Mathematics - 2016-12-02 (page 8 of 11)

Brody

09:00

@Axoren Give me an example.

Jasch1

notation

Axoren

@Brody Do you agree that $\{0, 2, 4, \dots 1, 3, 5, \dots\}$ is not a sequence?

Brody

$d(S)$ may denote the natural density of a subset $S$ in the natural numbers

@Axoren Sure

Axoren

@Brody So, to count one natural number at a time from the set of natural numbers, we need a sequence of natural numbers.

Brody

Let's roll back some

$A(n)$ denotes the count of $a\in A\subset \mathbb{N}$ at most $n\in\mathbb{N}$. Right?

Jasch1

09:03

can you write a simple fraction latex 1 over d

the special d

Axoren

\frac 1 d

Brody

It's just an ordinary d

Axoren

@Brody $a(n)$ is a count. $A(n)$ is a set.

Brody

@Axoren Sorry, right. It's the set of blah

Jasch1

what do I have to use

for the fraction

Brody

09:05

e.g. for the perfect squares, $A(5)=\{0,1,4\}$

Axoren

@Jasch1 \frac a b is $\frac a b$

@Brody No. $A(5) = \{0, \dots, 5\}$

$A(5) \cap S = \{0, 1, 4\}$

That's an important distinction.

Brody

@Axoren Oh, I was thinking of a different $A(n)$ then

Axoren

@Brody So, now back to the sequence argument.

Since we know that we're checking natural numbers in the set from a sequence, we can only randomly for a finite region of them.

Otherwise, we stop having a sequence.

Null

why can the absolute be dropped safely at $|\sqrt[n]{n}-1|$?

(n is some positive integer)

Jasch1

@Axoren If I were to use setbuilder, would x have to be greater than 0?

Axoren

09:09

@Jasch1 What is $x$ in that context?

Jasch1

{x | 2x }

Brody

@Null Question, when can we drop $|\;|$?

Null

@Brody if the term is always positive

Axoren

@Null It can't be. $n = 0$ or $n > 2$ results in a negative on the inside. Your application for that expression may not mind having a negative.

Wait, no.

Brody

@Null *nonnegative

Null

09:11

@Brody >0

Axoren

Strictly positive.^

Jasch1

k

Brody

@Null $|0|=0$

Axoren

I messed up, I think $n > 2$ still let's the equation stay positive.

Null

ah well

Jasch1

09:11

how do I use the includes sign in latex

Null

yep, 0 is ok too!

Jasch1

the is part of

Brody

@Null $\forall n\in\mathbb{N}: n^{1/n}\ge 1$

Axoren

wolframalpha.com/input/?i=n%5E%7B1%2Fn%7D+%3C+0

Null

@Brody so to say we can safely drop this, we have to show that the nth root of n is higher or equal to 1

Axoren

09:13

@Jasch1 \subset and \in

Brody

@Null If you want to prove it, go ahead. We can just "know" this, drop the bars, and be on our way :)

But it's not a difficult proof

Null

@Brody do i have to expand the root, or work with euler e?

Axoren

Really just a limit calculation and then a local monotonicity argument.

Brody

@Null Nope, it's pretty much one step

Null

ah well, i forgot that we can pull the limit and the condition for n out of the hat

Axoren

09:17

@Null $\sqrt[n]n$ converges to $1$ and is monotonic on $[e, \infty)$

Brody

@Null Um, why not raise both sides to the $n$-th power? Since $n$ is a positive integer, the applied function is monotonic and preserves the order

Axoren

$\sqrt[e]e$ is between $\sqrt 2$ and $\sqrt[3]3$

So, we know that at all points on this line, it's above $1$.

Brody

$1^{1/n}\ge 1$ follows from $n\ge 1^n = 1$

Axoren

@Brody Unless $n$ is even, then you can have $(-1)^n = 1$

Null

we only consider positives for roots

Brody

09:20

@Axoren Why is that relevant?

Null

(principal root)

Brody

Ah, I see

Axoren

It's not relevant in this case and I'm still cookieless

Brody

hands @Axoren a cookie

Axoren

Internet cookies don't count.

Brody

09:21

Darn... and I made them with love

Null

a cookie is a lot man

you first gotta eat the half

then the quarter

...

Axoren

Yeah, privacy policies are too loose with giving them out

Brody

I just block dem shits

@Null Whoah, a whole half to begin?

Axoren

If you want to dip a cookie for 2 seconds, how fast should you pull it out to avoid dipping it for too long while not pulling it out too quickly?

