Mathematics - 2014-08-22 (page 1 of 3)

I'll grow past it. I just need to finish Yu Yu Hakusho/Tokyo Ghoul and watch Hunter X Hunter until the end (which is in about 5 weeks). I might hold back on the Naruto episodes for a while so that I can marathon them. Then again, it might not be a good idea to clump the episodes together. Decisions, decisions ...

skullpatrol

Khallil

I've done that experiment before. It's to do with damping, isn't it?

I can't remember the name ...

skullpatrol

dunno, it doesn't look real

Khallil

My one didn't work, so I wouldn't know!

Balarka Sen

12:08 AM

wierd. it looks expanding instead of damping.

Khallil

It's really strange.

It must've been tampered with somehow.

Any of y'all heard of ZOIDS?

Mike Miller

It damps eventually.

They're all vibrating at different frequencies

Balarka Sen

@MikeMiller It should. But it doesn't look so.

Mike Miller

Which is why you see that change occuring

Khallil

Mike Miller

12:10 AM

@BalarkaSen Not in the video, because it doesn't go long enough :)

skullpatrol

Khallil

What's that? It looks awfully cool!

(Also, what time is it where you are @skullpatrol?)

Are you still here, @BalarkaSen?

Balarka Sen

hmgrmh, wha?

sorry dozed off

Khallil

It's about 04:50 where you are, right?

Balarka Sen

it's 5:50

Khallil

12:21 AM

:O

Ghrfeghl, night.

It's only 01:21 here and I feel like a zombie.

Balarka Sen

night haha interesting remark

Khallil

I promise I'll do set theory tomorrow!

Hold me to that promise, good sir @BalarkaSen!

Balarka Sen

light light everywhere. not a drop ... hick ... of darkness.

and i am out

Jasper Loy

3:16 AM

@TedShifrin Is it important to study infinite-dimensional Banach manifolds as covered in say Lang's Fundamentals of Differential Geometry? It is one of the few books I know that uses Banach manifolds and Hilbert manifolds.

Chris's sis

6:44 AM

Greetings

skullpatrol

Greetings

Chris's sis

@skullpatrol Hey there. How is it going?

skullpatrol

Fine thanks, how are you?

Chris's sis

@skullpatrol I'm admiring my last creation. :D

skullpatrol

icic

Chris's sis

6:48 AM

@skullpatrol $$\int_0^1\sqrt{x}\tan^{\large 1/2^2}(2\arctan(x))\tan^{\large 1/2^3}(2^2\arctan(x))\tan^{\large 1/2^4}(2^3\arctan(x))\cdots \ dx = \log(2)$$

skullpatrol

nice

Chris's sis

Yeah, it's crazy awesome. :-)

rehband

@Chris'ssis =O

Chris's sis

:-)

Chris's sis

7:03 AM

bbl (some work to do now)

skullpatrol

later

gideon

7:48 AM

hey folks. I was wondering, is it a good idea to start studying calculus with Apostols book? Volume 1?

I generally know calculus, right up to what differentiation and integration is, Although not intuitively.

Math (Especially calculus) doesn't come all to natural to me but I've been working on it. I've started studying right from trigonometry again.

Will getting apostols first book be useful? Or will it be too much?

skullpatrol

The best way to find out is to try. If you don't like it there are plenty of other textbooks to study from.

I would recommend going to the math section at a university library and start "browsing."

Chris's sis

8:06 AM

Prove that $$\sum_{n=1}^{\infty} \frac{1}{2n-1}\log\left(\sqrt {\frac{4n-1}{ 4n-3}}\right) = \arctan(1)\int_0^{ \arctan(1)} \frac{\tan(x)}{x} \ dx$$

Huy

8:22 AM

I'm slightly confused. How is
$$|\Delta u|^2 = \sum_{i,j} (\partial_i(\partial_i u \partial_j^2 u) - \partial_i u \partial_j^2 \partial_i u)$$?

Got it. I kept trying to get from the LHS to the RHS but the other direction is straight forward.

