Mathematics - 2018-11-22 (page 1 of 2)

Startec

01:05

So if a derivative is how steep a function is at point on the graph, what is the second derivative?

How much the slope itself is changing?

Prashin Jeevaganth

01:52

Does anyone here know what's the difference between a disconnected cake and connected cake? I need examples other than the technical definition because I can't appreciate it

Semiclassical

03:11

@Startec That's right. For example, take f(x)=x^2. At the middle of the graph of y=x^2, the curve is flat; to the left and right it becomes increasingly steeper, but in the opposite directions. So the slope is more and more positive/negative the further out you go, and the rate at which that slope changes happens to be constant.

(Another way to put it: f'(x) is the rate at which f(x) changes with x, whereas f''(x) is the rate at which f'(x)---the rate of change---changes)

user76284

Any ideas on how to prove this is always positive for $\alpha^2 < 1$ and $\theta > 0$?

Silent

05:03

Is the function given below, a counterexample to 'Suppose $f$ is a continuous bijection of a (not necessarily compact) metric space $X$ onto metric space $Y$, then inverse $f^{-1}$ continuous':
$$
g(x)=\left\{
\begin{array}{c}
\frac1y, \,\,\,1\le y<\infty \\
0,\,\,\,y=0
\end{array}
\right.
$$

I meant $$
g(x)=\left\{
\begin{array}{c}
\frac1x, \,\,\,1\le x<\infty \\
0,\,\,\,x=0
\end{array}
\right.
$$

can't edit now

Silent

05:22

@AkivaWeinberger, will you please look at my question above?

Akiva Weinberger

So it's a map from $\{0\}\cup[1,\infty)\to[0,1]$?

I think works, actually

You could also do $[0,1)\cup[2,3]\to[0,2]$ defined by $f(x)=x$ if $x\in[0,1)$ and $f(x)=x-1$ if $x\in[2,3]$

or $[0,2\pi)\to S^1$ defined by $f(x)=e^{ix}$ (i.e. rolling it into a circle)

@Silent

Silent

Wow, @AkivaWeinberger, thank you very much! I, in fact had only that, rolling into circle example, which seemed a bit sophisticated. Thanks for this elementary example.

Startec

Thanks @sem

@Semiclassical

Akiva Weinberger

Continuous functions keep what's connected connected. Homeomorphisms also keep what's disconnected disconnected.

Alessandro Codenotti

06:06

@Silent a very easy example: the identity from $\Bbb R$ with the discrete topology to $\Bbb R$ with the usual topology

Albas

06:23

@BalarkaSen My project advisor says that most M.Math students from ISI just end up knowing a whole lot of fancy words without knowing anything about them. Didn't know how true that was

Adam

06:35

awesome Pafnuty Lvovich Chebyshev fact of the day: he has family roots that trace back to a militant leader of the Tartar clan, who are more well known for an alliance they made with the Mongols lead by Genghis Khan

@WillHunting I hope global warming increases the iceberg separation and there is more shark food in the ocean for our prehistoric friends that are on the top of your food chain, basically the most necessary animal in the oceans ecosystem arguably

Secret

this post is soooo random

Silent

06:56

@AlessandroCodenotti Wow! Awesome :-)

Thank you

Alessandro Codenotti

07:57

I missed it but I guess you already solved the issue

Akiva Weinberger

08:12

I guess "open functions" (functions which map open sets to open sets) would be functions that "preserve disconnectedness"

and continuous functions are ones that "preserve connectedness"

Does that make sense?

No it doesn't. Never mind.

$f:(-2,-1)\cup(1,2)$ defined by $f(x)=|x|$ doesn't "preserve disconnectedness" in any meaningful way.

Nor does the constant function (the function to a one-point space).

Liad

@AlessandroCodenotti nope

i still haven't figured out $\{(z,e^z) : z\in \Bbb C\}$ :/

Alessandro Codenotti

Did you think about the hint I gave you?

Liad

that $e^z$ is periodic?

Alessandro Codenotti

08:45

Yes

Liad

hmm. im not sure how to use it. i think it can be shown that the set's closure is the whole space, but how to "project" (i know projection isn't good) it to $\Bbb A ^1$ is not clear to me @AlessandroCodenotti

maybe looking at $z$ in a circle of radius $2\pi$ ?

Alessandro Codenotti

The projection always works in the same, $(x,y)\mapsto x$ or $(x,y)\mapsto y$, you can think about this problem in terms of projections if you want, but which coordinate are you projecting on?

