Mathematics - 2016-11-19 (page 3 of 10)

Fargle

03:00

@DHMO If you're talking about making $f$ a function from $\Bbb Z$ to the power set of $\Bbb R$, no, it is not surjective.

For instance, the set $\Bbb Z \subset \Bbb R$ is not mapped to, nor is $[\frac{1}{2}, \frac{3}{2})$, or many others.

In fact no function from $\Bbb Z$ to $\Bbb R$ (or to its power set) can be surjective.

meow-mix

@Fargle wouldn't your definition of mapping always have to be injective?

Fargle

@meow-mix Not at all. $f(x) = x^2$ is a mapping under my definition, but it fails to be injective.

meow-mix

oh.

hmmmm

DHMO

$\Bbb R$ can be divided into two dense disjoint subsets, namely $\Bbb Q$ and $\Bbb R \setminus \Bbb Q$.

Do you guys know any nice or nicer examples?

meow-mix

idk, thats what i learned from my algebra book

DHMO

03:02

(It is not "nice" because the other subset is $\Bbb R \setminus \Bbb Q$ instead of having a straightforward definition)

meow-mix

that mappings are relations

sorry for the miscommunications

Fargle

A function must be a relation--i.e. a subset of $A \times B$--for which every $a$ is related to one and only one $b$.

Brody

Let $f:\mathbb{Z}\to X$ where $X=\{[i,i+1)\subset \mathbb{R}: i\in\mathbb{Z}\}$.

meow-mix

yeah i know

its just that

Fargle

An injective function is one where if $aRb$ and $cRb$, then $a = c$.

meow-mix

03:03

your restriction wasn't named

Fargle

@meow-mix I named it in different terms earlier, I thought.

Brody

And we have $n\mapsto [n,n+1)$. Then is it surjective?

meow-mix

well

DHMO

@Brody then it is bijective, but $X$ is not $\Bbb R$

meow-mix

every element of $\mathbb{R}$ is mapped to by an integer

so i'd say yes?

Fargle

03:04

@Brody No--this is not a function over $\Bbb R$, and if you're taking it over $\mathcal{P}(\Bbb R)$, there are infinitely many sets not of that form

@meow-mix Injectivity and surjectivity are only defined for functions, not for general relations.

meow-mix

ok

Fargle

@Brody Oh I see now. This example is in fact surjective, but kind of by construction.

meow-mix

well, i had 5 hours last night

i think im going to go to bed

sorry for the hassle, fargle

Brody

@DHMO Sure, but as written I was curious what it was

Fargle

It's no hassle, @meow!

DHMO

03:05

@Brody ok

Brody

Although it is confusingly like $\mathbb{R}$ in a way, in the way that it is a partition

meow-mix

good night

Fargle

G'night.

DHMO

@Brody it is a partition, but it is not like $\Bbb R$ in any other way

Brody

Cya meow

Fargle

03:07

@Brody Yeah, this is where a lot of people get confused about this sort of thing. $X$ is a set of sets of reals. In that sense it bears no resemblance to the reals.

But the fact that its elements are sets of reals will confuse people endlessly.

Brody

@DHMO Of course, but the partitioning may give beginning students a misguided feeling of $\mathbb{R}$-ness, especially if they're ignorant to the very fact it's a partition

DHMO

@Brody sure

Brody

The intuition being that the set is informally misinterpreted as a sort of union of its members, I guess

Fargle

Right.

Brody

But that's neat. I didn't know (or either forgot) that $\mapsto $ can involve sets

Fargle

03:14

@Brody One way to show that $|\Bbb R| > |\Bbb N|$ is to use the binary representations of real numbers to draw a one-to-one correspondence between $[0,1]$ and the power set of $\Bbb N$, mapping the elements of $[0,1]$ to subsets of $\Bbb N$ using their binary representation as a way to denote whether each natural is in the subset or not.

As an example, $\frac{1}{2} = 0.1$ in base 2 would be mapped to the subset $\{1\}$, and $\frac{3}{4} = 0.11$ would be mapped to the subset $\{1,2\}$.

DHMO

@Fargle hint: it is not an injection

Fargle

@DHMO There are ways to fix it. :P I'm doing this ad-hoc

Brody

I don't understand the mapping process. Why does $0.11_2 \mapsto \{1,2\}$?

