Mathematics - 2015-11-27 (page 1 of 2)

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12:02 AM

@Josh If M/L/K is a tower of Galois extensions, then Gal(L/K)->Gal(M/K)->Gal(M/L) is a pair of group homomorphisms. the first is one-to-one, the second is onto, and the image of the first is precisely the kernel of the second. when we have H->G->K (for some groups H,G,K) with those properties, we cannot say that G is a direct product of H and K (not even a semidirect product in general).

@evinda Make Y the subspace of sequences that are eventually bounded by r^n for some r<1, and make x(n)=1/n.

or you could make Y the subspace of sequences that are just 0s with finitely many exceptions, and make x any sequence with infinitely many positive terms

Imagine Dragons

Can someone help me with a quadratics question

12:24 AM

what is it

Don't ask to ask, just ask :)

3 hours later…

3:46 AM

it sure is a good feeling to finally finish all the assignments you have due in a really busy week

2 hours later…

5:31 AM

I have a map of spaces $f_t: X \to F_t$. The $F_t$ are all noncanonically isomorphic, and putting this together for all $t$, I obtain a map $X \times [0,1] \to F$, where $F$ fibers over $[0,1]$ with fiber $F_t$. This is an uncomfortable situation; I'd like the codomain to have dimension one lower. I have fiddled for a bit but can't see how to reduce the dimension of the codomain without just trivializing the bundle and projecting (which I would like not to do). Thoughts?

Would prefer something more natural/functorial, whatever that means. If you'd like, pretend everything here is a smooth map of smooth manifolds.

1 hour later…

user174558

6:59 AM

@robjohn How was the turkey?

@JasperLoy It was very nice. How was your day?

user174558

@robjohn I slept through most of it. I think about things, get tired, and then sleep.

user174558

It's almost time for SE hats again.

I'm mostly just an idiot

What textbook should I learn calculus from, from a rigorous viewpoint, given that I have already done much more 'advanced' mathematics?

7:46 AM

@MikeMiller Does "huh?" count as a thought?

4 hours later…

Overly Excessive

11:16 AM

Why is $\sqrt{n} = n^{\frac{1}{2}}$ ?

I'm mostly just an idiot

@OverlyExcessive $\sqrt{a}^2=a=a^{\frac12 \times 2}$

I am new here, so I am not sure if it is normal that we have pretty much no chat here?

Tobias Kildetoft

@I'mmostlyjustanidiot It varies a lot

I'm mostly just an idiot

11:31 AM

@TobiasKildetoft As in on a day to day basis, or there are what part of the year that we are in trends?

Tobias Kildetoft

@I'mmostlyjustanidiot Not sure

Overly Excessive

Can you elaborate on this please I am not 100% sure, $a^{\frac{1}{2}} \ne a^{\frac{1}{2} * 2}$

I'm mostly just an idiot

Try \ne for 'not equal'

No you missed my $\sqrt{2}^{\huge 2}$

So we pick $Y=\{ x \in l^2(\mathbb{N}): \sum_{j=1}^{\infty} |x_j|^2< r^n \text{ for some } r<1 \}$ and $x(n)=\frac{1}{n}$, right?
But how can we now find the distance $||x-Y||= \inf \{ ||x-y||: y \in Y\}$ ?

@robjohn
Yesterday i answerd this question http://math.stackexchange.com/questions/1547804/cauchys-theorem-prove-that-sum-n-1-infty-frac1-lambda-n2-f/1547878#1547878 and get a request how to justify the contour of integration i used. I did think about this a little while but don't get to a rigorous justification which goes beyond "because it works". Because i think u are an expert on this kind of calculations , i kindly ask you if you could give me a hint in the right direction.

user174558

11:45 AM

@I'mmostlyjustanidiot Yes, on weekends on holidays.

11:58 AM

Is there a relationship between the linear transformation associated to a matrix and the one associated to its transpose?

I'm mostly just an idiot

12:19 PM

@Alessandro Why don't you draw some lines and see?

user174558

1:01 PM

@I'mmostlyjustanidiot Are you a math student?

I'm mostly just an idiot

@JasperLoy I'm mostly an idiot.

user174558

@I'mmostlyjustanidiot Did you have turkey?

I'm mostly just an idiot

@JasperLoy Noone wanted to share one with me.

user174558

@I'mmostlyjustanidiot I eat chicken, not turkey.

I'm mostly just an idiot

@JasperLoy Is it normal to have to go back and relearn things all the time?

user174558

1:07 PM

@I'mmostlyjustanidiot Yes, very normal in mathematics. Like a normal operator, or a normal subgroup, or a normal extension.

I'm mostly just an idiot

@JasperLoy Yes it was sort of to do with the second.

user174558

@I'mmostlyjustanidiot Are you using any algebra text?

user174558

@I'mmostlyjustanidiot I am glad I made you happy twice, goodbye.