Chris's sis

$$\left(\frac{3}{1}\right)^{1/2}\cdot \left(\frac{7}{5}\right)^{1/6}\cdot \left(\frac{11}{9}\right)^{1/10}\cdot \left(\frac{15}{13}\right)^{1/14}\cdots =\exp\left(\arctan(1)\int_0^{\arctan(1)} \frac{\tan(x)}{x} \ dx\right)$$

2

(newly created)

rehband

@Chris'ssis Mindboggling

Chris's sis

@rehband This is just the beginning. I hope I'll do great things in the future, I'm full of ideas.

rehband

@Chris'ssis I'm sure you will do fantastic. And complete your book asap! :P

Chris's sis

:D

MickLH

8:34 AM

@Chris'ssis Great things, are just the beginning for you.

rehband

@Chris'ssis Your work ethic is contagious

MickLH

I hope you'll find more proper recognition in the future, maybe through converting some engineering problems into some of these magical closed form solutions

Chris's sis

@rehband hehe, a crazy amount of work is needed, and the pleasure you get after that is very high.

@MickLH At my last job I used a lot my knowledge and produced some great changes.

(being good is not enough for everybody)

AndrewG

@gideon I personally liked Apostols books, but different people have different opinions. He has a nice mix of calculus and linear algebra, and if I remember he doesn't shy away from the proofy aspects of things like some books.

@Chris'ssis That's a jaw-dropper :O

MickLH

8:56 AM

@Chris'ssis I appreciate your contagious work ethic as well, if it weren't for your inspiration I'd not have dug deep enough to find the same passionate interest in hypergeometric functions as I did. Without those insights, I would never have created my latest musical synthesizer which has brought me so much joy, and brought so much freedom of expression to my musical hobby.

2

Chris's sis

@MickLH Glad to hear that. :-)

brb

Huy

9:26 AM

@DanielFischer: I'm slightly confused, is $|\Delta u|^2 = |\nabla^2 u|^2$ for $u \in C_c^\infty$? I would have said yes, but then a lot of what I'm reading doesn't make sense.

Daniel Fischer

@Huy Depends on what is meant with $\nabla^2$.

Huy

@DanielFischer: I have no idea but would have assumed $\nabla \cdot \nabla$?

Daniel Fischer

@Huy Then $\nabla^2 = \Delta$. It would be different if $\nabla^2 u = \nabla(\nabla u)$. What that you're reading doesn't make sense?

Huy

Maybe context would help. We're trying to find a-priori estimates for the Dirichlet problem. For $u \in C_c^\infty(\Omega)$ we have
$$|\Delta u|^2 = \sum_{i,j} (\partial_i(\partial_i u \partial_j^2 u) - \partial_i u \partial_j^2 \partial_i u)= \sum_{i,j} (\partial_i(\partial_i u \partial_j^2 u) - \partial_j(\partial_i u \partial_i \partial_j u)) + \sum_{i,j} |\partial_i \partial_j u|^2,$$
but the last term is equal to $|\nabla^2 u|^2$. If this was equal to the LHS, then the first term on the RHS would be zero but I don't think that is necessarily the case? The lecture notes call that term t…

@DanielFischer

Daniel Fischer

@Huy There $\nabla^2 u$ is the matrix of second derivatives (Hesse matrix), not $\Delta u$.

Huy

9:40 AM

I have never seen that notion before. O_o

@DanielFischer: And I guess $\sum_{i,j} |a_{ij}|^2$ would be the corresponding norm?

Daniel Fischer

However, since the other sum consists only of derivatives of $C_c^\infty(\Omega)$ functions, the other sum integrates to $0$.

$\sqrt{\sum \lvert a_{ij}\rvert^2}$

Afk for a while, bbl.

Huy

Ah, right. Thanks.

gideon

9:55 AM

@AndrewG went and clicked buy! :)

I did enjoy linear algebra a LOT. However I've been on a quest to study calculus on my own for years now..

but haven't been able to grasp it to a level where I can natural think in terms of calculus.

Balarka Sen

@MatsGranvik This seems right up on your alley.

Bunch of determinants and recursions to calculate zeros of Riemann zeta.

It's somewhat efficient too.

WAT. I just noticed who the author is.

Khallil

Matiyasevich?

Also, good morning!

Balarka Sen

Yes. He solved the Hilbert's 10th problem.

When did he started working with RH?

Khallil

Woah.

He probably did it on the side, in secret.

Balarka Sen

I don't think so.

He is one of the many who proved Hilbert's 10th

Khallil

10:11 AM

How did Euler find all of those values of $\zeta(s)$ where $s$ is an even positive integer?

Balarka Sen

Sum manipulation.