Fargle

@Brody The nth bit after the decimal tells you whether n is in the subset.

If the nth bit is 1, n is in the subset. If the nth bit is 0, n is not.

But as DHMO points out this construction doesn't actually give an injection: $0.11 = 0.1011111...$ so that $\{1,2\}$ and $\{1,3,4,\dots\}$ are both mapped to by $\frac{3}{4}$.

Brody

Okay, I get the mapping now with that last remark :)

Fargle

03:21

@Brody Yeah, sorry, I would have given more examples to elucidate it but I'm lazy.

It's better to start with $\{0,1\}^{\omega}$ and go from there.

And then use some clever trickery to find a bijection between $\Bbb R$ and $\{0,1\}^{\omega}$.

Brody

@Fargle No worries. Your explanation was more than sufficient looking back now

Fargle

Woohoo! That connection blew me the hell away when I was first shown it.

Brody

@Fargle Sorry, what does $\{0,1\}^\omega$ denote?

Fargle

@Brody The set of all infinite sequences of $0$s and $1$s.

@Brody More precisely, it's the set of all functions $f : \Bbb N \rightarrow \{0,1\}$.

I guess I should have written $\{0,1\}^{\Bbb N}$...I'll have to remind myself why the way I wrote it is okay.

herp

Brody

Whoo, ok now :p

Interesting

DHMO

03:26

Why is $\{0,1\}^{\omega}$ invalid? @Fargle

Fargle

@DHMO I guess it's not--it's just not immediately clear to me why it's fine. But I guess most people take $\omega = \Bbb N$...

DHMO

@Fargle but it is the very definition of $\omega$

$\omega := \Bbb N$

Fargle

@DHMO That I didn't exactly know but a quick google search helped me there.

DHMO

but $|\Bbb N|$ is another thing

Fargle

@DHMO That I know to be $\aleph_0$.

Brody

03:28

sigh

DHMO

@Fargle yes

Brody

So omega can mean the set natural numbers?

Fargle

@Brody Whigh the sy?

Brody

@Fargle ^

Fargle

@Brody It is defined to mean that, yes.

DHMO

03:29

@Fargle but in a different context... it's quite confusing

Fargle

@DHMO I assume you're meaning ordinal arithmetic.

Brody

But that's different from the one seen in ordinal stuff

DHMO

$\omega$ is defined to be the set $\{0,1,2,\dots\}$ in the sense that $4 := \{0,1,2,3\}$

@Brody how?

@Fargle yes

Brody

@DHMO I just presumed. Are they really the same?

DHMO

1 min ago, by DHMO

$\omega$ is defined to be the set $\{0,1,2,\dots\}$ in the sense that $4 := \{0,1,2,3\}$

Fargle

03:31

@Brody It's the same notion. $\omega + 1$ is the set $\{0,1,\dots\} \cup \{\omega\}$ where $\omega > n$ for all natural $n$.

Brody

@DHMO @Fargle Hmm, ok. Thank you

Fargle

Where as $1 + \omega$ is just $\{a\} \cup \{0,1,\dots\}$ where $a < n$ for every natural $n$, but this has the same order type as $\omega$--that is, its order behaves the same--so $1 + \omega = \omega$.

DHMO

@Fargle this is too complicated for @Brody maybe

Fargle

That's possible. There's a Vihart video that does this concept much more justice.

DHMO

@Fargle I'm not sure if I follow your first statement

Fargle

03:34

@DHMO About $\omega + 1$ or $1 + \omega$?

DHMO

@Fargle "$1 + \omega$ is just $\{a\} \cup \{0,1,\dots\}$ where $a < n$ for every natural $n$"

and did you mean VSauce instead of ViHart?

How To Count Past Infinity

►

(warning: none of the above video is formal)

Fargle

@DHMO $S + T$ in ordinal arithmetic means "adjoin S to T as well-ordered sets so that every $t \in T$ is greater than every $s \in S$.

Brody

Reading the Wiki now...

Fargle

So more precisely, $1 + \omega$ is $\{0,0',1',2',\dots\}$, which has the same order properties as $\omega$ (one element has no direct predecessor, every other element does)...