I'm mostly just an idiot

@JasperLoy Dummit and foote

@JasperLoy Bye orange

I am trying to prove the first isomorphism theorem for groups, despite having done this a year ago

So we want $G/\ker(\phi) \cong \phi(G)$

Showing $\ker(\psi)$ normal in $G$ is alright

Now $\psi:G\to H$ is a homomorphism

Now I guess I need to think why this means that we have a homomorphism between the things we want to build the isomorphism

Tobias Kildetoft

@I'mmostlyjustanidiot There should be an "obvious" choice for the map

it is just a matter of showing that it is actually welldefined

I'm mostly just an idiot

1:17 PM

$\psi(g)=g+K$

(oops I changed between $\psi$ and $\phi$)

@TobiasKildetoft Oh yes, that works for me. That is indeed a homomorphism, thanks, I can get bijectivity surely from here.

Tobias Kildetoft

@I'mmostlyjustanidiot Yeah, being welldefined is actually the most tricky here (and it is not hard)

I'm mostly just an idiot

@TobiasKildetoft Well defined means that is indeed a homomorphism, or something else?

Tobias Kildetoft

@I'mmostlyjustanidiot Well, I am not sure what map you are considering. For now you seem to have one from $G$ to $G/ker(\phi)$, which is not what you want

I'm mostly just an idiot

Oh that's true

So well defined would normally mean the correct domain and range?

Tobias Kildetoft

@I'mmostlyjustanidiot No, the thing is here that you will be defining a map from $G/ker(\phi)$ to $H$

and you probably want to define it on each coset by picking a representative of that coset. Welldefined means that it does not depend on that choice

1:35 PM

yeah

hi @L33ter

so, what he means is that we pick two different elements of the same equivalence class and show that mapping is the same.

Hi @BalarkaSen

how r you ?

I heard from ted that you went to some seminar?

I'm good :)

I covered the fundmental group I will do this week covering spaces and fundmental group of the circle

I understood now why you considered loops

Have you done my star convex set problem?

1:38 PM

no not yet I will do it this weekend

haven't had time

ok.

I'm mostly just an idiot

@L33ter Oh okay I get it now, thank you

@L33ter yes.

you know I asked my professor the following question

it was a school on topology & condensed matter physics.

1:39 PM

Can you classify all topological spaces that give fundmental group that is abelian.

he told me no

oh cool

I would have loved to attend it

I learning atm cool stuff about hamiltonian mechanics

its also related to topology

I dunno about that yet, but I have heard that symplectic topologist study "phase space"

@L33ter A subcase of your question is to classify all topological groups.

Which is of course impossibly hard.

yeah

D'you know why a topological group has abelian fundamental group?

no why ?

1 sec let me think

It's an exercise in Munkres. You'd have to do it by yourself by following Munkres' hints.

It requires a trick, hard to come up by yourself.

1:43 PM

alright I will let you spoil it for me

But you can try. If you're smart, you can figure out the trick.

yeah

I will try it out

I will be doing some topology and algebra today so that is nice

I'm mostly just an idiot

What is the obvious map @L33ter?

from where to where ?

I'm mostly just an idiot

$G/\ker \psi \to \psi(G)$

For proving the first isomorphism theorem of groups

1:49 PM

what are the elements of G/Ker ?

what does it look like

I'm mostly just an idiot

Cosets $a+\ker(\psi)$

or better yet [a]

I'm mostly just an idiot

Yeah

how can we get an element of $\psi(G)$?

given this ?

I'm mostly just an idiot

From $H$(since $\psi:G\to H)$?

1:51 PM

so we have the elements G/Ker are of the form [a]

I'm mostly just an idiot

Yep

those are representative of equivalence classes

now we need to pass from representative of equivalence classes to normal elements

I'm mostly just an idiot

Just take $[a]\mapsto a$?

yeah

@I'mmostlyjustanidiot Yes. But you have to show it's well-defined, that is, no matter whatever representative of $[a]$ you choose and pushforward by $\psi$, the result is the same.

1:52 PM

yeah

So instead of picking $a$ in $[a]$, pick some $b \ne a$ in $[a]$ does not change things.

@I'mmostlyjustanidiot Also, typo. You mean $[a] \mapsto \psi(a)$.

I'm mostly just an idiot

@BalarkaSen Okay thanks, I will try that. Just for language sake I should ask if pushforward means map?

yeah

I'm mostly just an idiot

@BalarkaSen Okay I'll try that now

@I'mmostlyjustanidiot By pushforwarding $a$ by $\psi : G \to H$, I mean letting $\psi$ eat $a$ to get $\psi(a)$ :) so letting $a$ go forward by the arrow $G \to H$.

I'm mostly just an idiot

1:54 PM

:D

I like to think about algebra visually by arrows and stuff, you see.

hi @KarlKronenfeld

Karl Kronenfeld

hello

how's life?