Khallil

Were they all derived from $\zeta(2)=\pi^2/6$?

Balarka Sen

No, no.

But essentially all of them can be derived by a generic method of complex analysis.

There are real analytic methods, but I am not familiar with them. Ask @Chris'ssis

Khallil

Oh, that's something I'll eventually get to. I'll hold the questions on those until later.

It's all really mind-blowing stuff.

Balarka Sen

I don't really care about the special values though.

I am interested in the connection with primes.

Khallil

10:15 AM

Is the RH to prove whether or not there are zeroes of $\zeta(s)$ for $s$ strictly between 0 and 1?

Balarka Sen

Nah, man.

Khallil

Oh. What is it then?

Balarka Sen

RH can't be elementarily stated.

At least, not with the series definition of zeta

Khallil

Actually, did you write about it in your forum thread?

Balarka Sen

yes

it took 3 large introductory posts just to state the conjecture.

@Khallil There are some kind of elementary versions of RH but they are all unintuitive.

Khallil

10:21 AM

Oh, it turns out I already have an MHB account.

Balarka Sen

what's the username?

Khallil

Strawberry.

(Without the full stop.)

Balarka Sen

wat

Khallil

I joined on the 1st of August last year! :O

Balarka Sen

There you go.

Khallil

10:28 AM

^_^

I think the reason I joined was a result of looking at ZaidAlyafey's differentiation under the integral thread. That and his other advanced integration threads too.

Balarka Sen

Yeah, he has quite a few tutorials on integration.

Khallil

I might check a few of them out sometime soon.

Before all that, I've gotta make some breakfast.

Balarka Sen

have fun!

ah, just figured out how to complex plot in pari.

Alec Teal

Morning

Balarka Sen

Morning, @AlecTeal

skullpatrol

10:33 AM

hi everybody

Balarka Sen

hello to you too skull

Mats Granvik

10:53 AM

@BalarkaSen Thank you for the link @BalarkaSen

Balarka Sen

No problem

Khallil

11:07 AM

𝒞

What is that symbol?

Also, what's it's $\LaTeX$ code?

Khallil

11:21 AM

Happy birthday, @JasperLoy!

N3buchadnezzar

ahoy

Vrouvrou

hello please

skullpatrol

hello thank you

:-)

Vrouvrou

looooooooooooooool what is the theorem who tel us that the zero of an ivesible function is isolated

please

Balarka Sen

@Khallil err. $\mathcal{C}$?

ahoy @N3buchadnezzar

Khallil

11:32 AM

No, it's not that, @BalarkaSen. Do you just see a plain C in this symbol '𝒞'?

Balarka Sen

That seems like a cursive 'b'

Doesn't work.

Vrouvrou

@ChristopherA.Wong

Khallil

$\mathscr{C}$

I found it, @BalarkaSen.

It's a set of letters \mathscr{...}.

Balarka Sen

ah

Khallil

$\mathscr{a}, \mathscr{b}, \mathscr{c}, \mathscr{d}, \mathscr{e}, \mathscr{f}, \mathscr{g}, \mathscr{h}, \mathscr{i}, \dots, \mathscr{A}, \mathscr{B}, \mathscr{C}, \mathscr{D}, \mathscr{E}, \mathscr{F}$

Balarka Sen

11:42 AM

$\mathscr{A}$

It only works for upper case letters

Khallil

It does? It works for lower case ones too!

(See above!)

Balarka Sen

where?

$\text{a}$. $\mathscr{a}$.

Is there any difference?

I see a difference.

?

11:46 AM

wat

Khallil

Do they look the same to you?

This is puzzling.

Balarka Sen

??

Khallil

Hmm ...

It might be that I have a $\LaTeX$ program installed on my comp.

That, and perhaps it's included packages, might be responsible for the difference.

Alizter

@BalarkaSen Someone asked why $\Bbb Z/n\Bbb Z\cong \Bbb Z_n$. I showed it by showing that set theoretically they are exactly the same. Then I whipped out the identity aut for the iso :)

Balarka Sen

@Alizter intuitively they are isomorphic because you can relabel the elts of $\Bbb Z_n$ with the elts of $\Bbb Z/n\Bbb Z$ leaving the operations invariant. this is exactly what you did, but with complicated names =)

although, in fact, $\Bbb Z_p \not \cong \Bbb Z/p\Bbb Z$ in a blackboard.