DHMO

@Fargle but you cannot have $a<n$ for every natural $n$

Fargle

03:36

@DHMO Sure you can. $a = -1$

DHMO

@Fargle $-1$ is not an ordinal

Fargle

@DHMO Look at my second explanation.

DHMO

when I learnt it, $1+\omega := \bigcup\limits_{n \in \omega} (1+n) = 1 \cup 2 \cup 3 \cup \cdots = \omega$

Fargle

You define the order on $S + T$.

@DHMO But then $1 + \omega$ doesn't equal $\omega$, it's a subset of $\omega$.

@DHMO Unless you take "equals" to mean order-equivalent.

DHMO

@Fargle no, both sets are equal

Fargle

03:38

$\{0,1,\dots\} \neq \{1,2,\dots\}$

DHMO

@Fargle $\{0,1,\cdots\} = 1\cup2\cup\cdots$

Fargle

@DHMO That's true, but you're conflating sets with elements here.

DHMO

@Fargle everything is a set in ZFC

$1 := \{0\}$

Fargle

$a \cup b \cup \cdots \neq \{a,b,\dots\}$

DHMO

$2 := \{0,1\}$

Fargle

03:39

I am aware of this

DHMO

@Fargle I didn't say they are equal

Fargle

@DHMO Yes, you did.

DHMO

@Fargle when?

Fargle

Ah, you edited.

DHMO

yes

Fargle

03:40

Both our definitions turn out to be equivalent at any rate.

DHMO

sure

Fargle

When you take "equals" to mean order-equivalent.

I find it better, for me at least, to think of ordinal addition as adjoinment rather than unions.

For example, how would you define $\omega + 1$?

DHMO

@Fargle $\omega \cup \{\omega\}$

Fargle

How is that compatible with the definition of $1 + \omega$? Not trying to be dense, really asking

That is, how are these the same notions of addition?

DHMO

@Fargle in $\omega+1$, the right hand side is not a limit ordinal

in $1+\omega$, the right hand side is a limit ordinal

Fargle

03:43

Hrm. Yours works--I just prefer mine because you don't have to worry about limit ordinals.

DHMO

@Fargle I do appreciate questions

@Fargle how do you define order?

Fargle

@DHMO In $S + T$, you retain the orders on each of $S$ and $T$, and make $s < t$ for each $s \in S, t \in T$.

DHMO

@Fargle how do you define order?

Fargle

(Requiring that $S$ and $T$ be well-ordered)

Ah. An order is a relation $<$ such that:
1. Either $x < y$, $y > x$, or $x = y$,
2. If $x < y$ and $y < z$, then $x < z$, and
3. $x$ is not less than $x$.

DHMO

@Fargle what is order-equivalent?

Fargle

03:48

@DHMO $S$ and $T$ are order-equivalent if there exists a bijection $f : S \rightarrow T$ such that if $s_1 <_S s_2$, then $f(s_1) <_T f(s_2)$.

DHMO

@Fargle why isn't $\omega+1$ order-equivalent with $\omega$?

Fargle

$\omega + 1$ has a maximal element, namely $\omega$. $\omega$ has no maximal element.

DHMO

using your definition?

Fargle

Say we map from $\omega + 1$ to $\omega$. Then $\omega$ must get mapped to some natural, say $n$.

And every other natural must also get mapped to some natural.

Let $k \in \omega + 1$ be the element mapped to $n + 1 \in \omega$; then $k < \omega$ but $f(k) = n+1 > n = f(\omega)$.

Therefore no order equivalence can exist.

DHMO

what is --? did you mean $\implies$ or $\iff$?

Fargle

03:51

No, it's just a connector.

Sorry >_>

Brody

the vsauce guy mentions $\mathscr{P}(\aleph _0)$. anything wrong with this?

Fargle

@Brody Haven't seen that video.

DHMO

@Brody what is wrong with $\aleph_0$?

Fargle

@DHMO And no, I meant ViHart. Let me go dig it up.

How many kinds of infinity are there?

►

Brody

@DHMO Fixed it.

DHMO

03:53

@Brody where do you see it?