Liad

the second one would be better

because as we said $Im(e^z)$ is not closed in $\Bbb A ^ 1$

s.harp

Lie groups! Does every element of a connected Lie group admit a root? If the exponential map is surjective, then yes. But its not always surjective

Alessandro Codenotti

But the projection is not closed, so the image not being closed isn't very helpful

Liad

09:03

i know, that's why im stuck.

Alessandro Codenotti

09:15

So think about my hint instead

Another hint is that often the fibers of projections are more interesting than their images

Liad

what is fibers ?@AlessandroCodenotti

Alessandro Codenotti

Preimages of points

Liad

ok so it is enough to look at the preimage of $B_0(2\pi)$

maybe looking at the preimage of $0$ ? @AlessandroCodenotti

Alessandro Codenotti

I'm thinking of projecting on the first coordinate actually

Liad

that would give $\Bbb A^1$

Alessandro Codenotti

09:27

Sure, but look at the fibers

Liad

the fibers are $\{(z,e^z)\}$

Alessandro Codenotti

What's the fiber over 0

Liad

$\emptyset$

Alessandro Codenotti

If the image of the projection is the whole of $\Bbb A^1$ how can it have an empty fiber?

Liad

i know its wrong.

its only $\{(0,1)\}$ as i wrote earlier?

Akiva Weinberger

09:38

Whoa!

If $\epsilon_{12}=1$, $\epsilon_{21}=-1$, and $\epsilon_{11}=\epsilon_{22}=0$, then $P\epsilon P^\top=\det(P)\epsilon$

Is that just a weird coincidence?

Oh wait I see

Sort of

Alessandro Codenotti

Ah, wait, I got confused, projecting in the second coordinate was good, look at those fibers

mercio

that only works for size 2 matrices

Akiva Weinberger

Hold on, so the inverse of $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is $\frac1{\det(P)}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$

mercio

yes

something like that

Liad

@AlessandroCodenotti so here the preimage of $0$ is empty because $e^z$ is never zero

Akiva Weinberger

09:44

and that last one has a superficial similarity to $P^\top$

Alessandro Codenotti

Ok let's look at the preimage of $1$ instead

Liad

so its $\{(z,e^z) : e^z =1 \}$

Akiva Weinberger

Is that actually $\epsilon P^\top\epsilon$?

Alessandro Codenotti

@Liad right, how many elements does this have?

mercio

no you are missing a minus sign because $\epsilon^2 = -I$

Akiva Weinberger

09:47

Oh OK so hold on

Alessandro Codenotti

Hmm isn't that suspicious? Can that happen with polynomials?

Liad

it can't

because polynomials have finite number of zeros, this is how i proved it's closure is the whole space

(the closure of the set we started with)

Akiva Weinberger

So $P^{-1}=-\frac1{\det(P )}\epsilon P^\top\epsilon$?

mercio

for size 2 matrices only

Akiva Weinberger

Neat

Alessandro Codenotti

09:49

Wait if proved the closure is the whole space then you're done, you already know it's not a variety

Akiva Weinberger

@mercio Right, because for 3x3 and above, the entries of the inverse matrix are quadratic instead of linear

(divided by the determinant)

Liad

my bad. @AlessandroCodenotti but itsn't it easier showing the closure is the whole space?

Alessandro Codenotti

It's the same argument

Liad

maybe i did something wrong

the closure is the zero set of a finite number of polynomials right ?@AlessandroCodenotti

Alessandro Codenotti

It is

mercio

09:51

The entries of the inverse matrix still are degree (n-1) polynomials in the entries of the original matrix, divided by the determinant

Akiva Weinberger

Right, yeah

Alessandro Codenotti

So how did you show that the closure is the whole space?

Liad

i just going over it now, doubting my argument ^^

if $f$ vanish on the set then $f(z,e^z)=0$ for all $z$

that implies that $f$ has infinite number of zeros, doesn't it? @AlessandroCodenotti

Liad

10:14

if we fix $z_0$ and look at $f(z_0,.)$ we get that this is a polynomial in $1$ variable

@AlessandroCodenotti with infinite number of zeros, so it must be the zero polynomial

Alessandro Codenotti

Yeah that works

Balarka Sen

12:38

@Albas He's quite correct

M.Math in ISI is bullshit

Secret

13:02

One thing I really want to happen if dedekind finite sets exists: Once a fool is smacked by them, they took a segmented shape that increasingly get segmented and then vanished like seeing an abstract artwork thining out of sight

Sounds like M.Math is a good test subject > : D

Secret

13:14

in The h Bar, 2 hours ago, by Physics Meta

0

Q: Is the demise of Stack Exchange (as we know it) ineluctable?

