Mathematics - 2018-03-30 (page 1 of 3)

Faust

00:45

Pants

pantsu

Xander Henderson

mmm... pants?

Secret

Xander Henderson

01:10

What is that, @Secret?

Basins of attraction for some dynamical system?

Secret

It's the following complex graph:

davidbau.com/conformal/#z%2B1%2Fsin(1%2Fz)

zoomed towards the essential singularity

Xander Henderson

Ah... okay! That explains it

nifty

so the black bits are images of squares in $\mathbb{C}$, and the colored bits are images of different squares, with the hue and value representing argument and modulus, respectively.

that is a relatively nifty implementation

I like it!

hash tag bookmarked

OH! It knows about IFSes!

davidbau.com/conformal/…

Abra001

that looks like my laptopscreen i broke yesterday.

Xander Henderson

ha!

it take some time to render

then monkey about with the slider

Isa

01:28

Could someone check if my solution to pde is correct ? I answered my own question :D since I didn't receive detailed help.. math.stackexchange.com/questions/2709268/… Thanks!

Secret

wow I did not knew you can put sliders, that opens up a whole new world

Xander Henderson

heh

inorite?

Leaky Nun

01:45

0

Q: Are they any algebraic expressions that are made of irrational numbers that equal a rational number?

I know about the non-proven Euler equation, but I'm wondering if there are any products of irrational numbers or sums which multiply or sum to a rational one?

hi, theDoctor :) — Kenny Lau 13 secs ago

theDoctor returns

Abra001

@LeylaAlkan if the order of permutation is what i think it is, there is $(\binom(3)(12)*\binom(3)(9)*\binom(3)(6)*1)3^4$ permutations

ps: for the first question.

Leyla Alkan

Hi @Abra001, I asked it on the site. Can you check it and leave comments there please ?

3

Q: Finding the number of permutations of $[12]$ of given orders

Find the number of permutations of $[12]$ whose order is $a)3 \\b)4 \\c)12$ As @астонвіллаолофмэллбэрг stated in the comments, it's not an easy process and I missed counting some possible permutations already.. My solution: a)I started choosing $3$ elements out of $12$ that are form...

combinatorics permutations

Abra001

hmmmm the answerer multiplied by 2 not 3 wonder who's right.

butsince i'm always self condemnatory i'llrecheck my calculations in prior.

oh well never mind i did a mistake of counting some tuples twice @LeylaAlkan the answer in main is fine and dustless.

Leyla Alkan

02:04

I was editing the post and still havent read the answer

That answer is hard to follow actually @Abra001

Abra001

well you need just to figure how many non duplicating combinations of 3 elements among 12 , and multiply them by the count of configurations constituted by a permutaion of class 3°

i'm trying to fathom the other case of order=4

Xander Henderson

@LeakyNun Is theDoctor a well known person?

did I just get drawn into some kind of troll?

Leaky Nun

@XanderHenderson he's a crank

he sometimes makes trouble here

Xander Henderson

Oh, joy.

That question seriously needs to be put out of its misery. :\

Leyla Alkan

As I first skimmed the answer , I said hell no :D But now it seems fine as you also said @Abra001

Hmm, I didnt get this part :"but realize that we've chosen these two cycles to appear in that order, so we've over-counted by a factor of 2!." @Abra001

Abra001

02:19

it's ecause he overcounted things when exponentiating 2 and 3 by itself

any permutation of that order can be its adjacent, so he divided by 3! possible permuations of permutations.

Leyla Alkan

does not ring a bell :(

Abra001

if we call a,b,c permutations of order 3, abc can be equal to cba, since the same patterns are mapped to themselves, in other terms (145)(267) is the same (267)(145)

sorry i'm not a very good explanatory instrument.

Leyla Alkan

Ohh, okay got it now. So for the b part if I take the permutation as this one (1234)(56)(78)(9 10) in this case I should divide it by $4!$ right ?

Abra001

yes, the orderings differ, while the permutation yielded is same at the end.

Leyla Alkan

Okay thanks! Your help is much appreciated than you think, thanks a lot :)@Abra001

Abra001

02:27

yw, need to call it a night, tc.

Leyla Alkan

You too..

Secret

02:53

I am currently thinking about some geometry, but nothing comes out yet

Faust

03:36

@Secret what exactly does that mean?