Brody

@DHMO 8:58 in the vid linked earlier

Just wondering if you can take the powerset of a cardinal number (I suppose if it's identified with the set it describes?)

DHMO

@Brody you're right, $\aleph_0$ is not a set so you can't take the power-set of it.

Brody

@DHMO Gotchya. Thanks

Phew. Palpitations begone

DHMO

@Brody wait. According to wikipedia, cardinal numbers are ordinals which are sets by definition

> Formally, assuming the axiom of choice, the cardinality of a set X is the least ordinal number α such that there is a bijection between X and α.

but I have some problems with that definition

Brody

@DHMO I suppose that makes sense. Like $\{0,1,2,3\}=4$ can define the fourth cardinal, and the ordinals and cardinals coincide for finite collections

DHMO

04:01

isn't it true that under that definition, $\aleph_1 = \omega + 1$?

Brody

@DHMO What does $\aleph_1$ mean to you here?

DHMO

@Brody well usually $\aleph_1$ is equal to $2^{\aleph_0}$

Fargle

@DHMO EHHHHHHHHHHHH

Only if you take the continuum hypothesis to be true.

Though there's no reason not to, as far as I know.

DHMO

I know

well, if we define $\aleph_0 = \omega$, then $2^{\aleph_0} = 2^\omega = \omega$...

Brody

@Fargle Had the same objection but that doesn't matter insofar as DHMO's question, huh?

Fargle

04:09

@Brody Not particularly.

I'm just a pedant.

Brody

The "EHHHHH..." speaks for itself :p

Really want to star that too... but I'll abstain for the greater good

DHMO

what does $2^{\aleph_0}$ mean?

Fargle

@DHMO The cardinality of the power set of a set with cardinality $\aleph_0$.

Brody

$2\underbrace{\times 2\times 2\times \cdots}_{\aleph_0-1\text{ times}}$

(jk)

Fargle

pukes

Brody

04:15

laughs trollingly

Fargle

This comes from the fact that you can identify the power set of a set $A$ with $2^A$ (i.e. $\{0,1\}^A$).

And for finite sets $|\mathcal{P}(A)| = 2^{|A|}$

Brody

04:37

@DHMO The vsauce video wasn't bad imo. Albeit, even for an informal / popularization-type vid, still a mindf**k

Fargle

@Brody Yeah I don't ever treat Vsauce as a place for formal or complete explanations, and in that regard, it does well.

MonsieurGalois

Can I ask for the link of the community wiki?

Brody

@Fargle So it looked like the guy was constructing $\aleph_0$-sized collections of ordinals like $\{\omega, \omega+1, \omega+2,\ldots\}$ and $\{\omega, \omega\cdot 2,\omega\cdot 3,\ldots\}$ and $\{\omega,\omega^\omega, \omega^{\omega^\omega},\ldots\}$

Fargle

Yeah, I seem to remember that. It's been a while since I've seen that video though.

Brody

also, going in order I missed $\{\omega,\omega^2,\omega^3,\ldots\}$

but point remaining, it appears that these sets have an associated hyperoperation. And the video gave the impression that $\varepsilon_0$ is the solution to not having a standard way to write whatever ordinal follows all the power towers of $\omega$

@Fargle though non-standard, I'm thinking Knuth's up-arrow notation or some similar format would be equivalent, e.g. $\sup\{\omega,\omega^\omega,\omega^{\omega^\omega}\}=\omega\uparrow\uparrow\upar‌row 2$ or $\omega\,^3\uparrow 2$

Fargle

04:54

That I'm not sure of. I'm by no means an expert on ordinal arithmetic.

Brody

blasted TeX bug

Corrected: $\sup\{\omega,\omega^\omega,\omega^{\omega^\omega},\ldots\}$ $=\omega\uparrow\uparrow\uparrow 2$ $=\omega (^3\uparrow )2$

DHMO

@Fargle yeah, so $2^{\aleph_0} \ne 2^\omega$

@Brody indeed it is

@Brody yes, the axiom of replacement

@Brody also $\omega\uparrow\uparrow\omega$

Brody

@DHMO $\omega\uparrow\uparrow\omega$ is what?

