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12:01 AM
@copper.hat I meant it was about time you yelled surrender. :P
a rare moment when i am caught for words :-)
@user863565 are you asking why if region $A$ fully contains region $B$ that the area of $A$ is greater or equal to area of $B$?
@user863565 Do you see that the areas of the triangles and the wedge are contained in a chain?
which requires a yes to Quin's query, or is it a no that they are not asking about that?
my question is not meant to be rhetorical.. im honestly just curious. the proof seems very transparent to me
at least in the first quadrant, which is all we need.
12:16 AM
transparent meaning you can't see it, or transparent meaning you can?
!!! maybe both! this question has me second guessing myself
but i meant it in that it is not hard to see the main idea
transparent things should be hard to see. you'd go out of business selling glasses if all people saw when they put them on was glasses.
i'm going to write the OED about this.
maybe i should go into sunglasses?
i do remember getting to that limit when i first took calculus and thinking things were getting unusual. it had been all polynomials and factoring and canceling x - a and suddenly we're drawing pictures of a circle. the book did say something like, this is because we want to use sin x before we have more algebraic ways of expressing it.
12:25 AM
i love the comment, use l'hopital's rule. right, we know sin x is differentiable at 0 because why?
One of my pettest of peeves is the use of L'Hôpital when it's in fact the definition of the derivative.
wow, we even agree on a pet peeve.
we take the limit of application of L'Hop to the definition of the derivative. maybe some category theorist can make sense of this?
just kidding. im spewing total nonsense
then a category theorist will definitely be able to make sense of it
was there a name for matrices of the form $\begin{pmatrix}a&b\\-b&a\end{pmatrix}$ with determinant $1$ (i.e. those representing unit complex numbers)?
12:35 AM
$SO(2)\times\Bbb R^*$?
oh, it's just $SO(2)$, right
think so
patrick if you've known him as long as we have
12:44 AM
yeah, patrick rotations. of the cork rotations
1:41 AM
Oh, I missed the determinant $1$ in there. Duh.
1:59 AM
@robjohn I am saying that because the area in your proof is redundant
u can compare height directly
and area doesn't make sense
sorry for writing too fast I am at school lol
not allowed to use 9hone
@user863565 If you don't use area, you have to deal with arclength of the arc of the circle. Area is much more basic, going back rigorously to the early Greeks.
actually, hamilton would be a much more apropos name for a rotation
Is Alexander spinning around in his grave?
2:53 AM
@robjohn, I cheated, looked up the solution and dont get it.
@TedShifrin I was thinking William.
Correction, i get the explaination, just not the specific notation used. And also I misunderstood the problem (ie recalled it incorrectly)
@AndrewMicallef which solution did you look up?
not all solutions are created equal
Or rather the linked pdf
Are those fractions in the braces?
(Moot anyway my alarm just rang and i have to go(
math mag really committed to those 70s-ass fonts for titles and author names
looks like some variant of optima for the author
2:57 AM
@AndrewMicallef Try this one
sorry that's as far as i'm reading in this
now that's what i call an answer!
does math.se support optima as a font
That answer looks pretty much like mine
you might track down that guy and sue him
@leslietownes I'm sure I wasn't the first
don't need to be first as long as you can prove that they copied you
there's a highly suspicious similarity between that poster's moniker and representation and your own
3:06 AM
@leslietownes Mark Noy?
oh i'm sorry i was looking at the link to your own answer
my remarks will make more sense now
Ah, yes. He did copy my answer! Get my lawyer!
I wonder if I'd be the first to sue oneself.
i'll take the case on contingency. you cannot lose
3:45 AM
have you met my legal friend murphy?
really hoping this is the setup for a joke. no, i haven't
sry, no joke. murphy's law
you cannot lose is tempting the non existent gods
the weird thing is, i even thought 'murphy's law' but didn't make the connection because i was actually thinking about the substance of murphy's law
i stop drinking caffeinated tea around noon
:-). i'm in a murderous mood.
if i were to explain at the moment i might damage my laptop so i will refrain.
will look for PSQs to answer to taunt the mse gods
4:20 AM
hides from @copper
no need :-), it is myself who is getting beaten up by myself :-0
you're just tired of looking at that PSQ in the mirror when you wake up in the morning
how can i improve the question that is me
i just look for sand to bury my head under
i used to scour the news daily, then we had the last four years of overload and now i sometimes reach dinner without having checked the headlines.
i need to find a balance.
the balance may be not checking the headlines at all. remember, they're just your CPU come to life
patent waiver on vaccines!