Alizter

11:54 AM

@BalarkaSen why is that?

Balarka Sen

@Alizter heh, i am just kidding. look at p-adic numbers.

Alizter

I never understood p-adic I have read that wiki artile a coupe of time still no significance for me for some reason

Balarka Sen

read a decent book on algebraic number theory. wiki articles are worthless.

@Alizter Do you have a copy of Dummit-Foote?

Alizter

@BalarkaSen Unfortunately no.

I would like to learn topology and some number theory soon

Balarka Sen

well, now that i think about it i am not entirely sure if it's explained in there. read about inverse limits from somewhere.

@Alizter p-adics are related to both.

Alizter

11:58 AM

@BalarkaSen Ok.

Balarka Sen

it's a kind of a entrance to the topological essence of number theory.

Alizter

I know very little topology

Balarka Sen

me too =(

Vrouvrou

tel me please what is the theorem who tel us that the zero of an ivesible function is isolated ?

Alizter

@BalarkaSen I just realised that I don't have to waste much time at school with powers trig integrals. I just read about the tangent half-angle :D

Then i can get a rational function

And be like

Martin Sleziak

12:06 PM

@Vrouvrou What is ivesible function?

Alizter

Contours beach

Balarka Sen

@Alizter tangent half-angle?

Alizter

Because nobody will understand anything I say when i whip out residues

@BalarkaSen en.wikipedia.org/wiki/Tangent_half-angle_substitution

Balarka Sen

ah

Alizter

No more chasing trig identities

Balarka Sen

12:08 PM

@Alizter are you allowed to use complex analysis though? in school?

Alizter

@BalarkaSen Usually in exams you can argue that if you are mathematically correct it is ok

but I might use this in hw then do the standard exam technique

too risque tho

Also being a smart-ass never goes down well.

Balarka Sen

haha, right

i used modular arithmetic in this exam on a problem.

i guess i won't get marks on that

Alizter

Yeah explaining to an exam board why you have residues integrating sec x won't go to well

@BalarkaSen Although I do know that somebody used FLT in an olympiad problem

2

Balarka Sen

oof

Alizter

@BalarkaSen I have some trig integrals to take care of

Balarka Sen

12:16 PM

integrals. meh.

heya @Nick

rehband

Whats an alternative way of expressing $$\Gamma(\frac{1}{n+1})$$?

Without the Gamma function

Balarka Sen

@rehband $\Gamma(1/5)$, for one, has no closed form.

Nick

@BalarkaSen: Ahoy. Great news, I do NOT have an open problem for you today :D

Balarka Sen

On the other hand you could use the multiplication formulas for some special cases.

rehband

@BalarkaSen Hm okay

Balarka Sen

12:19 PM

I dunno.

rehband

@BalarkaSen Sorry, nevermind :)

Balarka Sen

@rehband Hmm, I think I spoke too soon. I think there might or might not be a closed form for $\Gamma(1/5)$ in terms of periods of genus-2 elliptic functions. Hmm.

yes, @Nick.

rehband

@BalarkaSen Ok, I'm wondering if we put $$x_n = \frac{ \Gamma(1/1) \Gamma(1/2) ... \Gamma(1/n) }{n^n}$$, what is $$\frac{x_{n+1}}{x_n}$$ ?

Balarka Sen

I am absolutely not the person who can give you the answer.

rehband

Okay no worries :)

Balarka Sen

12:26 PM

i googled and found that most sites are expressing gamma of inverted integers in terms of periods of trig and elliptic functions instead and saying that "nothing like that is known for $\Gamma(1/5)$". I guess a genus-2 (2? what's the genus of $x^5 + y^5 = 1$?) hyperelliptic function should do the trick.

this, by the way, is the reason i am referring to hyperelliptic curves

Vrouvrou

12:52 PM

@MartinSleziak invertible sorry

Chris's sis

@rehband By Euler's reflection formula, when $n$ large, we have that $$\Gamma\left(\frac{1}{n+1}\right) \sim n+1$$

AndrewG

1:15 PM

@TedShifrin what is the genus of Balarka's example, $x^5 + y^5 = 1$? I thought the degree and genus were related by $d = 2g +1$ or $d = 2g + 2$, which would suggest it's genus 2, but wiki's genus-degree formula seems to suggest it's 6, which sounds quite absurd. What gives?