DHMO

@Brody the same as $\omega\uparrow\uparrow\uparrow2$

> $$a \uparrow \uparrow \uparrow \uparrow b =
\underbrace{
\left.\left.\left. \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\}
\underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\}
\cdots \right\}
a
}_{b}$$

Brody

05:09

@DHMO I disagree, but I'm probably wrong

DHMO

@Brody why do you disagree?

Brody

at least as far as finite operands are concerned, not sure in the ordinal case

DHMO

well you already broke the rule of finiteness by applying uparrow on transfinite ordinals

Brody

@DHMO $x\uparrow\uparrow\uparrow 2\;=\;^x x$ in tetration notation

@DHMO OHHH, I misread

Sorry, I kept thinking you meant $\omega \uparrow\uparrow 2$ is the same as $\omega\uparrow\uparrow\uparrow 2$, even as I rewrote your statement above. Big brain fart

DHMO

lol

Sophie

05:16

if you have a continued fraction of the form $\sqrt{n}=[a_0;a_1,a_1\dotsb,a_n]$ and $\frac{p}{q}=[a_0,\underbrace{a_1\dotsb, a_n}_{\text{this k times}}\dotsb]$ then $p^2-nq^2=\pm 1$

I suspect this is true, no proof yet

DHMO

@Sophie rational numbers always have finite continued fraction

I also do not know what you mean by $\frac{p}{q}=[a_0,\underbrace{a_1\dotsb, a_n}_{\text{this k times}}\dotsb]$

Sophie

yes, I meant that the period is repeat a number k of times

for a finite k of course

DHMO

also, $\sqrt{n}$ should have an infinite continued fraction

Sophie

yes it does and it's periodic

DHMO

but your notation of it terminating at $a_n$ tells me otherwise

could you give us an example?

Sophie

05:19

$\frac{1+\sqrt{5}}{2}=[1]$

that is, all the terms in the continued fraction are 1

Brody

@DHMO an aesthetic disadvantage of the up-arrow is that the $n$-th hyperoperation is identified with $n-2$ arrows or $^{n-2}\uparrow$, so the notation $a[n]b$ may be preferable if we want to emphasize this replacement hierarchy's ascendancy as identified with the natural numbers

DHMO

I mean an example of your conjecture

@Brody on the other hand I do not prefer using hyperoperation lol

Sophie

@DHMO I've proved it for n=2,3,5,6

DHMO

@Sophie can you give me an example of $n$ and $p$ and $q$?

Brody

@DHMO lol, just food for thought. It's a nice alternative to introducing a whole new letter (i.e. $\varepsilon$) for conceptualization and pedagogy's sake

just im(humble)o

DHMO

05:22

@Brody what is $\epsilon$?

is it in the video? if so, where?

Sophie

okay for $n=6$, $\sqrt{6}=[2;2,4]$ then if instead of $p,q$ I had $p_k,q_k$ where k is the amount of times the period is repeated, then $\frac{p_{k+1}}{q_{k+1}}=\frac{12q_k+5p_k}{5q_k+2p_k}$ then it can be proved by induction

DHMO

I see

Brody

@DHMO Referring to the epsilon numbers. $\varepsilon_0=\sup$ $\{0,1,\omega,\omega^\omega,\omega^{\omega^\omega},\ldots\}$

DHMO

I see

heh, $\epsilon = \omega^\epsilon$

Sophie

this implies a connection between continued fractions and Pell equations that I find interesting. It also related the smallest solution of the Pell equation to the period of the continued fraction of $\sqrt{n}$

DHMO

05:26

well obviously this does not hold if $n$ is a perfect square, in which case $p^2-nq^2=0$

Sophie

that explains why the minimal solution of $x^2-61y^2=1$ is so big, because $\sqrt{61}=[7;1,4,3,1,2,2,1,3,4,1,14]$ which is an abnormally long period

Brody

@DHMO In the vsauce vid, this bit starts around 16:02 and epsilon-null introduced about 20 seconds later

Sophie

@DHMO I've been ignoring these cases. It's 3am and I can't sleep because of this

DHMO

@Brody right

@Sophie you meant $\sqrt6=[2;\overline{2,4}]$

it fails for $\sqrt8=[2;\overline{1,4}]$, where $[2;1,4]=\dfrac{14}5$ and $14^2-8\times5^2=-4$

Sophie

@DHMO I'm used to the notation with only the ; and , and no overbar, but that's ok

DHMO

05:30

and then you would argue "hey $n$ must be prime"

Sophie

it works for 6 too

DHMO

in which case I say $\sqrt{11} = [3;\overline{3,6}]$ and $[3;3,6]=\dfrac{63}{19}$ and $63^2-11\times19^2=-2$

Sophie

jeez do you always have that many counterexamples up your sleeve?