Null

i find it hilarious that sites now have to say it to us, that they use cookies

because that sucks on mobiles

Axoren

09:23

I can eat an entire box of Chips Ahoy in one bite.

DHMO

Concept check: explain why (-1)^(1/3) = (-1) while e^(ipi/3) not equals (-1)

Brody

It's harder starting from nothing and nibbling twice larger portions. Certainly less fulfilling

Axoren

I avoid Zeno's Paradox.

Null

@Brody you sure can start with a quarter, if you want so

Brody

@DHMO Something something complex exponentiation yadda yadda

Axoren

09:25

I'm disappointed in the nature of mathematics.

DHMO

@brody correct

@Axoren which nature?

Axoren

You can't reverse the order of Zeno's paradox

DHMO

what means?

Null

xodarap s'oneZ

Axoren

To start with $\varepsilon$ of the cookie, then $2\varepsilon$, and then $4\varepsilon$

Brody

09:26

lol

@Axoren sure you can

DHMO

@brody you can't. there is no (infinity-1)

Axoren

In a transfinite process

Secret

@DHMO $(exp())^n=exp(n )$ don't hold for n not an integer due to the complex exponential is not injective over all angles

Axoren

But I don't think there's a way to define a sequence in a stable way on the transfinite numbers.

DHMO

@axoren even in transfinite. you can't have omega-1

Secret

09:27

e.g. $exp(x)=exp(x+m2\pi)$

Axoren

@DHMO I think I agree, but I don't know that that means.

Now I do

I agree, but that's not what I mean

Brody

@DHMO Huh?

Why not?

Axoren

$\varepsilon = \frac 1 \omega$

It's still in that vein of why that sequence doesn't make sense.

DHMO

@brody because infinity is not a number

omega is not an integer either

Axoren

It doesn't follow fundamental properties of arithmetic that allows us to do the same thing.

DHMO

09:29

@Secret I don't like that explanation.

Brody

No arithmetic on infinitesimals?

DHMO

@brody just addition and multiplication

Brody

So then the issue is...

Axoren

@Brody $\omega + 1 = \omega$, to say $\omega + 1 - 1$, you need to justify what it means to be $\omega - 1$

DHMO

and omega isn't infinitesimal

Axoren

09:30

If I remember correctly.

Is that correct, DHMO?

DHMO

if tou are talking about hyperreals, then 2^n epsilon is still close to 0

Axoren

@DHMO I agree.

DHMO

@Axoren no. 1+omega=omega but omega+1 not equals omega

Brody

Is $\varepsilon$ not an infinitesimal?

Axoren

@DHMO Right, I had it reversed. $\{1, 2, \dots, a\}$ is a set with cardinality $\omega+1$

DHMO

09:32

subtraction is not well-defined because subtraction is a short-hand for addition of additive inverses, and additive inverses are not defined for ordinals

Axoren

@Brody It is, but that's precisely why we can't start with that size of a bite.

DHMO

@axoren no. the cardinality is still omega

Brody

@Axoren Ah, so we can't simply choose a starting point for our sequence

Axoren

@DHMO You're right, I could just provide a different ordering which puts $a$ at the beginning. Do you have an example of an $\omega+1$ set?

DHMO

@Axoren it does not exist

omega+1 has the same cardinality as omega

Brody

09:34

What is the union definition of addition on $\omega$s?

Axoren

Now, I'm lost and will have to go back to my texts for this

The ordinals and transfinites are separate, that much I've gathered.

DHMO

@brody go to proofwiki and search Definition:Ordinal Addition

@Axoren "transfinite ordinal". transfinite describes a class of ordinals

Tobias Kildetoft

Awesome. My latest paper has just been accepted in Algebras and Representation Theory, pending some minor changes.

Secret

what's the topic of the paper?

DHMO

@Axoren any ordinal you can come up with which has omega has the same cardinality with omega

Tobias Kildetoft

09:36

The reviewer really liked the paper and gave a ton of detailed feedback.

Axoren

@DHMO So those ordinals just aren't used as cardinalities for any set.

Brody

So what's wrong with starting with an infinitesimal chunk of a cookie and scaling it?

DHMO

@Axoren yes

Tobias Kildetoft

@Secret Representations of algebraic groups (in broad terms).

Axoren

@Brody You'll never leave the infinitessimal chunk

DHMO

09:37

@Brody nothing's wrong, but you can never get to 1

Axoren

If anything, you'll always be at the same infinitessimal chunk forever.