Secret

nothing comes out yet = I cannot think of any interesting examples to play with yet

Faust

try fiugre out what the length of a diagonal connecting two vertices is of a regular pentagon is

WDUK

Can some one explain how the eigenvector for an eigenvalue of 2 is not the vector i got but the one below it: i.imgur.com/HPB6zOB.png

i just do not see how they got that vector

do they just choose 1,0 because 0,0 is not an acceptable choice ?

Secret

03:52

zero vector is not an eigenvector because it will stay the same under any linear map

WDUK

right so they just picked some other random valid vector then

Secret

yup

Rumplestillskin

Can anyone help me with a quick sanity check regarding vector calculus. I have an expression (V^jV^j) and I want to take it's derivative with respect to V^i. The coordinate system is Cartesian so I'm not worried about index placement. Is the answer 2V^i? Like does the index change or stay as j?

Secret

d(V^j)/d(V^i)=delta^j_i?

Rick

04:12

How do I find the coefficient of a general $x^r$ in $(1-2x+2x^2)^n$?

Faust

factor it

use binomail

?

Rumplestillskin

@Secret of course! Many thanks!

Zee

Silence and math , is all we need

Kanwaljit Singh

04:50

Hello can anyone help me with this question. Present population is 125000. If annual birth rate 5.5% and death rate 3.5%. Then population after two year?

Zee

05:04

37

Daminark

06:57

Hey @Astyx, @Akiva, and @Alessandro!

Astyx

Hi

Alessandro Codenotti

Hi @Daminark @Astyx

Daminark

How's it going?

Fargle

wow

Astyx

wow

Daminark

06:58

Hey @Fargle!

wow

Astyx

It's going wow

Eddy Wally wow meme

►

Fargle

lol, hey @Daminark. What's good?

Daminark

Not too much man, how about you?

feynhat

Hello

Daminark

Hey

feynhat

07:08

I am trying to evaluate this sum: $$\sum_{n=5}^{\infty}na^i b^j c^k d^l$$ where $i+j+k+l = n$ and $0 < a, b, c, d < 1$.

Now, of course this will be convergent because $na^i b^j c^k d^l \leq n \max(a, b, c, d)^n$ and $\sum nx^n$ is convergent for 0<x<1.

Now, can I re-write the sum as: $$\sum_{n=5}^\infty \sum_{i=n-5}^{n} \sum_{j=n-5}^{n-i} \sum_{k=n-5}^{n-i-j} na^ib^jc^kd^{n-i-j-k}$$ ?

Zee

May god have mercy on our soul

feynhat

@Zee huh?

Zee

You know , a fields medalist came to my school today to give a talk , and everybody wanted to go see his talk , even though they know nothing about him, just strictly couse he is a field medalist , if you ask me , that is disgusting

Astyx

07:23

Who was that fields medalist ?

Zee

I don’t even know

I forgot the name , Harris or something

Astyx

@feynhat Why $n-5$ ?

feynhat

@Zee Did he talk about his research or was it just pop-math stuff? Because, if its the latter then its fine if people go to his talk even though they know about him?

Zee

Martin hairer

feynhat

@Astyx Actually I realized that it should be this: $$\sum_{n=5}^\infty \sum_{i=0}^{n-5} \sum_{j=0}^{n-5-i} \sum_{k=0}^{n-5-i-j} na^ib^jc^kd^{n-5-i-j-k}$$
but its too late to edit.

Secret

07:26

@Zee The best way to knew who in your social circle does not care about you, is to first become world famous, and then check which of the guests that appeared are from those that have no contact for a long time

Zee

He gave specific talk about a theorem

Astyx

@feynhat Still, why $n-5$ ?

Liad

any idea how to find the minimal polynomial for $\sqrt(3) + \sqrt(7)$ over $\Bbb Q$ ?

Zee

I just don’t understand how mathematician can be so smart and moral and yet act like teenage girls at a Justin Bieber concert

Astyx

Not all are moral

Some do science for the fame and do anything they can to get recognition

Well, maths instead of science in this context

feynhat

07:29

@Astyx the sum of the powers should be $n-5$.

@Astyx again, I didn't mentioned that in the first post. My bad.