Martin Sleziak

@Vrouvrou BTW I am not sure I understand the question.

1 hour ago, by Vrouvrou

tel me please what is the theorem who tel us that the zero of an ivesible function is isolated ?

If a function is bijective, then it only has one root.

If it is injective, it has at most one root.

BTW if there are some fans of hyperbolic geometry here:

in Geometry, 20 mins ago, by rschwieb

I would be really interested in talking about this hyperbolic plane question I have: http://math.stackexchange.com/q/905123/29335

rehband

2:06 PM

@Chris'ssis That makes things clearer! Thanks Ramanujan

Chris's sis

@rehband :D

robjohn

@Khallil It's not damping, it's due to the differing periods of the various pendula

@Chris'ssis It also follows from $x\Gamma(x)=\Gamma(x+1)$ and because $\Gamma(1)=1$.

Chris's sis

@robjohn Yeap.

Mats Granvik

@robjohn Is there a name for matrices with non zero entries, that sum to zero?

robjohn

We can get a better approximation noting that $$\Gamma\left(\frac1{n+1}\right)=(n+1)\Gamma\left(1+\frac1{n+1}\right) =(n+1)\left(1-\frac{\gamma}{n+1}+O\left(\frac1{(n+1)^2}\right)\right) =n+1-\gamma+O\left(\frac1{n+1}\right)$$

rehband

2:22 PM

@robjohn Very nice. What happened in the second equality?

Rifaz Nahiyan

Hello all.
Easy question here, but cant solve it. :|

"Mary's income is 60% more than Tim's income. Tim's income is 40% less than Juan's income. What percent of Juan's income is Mary's income?"

robjohn

@rehband The second equality is the expansion of $\Gamma$ about $x=1$

rehband

@robjohn Great, got it

Mats Granvik

Zero-sum square matrices seems to be the name.

robjohn

@MatsGranvik That describes them; is it a common one?

Chris's sis

2:29 PM

@rehband robjohn is an exceptionally clever person. There is much to learn from him.

Mats Granvik

@robjohn it is a variant of the von Mangoldt function matrix.

United States Excel spreadsheet formula: =IF(OR(ROW()=1, COLUMN()=1),1,IF(ROW()>=COLUMN(),-SUM(INDIRECT(ADDRESS(ROW()-COLUMN()+1,COLUMN()‌, 4)&":"&ADDRESS(ROW()-1, COLUMN(), 4), 4)),-SUM(INDIRECT(ADDRESS(COLUMN()-ROW()+1,ROW(), 4)&":"&ADDRESS(COLUMN()-1, ROW(), 4), 4))))*(IF(AND(ROW()=1,COLUMN()=1),0,IF(ROW()=COLUMN(),2,1)))

rehband

2:45 PM

@Chris'ssis Awesome, I'll study some of his MSE answers

Chris's sis

@rehband Great. And don't forget to do your own research that is extremely precious.

Without research I would have never obtained a result like the one below $$\int_0^1\sqrt{x}\tan^{\large 1/2^2}(2\arctan(x))\tan^{\large 1/2^3}(2^2\arctan(x))\tan^{\large 1/2^4}(2^3\arctan(x))\cdots \ dx = \log(2)$$

(and many others of this type)

Or this one ...

@robjohn What do you think about the tan(x) integral, is it good enough for Putnam exam? :-)

I can already see and smell the mushroom cloud.

:-)

Balarka Sen

4:19 PM

@MatsGranvik

Mats Granvik

@BalarkaSen

Balarka Sen

@Mats Is it me or the Gram points are always near the extremas of Riemann-Siegel? Or am I being too stupid to realize why that is happening?

Mats Granvik

@BalarkaSen I don't know. I have forgotten what a Gram point is.

Balarka Sen

places where the theta function is multiple of $\pi$

i.e., points of the critical line where zeta takes real values.

@Mats Take an arbitrary gram point. Wiki says 60.3518119691... is (imaginary part of) one. I can be that the Riemann-Siegel has an extremum near this point somewhere.

sure enough, Z'(t) sign changes somewhere between t = 63 to 64.

O_O

Mats Granvik

@BalarkaSen I am not competent to comment.

Balarka Sen

4:31 PM

@MatsGranvik at least you can numerically check this phenomenon for me with mathematica.