I feel concerned about a thought that I had recently. I am wondering if the current policy about duplicate questions may become problematic in the (very) long term for the health of Stack Exchange. Let me explain: as far as I understand, questions are marked duplicate when they have already bee...

discussion

I don't think questions can ever be exhausted

Balarka Sen

13:31

Hi @Eric

Jake Rose

Can anybody explain how a function can be represented as a weighted sum of Dirac delta functions?

Érico Melo Silva

hlo

Albas

14:12

@BalarkaSen She's*. Anyway, whats up

Balarka Sen

Came back home today, winter break for the next 1.5 months

Albas

Oh you got your winter break already... I still have exams

Balarka Sen

rip

Good luck!

Albas

I have to study biology man, it's just crazy. Why do I have to care about how many offsprings a fruit fly has ...

Balarka Sen

RIP

Astyx

14:38

I don't get a break after my exams :(

Balarka Sen

Sad

Prashin Jeevaganth

15:00

Does anyone know how to calculate the number of ways to put 12 students into 4 groups of any size(excluding size 0)?

vzn

15:11

in theory salon, 24 secs ago, by vzn

The Greatest Mathematician You've Never Heard Of — Robert Langlands’s work may have no practical applications—and he doesn’t mind at all / the walrus

Akiva Weinberger

I've heard of him but I don't really know what he does

There's something called the Langlands program, which is a series of conjectures(?), that's all I know

Mats Granvik

@Secret Yes that is right. I plotted $n/k$ for $n=1,2,3,4,5,6,7,8,9,10,11,12$ and $k=1,2,3,4,5,6,7,8,9,10,11,12$.

Astyx

Is there a way to make the Peano axioms complete ?

By adding more axioms I mean

Leaky Nun

@Astyx yes

Astyx

By adding finitely many axioms ?

Leaky Nun

15:24

Let $T$ be all the sentences $\varphi$ in the language of PA such that $\Bbb N \vDash \varphi$

then no

by Godel's incompleteness

Astyx

Doesn't a theory need to have countably many axioms at most ?

Oh wait silly question

Leaky Nun

not really, but it doesn't change anything in this case

Astyx

Yeah I figured

Leaky Nun

so Godel's incompleteness tells you something stronger

Astyx

Cause there are only countably many propositions anyways

Leaky Nun

15:27

you can't add a $\Delta_1$ set of axioms to form a complete extension of PA

Astyx

$\Delta_1$ meaning recursively enumerable ?

Leaky Nun

@Astyx sorry, slow internet

Astyx

No issue

Leaky Nun

@Astyx $\Sigma_1$ = recursively enumerable, $\Pi_1$ = recursively co-enumerable, $\Delta_1 = \Sigma_1 \cap \Pi_1$

Astyx

Oh cool

Leaky Nun

15:36

and we know that my $T$ is $\Sigma_1$ so it can't be that one

Astyx

Yeah that's the proof right ?

You show that $T$ is $\Sigma_1$ and that $Th(\Bbb N)$ isn't

So they can't be equal

Leaky Nun

no, I'm stupid, my T is Th(N), and neither of them is $\Sigma_1$, sorry

Akiva Weinberger

The whole point of Gödel's incompleteness theorem is that arithmetic can never be complete (as long as we can describe the axioms)

Astyx

What does the thing in parenthesis mean ?

Akiva Weinberger

The whole point of Gödel's incompleteness theorem is that arithmetic can never be complete (as long as we can describe the axioms)

Astyx

15:48

(And just to make sure, the proof of Gödel tells you that the set of things you can prove is recursively enumerable and that the set of things that are true/false in $\Bbb N$ is not right ?)

user193319

If $X$ is a connected topological space, does it follow that $X$ has a single connected component (namely, itself)? This feels true, but I can't seem to prove it.

Astyx

What definitions do you use ? @user193319

But yes this is true IIRC

user193319

A space is connected if it cannot be written as a union of two nonempty, disjoint, open sets.

And connected components are maximal connected subspaces.