Astyx

The media like to idolize some people

Zee

True but I noticed mathematicians tend to be more people of principle than usual people are but perhaps that’s just my shallow understanding

Astyx

Then yeah, it's correct @feynhat

@Zee Maybe your usual mathematician, but imo not the famous ones

I mean, the only ways to become famous is either to be really really, really good, or do everything you can to become famous

Alessandro Codenotti

@Liad square it, isolate the root, square it again, check that what you've got actually is the right polynomial

Astyx

(with some exceptions, for instance Hazking became famous thanks to his sickness imo)

feynhat

07:31

Actually, I arrived at this sum while finding the probability function of a discrete random variable (say $X$). It turns out that the variable will never attain any value less than or equal to 4, $P(X \leq 4) = 0$.

Liad

@AlessandroCodenotti i can see why this will work but how can you justify that the degree would be 4?

i can see why it is at most 4

Zee

Its a dog eat dog world

Alessandro Codenotti

Well how's the minimal polynomial defined?

Liad

the degree of the polynomial with minimal degree that vanishes at the point

Alessandro Codenotti

There are many polynomials of minimum degree vanishing at a point, if $p(x)$ is one so is $2p(x)$

Liad

07:36

with $a_n=1$ ... the degree wouldnt change

Alessandro Codenotti

What I'm trying to get at is that there are some properties that will tell you a polynomial is the minimal polynomial of its roots

Liad

if its irreducible for example

feynhat

Now, I know the exact values of $a, b, c, d$ and I only want to evaluate the sum upto 2 decimal places. I wrote a python script to do this. On my machine, I ran it upto n=400, the value upto 3 decimal places does not change at all after n=91. So can I safely assume, that this sum is approx. 297.309? Can I say, that further terms will never add up to more than 0.001?

Alessandro Codenotti

@Liad Right, so after doing this calculation you can check whether you got something irreducible, if that's the case it must be the minimal polynomial

Liad

ok. something that bothers me - we have that $x \ ^ 2 -2$ is the minimal polynomial of $\sqrt(2)$ over the rationals. but if a polynomial have roots(with degree<=3) it is reducible, and the minimal polynomial cant be reducible.. how is that?@AlessandroCodenotti

Astyx

07:40

@feynhat You can evaluate it precisely i think

$$\sum_{n=5}^\infty \sum_{i=0}^{n-5} \sum_{j=0}^{n-5-i} \sum_{k=0}^{n-5-i-j} na^ib^jc^kd^{n-5-i-j-k} = \sum_{n=5}^\infty \sum_{i=0}^{n-5} \sum_{j=0}^{n-5-i} na^ib^jc^k \sum_{k=0}^{n-5-i-j} d^{k}$$

Alessandro Codenotti

@Liad it has no roots over the rational, so it is irreducible over the rationals

Liad

right right. dont know why it bothered me ^^

Alessandro Codenotti

It can be reducible over a bigger field, but you only need irreducibility over $\Bbb Q$

Liad

@AlessandroCodenotti thanks!

Astyx

@feynhat And you'd have to assume $a\le b\le c\le d$ or something for things to work out nicely

$$5 - 4 a - 4 b + 3 a b - 4 c + 3 a c + 3 b c - 2 a b c - 4 d +
3 a d + 3 b d - 2 a b d + 3 c d - 2 a c d - 2 b c d +
a b c d\over (a-1)^2 (b-1)^2 (c-1)^2 (d-1)^2$$ if you ask mathematica

$$5 - 4(a+b+c+d) + 3(ab+ac+ad+bc+bd+cd) - 2(abc+abd+acd+bcd) + abcd \over (a-1)^2 (b-1)^2 (c-1)^2 (d-1)^2$$

Fargle

08:05

@Daminark lol sorry, I had to step away. I've just been reading and reading and reading.

feynhat

@Astyx: Thanks a lot.

Astyx

Glad to help

Liad

@AlessandroCodenotti the polynomial is $x \ ^ 4 -20x \ ^ 2 +16$, i hoped that i could use Eisenstein's criterion but i cant.. any idea how to show it is irreducible?

Daminark

It's aight, and nice

Astyx

08:21

@Liad None of its roots are rationnal, thus if it is reducible, it has to be the product of two degree 2 polynomials

Daminark

Hmm, is there some slick thing we could do instead though? Maybe reduce it mod some $p$?