DHMO

have fun passing through 0.000001

Axoren

At least Zeno made progress.

Brody

@Axoren Exactly. Sounds like Zeno's paradox going the other way

Axoren

You wouldn't even

But if you COULD start with $\varepsilon$ and make it to $1$, you'd be eating twice as much cookie every second.

Brody

09:38

Of course, rather than taking partial sums of $2^{-n}$, we half our cookies but never reach nothing. The reverse is starting with almost nothing and never reaching a half

Tobias Kildetoft

Well, you could scale by an infinite amount assuming you work in a suitable field for that.

DHMO

@TobiasKildetoft which field?

Axoren

I think I've found a way to do the reverse Zeno

Tobias Kildetoft

@DHMO The surreals would be one choice

DHMO

@TobiasKildetoft not really. you'd never leave 0

@Axoren how?

Tobias Kildetoft

09:40

@DHMO I thought you wanted to start with an infinitesimal

Axoren

@DHMO First, you perform Zeno's paradox but you don't eat halves of cookies. Instead, you leave the smaller sections off to the side Then, "after you're done", you eat the sections in reverse order.

DHMO

@TobiasKildetoft i mean you will never leave the infinitesimals

Axoren

At some point, you accept your lasted smallest chunk as a good enough estimation for the rest of the process.

Brody

@Axoren Heh, discretization

Alessandro

Don't confuse cardinals and ordinals, there are sets of order type omega+1, but it's still a countable ordinal, it's probably better to say that it has cardinality aleph_0 though because if you use omega for both as a cardinal and as ab ordinal stuff like 2^omega is ambiguous

Axoren

09:42

@Alessandro Could you describe a set with that order?

DHMO

@Axoren {1,2,3,...,0}

Axoren

@DHMO So the one I subscribed and you said no.

DHMO

@axoren you said it has cardinality omega+1

Alessandro

$\omega\cup \{omega\}$ :P jokes apart N with a bonus last element after all the integers

Axoren

@DHMO So, the order of the set is separate from cardinality?

Sophie

09:44

what is the Fourier transform of $x^2$? I'm getting something bizarre $$a\sum_{n=-\infty}^\infty\frac{e^{inx}}{i\pi n^3}$$ where $a=e^{2\pi}(2\pi^2-2\pi+1)-2$ is a bizarre constant

Brody

So any bunch of matchsticks + a matchstick

Alessandro

Yes @Axoren

DHMO

@Axoren it has order type omega+1 but cardinality omega (or aleph_0)

Axoren

Intuitively, I get the distinction now.

DHMO

@brody hmm... wut?

Axoren

09:45

I'll have to read about the distinction between order type and cardinality to understand the difference formally.

Brody

@DHMO an $\aleph_0$ bunch of matchsticks, then a matchstick

DHMO

@axoren cardinality is how many things there are; order type is how they are ordered

@Brody yes...

Secret

(rant) I really don't like to work with indexed arrays, so hard to keep track of entries when you cannot see the entire array all at once

Axoren

@DHMO Oh, so it's not as in an "order of magnitude", it's an actual result of the ordering of the sets.

Alessandro

Order type of $X$ is, if it exists, the only ordinal $\lambda$ for which there is an order preserving bijection $X\to \lambda$

DHMO

09:47

@axoren yes

Tobias Kildetoft

@DHMO If you scale an infinitesimal by its inverse then you leave the infinitesimals.

Brody

@DHMO

This sort of graphic

DHMO

@brody that would be omega times omega

Brody

I know, just an example of the visual

DHMO

sure

Axoren

09:49

I feel like it would be some sum where each other summand of $\omega$ is shrunk

Can't you see them getting smaller?

$a_1\omega + a_2\omega + \dots$ for $a_i > a_{i+1}$

DHMO

@axoren only to make it span a finite space

technically you cant make them to infinity

Axoren

@DHMO I know. It was a joke.

Also, was that last bit tongue-in-cheek?

DHMO

Exercise: prove that there does not exist an ordinal x such that x+1 = omega

Sophie

@DHMO do you know anything about Fourier series?

DHMO

@sophie x^2 is not like a wave at all

Axoren

09:53

Assume there was such an ordinal. There's some definition which trivially makes it a contradiction.

Sophie

can't I decompose anything into waves?