Astyx

Well suppose it has two connected component

And see it's not connected

Akiva Weinberger

That wasn't meant to be sent twice

user193319

15:53

Well, two connected components is easy. But a priori $X$ could be an arbitrary union of connected components.

Akiva Weinberger

Sorry

Internet was laggy

Leaky Nun

@Astyx the "proof" of Godel doesn't say that, Godel itself says that.

Akiva Weinberger

@Astyx There's an algorithm that, given a statement, tells us whether or not it's an axiom

is what it means

Astyx

Ok right

Akiva Weinberger

An algorithm that halts and tells us "yes this is an axiom" or "no it's not"

Leaky Nun

15:54

so basically $\Delta_1$

Akiva Weinberger

That's why just adding "all true things" doesn't count

There's no axiom that tells us whether or not a given statement is true

Leaky Nun

but wiki says $\Sigma_1$ so I don't know

maybe $\Sigma_1$ suffices

i.e. the algorithm says "yes this is an axiom" or times out

Astyx

I was taught $\Sigma_1$ IIRC

Leaky Nun

yeah sorry

Astyx

@user193319 Suppose it has at least two then

Akiva Weinberger

15:56

@user193319 What does "maximal" mean

Astyx

And take one of the two and the union of the rest

user193319

@Astyx But are the connected components open? I know they are closed.

Akiva Weinberger

Not necessarily

In $\Bbb Q$ the connected components are points

and points aren't open in $\Bbb Q$

Astyx

It depends on the topology though

user193319

@AkivaWeinberger $C$ is maximally connected if whenever $K$ is connected and $C \subseteq K$, then $C=K$.

Akiva Weinberger

15:58

So if the entire space $X$ is connected, then $X$ is the only maximally connected set

so there is only one connected component

Bye

Astyx

There's path connectedness and connectedness though

I'm pretty sure $\Bbb Q$ is connected

user193319

@Astyx No. If $r$ is any irrational, $(-\infty, r) \cap \Bbb{Q}$ and $(r, \infty) \cap \Bbb{Q}$ separate $\Bbb{Q}$.

Astyx

Huh

Fair enough

Alessandro Codenotti

@Astyx $\Bbb Q$ is totally disconnected

Leaky Nun

@AlessandroCodenotti maybe $\Bbb Q$ is a singleton :P

Astyx

16:05

Yeah my brain just broke for a second

user21820

@LeakyNun Th(N) cannot be Σ[k] for any natural k, otherwise it can be decided by the k-th Turing jump, which can be captured by a Σ[k+1] arithmetical property, which is impossible by the relativization of the incompleteness theorem.

Just in case you wanted a quick justification for that claim.

Leaky Nun

thanks

I'm also wondering, if r1 and r2 are real numbers such that {q rational | q<r1} and {q rational | q<r2} are Δ[1], does it follow that {q rational | q<r1+r2} is also Δ1?

it would be very stupid if it wasn't

but I can't prove that it is

@user21820

Astyx

In my exam we proved that it was but we assumed operations to be computable in the coding of rationnals we had

Leaky Nun

@Astyx how does the proof go, briefly?

I mean, rational addition is obviously computable

Astyx

We proved that for a number to be $\Delta_1$ you have to be able to compute a sequence of rationnals of which it is the upper bound

user21820

16:16

@LeakyNun I didn't think through carefully, but... Δ1 is equivalent to computable, right? Computable reals are closed under addition. And given any real r we have { q : q∈Q and q<r } is recursive iff r is a computable real.

Astyx

Same for lower bound

Computable reals form a closed field

Leaky Nun

@Astyx inverse...

Astyx

So once you have that it's easy to compute the sum of two rationnals, so it's easy to bound below and above

inverse ?

Leaky Nun

what do you mean for a number to be $\Delta_1$?

user21820

@LeakyNun Also there. If a computable real is nonzero, its inverse is computable too.

Astyx

16:19

The definition we had is that you can enumerate $\{q\in\Bbb Q, a\le r\}$

Leaky Nun

@user21820 how do you prove your last statement?

Astyx

And $\{q\in\Bbb Q, a\gt r\}$

Leaky Nun

I've heard that it's very far away from the truth

great @Astyx

so there isn't like a simple formula?

Astyx

What for ?

Leaky Nun

to express q < r1 + r2 in terms of ?q < r1 and ?q < r2

where ?q means a rational number (generally distinct from q)

Astyx

16:23

Sure

Oh wait

No

Not that I can think of at least

Leaky Nun

how do you go from sequence to dedekind cut?