Oh here's an idea actually

Liad

maybe reverse the polynomial?

Daminark

Factor it in $\mathbb{R}$

Liad

nvm ^^

Daminark

$y = x^2$, so this is $y^2 - 20y + 16$, giving us the factorization $(x^2 + 10 + 2\sqrt{21})(x^2 + 10 - 2\sqrt{21})$

Can that be reduced further?

No it can't!

Liad

08:30

it cant the determinant is negative

Daminark

So now that's our factorization in $\mathbb{R}$. But this factorization is unique in $\mathbb{R}$ and it's not in $\mathbb{Q}$. So we're done

It's not the most satisfying way but it works

Liad

yea if i wont think of something else i will use that

thank you!

Fargle

Huh, by coincidence I'm doing a problem right now that has to do with irreducible polynomials.

Daminark

No problem!

Oh nifty!

Is it like, a problem in algebra? Or is it a problem that comes up just in life which happens to involve it?

Fargle

Namely, showing that $\sqrt{2} + \sqrt{3} + \sqrt{5}$ is irrational. The tactic I chose was to assume it's rational, find a polynomial in $\Bbb Z$ it satisfies, then use the rational roots theorem and the fact that the polynomial has no rational roots between $-10$ and $10$.

(10 being because clearly $\sqrt{n} < n$ for $n > 1$)

The polynomial turns out to be even, so that simplifies things, but it turns out that the minimal polynomial is $x^8 - 40x^6 + 352x^4 - 960x^2 + 576$, so there are a lot of numbers to check, even after bounding (1,2,3,4,6,8,9). Still, brute force does suffice.

Daminark

08:37

At least it's just a question of roots and it's not irreducibility. Since the constant term is a square so press F to pay respects to Eisenstein

Fargle

Yeah, unfortunately. Would have loved to use that shortcut.

Daminark

Anyway I should head over to sleep now but yeah, we're actually getting ready to do Galois theory in algebra which is pretty neat. Our professor spends a lot of time talking about how one should think about things (partially from a cognitive standpoint, also he gets philosophical sometimes)

Fargle

I guess one could also proceed by evaluating the polynomial at 1, 2, 3, 4, 6, 8, and 9, but I don't want to do eighth powers of that.

Daminark

We aren't getting that far because of it though

Fargle

What text are you using for Galois?

Liad

08:40

just a thought - maybe we can use the fact that $deg(\alpha +\beta) \le deg(\alpha) deg(\beta) $ ?

in the case $\alpha = \sqrt(3) , \beta = \sqrt(7)$

Daminark

But yeah Galois actually is quite interesting, from what our prof said it basically involves relating the ability to solve polynomials via radicals to the structure of fields

Dummit Foote

Fargle

Ah nice. And yeah, Galois theory is super elegant IMO. It's the natural step of applying group theory once you recognize that there are natural symmetries in the solutions of polynomial equations.

And then the insight that properties of the group correspond to solvability by radicals is honestly just some next-level shit.

Daminark

Nice

Fargle

...just thought of a way I could have avoided brute force.

Daminark

How?

Fargle

08:49

Bound $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$ by their tenths-place digits, and you find that the sum is between $5.3$ and $5.6$, so is clearly not $1,2,3,4,6,8,9$.

Or any integer for that matter.

Daminark

Ah, true

I like that

Well, anyway I've gotta go to bed since it's almost 4 and I've got class tomorrow so see you around!

Fargle

lol take care.

Astyx

4 !?

That's a big number

Fargle

@Astyx 4 is the biggest number.

Astyx

That's what I thought

Mats Granvik

09:32

HerrWarum

10:22

hello

I want to show that if N is a submodule of M such that M/N is free, the N is a direct summand of M

I should probably get M = N \oplus M/N

probably have to use the equivalent conditions for when an exact sequence splits

but not sure how

Ropstah

@Secret chat.stackexchange.com/transcript/message/43701032#43701032

Jannik Pitt

10:43

I'm a bit confused: In a general topological space, does continuity always implies sequential continuity? I wanted to prove that in a first countable space the two are equivalent but it seems that for the above implication first a countability isn't needed.

Alessandro Codenotti

It's needed only for the other direction, that's right

Jannik Pitt

Okay, thanks!