4:28 AM
i hope it's just on the vaccine compounds themselves. they should leave a backdoor, like a patent on a method of preparing the compound, or making a machine that prepares the compound
give us something
:-) new socialism
i spent a lot of money having attorneys prepare patents on stuff we read about, they've got to leave me in peace
maybe change field to divorces for billionaires? seems to be a popular industry at presnt
i wonder if 640k will be enough for everybody
high end family law is a good racket
or high end trusts and estates
most of it is not high end :(
i should do our trust, i have deferred it for a long time.
4:37 AM
and everybody wants to kill each other even at the high end, for varying reasons
i was involved in one dispute over a trust, worst behavior i'd ever seen from human beings
i want the simple life :-).
think of a trust as a little time bomb you can set off to destroy your family after you die
i am torn about it for a variety of reasons.
there isn't a huge amount of law on trusts. it will surprise you to learn that california has a civil code on this subject, and judges just cut-paste lines from that and try to do whatever seems right. not a lot of 'precedent'
i have huge trust issues :-).
4:40 AM
all of the main cases have names like getty in the caption, too, so if there is precedent it's hardly applicable to normal folks
i don't like things without a contractual end.
your honor, by the case of hearst v. hearst, where the trustee had to give over the castle and the personal zoo and the yachts to the other hearst, we have to ---
we do still have a sunfish in my mother's garage, but i suspect no one will care if it disappears from the estate...
and the castle, that was many generations ago
one of my ancestors mowed the lawn near a castle in killarney
there's probably a way of setting up an LLC or something and having people contract with it in a way that achieves the same end as a trust, but that seems like a tightrope act
unless it is in common use it adds burden to the next round.
i will probably remain on the fence long enough that it becomes a non issue :-)
4:45 AM
sounds like a next round problem to me
time settles all things
yes, unfortunately the next round is my issue currently (hence mood)
i'd rather deal with PSQs
i think i found a brainless therapeutic problem to solve...
Brainless therapy?
about all i can handle right now :-)
5:02 AM
Is minimal element of a set same as minimum element of the set?
minimal is a weaker notion.
i'm assuming e.g. a partially ordered set
think of a graph
In poset only I saw that term for the first time
if we know more about the order it would advantage all of us to learn what that might be.
5:03 AM
a set may have a minimal element but no minimum
Minimal of a set is like : There is nothing in that set less that that element.
that is not a usual definition
Of course, in poset it will change mutatis mutandis
i might step back here, if you want to start with a precise definition i am good, otherwise this ends up chasing tails.
in many ordered sets of varied hypotheses a minimum of a set S is, at a minimum (ha!) comparable to every other element of S.
this is not automatic in many situations where there may still be things that are in the set and not greater than any other thing in the set.
but we're just floating in the sea right now without definitions and hypotheses.
5:08 AM
Let (P, $\le$) be a poset and there exists $p\in P$ such that for all $x\in P$ we have $p\le x$ then p is minimal
Is this definition good enough?
i like that, if that's what you're working with and you know what properties $\leq$ enjoys by virtue of $(P, \leq)$ being a "poset."
$\le $ is partial ordered relation.
i misread. i would call that a minimum element.
i think the usual definition is that $p$ is minimal if there is no element $q$ such that $q < p$.
this is the one i am most familiar with.
you're a minimum if you're in the society and below everybody else in the society. think of being the only serf in a society of noblemen. you're minimal if you're in the society and not better than anyone else in the society. think of people under communism.
for more about communism, please consult my blog
5:14 AM
That makes sense.
the poset pictures here would be a dot connected via lines to a bunch of dots above it (the serf) and a long row of millions of dots, all on the same level, nobody connected to anybody else. forget about party members for a moment.
extreme examples are helpful. if you're not comparable with any other member of the poset, you're minimal by definition and will fail to be a minimum unless you're the last person on earth.
think a finite (or infinite, probably better) directed graph . any leaf element is minimal.
If $\le$ is less or equal in its usual meaning and let’s stay in the set of reals for example then minimal element of a subset of R is same as minimum element of the subset. Is this statement correct?
if its usual meaning is the real numbers, sure, a minimal element (if it exists) will be a minimum.
@copper.hat: If you are referring to graph theory, I’m afraid I don’t know it yet :’(
@leslietownes: Yes!! That’s what I wanted to know :) Thank you!