Astyx

One direction is obvious

Leaky Nun

neither direction is obvious, realy

Astyx

Well if you can enumerate $\{q\in \Bbb Q, q\le r\}$ you already have a sequence of which $r$ is the upper bound

Leaky Nun

how?

Astyx

16:25

Take the enumeration

user21820

@LeakyNun If { q : q∈Q and q<r } is recursive then you can compute arbitrarily precise approximation of r, so r is computable. But if r is computable, I might be incorrect in claiming { q : q∈Q and q<r } is computable.

Leaky Nun

ok

@Astyx I’m stupid

Astyx

Neh, these things drive one crazy

user21820

@Astyx I don't know why people want to use Dedekind cuts... =P

Astyx

I guess it makes life easier

user21820

16:27

No?! Cauchy sequences are much easier in my opinion.

Astyx

We used Dedekind cuts to prove that it's arbitrarily approximable

user21820

@Astyx I mean we can just never introduce Dedekind cuts, and hence never have to bother about them.

But back to LeakyNun's question, I don't know how to make a counter-example.

Astyx

For the other way around, you take you're computable sequence $u_n$ and compute $M_n = \sup(u_1,\dots, u_n)$, then a rationnal if it's $\le M_n$, then $M_{n+1}$, etc. And everytime $M_n$ gets bigger you start over for the rationnals

Balarka Sen

Ultrafilters are the best way to construct reals :3

Astyx

We're not trying to construct reals

We're trying to compute them !

Answering this is giving me anxiety because I feel like I failed explaining this in my exam ..

Akiva Weinberger

16:32

I think Turing, in his original famous paper, said that if $\gamma$ turned out to have a finite decimal expansion (which it might, we don't know), then the algorithms we have for determining its digits might never halt

because they'd work by giving tighter and tighter bounds on the number, which will eventually be $(0.57\dots 39999,0.57\dots 40000)$

(assuming the last digit is a $4$, which I chose randomly)

so the algorithm will never be able to find that last digit

$\gamma$ would still be computable in that case anyway - just have the algorithm write out the finitely many digits - but it's still weird

Like, we don't have a constructive proof that $\gamma$ is computable.

This is weird because it's dependent on the base we're using

user21820

@LeakyNun Oh so I see my claim is true non-constructively. If r is rational, clearly { q : q∈Q and q<r } is computable. If r is irrational, then given any rational k/m we can compute m·r until it clearly falls between two integers, so we know whether k<m·r or not.

Leaky Nun

@Astyx does the exam assert my statement or ask you to prove or refute it?

Astyx

Assert it

Leaky Nun

great

user21820

@AkivaWeinberger My above comment doesn't seem to rely on bases.

mercio

16:37

o..o

Astyx

It asserts the set of computable reals is a closed field

Leaky Nun

using my formulation?

mercio

a closed field where equality is not decidable

Leaky Nun

Cauchy or Dedekind?

Astyx

Using the formulation $\alpha$ is computable iff $\{q\in \Bbb Q, q\le\alpha\}$ and $\{q\in \Bbb Q, q\gt\alpha\}$ are recursively enumerable

mercio

16:38

actually how are you going to do divisions if you don't know if you are dividing by 0 or not

Leaky Nun

you must be kidding me

Astyx

Or rather the sets of coding for those rationnals

Leaky Nun

@mercio make it a partial function

(I don’t know how deeply you want to dive into this, mercio

how about an easier question

user21820

@mercio Given a program encoding r (as Cauchy sequence), you can computably construct a program encoding the sequence of the reciprocals of the approximants. If r was nonzero, that program would encode the inverse of r. But the axioms for RCF don't have a division function-symbol.

Leaky Nun

@Astyx how to prove that if r is computable then so is -r?

mercio

16:41

by making a special case for when $r$ is rational

Astyx

My guess is you have 0

And it is assumed substraction (of rationnals) is computable

mercio

do the operations have to be computable ?

Astyx

So if you have a superior sequence of an irrationnal (the rationnal case is obvious) you take it's negation and you get an inferior sequence

Leaky Nun

well this is not very un-disappointing

Astyx

(by inferior/superior sequence i mean enumeration of things above/below that real)

Nothing transcendental tbh

Leaky Nun

16:44

So this isn’t a computable function R -> R

Astyx

Just technical stuff

user21820

@mercio As a field, no, because all we are asking is whether the set of computable reals under normal arithmetic forms a field or not. But it is clear (at least for Cauchy computable reals) both the addition and multiplication operations and additive inverse are themselves computable.