Kasper

MathFoundations80: Inconvenient truths about sqrt(2)

►

Got to say, since I began working as a programmer and teaching applied mathematics next to it, I begin to appraciate more and more the points that Wildberger makes.

if you teach applied mathematics. There is little use of talking about real numbers. Why should I talk about real numbers? Why should I just say, hey, with \sqrt{2} we mean that rational number or decimal that comes as close as it may be useful for your computation. You don't actually need to talk about real numbers, it is just of no use.

Constructing the real numbers is just awkward, not intuitive, and feels completely artificial.

It has nothing to do with how the "real" world works.

And not even a computer can grasps this idea. The computer only understand rational numbers.

Fargle

11:10

By the same token, this would be saying that 1/3 isn't sensical, since it has an infinite base-2 expansion.

In fact, only the dyadic fractions would be worthy of reasoning about if we took that kind of lens.

Kasper

You can compute with rational numbers. The computer understand that kind of arithmetic.

Fargle

Computer algebra systems understand how to work with logs and roots just fine, too.

Well, modulo a sufficient definition of "understand".

Kasper

Only if the view it in a algabraic way, as an extension field, not if you see it as an infinite decimal.

That is the point he also makes in the video.

Fargle

I mean, I certainly think that fields like numerical analysis have to be concerned with the limitations of computer architecture and so on.

But the rules of logic and arithmetic permit the construction of the reals, so they are by definition reasonable.

anon

he claims "completed infinity" is nonsense, and so anything built with infinite sets (e.g. the real number system with dedekind cuts or cauchy sequences) is nonsense in his view.

Fargle

11:14

My main problem with Wildberger is that he never gives a truly compelling reason to reject the premise that infinity is a reasonable concept.

At least, compelling to my ear.

Alessandro Codenotti

@Kasper Computers don't really work with rational numbers, not even a finite subset of rational numbers, floating point addition isn't associative for example

anon

it's also unclear any computer at present "understands" anything at all. (see: chinese room argument.)

Kasper

@AlessandroCodenotti Well, we can teach computers how to calculate with rational numbers. We can not teach a computer how to interpret the real numbers system with dedekind cuts. Because a computer cannot grasp a infinite procedure.

Semiclassical

Computers do work in terms of discrete math, though

Kasper

But my point of view, more practical. Why would I teach my students real numbers? Who are interested in Mathematics in applied sense.

Alessandro Codenotti

11:18

Because a lot of real problems have solutions which aren't rational?

Fargle

Because many models of the world make the assumption that things vary continuously and that we're working in space that's locally Euclidean.

anon

you shouldn't teach applied math students about formal constructions of the real numbers (dedekind cuts, cauchy sequences) unless they choose to take a formal / foundations class from you in which that is appropriate. this is different from using real numbers. most programmers don't need to understand machine language, after all.

Fargle

At that point, you must confront the idea of continua, and of uncountability.

Kasper

@AlessandroCodenotti Why should I not just say, hey, this square root of 2 here, what we mean with that is, the fraction or decimal that comes as close as it may be useful for your computaiton?

Semiclassical

I’d nevertheless say that real numbers still matter on the grounds that the continuous is a very convenient approximation of said discrete math

Kasper

11:20

Why would I introduce dedekind cuts?

Alessandro Codenotti

@Kasper How do you know how close is close enough?

anon

Who told you to introduce Dedekind cuts in an applied mathematics class to applied mathematics students?

Alessandro Codenotti

@Kasper You don't need to, to solve real problems an intuitive understanding of the reals is plenty enough

Fargle

I mean, for an applied math class, don't. Just like you wouldn't teach how to glaze a ham in an applied math class. It's not relevant.

Alessandro Codenotti

Do you just work with the rational or do you think about them as $\Bbb Z\times (\Bbb Z\setminus\{0\})$ quotiented by the usual equivalence relation? Do you write $\frac 13$ or $[(1,3)]$?

Kasper

11:22

@Fargle Exactly, and that is I think the point that Wildberger makes. Let's not talk about real numbers, but talk about rational numbers. And if write \sqrt{2} we don't mean a real number, we mean a close rational approximation which square is 2.

anon

Formally constructing the reals is different from merely using reals, of course. We get limits -> derivatives -> calculus -> differential equations which are extremely useful in physics. I don't know how to build a computer using circuits and programming machine language et cetera but I'm perfectly fine with using a computer.