5:18 AM
note that any two elements of R are comparable to one another. this is a big part of what makes R and other "linearly ordered' sets different from general partially ordered sets.
$\varepsilon$ $\epsilon$
$\in$. i had to do it.
@leslietownes: I understand that. R is an ordered set.
Thank you!
ah no mathjax doesn't support this character
menya zovut ne koro
5:25 AM
Sorry I deleted my message as I thought it was not relevant here.
Thanks for letting me know the negation of the sentence . @Euler2
time goes fast but it also goes slow
can someone make a time machine
i would pay any amount blyat
If we're doing Russian in here, why not do it in the Cyrillic alphabet?
Да почему бы и нет?
kirillitsa :)
partially derived from the Greek alphabet
that D in Da is just a delta with a fedora
5:36 AM
Я этово не знаю!
So, Balarka's not been here over a week.
seems like leslie is omnipresent
he is talking in two rooms
Leslie thinks he's omnipotent and omniscient and omnivorous.
pronounced dhelta
2 hours later…
7:54 AM
@Quin I am just saying that you can just compare height sin(x)<=x<=tan(x) and comparing area is redundant in that proof
and still get sin(x)/x\to 1 as x\to 0
@TedShifrin I think you understand what I mean. But arclength was already known to early Greek.
also I don't think someone will make sense out of area for that inequality
8:15 AM
@Euler2 $\unicode{xa792}\unicode{xa793}$
8:26 AM
n i c e
You can always use unicode characters.
@user863565 how do you compare the non-line arc for $x$?
there is an argument for why it is $\ge\sin(x)$, but it is not clear why it is $\le\tan(x)$
lemme try some random number in uni $\unicode{xa843}$
something tells me that I've made a mistake.
why does reducing to the identity mod a square-zero ideal imply being an isomorphism?
1 hour later…
10:03 AM
@robjohn I think I kinda get it why area is used thanks
I was using drawing as proof
now I see the motivation of this proof
10:21 AM
@robjohn wait a minute then it is also not clear why inequality of area is true 😅
I mean we are working with geometric intution only
10:45 AM
Let $f$ be some function $f:\mathbb{R}^d\to \mathbb{R}$, and $\nabla f$ its gradient
let $A \in \mathbb{R}^d\times \mathbb{R}^d$ a symmetric invertible matrix.
Can anyone help me study $A \nabla f $
for instance is $A \nabla f (x) = \nabla g(x) $ for some function $g$
1 hour later…
12:13 PM
@TedShifrin I'm lurking
12:48 PM
Hi, I have a doubt
Suppose $F$ is a field. Suppose $\sigma_1, \sigma_2 \dots \sigma_n$ are distinct elements of finite group of automorphisms of $K$, then lang proves that they are algebraically independent when $K$ is infinite. What about finite field $K$?
I cannot find any counterexample. Can anyone help?
12:59 PM
@permutation_matrix it still works
@LeakyNun, Okay. How?
I am talking about algebraic dependence. Not linear dependence
what is the proof?
1 hour later…
2:33 PM
oh, i missed that
if \sqrt is the principal square root function on C, then sqrt(e^z) should be an entire function, because e^z does not vanish anywhere, and so we have no branch points. However, in the function plot on wolframalpha there appears to be a cut at z=x+iy for y=pi; see the contour plots here: wolframalpha.com/input/?i=plot+sqrt%28exp%28z%29%29
can someone explain this to me please?
maybe wolframalpha just messes something up... but from my understanding sqrt(e^z) should be analytic everywhere
I suppose wolframalpha doesn't understand that there exists an analytic continuation, and thus messes up
@permutation_matrix every automorphism is a power of Frobenius (x |-> x^q)
3:35 PM
hi all
$$y"-(2y/(xlnx)) + (2+lnx-ln^2x)y/(x^2ln^2x)= lnx$$
i wanted to solve this DE
I did, but my answer is not matching
its off slightly, and wolfram gives no solution
as in , it does not claim that there is no solution, it simply didnt give the solution
4:01 PM
@BalarkaSen Wonderful!
@satan29 what's your solution?
@courge9 you seem to be mixing up $\sqrt{e^z}$, which has two entire branches, with $e^{\sqrt{z}}$, which still needs a branch cut.