Axoren

I'm too tired for this at 5am

DHMO

@axoren you don't say

Sophie

well in the $[0, 2\pi)$ interval

Axoren

@DHMO Is it some open problem?

Or is that the actual gist of the proof?

Sophie

09:55

@DHMO well I wanted to Fourier transform something, and $x \mapsto x$ was trivial, so polynomials were the next best choice

DHMO

@axoren it is a beginner problem.

you laid out the framework of the proof

Axoren

Well, isn't $\omega$ the smallest ordinal larger than every natural number?

I'm trying to see what I'm missing.

Oh, I see

It's not directly from a definition

But it's a result of it.

Because it's the smallest infinite ordinal, any ordinal less than it is a natural number.

And x+1 is natural

That's sneaky, @DHMO

That's REALLY sneaky

DHMO

that's rather unrigorous

Brody

Assume $\exists x:\, x+1=\omega$. Then, $(x+0)\cup (x+1)=0\cup 1\cup 2\cup\cdots$

Dunno, lol

Axoren

@DHMO I'm hungry. Do you want me to write it in logic notation?

You'll have to pay for that in food.

DHMO

10:02

@Brody Well, firstly you need to know the definition of "+1"

Brody

Why am I up at 5?

Sophie

@DHMO and "="

Brody

Why are you up at 5? @Axoren

DHMO

@Sophie heh, and "omega"

Sophie

@Brody solving math problems, probably. That's why I am

Axoren

10:03

@Brody Because I've lost control of my life.

That reminds me, I should have some pudding mix.

DHMO

@Secret yes, but she just proved $\beta(n_1)=\beta(n_2)$ implies $n_1=n_2$

Secret

But she is not asked to prove $\beta$ being a bijection, only an injection?

> this is a proof I wrote to try to prove that β:Z→Zβ:Z→Z is one to one...does it look alright?

DHMO

@Secret oh, I thought one-to-one means bijection

thanks

Axoren

Bijection

In mathematics, a bijection, bijective function or one-to-one correspondence is a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set. There are no unpaired elements. In mathematical terms, a bijective function f: X → Y is a one-to-one (injective) and onto (surjective) mapping of a set X to a set Y. A bijection from the set X to the set Y has an inverse function from Y to X. If X and Y are finite sets, then the existence of a bijection means...

You need both injection and surjection

Secret

[Abstract algebra] Meanwhile, trying to represent the axiom of associativity as a geometric feature of the cayley table...

DHMO

10:08

@Axoren after all the only debate is on what one-to-one means

Null

has $e^x$ a root in some field?

Axoren

Then one-to-one is just injection, as Secret said. I'm going to depart as I can only pay half attention and barely make coherent math things

DHMO

@Null so you are asking if $f(x)=\displaystyle\sum_{n=0}^\infty \dfrac{x^n}{n!}$ has a root in some field

Secret

As for surjection, I often attempt to not prove by contradiction by checking the following condition $\exists S: f(S)=\emptyset$

By actually writing down the emptyset, suddenly morphisms became a little bit more straightforward (at least that's what happen when I was working through Munkres)

Sophie

Is it essential to read the topology chapter on Rudin? I'm not getting the motivation for any of this...

Null

10:11

@DHMO eh, this sum is 0 for x=0. But $e^0=1$ :s

DHMO

@Null no, 0^0 is taken as 1 here, so the first term of the summation is 1 and the other terms 0

Null

@DHMO ah

yep yep

Sophie

Well you can also define it by the limit, because you know it is a removable singularity

Null

actually, in a field where $-\infty$ is a thing

Sophie

I don't think this thing would obey the field axioms

Null

10:13

@Sophie me neither

DHMO

@Sophie what limit?

Sophie

$\lim_{x\to 0} e^x=1$

DHMO

well, firstly your field has to have the natural numbers, or else what is $n!$.

Null

why do we have to show both sides for a limit. But when we talk about the limit as x approaches infinity we only check one side.

DHMO

@Null because you can't approach infinity from the other side

Sophie

10:16

what is on the other side of infinity?

DHMO

"the other side of infinity" is an ill-defined concept

Brody

I've seen a few who'd argue it'd take you back down the real numbers. What are they?

DHMO

@Brody hyperreals

Brody

If only "God" were a finitist

DHMO

@TobiasKildetoft oh, sorry, I confused "surreal" with "hyperreal". I'm not familiar with the surreals. What is $\varepsilon \times \omega$?

Brody

10:21

Could there be a Supreme Being in a finite world?