Fargle

But the rationals don't even have a square root of most numbers. It is only by acknowledging the existence of the real numbers that we even have an approximation.

It could just as easily have been true (or, well, I guess not, but bear with me) that the analytic completion of $\Bbb Q$ had no such square root.

Semiclassical

I think focusing on the rationals is a bit of a distraction

Alessandro Codenotti

@Kasper So does $x^2=2$ have any solution? Because if it doesn't we can also find circles in the plane and lines through their centers that don't intersect the circumference

Semiclassical

The point is that the workings of a computer ultimately all rest on discrete structures not continuous ones

anon

11:26

Except for error-correcting codes need to be in place because errors occur, and the rate at which they occur and under what conditions can be predicted using physics that rely on real numbers. (:

Semiclassical

Hah, point

Kasper

@AlessandroCodenotti Well, he explains that in the video. No exact solution, but an approximation exists.

Fargle

I guess the important distinction I want to draw is that infinite structures don't require infinitely large representations.

anon

Not in the rationals, but in the reals yes there is a solution. There's no reason to censor ourselves in teaching.

Fargle

You don't need infinite space to define the square root of two--it's the length of the diagonal of a unit square.

Just because the way we approach computing means that we can't exactly represent it numerically doesn't mean it's unreasonable, or unworthy of discussion in an applied context.

Kasper

11:28

@anon In the applied world, it doesn't exist. So why would we talk about it?

Fargle

That's a very bold claim.

Semiclassical

@anon though even there I presume the error codes ultimately work on a PC by picking some finite truncation of said probabilities

Though that’s a bit of a guess on my part

anon

@Kasper When you teach equations, they have solutions, and there's no reason to censor ourselves in talking about the solutions to the toy equations we teach. If your applied math uses any calculus, then of course you're using stuff made with real numbers. This is silly.

Alessandro Codenotti

@Kasper So you're fine with the idea that there are non parallel lines on the plane that don't intersect?

Fargle

In the applied world, I've almost never seen a case where someone would rather work with a kajillion Planck lengths than with an assumption of continuity.

Semiclassical

11:31

But to me the point is not so much that you “need” the real numbers in order to understand the operation of a computer. But excluding them usually gives no advantages and opens up a lot of problems

Kasper

@anon Yes, but you just "word" yourself different. You don't say, it has a real solution, you say it has an approximate solution.

Because that is the way, you are going to use it later in your applied work.

anon

Self-censorship.

Fargle

Or, you could say, "This has a solution, which certainly exists in the same fashion in which any other number exists, and we approximate it by 1.4142."

Semiclassical

@Fargle ehhh. You do run into people talking about finite size effects, eg how a crystal lattice is ultimately a discrete object with finite but large numbers of particles

Fargle

@Semiclassical Fair enough. I just mean, in my physics education, for example, we did integrals for calculating moments of inertia rather than finite but large sums over all the molecules.

Or, as another example, differential equations are often solved numerically in practice. But the only reason we can be justified in doing this is because of existence and uniqueness theorems that use the topological properties of intervals and the reals.

Kasper

11:35

@AlessandroCodenotti There is point where they come very close to intersecting. If we would zoom in, the computer, it looks like they are intersecting, but there is actually no point you could draw with the computer where they actually intersect.

Fargle

It might not be immediately relevant to the applied mathematician actually doing the work, but it is certainly relevant to the epistemology of it all.

Kasper

That is the applied way of seeing it. That is how a programmer has to work with this.

Alessandro Codenotti

Since you're so fixated on computers then you agree that addition of rational numbers is not associative? Since that's how computers do it, (0.1+0.2)+0.3 isn't the same as 0.1+(0.2+0.3)

Kasper

@anon It is the same that in primary school or highschool you will say that x^2=-1 has no solution.

Because you are not concerned yet with the solution of x^2=-1.

In the same sense that an applied mathematician is not concerned with \sqrt{2} as an exact real number.

Fargle

@Kasper But there's nothing logically inconsistent about asserting that it does have a solution, and constructing a set which does meet that requirement.