@courge9 there is no principal square root function on all of C, at least not a continuous - let alone holomorphic - one, regardless of whether you remove 0 or not
@robjohn I don't think I'm mixing that up... z\mapsto e^z is a holomorphic function on all of C, which takes values in C\{0}. Hence this function has a holomorphic logarithm, and thus holomorphic n-th roots for all n.
where's the mistake here?
taking n = 1/2 ?
n=2 for the square root
4:09 PM
The square root is $e^{z/2}$ not $e^{\sqrt{z}}$
sqrt( ) is its own thing. you can do stuff with e^z by popping stuff into the exponent
@robjohn exactly, I never claimed something else
e^(z/2) is not sqrt(e^z)
@courge9 Sorry, I misread the question
defining a sqrt requires slicing the plane from 0 to infty in some fashion
4:11 PM
@leslietownes why not, if we choose the principal square root function?
there is no principal square root function
the principal square root function, if you mean the one where the slice is the nonpositive real axis, is not holomorphic on C \ {0}
or continuous there
i assume WA interprets sqrt(e^z) as "choose a branch of sqrt and apply that to e^z"
yes, exactly
thorgott many textbooks do refer to a specific sqrt as "the principal square root," it is not nonsense
it's just not continuous or holomorphic either
on C \ 0
I don't believe in discontinuous functions
but yeah, I already specified that part in my first message
4:14 PM
@TedShifrin Had my graduate school admission interview. Did well there. Want to hear what they asked?
I'm confused
courge9 the principal square root in many books is given in polar form by sending r e^(it) to r^(1/2) e^(it/2). the problem with this is that it treats points near the negative real axis in very different ways.
points slightly above i get pulled back toward the positive y axis. points slightly below i get pulled down toward the negative y axis
there's no way of defining this at i to make it continuous let alone differentiable there
and you encounter a similar problem if you use other formulae
I thought that since sqrt has a branch point at 0, but e^z never vanishes, sqrt(e^z) has no branch point and hence everything is fine
oh wait a moment
it's fine in a strip around the real axis, but not where the exp takes negative values, which is at $i$
If $X$ is path connected, locally path connected, semilocally simply connected then there is a covering space $\tilde{X}$ such that $p:\tilde{X}\to X$, $p_*(\pi_1(\tilde{X})) = H$ for given $H<\pi_1(X)$. But we still don't know such $\tilde{X}$ is path connected right?
well, it may or may not be if that's all you're going off
but just take the connected component containing the point at which you compute the fundamental group and you have a path connected one
4:21 PM
@leslietownes is that correct? so the problem here is that while 0 is not a branch point of sqrt(e^z), we still have a branch cut on the negative real axis in the image of e^z (that is, at z=i) where sqrt(e^z) becomes discontinuous
Follow $\sqrt{e^z}$ along the line from $z=-\pi i$ to $z=\pi i$. That goes around the unit circle counter-clockwise from $-i$ to $i$. We cannot continue along this path since it would take us out of the range of the principal square root, so we need to jump back to $-i$. Thus, the discontinuity
The range of the principal branch of the square root is the right half plane
ok, but \sqrt(e^z) is continuous (in fact, analytic) inside an open strip around the real axis with imaginary part between (-i,i), right?
and did I put it right with what I've written above: " so the problem here is that while 0 is not a branch point of sqrt(e^z), we still have a branch cut on the negative real axis in the image of e^z (that is, at z=i) where sqrt(e^z) becomes discontinuous"
$z=\pi i$? since that is where $e^z$ meets the branch cut for sqrt
4:26 PM
oh yes of course, sorry
I was thinking e^(pi*z), that's why I wrote $i$
that comes from simplifying my notes when asking a question, haha
@courge9: are you running ChatJax?
okay, so the misconception on my side was that I thought it suffices that 0 is not a branch point to conclude that the function is entire, while in reality one still has to consider the cut along the negative real axis
@robjohn I'm not, sorry
@courge9 yes
4:28 PM
thanks, that clears it up a lot...
@courge9 It might help reading this room.
Look at the "LaTeX in chat" link at the upper right of this page
robjohn gets 5 cents every time someone clicks on that
I'm doing research, but my analysis knowledge is really bad and too long ago... I'm not an analyst at all, but sometimes there's no way around, haha
@leslietownes I wish!