Astyx

@LeakyNun What isnt ?

Akiva Weinberger

@user21820 I wasn't talking about what you said, I was talking about what I said

Leaky Nun

@Astyx negation

Astyx

16:45

I think it is

user21820

@AkivaWeinberger Yes I was saying that my comment seems to be the same flavour as your comment, just without the bases.

Leaky Nun

@Astyx i think it isn’t because you need to decide whether the number is rational

Astyx

Not really

That's only for simplicity of the proof

Leaky Nun

then how?

mercio

I agree with Leaky on this one

Astyx

16:47

But otherwise you just need to remove the supp from the upperbound (which you can do by keeping the sup of values already seen and enumerating it everytime you come accross something bigger)

user21820

@Astyx Are you saying additive inverse of Dedekind computable reals is computable? I don't think so...

Astyx

For the otherway around though it might be trickier

@user21820 Assuming all operations are computable that is

user21820

@Astyx What is the meaning of that? We of course use a standard encoding of rationals, but what is your program that computes the additive inverse?

Astyx

In my exam they didn't talk about a specific encoding, they just said we had an encoding for which +, -, * and / are computable.

mercio

yes but only for rationals

Astyx

16:50

But isn't the additive inverse just negating the first bit in standard representation ?

mercio

can you re explain what your set R is ?

Akiva Weinberger

Would you say that Moore's law is descriptive or prescriptive?

user21820

@Astyx Choose your favourite encoding and demonstrate a program that does it.

Astyx

I don't know much about that

I'm not saying it's possible, I'm saying we assumed it

mercio

an encoding of rationals for which + - * and / are computable doesn't mean you can assume - is computable for all reals

user21820

16:52

@Astyx Are you sure the exam assumed that the Dedekind cut encoding of computable reals supports this?

Semiclassical

@AkivaWeinberger it's a little bit of both. As an observation about how computational power was increasing over time, it's descriptive. But people then went and used it to set growth targets, in which case it's prescriptive

user21820

As I stated above, the Cauchy sequence encoding supports +,−,· and division by nonzero, but it's still impossible to determine equality to zero so you cannot get computable division in general.

Astyx

@mercio The set of reals such that the sets of encodings of the rationnals of the sets $\{q\in\Bbb Q, q\le\alpha \}$ and $\{q\in \Bbb Q, q\gt \alpha\}$ are recursively enumerable. But you might not want to dive into that

mercio

I want to dive into that

recursively enumerable means ?

user21820

@mercio A program enumerates it (but may have to run forever to do so).

mercio

16:55

so we have a program that takes a rational number and tells us in finite time if $q \le \alpha$ or if $q > \alpha$ ?

Astyx

@user21820 What do both those terms mean ? I'm very new to this subject. The goal of the exam was to construct the field of computable reals over 4 or 5 questions, assuming we had a computable coding for rationnals for which addition, substraction, multiplication and division are all computable

No

We have a program that spits out $q_1\le \alpha$, then $q_2\le \alpha$, ...

mercio

it's basically the same thing because your two sets partition $\Bbb Q$

Astyx

And such that $\{q_i, i\ge 1\} = \{q\in \Bbb Q, q\le \alpha\}$

It's not the same thing at all

user21820

@Astyx I simply don't believe the claim you made, whether it is an accurate representation of your exam question. @LeakyNun said he doesn't think the negation of Dedekind cut computable reals is computable. I also don't.

mercio

I also don't

Akiva Weinberger

16:57

There are more transistors than leaves

mercio

besides, it is also pretty hard to tell if a Turing machine will describe a real number or not

Akiva Weinberger

on the planet

mercio

that's scary

Astyx

I'm saying you only need a finite time to know if a rationnal below $\alpha$ is below $\alpha$. You might never know wether a rationnal above $\alpha$ is not below $\alpha$

mercio

but isn't $\{q \in \Bbb Q, q > \alpha\}$ recursively enumerable ?

Astyx

16:58

Yes

That's why you need both

Oh my bad

I misread what you meant

Yes you're right

mercio

and I suppose you want your field to be a quotient of the set of turing machines that do that

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$Mathematics$ Mathematics