DHMO

@Sophie nah, i just used wolframalpha

Brody

In elementary calculus, when we speak of "some small" $\varepsilon >0$, are we actually considering a literal infinitesimal?

DHMO

05:34

@Brody heh, conflation

Sophie

no, it's a real number that can get arbitrarily small

DHMO

^

Brody

@Sophie Thanks

Sophie

the whole point of epsilon-delta definitions is to formalize calculus without infinitesimals (that can also be done with nonstandard analysis but that wasn't available at the time)

@DHMO you're stopping the period at the wrong point

I'm sorry I've been imprecise, let me reformulate

Brody

@Sophie I see. I was reading the MathWorld page for "Epsilon" which says: In mathematics, a small positive infinitesimal quantity, usually denoted $\epsilon$ or $\varepsilon$, whose limit is usually taken as $\epsilon\to 0$

Wasn't sure if this meaning worked (albeit tacitly) in ordinary calculus too

Sophie

05:40

@Brody this page is in the "Mathematical Humor" category, kind of fishy

Brody

@Sophie Haha, saw that too. But there are other serious pages in there as well

Probably for the bit about Erdős, while they couldn't/didn't append a normal category for the dry stuff

Sophie

@DHMO You're stopping at the wrong point! The notation is kind of getting in the way. $2+\frac{1}{1}=\frac{3}{1}$, $3^2-8\times 1^2=1$, $1+\frac{1}{1+\frac{1}{4+\frac{1}{1}}}=\frac{17}{6}$, $17^2-8\times 6^2=1$

DHMO

@Sophie so basically for $[2;\overline{1,4}]$ we consider $[2;1]$ and $[2;1,4,1]$

Sophie

and then add more cycles of 1,4. Yes

I have proven it by induction for 8 too

DHMO

05:56

It fails for $\sqrt{61} = [7; \overline{1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}]$ where $[7; 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14, 1] = \dfrac{469849}{60158}$ and $469849^2 - 61\times60158^2 = -3$

Brody

@DHMO Thanks for the help earlier. Also had fun with the transfinite stuff, though I'll be exiting with more questions than answers. Indeed, it unfortunately seems that (with the appropriate definitions) $\mathbb{N}=\omega=\aleph_0=|\mathbb{N}|$, unless any two of the definitions I've taken are incompatible

DHMO

@Brody it is quite unfortunate.

but fortunately I was wrong in stating that $\aleph_1=\omega+1$ which is not true

because $\omega$ is equinumerous with $\omega+1$

I believe $2^{\aleph_0}$ equals $\omega^\omega$ instead of $2^\omega$ since $2^\omega=\omega$

then $\Bbb N = |\Bbb N|$

Sophie

@DHMO if you stop at $\frac{29718}{3805}=[7;1,4,3,1,2,2,1,3,4,1]$ you get $29718^2-61\times 3805^2=-1$

which is just before the biggest number in the fraction, which means it has a good precision in the approximation, relative to the size of $29718$ and $3805$ themselves. And if $x^2-ny^2=1$ then $\left| \sqrt{n}-\frac{p}{q }\right|$ is approximately $\frac{1}{\sqrt{n}q^2}$

doesn't anyone find it spooky that the continued fraction of $\sqrt{61}$ is almost symmetric?

DHMO

06:18

@Sophie planetmath.org/tableofcontinuedfractionsofsqrtnfor1n102

so is every other continued fraction

Brody

Anyone here familiar with "fractional flowers"?

DHMO

@Brody what is it?