@robjohn thanks, I'll do that
4:35 PM
This isn't analysis, it's pure topology. $\mathbb{C}\setminus\{0\}\rightarrow\mathbb{C}\setminus\{0\},z\mapsto z^2$ is a non-trivial, connected cover, so it doesn't admit a global section, i.e. there is no global square root. The image of $\pi_1(\mathbb{C}\setminus\{0\})\cong\mathbb{Z}$ under this covering map is identified with $2\mathbb{Z}$, so any map $X\rightarrow\mathbb{C}\setminus\{0\}$ which maps $\pi_1(X)$ into $2\mathbb{Z}$ admits a lift through the covering.
The covering map is holomorphic, whence a local biholomorphism, so a lift of a holomorphic map through it will be holomorphi
@Thorgott rofl
@BalarkaSen glad to see your doing fine.
yeah, things are not looking good
@Thorgott : but $z\mapsto e^z$ is a nowhere vanishing holomorphic function $\mathbb{C}\rightarrow\mathbb{C}\setminus\{0\}$, so why doesn't it admit a holomorphic square root?
4:40 PM
@courge9 it does, $e^{z/2}$ (or $-e^{z/2}$)
@courge9 it's just not $\sqrt{e^z}$
or rather, $\sqrt{e^z}$ is simply not a well-defined expression
ah, so it's a 2nd root in the sense that its square gives the original function e^z, but it's not THE square root in the sense of $\sqrt{e^z}$
@Thorgott it is if you take the principal branch of square root
@courge9 exactly
I am making youtube videos on STEM youtube.com/watch?v=vTWtzZ6glJk and advanced math and basic math just for myself
if anyone is interested
4:43 PM
robjohn, thorgott doesn't believe in that. :)
the principal branch DOESN'T exist?
@robjohn but if we take the principal branch, is it still true that $\sqrt{e^z}$ becomes discontinuous at $\pi i$? How does that fit together with what you said before?
24 mins ago, by robjohn
Follow $\sqrt{e^z}$ along the line from $z=-\pi i$ to $z=\pi i$. That goes around the unit circle counter-clockwise from $-i$ to $i$. We cannot continue along this path since it would take us out of the range of the principal square root, so we need to jump back to $-i$. Thus, the discontinuity
yes, it is discontinuous, or rather it cannot be continued across $\pi i$ continuously
but $e^{z/2}$ is continuous everywhere
it's a bit like why i don't like the classical notation sqrt(-1) for i. it invites "power arithmetic" like -1 = (sqrt(-1))^2 = sqrt((-1)^2) = sqrt(1) = 1
4:47 PM
@courge9 it is.
fine, go enjoy your discontinuous branch if you like it that much
@leslietownes I agree. I don't like the $\sqrt{z}$ notation, unless it is clearly stated what that means.
@robjohn $lnx (c1x^2 + c2/x + x^2ln^2x/6 -x^2lnx/9)$
@robjohn ok, but $e^{z/2}$ is NOT the principal square root of $e^z$, right?
(at least not everywhere)
similarly (e^a)^(1/2) invites us into a nightmare world while e^(a/2) is just fine
4:49 PM
@Thorgott it's a side-effect of composing a function with a function that has a branch cut
@leslietownes I don't like raising complex numbers to non integer powers, unless the definition is clearly stated.
for the same reason. yes
raising positive reals to complex powers is fine
i don't want to be invited into a nightmare world unless i know the definitions
I think I have to deal with $\sqrt{e^z}$ as I want to consider an analytic continuation of $\sqrt{e^x}$
4:53 PM
@courge9 extending from the reals?
then why not extend $e^{x/2}$ which is the same on the reals?
and has a much nicer continuation
at the moment I was writing this comment, I was wondering the same
The bill is in the mail ;-)
oh joy
4:56 PM
does that help some? that is, can you do that in the context of what you're doing?
it potentially does, thanks
Is it generally so that taking the power 1/2 is better than sqrt
concerning analytic continuation from the reals
in this case. Not all cases
for example, $\sqrt{\sinh(x)}$ still suffers from the same problems due to the zeros of $\sinh$, right?
4:58 PM
post-composing with some branch of sqrt( ) with some random function is unlikely to give you the pleasant result that precomposing e^z with z/2 does
@courge9 due to the fact that you're taking the square root of negative reals, so this is not a function on the reals to the reals
@robjohn right, bad example...
@robjohn did you solve it?