Null

@DHMO i'd say 1. but i dunno

DHMO

Surreal number

In mathematics, the surreal number system is a totally ordered class containing the real numbers as well as infinite and infinitesimal numbers, respectively larger or smaller in absolute value than any positive real number. The surreals share many properties with the reals, including the usual arithmetic operations (addition, subtraction, multiplication, and division); as such, they form an ordered field. If formulated in Von Neumann–Bernays–Gödel set theory, the surreal numbers are the largest possible ordered field; all other ordered fields, such as the rationals, the reals, the rational functions...

@Null have a look to get some intuition.

Null

@Brody that's part of the definition of supreme being to some. (bla bla, god can build a rock he can't lift, but if he can't lift it, he's no god)

DHMO

@Brody ill-defined questions are ill-defined.

Brody

@DHMO Then what's an ill-defined answer

DHMO

10:23

::facepalm::

Sophie

define "define"

DHMO

stop messing with my brain

Brody

v.transitive : to define

$x\cup \{x\}=\omega$. Not getting very far...

DHMO

@Brody nice.

Null

@Brody can you help me? i want to bound $\sqrt[n]{n}-1$ by above. So conjugate to get a fraction or how?

Brody

10:28

@Null No calculus?

Sophie

@Null How good a bound do you need?

Brody

1000

Null

@Sophie such a good one to proof the limit of the nth root of n ;)

i have difficulty with N, because $\sqrt[n]{n}\text{?}\sqrt[N]{N}$, where n>N

Sophie

$\lim_{x\to\infty}x^{1/x}=\exp(\lim_{x\to\infty}\frac{\ln(x)}{x})$

Brody

Can you prove $\sqrt[3]{3}>\sqrt[n]{n},\,\forall\,n>3$? @Null

Null

10:35

@Brody out of my hat no.

Brody

@Null You can assume $3^n>n^3,\,\forall\, n>3$ and call it a day.

Null

@Brody procrastination expert haha

DHMO

:o $\omega-1$ is a thing in the surreals

What the hell is $\sqrt\omega$

Brody

@DHMO That's surreal

Null

@Brody $\sqrt[3]{3}>\sqrt[n]{n}$
$\sqrt[\frac{3}{n}]{3}>n$
$\sqrt[\frac{1}{n}]{3}>n^3$
$3^n>n^3$

Tobias Kildetoft

10:42

@DHMO I don't think you have squareroots in general, but subtracting 1 works as "usual". And indeed $\omega\varepsilon =1$ if one picks the correct ones (the $\omega$ being left choosing between all naturals and $\varepsilon$ being right choosing between all negative powers of 2).

DHMO

@TobiasKildetoft $\sqrt\omega$ is a thing in the surreals, according to my source

Tobias Kildetoft

@DHMO I see. I never worked that much with them.

Brody

@Null Mhm. So what can you say about $n^{(1/n)}-1$?

DHMO

@TobiasKildetoft well, one could start by working with $\dfrac\omega2$

Null

@Brody actually i'd start at $\sqrt[n]{n}-1< \sqrt[n]{n}$ to avoid the -1

well, that doesn't work, because then it's certainly bigger then epsilon

Tobias Kildetoft

10:49

@DHMO That one is not so bad since 1/2 is easy enougv to write down

Null

so i have to keep the 1

DHMO

@TobiasKildetoft but multiplication isn't

Fargle

Just realized something I should have realized a while back: if $f : A \rightarrow B$ is injective, then there exists a surjection $g : B \rightarrow A$, and vice versa.

Tobias Kildetoft

@DHMO Depends on what you mean by "easy". At least it has an explicit formula, however horrible it is.

Fargle

I knew the whole inverse business but I'd never put two and two together, so to speak.

DHMO

10:52

@Fargle that is not true.

Tobias Kildetoft

@Fargle The converse actually uses a bit of Choice

Fargle

@TobiasKildetoft Fair.

Makes sense, actually.

@DHMO A function is injective iff it has a left inverse $g$, and $g$ certainly has $f$ as a right inverse.

Tobias is right that AC is involved here, though--my textbook is just getting around to noting that.

Tobias Kildetoft

Though not the full power of choice.

Fargle

How much does it use?

Tobias Kildetoft

I think it is to some extend open precisely how strong a statement it is

Fargle

10:59

Ah, I see this has been asked on MSE before.

I'd rather have this result and Banach-Tarski be true than have this result be false, at any rate.