Sophie

$\frac{1766319049}{226153980}=[7;1,4,3,1,2,2,1,3,4,1,14,1,4,3,1,2,2,1,3,4,1]$ and $1766319049^2-61\times 226153980^2=1$

Brody

@DHMO It's a way to visualize (i.e. graph) the periods of the decimal expansions of rational numbers. Take the base-10 digits and arrange them in order in a circle, like the numbers on a clock. And draw line segments between the digits as given by the period of a repeating expansion, e.g. $1/7=0.\overline{142857}$

DHMO

@Brody what is special about it?

Brody

06:29

@DHMO Many of them exhibit a certain symmetry, and I'm curious why certain ones have it and others not. Also, they're pretty

DHMO

@Brody for example?

Brody

@DHMO Wolfram has a page on them, but I don't like the in-site app too much demonstrations.wolfram.com/FractionalGraphsAndFlowers

Hmm, might draw a picture

DHMO

I see

Brody

06:44

@DHMO Here's a working format for radix 10. Look at cycles in the expansions and connect the digits in order, e.g. for 1/7 do $1\to 4\to 2\to 8\to 5\to 7\to 1$

Balarka Sen

what's new

Brody

@BalarkaSen Hello :) Not much

Spending a Friday night doing homework and wasting time in a math chat, I'm so fun/busy/cool :p

Balarka Sen

nah, doing what you like is not wasting time. but don't get used to this chat too much, otherwise you'll end up procrastinating most of the time like me :P

(and many other people)

holy hell the transcript is huge. i give up.

Brody

07:00

This is the graph for the digits of the period in $\frac{18}{19}=0.\overline{947368421052631578}$. I'm thinking it's the same for all $\frac{n}{19}$ with $1\le n\le 18,\,\text{gcd}(n,19)=1$. As immediately noticeable, the graph shows vertical symmetry @DHMO

DHMO

@Brody nice

Brody

You can trace the path of digits with your fingers, starting from 9 to 4, then 4 to 7, etc. etc.

I meant cycle for the above, the period is of course $19-1=18$

DHMO

Question: comment on the name "math underflow" which certain people in mathoverflow call us

Balarka Sen

I think it's a great name

Brody

Might be taken as a little demeaning, lol

DHMO

07:10

Quite beautiful

Brody

@DHMO It is

$\frac{1}{21}=0.0\overline{476190}$ (or any related multiple thereof) is odd. Period 6, graph shows no symmetry

DHMO

07:28

@Brody you meant $0.\overline{047619}$

Brody

@DHMO Either is fine. I think the convention is to not permit leading zeroes in the repeating string, for whatever reason (though yours is slightly better imo)

Sophie

@Brody doesn't it mean which digits are in the period? Then you need to include the 0

because otherwise you'd think $\frac{1}{21}=0.0476194761947619...$

Brody

@Sophie Look again, it's just trailing

Wolfram documents have it this way too, not to say it's better or more correct

Balarka Sen

07:55

I have to decide what I want to do today

Brody

@BalarkaSen Looking at the starred comment. Is the Warsaw circle a circle?

Balarka Sen

It's not topologically equivalent to the standard circle, no.

DHMO

@Brody now that i learnt ordinals formally it makes more sense when i watch that video again

Balarka Sen

Look at a small neighborhood of some point on the vertical bit (i.e., one of the "bad points"). That's a topologist's sine curve which, as we already discussed, is not path connected.

So there are points which have non-path-connected neighborhoods. The terminology for this is "not locally path connected".

The circle, however, is locally path connected.

DHMO

So $\omega^\omega$ is still equinumerous with $\Bbb N$

Brody

08:01

@BalarkaSen Okay, was just wondering if there was more to the name

@BalarkaSen I see

@DHMO True for any ordinal less than $\omega_1$, no?

Actually, that might assume CH (edit: nevermind, it doesn't)

DHMO

but I have no idea how to prove it

Brody

@DHMO Doesn't $\omega^\omega=\{\omega,\,\omega\cdot 2,\,\omega\cdot 3,\,\ldots\}$? If so, the correspondence is visible

DHMO

no it doesn't

is that a result of being misled by VSauce?

