@Ted 50 years ago, what did you think life would be like now?

Another forum, but still a math problem. How would one solve this partial differential equation? physicsforums.com/threads/…

So, looking for confirmation. Is it true that $\prod\limits_{n\in\mathbb{N}}\mathbb{Q}$ has a subgroup isomorphic to $\mathbb{R}$? Namely, the subgroup of sequences that converge to one particular limit or another.

@Rithaniel, as a subgroup? How do you make sense of convergence? It would make more sense as a topological space, wouldn't it?

Also you could have two different sequences converging to the same point? One from the left, the other from the right(and various others). Will it be a bijection then?\ (if you are looking for an isomorphism that is)

Hmmm, so it's a bit more complex than I thought.

So, I'd need to define what kind of sense of convergence I'm using? Also, it would probably be better to say that this subgroup would be homomorphic (epimorphic?) to the reals.

@Rithaniel It has a quotient isomorphic to $\Bbb R$, surely, if you quotient out by the group of sequences that go to zero

Oh, wait

No that allows for (the equivalence classes of) nonconvergent sequences

OK - the set of Cauchy sequences quotiented out by the group of sequences that go to zero is $\Bbb R$

which is a bit worse than what you wanted

On the plus side, this is exactly Cauchy's construction of the reals

(I can't remember if you know what a Cauchy sequence is. The short story is that, you want a way of saying "This sequence in $\Bbb Q$ converges to something in $\Bbb R$" without having defined $\Bbb R$ yet. Something's Cauchy if, roughly, eventually all the elements are really close to each other.)

Yeah, isn't that a sequence is Cauchy if after a certain point the gaps between adjacent entries are less than a given N?

Or is it that, for all $n,m$, if $N<n,m$ then $\vert x_n-x_m\vert<\epsilon$?

Either way, I'm always looking for the reals in the infinite product of the rationals, so I feel that I should get a better grip of when/how they crop up in there.

@Rithaniel $\forall\epsilon\exists N\forall m,n>N,|x_m-x_n|<\epsilon$, yeah

@Rithaniel I don't know the answer, but note that with the axiom of choice, $\Bbb R$ is group isomorphic to $\Bbb R^2\simeq\Bbb C$

and that's not necessarily true without it

so maybe you can use that somehow?

Yeah I dunno anymore

With the axiom of choice I think $\prod_{n\in\Bbb N}\Bbb Q$ is group-isomorphic to $\Bbb R$, if only because, if you think of them as vector spaces over $\Bbb Q$, they have bases of the same cardinality

that's right

I know $\Bbb R$ doesn't necessarily have a basis over $\Bbb Q$ without the AoC. The same might be true for $\prod_{n\in\Bbb N}\Bbb Q=\Bbb Q^{\Bbb N}$? Not sure

(Fun fact: m comes before n in the alphabet, and this always catches me by surprise)

anything after L is a mystery

At some point I memorized the number of the position of any letter in the alphabet

so if you say like "Q" I can immediately tell you "17"

and the reverse

nise

14 9 19 5

that's like

word tablature

My mental image of "what are the reals" is based in topology, I suppose. I always think of them as the rationals unioned with all accumulation points of the rationals.

apparently tablature is absolutely awful as a musical notation system

or the Dedekind completion of any ordered set?

(Whichever one, they're both generalizations of turning $\Bbb Q$ into $\Bbb R$)

I don't believe I have ever learned either one, no.

If I have, it either wasn't called that, or wasn't given a lot of attention during the course.

A metric space is incomplete if there are Cauchy sequences that don't converge. (Like $\Bbb Q$)

An ordered set is incomplete if there are sets with no supremum. (Like $\Bbb Q$)

For Dedekind: what nice properties does $\{x\in\Bbb Q:x<\pi\}$ have?

Or in general $\{x\in\Bbb Q:x<\alpha\}$ for some real $\alpha$

How would you specify those sets without being able to talk about any real numbers

Well, $\mathbb{Q}$ is dedekind, so those sets have the same cardinality with all of $\mathbb{Q}$, which is the first thing that comes to mind. Don't know if it's useful, though.

I can't think of how to specify those sets without setting up some sense of a sequence that would converge towards that particular real.

I have a random subset of $\Bbb Q$ and I claim it's of that form. What questions would you ask to make sure I'm not lying and don't actually have something like $\{1,2\}$?

Why is $\{1,2\}$ not a set of that form?

Well, first, the set must be infinite.

OK. What about $\{\dots,-2,-1,0,1,2\}$?

Well, it doesn't contain $\frac{1}{2}$ and that'd be the next issue.

Why should $\frac12$ be in it?

Because $\frac{1}{2}\in\mathbb{Q}$ with $\frac{1}{2}<1<\alpha$ and our set is a subset of rationals $x$ such that $x<\alpha$.

hi :)

Ah OK so it should be downwards closed

If $a$ is in our set and $b<a$ then $b$ should be in our set

Is every downwards closed set of that form?

Well no, because there are downwards closed subsets of the rationals that contain $x>\alpha$. Though, we're still talking in terms of a real $\alpha$, aren't we?

I mean, for some $\alpha$

I'm watching a talk on random walks on graphs

Ah, wait, so are we going to talk about all downwards closed subsets of the rationals and define reals based on them?

Yeah :)

Okay, that's a nifty approach of defining the reals.

We need to exclude $\emptyset$, the universal set $\Bbb Q$, and things like $(-\infty,1]\cap\Bbb Q$

Alright, so any ordered set can be completed this way? What about partially ordered ones?

