mean "$x$ is contained in a finite definable set" (that way $i$ is almost definable 'cause it's in $\{i,-i\}$ which is finite and definable)

then there's only countably many almost definable things

Is this smaller set "equivalent" to $\Bbb C$ in some way? My intuition says "yes" but I don't know how to prove it

I suppose it's possible that there exists an infinite definable set with no finite definable subset, which breaks this probably

Actually, you know what, conjecture: a definable set is either $\Bbb C$ or finite.

Secret

11:56 AM

^ Some observations: Not all transcendentals are computable as it is uncountable while computable things must be at most countably infinite since we have a finite alphabet

Akiva Weinberger

Oh that conjecture's false, you could also get the complement of a finite set to be definable

$\Bbb C\setminus\{i,-i\}$ is definable

Secret

that's the set where all conjugates are identified pairwise?

Akiva Weinberger

$\setminus$, not $/$ @Secret

Secret

oops

Akiva Weinberger

It's the set of everything that's not equal to $i$ or $-i$

$\lnot(x^2+1=0)$

I suppose that it's true for quantifier-free formulas. I don't know how to deal with quantifiers, though

Secret

12:02 PM

btw, your more general idea can be analysed with the constructible universe, I guess...

Constructible universe

In mathematics, in set theory, the constructible universe (or Gödel's constructible universe), denoted L, is a particular class of sets that can be described entirely in terms of simpler sets. It was introduced by Kurt Gödel in his 1938 paper "The Consistency of the Axiom of Choice and of the Generalized Continuum-Hypothesis". In this, he proved that the constructible universe is an inner model of ZF set theory, and also that the axiom of choice and the generalized continuum hypothesis are true in the constructible universe. This shows that both propositions are consistent with the basic axioms...

Akiva Weinberger

But the constructible universe need not be equivalent to the original model

The original model might satisfy the negation of the axiom of choice $\lnot AC$ but I think the constructible universe always satisfies $AC$

GFauxPas

good morning frands

Avantgarde

Hi frand

Akiva Weinberger

Gwudmornurnlg

Gukmairns

Secret

Ah yes, $L$ is a model of AFC, thus there is axiom of choice in it

Akiva Weinberger

12:06 PM

Hm.

Avantgarde

'frand' is what some of the jerk guys here use to pick up women

Secret

[Unrelated] An example of a bypass (More on this later once I set up the chat room known as the Repository of Unnatural Algebraic Structures)

(Some Proposed definitions) Let $S$ be a structure that consists of formulae formed by the language of the underlying algebraic structure A of interest. These formulae can be themselves elements, or operators.

Using the jargon of universal algebra, we have identities taken from A. In S identities can be both elements and maps that take sets of identities and produces a set of identities

Let some identity $K_A$ which we want to discard from $A$, and let $B$ be the smallest set of identities we need to discard to discard $K_A$. $B$ are thus the necessary (and sometimes sufficient conditions) for $K_A$ to be proven

Now, a bypass is a set of identities $M\neq B$ which are not part of $A$ that allows $K_A$ to hold even after $B$ is discarded

Proof structures and semigroup actions are the reasons why I found abstract algebra quite visual. I mean what's the superficial difference between these (and possibly category theory diagrams) with drawing mechanisms in organic chemistry?

GFauxPas

is there a theorem that if a sequence is cauchy in one metric it's cauchy in another?

Secret

12:22 PM

0

Q: Sequence is Cauchy in one metric and not in another metric which both determine the same topology on a set $X$

I'm trying to figure out a case where for some set $X$ and two metrics $d_1$ and $d_2$ which determine the same topology, but a sequence is Cauchy in one but not the other. My idea was to take the interval $X= (- \pi, \pi)$, with $d_1$ being the euclidean metric. Then there exists a natural hom...

general-topology

GFauxPas

thanks Secret, so the ansewr is no

robjohn

yes, the answer is no

Alessandro Codenotti

@GFauxPas consider that with the trivial metric the only Cauchy sequences are the eventually constant ones

GFauxPas

what's the trivial metric?

Alessandro Codenotti

$d(x,y)=1$ if $x\neq y$ and $0$ if they are equal

discrete metric sorry

(which makes sense as a name since it induces the discrete topology on the underlying set)

user685272

12:29 PM

hi @robjohn long time no see

Leaky Nun

@AkivaWeinberger what if I make this conjecture instead: every definable set is either finite or has a finite complement

would this be true?

robjohn

@user685272 how are you?

user685272

fine thanks @robjohn and you?

robjohn

@user685272 good, but busy

Secret

in The Periodic Table, 4 mins ago, by Secret

A proof is then basically a question on how you can synthesize the product (what you try to prove) from a starting material (the starting point of a proof) using the reagents given (theorems, lemmas and axioms of the algebraic structure)

in The Periodic Table, 3 mins ago, by Secret

One major difference, however is the reaction yield is 100% for the maths, but not in chemistry

Now, to cross pollinate this back to maths:

Define an algebraic structure with some kind of fuzzy logic such that given a formula, another formula which rewrites the contents of said formula to get a new formula

such that there is a x% success rate in rewriting said formula

(Need to dig a short proof as an example... please continue whatever you are doing)

user685272

12:37 PM

x% = a/b

Secret

12:49 PM

For example: Prove that $\sqrt{2}$ is irrational

Original proof:
1. Suppose $\sqrt{2}=\frac{p}{q}$
Square both sides:
2. Then $2=\frac{p^2}{q^2}$
Rearrange
3. Thus $2q^2=p^2$
even*even=even, even*odd=even, odd*odd=odd
4. Thus $p^2$ is even, thus p is even, thus $q^2$ is even thus $q$ is even, contradict the fact that $p,q$ have no common factors

Proof after applying that crazy proposal:

Proof:
1. Suppose $\sqrt{2}=\frac{p}{q}$
Square both sides: (50% chance of success)
2. Then 50% of the time, $2=\frac{p^2}{q^2}$
Rearrange (40% success rate)
3. Thus 50%*40% of the time $2q^2=p^2$
even*even=even, even*odd=even, odd*odd=odd (100% success rate)
4. Thus $p^2$ is even, thus p is even, thus $q^2$ is even thus $q$ is even, contradict the fact that $p,q$ have no common factors
Thus 20% of the time, there is a contradiction

Therefore, $\sqrt{2}$ is irrational 20% of the time

Fuzzy logic

Fuzzy logic is a form of many-valued logic in which the truth values of variables may be any real number between 0 and 1. It is employed to handle the concept of partial truth, where the truth value may range between completely true and completely false. By contrast, in Boolean logic, the truth values of variables may only be the integer values 0 or 1. Furthermore, when linguistic variables are used, these degrees may be managed by specific (membership) functions. The term fuzzy logic was introduced with the 1965 proposal of fuzzy set theory by Lotfi Zadeh. Fuzzy logic had however been studied...

Now what happens if we have stochastic formal systems made from replacing classical logic in a deductive system with fuzzy logic...

(Pretty sure the artificial intelligence industry did similar things all the time...)

Transcript for

$Mathematics$ Mathematics