@courge9 Wolfram Alpha sees that you explicitly wrote sqrt(e^z), so it assumes you want the ugly version. If you really wanted $e^{z/2}$ then you would have writtten e^(z/2). It's kind of similar to $\sqrt{x^2}$ vs $\left(\sqrt x\right)^2$ for real $x$.
and Wolfram Alpha generally assumes the principal branch for sqrt, I suppose?
5:01 PM
at least that fits to the indicated cut at $\pm\pi i$ in my example
@satan29 no, i haven't looked at your solution yet.
@courge9 if you give the function as Sqrt[Exp[z]] there is not really anything else it can do.
@robjohn @leslietownes @Thorgott thanks for all of your help, I'll have to think about everything, but it's much clearer now
5:11 PM
Yesterday, I was messing around with $(1+1/n)^n \le e \le (1+1/n)^{n+1}$, and I noticed that $\pi$ is very close to the fixed point $n=(1+1/n)^{n+1}$.
In fact, $(\pi+1)^{\pi+1} \approx\pi^{\pi+2} \approx 360$. I'm almost tempted to post: Prove that $(\pi+1)^{\pi+1}< \pi^{\pi+2}$ on Puzzling.SE. :)
$(\pi+1)^{\pi+1} \approx 359.7956379$ and $\pi^{\pi+2} \approx 359.8670909$
I suppose there might be some deep analytic significance to this, or it might just be an interesting coincidence, in the vein of en.wikipedia.org/wiki/Strong_Law_of_Small_Numbers
5:27 PM
I have a quick follow-up question regarding what Thorgott wrote: any nowhere vanishing holomorphic function $f:\mathbb{C}\rightarrow\mathbb{C}\setminus\{0\}$ admits a holomorphic square root. I understand that this "square root" is (in general) not literally $\sqrt{f}$, but a function whose square gives the original function. But is it unambiguous to write $f^{1/2}$ for that function, in particular when trying to define an analytic continuation of $\sqrt{f}$ from the reals?
put differently: is it meaningful to regard this "holomorphic square root" as the proper analytic continuation of $\sqrt{f}$ on the reals?
I think it is still ambiguous because fractional powers of complex numbers are (usually) defined in terms of taking a branch of the logarithm and expressing the fractional power in terms of exp and log.
good point, so $f^{1/2}$ suffers from the same problems as $\sqrt{f}$ (at least in general -- there are exceptions, like $f(z)=e^z$ we discussed)
e.g., if you write $z^{1/2}$, usually this means that you need to choose a region in which to define $\text{Log}$ (normally $\Omega=\mathbb{C}\backslash (-\infty,0]$) and then define the branch $\text{Log}_\Omega$ to be the inverse of exp on $\Omega$. Then you can write that $z^{1/2}=e^{1/2\text{Log}_\Omega(z)}$ in this region.
but all this is still local and what thorgott was saying is that there are theorems which allow you to avoid this choice of branch under certain circumstances (choosing a branch is similar in my mind to choosing a basis!)
yeah, it's just gonna be nicer to work with $e^{z/2}$ as an extension of $e^{x/2}$
i wouldn't think of e^(z/2) as an exception to the case of f^(1/2) being a problem. post composing with ^{1/2} is the problem because you need to know what ^{1/2} means.
it's clear what z/2 means, so no trouble in precomposing it with anything.
i may be reversing post and pre because i'm left handed but we all know what i mean.
yeah exactly
but there is notational ambiguity when you do fractional powers of complex numbers unless you really nail down stuff explicitly. its annoying but such is life
okay back to (actual!) work :)
5:38 PM
@leslietownes I get your point, thanks... This is all far more involved than I anticipated, wow
i had to deal with this once and i forget the details what i did. i think i really needed a log, the formula really only made intuitive sense as a log, but i got rid of it by differentiating and then working in the world of rational functions of what had been stuffed into the log. the formulas weren't intuitive but i didn't have to worry about nightmare worlds.
Considering additive groups R (set of reals) and Q(set of rationals), I know that they can’t be isomorphic because R is uncountable and Q is countable so no bijection will exist between the two. But I wonder if there’s any element in R which has an order that no element in Q posseses.
Hi Leslietownes
what are the orders of the elements of Q, and why, would be a good starting point
5:55 PM
@Koro On a related note, see en.wikipedia.org/wiki/Beatty_sequence
on an unrelated note, floor(nx)/n is a sequence of rationals that converges to x. one way of showing Q dense in R. maybe not the simplest way.
as long as we're flooring multiples of things
@leslietownes: Every element in Q has infinite order
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