Alessandro Codenotti

11:40

@Kasper except that a lot of applied mathematicians are concerned with real numbers (and complex numbers while we're at it)

Fargle

And I disagree that an applied mathematician is not concerned with $\sqrt{2}$ as an exact real number, because if he's trying to make a 45-45-90 triangle as a support, he's going to calculate its hypotenuse's length algebraically and then approximate it.

If there were no exact solution, what would be the point of an approximate one?

anon

In applied math, the approximations that come from models are approximations of real numbers. Our models tend to use real numbers and continuity in fundamental ways, and it is important to understand how and why these models work. And there is no intuitive hurdle to talking about real numbers like there is with talking about imaginary numbers with schoolchildren.

Semiclassical

The continuous is usually a convenient approximation of the discrete

Kasper

I think I agree with that. My point is, why should I construct this set with my applied students? You guys actually say, hey, don't teach them what real numbers are, but still do math with them.

Wildberger approach is hey, be more rigorous and formal, and just explain what you are teaching.

anon

proximity is a symmetric relation :P

Fargle

11:45

@Kasper Again, don't. It's not relevant that they know how to construct it.

anon

@Kasper do you construct negative numbers or rational numbers with ordered pairs and equivalence relations, or do peano arithmetic and successor ordinals with whole numbers? Yet you still talk about whole numbers, integers, rationals because they are useful and because they are intuitive regardless of a formal construction.

Fargle

It's just relevant that they understand the properties of the continuum.

Namely, that the real numbers are an ordered field with the least-upper-bound property, which contains the rational numbers as an ordered subfield.

anon

(but not necessarily knowing the word "field" - replace with "number system" as necessary)

Fargle

But those things are generally either intuitively understood, or are explained in elementary school. (I'm thinking here of my math education, which taught me those "properties" like commutativity, associativity, etc. early on.)

Right.

Alessandro Codenotti

@Kasper Do you teach them what the natural numbers are via some axiomatic framework and then construct the integers and the rationals as quotients modulo some equivalence relation or just work with them?

Semiclassical

11:48

My attitude is that, even if one accepts the premise that computer algorithms work using discrete math, it still doesn’t follow that refusing to use continuity is a good idea

It hobbles the reasoning while usually not providing any compensating insight

Fargle

@Semiclassical Agreed, especially since we have ways of representing continuous reasoning within discrete systems.

Or at least, the results of that reasoning. (e.g. symbolic differentiation)

Secret

well, as we have discussed earlier, most of the time, both actual and potential infinities are convenient shorthands for otherwise hard to managed concepts.

Even if continuity is ultimately a fiction or at least an emergent concept, it has proved time and again to be useful

Semiclassical

That can still mean that there are scenarios where it is very smart to insist on s discrete description rather than blithely saying “oh it’s just infinite”

Secret

And I really REALLY love infinities, even if it is shown to be fiction, does not deter me from using them

Semiclassical

But in many practical cases it really is enough to assume continuity

Fargle

11:52

The example that springs to my mind is the catenary curve. If you take a wire of uniform density, hold it at both ends and let it hang in a gravitational field, its shape is a catenary. However, the function of the catenary curve, cosh x, takes on irrational values at almost all rational inputs.

Secret

@Ropstah the derivative of the identity map wrt to any variable will be the zero map, since the identity map does not have any parameters to depend on, hence unchanged

meanwhile the derivative of a conformal map depends on whether it has some variable in it, otherwise derivative of any fixed linear map is going to be the zero map

You can argue that the derivative of some nonlinear map may be conformal, similar to how the derivative of $x^2$ is the linear function 2x

Fargle

@Kasper I'm sorry if all this comes off as harsh, I don't mean it to be. I don't think there's even necessarily anything inconsistent about a finitist perspective, I just don't personally take it.

Secret

@Fargle so I am guessing lacking irrationals, the image of the caternary will be mostly empty?

Fargle

Again, for things like numerical analysis, the shortcomings of computers in handling continuous models absolutely has to be addressed, but rather than concluding that the real numbers aren't useful, I conclude that computers aren't perfect.

@Secret I mean, this is true of most continuous functions. If you restrict the identity function to only rational outputs, you've removed all the measure.

Secret

right

Fargle

11:59

The thing that's wonky about the catenary (or for that matter, sin and cos) is that rational inputs almost never correspond to rational outputs.

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