Alessandro

"Every surjective function has a right inverse" is equivalent to AC over ZF

DHMO

Is $f:\mathbb N\to\mathbb N$, $f:x\mapsto2x$ injective?

Alessandro

Do you put 0 in N?

DHMO

yes

@Alessandro so?

Sophie

@DHMO $f(x)=2x$, then $f(x)=f(y)\implies 2x=2y\implies x=y$

so yes, it is injective

DHMO

11:11

yet it does not have an inverse @Fargle @TobiasKildetoft

Sophie

it does have an inverse

DHMO

what is its inverse?

Sophie

$f^{-1}(x)=\frac{x}{2}$

DHMO

@Sophie Did you read $f:\mathbb N\to\mathbb N$?

Sophie

oh no I didn't. It doesn't have an inverse because it isn't surjective

DHMO

11:13

exactly

Fargle

@DHMO It does not have a right inverse, but it does have a left inverse. Let $g : \Bbb N \rightarrow N$ be $g(n) = \frac{n}{2}$.

DHMO

22 mins ago, by Fargle

Just realized something I should have realized a while back: if $f : A \rightarrow B$ is injective, then there exists a surjection $g : B \rightarrow A$, and vice versa.

@Fargle that isn't what you said ^

Secret

Suppose $f$ is not injective, then there exists $y1,y2$ such that $2y_1=2y_2$ Since $\mathbb{N}$ is an integral domain thus the set of all morphisms $g$ of the form f:x\mapsto ax, (where $a,x \in \mathbb{N}$) on $\mathbb{N}$ is injective. Therefore cancellation property holds and $y_1=y_2$ Therefore $f$ is injective.

$f$ however is not surjective since $f^\overleftarrow{}(n)=\emptyset$ for all odd integers $n$. Thus $f$ does not have a two sided inverse

Fargle

@DHMO Surjectivity is equivalent to there being a right inverse. Injectivity is equivalent to there being a left inverse.

DHMO

@Fargle fair enough

Sophie

11:15

if $\frac{1}{2\pi}\int_0^{2\pi}x^2dx=\frac{4\pi^2}{3}$, why is the constant term in the Fourier series of $x^2$, $\frac{2\pi^2}{3}$?

Fargle

And actually, to clean my definition up: let $g(n) = \frac{n}{2}$ when $n$ is even, and $\frac{n+1}{2}$ when $n$ is odd.

DHMO

Theorem: $\omega$ is odd.

Null

11:42

$f(x)=\frac{1}{f(x)+\frac{1}{f(x)+\frac{1}{f(x)+\cdots}}}$

^ can this be solved?

Fargle

@Null We have that $\frac{1}{f(x)} = f(x) + \frac{1}{f(x) + \frac{1}{f(x)+\cdots}} = 2f(x)$, giving $f(x) = \pm \frac{1}{\sqrt{2}}$.

Sophie

$f(x)=\frac{1}{2f(x)}$

Fargle

Er herp.

Secret

$2f(x)=f(x)+\frac{1}{f(x)+\frac{1}{f(x)+\frac{1}{f(x)+\cdots}}}$
Let $f(x)+\frac{1}{f(x)+\frac{1}{f(x)+\frac{1}{f(x)+\cdots}}}=2f(x)$.
Then $2f(x)=f(x)+\frac{1}{2f(x)}$

Fargle

To clarify, $f$ can be any function that takes the values $\pm \frac{1}{\sqrt{2}}$ on its entire domain.

That is, you could define $f(x) = \frac{1}{\sqrt{2}}$ if $x$ is rational and $f(x) = -\frac{1}{\sqrt{2}}$ when $x$ is irrational.

Null

11:48

so a nonconstant continuous function is impossible?

Sophie

it is possible, but if $f$ is continuous then it is constant

Fargle

Indeed. There are only two possible continuous functions.

Null

does there exist a continuous function that mimics your f(x) close enough? (thinking of the saw-function i saw yesterday) @Fargle Something like this: en.wikipedia.org/wiki/Sawtooth_wave#/media/…

Brody

What is $\frac{1}{u+\frac{2}{u+\frac{3}{u+\frac{4}{u+\cdots}}}}$ with $u=\cos x$

Null

@NaCl hi

Fargle

11:57

@Null Unfortunately, no--that function is continuous almost everywhere but the function I named (a linear transformation of the Dirichlet function) is in fact nowhere continuous.

Null

@Fargle ah, because of density right?

Fargle

@Null Yep.

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$Mathematics$ Mathematics