Brody

Probably, lol

Sorry, those should be exponents, not multiples. My brain doesn't work at this hour

DHMO

it is still wrong

Brody

08:15

Not surprised tbh

DHMO

don't feel bad lol

Brody

$\omega=\sup\{0,1,2,\ldots\};\;\;$ $\omega\cdot 2=\sup\{\omega,\omega+1, \omega+2,\omega+3,\ldots\};\;\;$ $\omega^2$ $=\sup\{\omega,\omega\cdot 2,\omega\cdot 3,\dots\}\;\;$

@DHMO Too late :p ^See what I'm doing?

DHMO

yes

Balarka Sen

"I once talked to the late Soviet physicist Landau on this subject. The setting was a shingle beach in the Crimea. 'What do you think,' I asked, 'does God exist or not?' There followed a pause of some three minutes. Then he looked at me helplessly. 'I think so.'At the time I was simply a sunburnt young boy, entirely unknown, son of the distinguished poet Arseniy Tarkovsky: a nobody, merely a son. It was the first and last time I saw Landau, a single, chance meeting; hence such candour on the part of the

DHMO

those statements are correct but useless

@brody

Brody

08:25

@DHMO sigh At least the thought train wasn't as off-the-tracks as I thought

DHMO

How does the cardinality of $\Bbb N^{\Bbb N}$ look like?

Brody

At this point I don't even know what that notation means anymore

@BalarkaSen Remarkable excerpt. From where is it?

DHMO

It means $\displaystyle\times\limits_{n\in\Bbb N}\Bbb N$

Balarka Sen

"Sculpting in Time" by Andrey Tarkovsky @Brody

DHMO

where $\times$ is cartesian product

treat it as vectors of natural numbers of dimension $\Bbb N$

Brody

08:30

@DHMO Not rendering on my side

DHMO

never mind

use my second description

Brody

So you mean the set of all vectors whose components are natural numbers, with dimension countable infinity?

DHMO

yes

Brody

I want to say $\aleph_1$, but anything I type is a wild swing

@BalarkaSen Ironically, he turned out to be quite renowned

DHMO

lol

Balarka Sen

08:34

Yep

Alessandro

No, it's $2^{\aleph_0}$

DHMO

@Alessandro why?

Alessandro

If you have $2\le\kappa\le\lambda$ cardinals with at least one of them infinite then $\kappa^\lambda=2^\lambda$

In your case $\kappa$ and $\lambda$ are both $\aleph_0$

DHMO

so?

Alessandro

So $2\le\aleph_0\le\aleph_0$ and $\aleph_0^\aleph_0=2^{\aleph_0}$

DHMO

08:41

ok thanks

Alessandro

Interesting quote @Balarka, I'll see of I can find the book since I already have to go to the municipal library later

Balarka Sen

It's a very beautiful book.

DHMO

@Alessandro then what does $2^\omega$ intuitively mean?

Alessandro

@Balarka should I see some of his movies before?

Balarka Sen

Yes. He talks mostly about his works in that book, but it has more.

Brody

08:48

@BalarkaSen What translation do you use?

Balarka Sen

So I think watching some of his movies before would give a complete understanding of the man and his ideas.

@Brody Oh, the English one. I don't know any of the other major languages :)

Alessandro

@DHMO Ordinal exponentation isn't very intuitive, $2^\omega=\omega$

DHMO

I know

Brody

@Alessandro In what undergrad course might one learn this subject?

DHMO

@Alessandro but how would you intuitively explain that?

Fargle

08:53

Hello chat.

Alessandro

@Brody where I study a little bit of this stuff is thaugth in a course called "introduction to mathematical logic" (despite having little to do with logic), but you usually need a set theory course (and where I am those aren't available for undergrads)

@DHMO it's the set of functions $\omega\to 2$ with finite support, ordered lexicographically with least elements first, I don't think there's a particular intuitive way to make sense of this

Brody

That makes sense. Thank you

Fargle

Yeah, you'd really only see it in an intro-to-proofs class in undergrad (depending on the level of rigor of your university), and at higher levels in a class on axiomatic set theory/order theory.

Alessandro

@Balarka I see, what would you suggest to begin with? I've had solaris on my list for a while

DHMO

@Alessandro I don't understand

Alessandro

08:59

Well actually you can make sense of this, it looks like binary expansions of the integers in the usual order, I was referring to cardinal exponentation in general

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$Mathematics$ Mathematics