Dunno, I think so

Also by the way how do we order our reals

so if $\alpha$ and $\beta$ correspond to these sets when is $\alpha<\beta$

I'd say "by the euclidean metric" but that just refers to a function to the positive reals.

Is it inherited from the order on $\mathbb{Q}$? If so, then that begs the question: How do we order the rationals?

Multiply out denominators to get integers and compare those?

Ah, yeah, there you go.

And the integers are ordered by the peano axioms?

If you define them that way yeah

What I was getting at was that $\alpha$'s set would be a subset of $\beta$'s set

Well, I would define the integers based around the natural numbers, and define the natural numbers from the peano axioms.

Ah, I like that, actually.

We're taking nonempty, bounded-above, downwards-closed sets without maxima, and ordering them by inclusion

yes

Aa can anyone suggest me a math project, on algebra,

@yuvrajsingh collage/school?

Hey recently I came across this weird solution to find total number of possible integer (satisfying condition $a_k=a_i+a_j$ which used Fibonacci series to find it. I think there is some flaw in it which I am not able to find If it is correct then it's really beautiful proof. If you've time please find the flaw in that answer.

College

@yuvrajsingh Chapters?

@VladimirReshetnikov Привет

@VladimirReshetnikov Привет

@AbhasKumarSinha are you preparing for jee

@yuvrajsingh yes

@AbhasKumarSinha will you help me in understanding p and c

I am struggling a lot

@yuvrajsingh I'm not an expert at it either, but I'll try, show me your question

@AbhasKumarSinha I have a 8 distinct object I want to distribute it among 4 child so that one particular child - 4,if he receives object 3 then he will not receive object 7 total cases

@yuvrajsingh I'm afraid, I don't understand your question

@AbhasKumarSinha first I have to distribute 8 distinct object to 4 children,

okay

8c4

I think it is 4^8 @AbhasKumarSinha

@yuvrajsingh No, because the order in which you give objects aren't important here, because you can give object A first with object B that'd be same as object B given first after object A.

In case the order in which the object given matters, then your answer will be right!

Ok

@yuvrajsingh I can understand your confusion. Do you've class 11th NCERT Mathematics? You'll easily master P&C if you read line by line NCERT and solve each and every question in examples and exercises without any help (solutions)

NCERT questions of P&C conver some part of Advanced too

@AbhasKumarSinha I doesn't have any notes or something

@yuvrajsingh That's a warning, don't read notes Read NCERT only, solve the examples, text questions and exemplar, I'm sure you won't regret that, that's worth being read

@yuvrajsingh here's the link to NCERT: ncert.nic.in/textbook/pdf/kemh107.pdf

@AbhasKumarSinha thanks will it solve my basic confusion

@yuvrajsingh Forget basic, This chapter in NCERT will even trail you some parts of Advanced too

@yuvrajsingh Don't forget to get this too: ncert.nic.in/ncerts/l/keep207.pdf

@yuvrajsingh some questions are even harder than few of advanced but they are important, don't leave them

Should I solve them line by oine

@yuvrajsingh no, you've to read the theory line by line and you don't have to solve them line by line, you can solve as you wish!

@AbhasKumarSinha are you in allen

@yuvrajsingh Bansal

@AbhasKumarSinha I trine his tomorrow and let you know thank you

@yuvrajsingh If you need additional materials/help/questions to solve, you can ask me here anytime, I'll help u

@yuvrajsingh you are welcome :)

TIL that the braid group relation $\sigma_i\sigma_j=\sigma_j\sigma_i$ when $|i-j|\ge2$ is called "far commutativity"

@AbhasKumarSinha thanks a lot I will

er - maybe only one person used that phrase

Oh never mind it's good I found multiple people calling it that

@yuvrajsingh no problem, don't get disheartened if you can't solve P&C questions, you are not alone, very few people excel here, but with efforts you'll definitly do it :)

@AbhasKumarSinha problem is that I belong to a poor family and cannot afford big coaching fees zirrad I article from net and then try to solve

@yuvrajsingh You can apply for a 10% reservation, the government will bear your expenses and scholarship. Good Luck, I can assure you that you can easily make up to NIT if you cover NCERT and Exemplar ones. (I'm also self-taught, I don't go to coaching)

@yuvrajsingh Have you tried Official IIT Coaching Videos?

@AbhasKumarSinha I read it and let you know tomorrow thanks good night

@yuvrajsingh Good night :)

Would a global HR conspiracy be called Big People

say you have two orthogonal vectors $\vec a$ and $\vec b$ and you add the vectors together to get $\vec a + \vec b= \vec c,$ s.t. $\vec c$ points in the positive vertical direction

and say instead of just one pair of orthogonal vectors being added, each point has a pair of orthogonal vectors

so you have a set of vectors pointing in the positive vertical direction

good so far?

so would this just be a combination of two vectors spaces via addition to create a third vector space?

@Ultradark Could you draw a picture

I don't understand

Yeah

the picture is arriving in 2 minutes

reminds me of a bunch of reference frames

I guess I'm doing a direct sum of two vector spaces $A$ and $B$ for $a_n \in A$ and $b_n \in B$

and the direct sum of two vector spaces is a vector space

is anyone there??

So, a question about axiom of choice: If you don't assume axiom of choice, is it possible for there to exist a set of cardinality greater than $n$ which does not have a subset of cardinality $n$?

Or, in other words, if I say "this set is infinite, thus it has a subset of 5 elements," have I used the axiom of choice?