Mats Granvik

@BalarkaSen Ok, I will give it a try.

Balarka Sen

oops, i meant 63.101867982...

sorry.

however, same would work for 60.351811969..., i am sure of that.

yes! Z'(t) sign changes in the interval [60, 61]

Mats Granvik

Balarka Sen

which ones are the gram points?

Mats Granvik

red ones

Balarka Sen

4:37 PM

no, they are the extremas of Z(t). i.e., where Z'(t) vanishes.

@Mats en.wikipedia.org/wiki/Riemann-Siegel_theta_function#Gram_points

gram points are real 't's where Arg(Gamma(1/4+I*t/4)) - t/2*log(Pi) is a multiple of Pi.

rehband

4:54 PM

@Chris'ssis It's incredible. What methods do you use to find these results? Could an undergraduate student understand what your steps were?

Balarka Sen

Tremendously ugly.

Chris's sis

@rehband I do nothing special, I learn, study and do research. The rest naturally comes to mind. Actually, I have no background in mathematics, I'm self-educated. I suppose that an undergraduated that also takes some courses with a professor should have some odds there (or not) ...

Balarka Sen

@Chris'ssis I think he asked how did you solve the integral.

Ah, that was the second question. Right.

Nevermind.

Chris's sis

undergraduate

Balarka Sen

@blue where the devil are you?

the last time you talked was 3 days ago.

in any case, i got a problem @blue help me out.

rehband

5:03 PM

@Chris'ssis I will keep working through Furdui's book, and hopefully one day i'll be able to find some result like that as well hehe

Chris's sis

@rehband by the way, have you seen 1.57.?

rehband

@Chris'ssis No, not yet. looking at it now.

Looks crazy

Have you worked on it?

Chris's sis

@rehband That solution seems nice enough. I was thinking to create something that uses that result.

rehband

@Chris'ssis There's a solution to it in the book?

Chris's sis

@rehband I also think it can be done by the Dominated convergence theorem.

rehband

5:07 PM

@Chris'ssis I'm not sure what dominated conv. theorem is. I've only heard the name before a couple of times :)

Chris's sis

Dominated convergence theorem

In measure theory, Lebesgue's dominated convergence theorem provides sufficient conditions under which almost everywhere convergence of a sequence of functions implies convergence in the L1 norm. Its power and utility are two of the primary theoretical advantages of Lebesgue integration over Riemann integration. It is widely used in probability theory, since it gives a sufficient condition for the convergence of expected values of random variables. == Statement of the theorem == Lebesgue's Dominated Convergence Theorem. Let {fn} be a sequence of real-valued measurable functions on a measure space...

rehband

@Chris'ssis Ah okay

Chris's sis

Back to my work, trying to finish some proofs.

rehband

@Chris'ssis Good luck, I will continue doing exercises

Chris's sis

@rehband Thanks! The same to you! ;)

Khallil

5:21 PM

This set theory is neat stuff, @BalarkaSen.

Balarka Sen

what made you think that?

it is neat, in any case.

Khallil

Some of these questions are just so cool that I can't help but admire set theory (or at least the minority I've seen).

I've got a question that just seems so clear but I can't prove it's truth. Unless I draw it in $\mathbb{R}^2$, that is. Would that be acceptable?

Balarka Sen

depends on what the question is

Khallil

I'm one step ahead of you! The question is to determine the truth of the equality $(\mathbb{R} - \mathbb{Z}) \times \mathbb{N} = (\mathbb{R} \times \mathbb{N}) - (\mathbb{Z} \times \mathbb{N})$.

Balarka Sen

Ah

Khallil

5:25 PM

It's to do with the Cartesian product being distributive across multiplication, right?

Balarka Sen

well, it's easy to do that by just explicitly writing the elements and using that one includes another and the other includes that proof for equality.

Khallil

I think I get what you're saying. I'm trying to use the fact that $\mathbb{Z} \subsetneq \mathbb{R}$ in my explanation.

Would it not be hard to explicitly write down the elements considering that $\mathbb{R}$ is included?

Balarka Sen

Let $(x, y) \in (\Bbb R - \Bbb Z) \times \Bbb N$

Can you show that it's in the other one also?

Khallil

Urm.

Well.

Negative, captain.

Balarka Sen

Can you prove that $(x, y) \in \Bbb R \times \Bbb N$?