Karl Kronenfeld

ah, the life assessment. I'd give a somewhat positive assessment right now.

I'm mostly just an idiot

So $\phi: [a] \mapsto \psi(a)$ and $b+K=a+K$ and $b\ne a$ gives $\phi(a+K)=\phi(b+K)=\phi(a)=\phi(b)=\psi(a)=\psi(b)$

Wait I used the homomorphism property in the wrong place

Wait is that map also a homomorphism?

1:59 PM

@KarlKronenfeld been doing any math?

Karl Kronenfeld

@BalarkaSen My criminal past is to remain a secret

I'm mostly just an idiot

Oh it is also a homomorphism $\phi([a]+[b])=\phi([a+b])=\psi(a+b)=\psi(a)+\psi(b)=\phi([a])+\phi([b])$

@I'mmostlyjustanidiot Not sure what you're doing up there. If $m, n \in [a]$, you have to prove that $\psi(m) = \psi(n)$. Where have you done that?

I'm mostly just an idiot

Above?

Let $a,b\in[a]$ then $a+K=b+K$, $a\ne b$

$(\phi([a])=\phi([b])=)\phi(a+K)=\phi(b+K)=\psi(a)=\psi(b)$

Why is $\phi(a + K) = \phi(b + K)$? That's what we are trying to prove in the first place!

Your proof is thus circular, right?

I'm mostly just an idiot

2:03 PM

Yes sorry

One sec

Try to be explicit there. Can you tell me, concretely, what $m, n \in [a]$ tells me about $m$ and $n$, without invoking fancy coset stuff?

I'm mostly just an idiot

They differ additively by an element in the kernel?

Exactly!

I'm mostly just an idiot

Okay I think I see, don't spoil for a sec :D

Sure.

I'm mostly just an idiot

2:06 PM

$\phi([a])=\psi(a)=\psi(b+k)=\psi(b)+\psi(k)=\psi(b)$

where $k\in \ker(\psi)$

Is that good?

Done, good.

I'm mostly just an idiot

Yay!!

You are a great tutor!

Most of the time I know what "mostly idiots" are stuck on/confused about because I am mostly an idiot too and I have gone through the struggle with precisely the things you have. I don't think I am good at teaching though.

But thanks.

I'm mostly just an idiot

I hope one day to be only partially an idiot :D

It's a bold dream

Me too.

2:15 PM

What does "comes to you" mean? "Generally speaking we think about periodic phenomena according to whether they are periodic in time or periodic in space. In the case of time the phenomenon comes to you."

Oh, I got it.

I'm mostly just an idiot

@Kartik What was the answer?

@I'mmostlyjustanidiot I had not read the full paragraph

It was

"In the case of time the phenomenon comes to you ........ In the case ofspace, you come to the phenomenon. "

I'm mostly just an idiot

Oh that is more sensical

The examples given were For time: "“sound” reaches your ear as a longitudinalpressure wave" and fYou take a picture and you observe repeating patterns.or space ".

Overly Excessive

What is the opinion on KA around here?

2:27 PM

@PVAL: Yes, but not a helpful one.

morning @MikeMiller

I figured out a resolution for my situation anyway.

Morning.

I'm mostly just an idiot

@BalarkaSen here is my proof $\phi$ is injective:

$$\phi([a])=\phi([b])\implies \psi(a)=\psi(b)$$$$\implies \psi(a-b)=0\implies a-b\in \ker(\psi) \implies [a-b]=[0]\implies a=b$$

Typo there. $[a - b] = [0]$ does not mean $ a= b$, but $[a] = [b]$

I'm mostly just an idiot

Oh ok, and then well-definedness comes in from above

2:29 PM

But yeah, the rest is fine.

I'm mostly just an idiot

Yay I have proved the first isomorphism theorem for groups again :D

Overly Excessive

What's the opinion of Khan Academy here?

I'm mostly just an idiot

It's good for really low level math @OverlyExcessive

Overly Excessive

@I'mmostlyjustanidiot What about entry-level Calculus? Is it adequate?

I'm mostly just an idiot

It would be cool if it covered higher level math, but it would be hard to implement automated proof verification(since proofs should be written in English more than symbolically in my opinion)

@OverlyExcessive Should be great

Overly Excessive

2:39 PM

Cool :)

I'm mostly just an idiot

I had no idea what KA was in acronym form for the record :D

3:16 PM

@evinda that is not what I defined Y to be

what you wrote doesn't even make sense

3:27 PM

@OverlyExcessive There are some ebooks which I found much better than KA for calculus. Unfortunately I dont have the links.

Overly Excessive

@Kartik For me I found the format of KA to be very good for self-study.

@Kartik but if you can find them please do share links

@OverlyExcessive I will try to find them, but actually they were left on my previous computer and I had found them after hours of searching.

Overly Excessive

Anyhow, how hard is it to learn Calculus if you have a decent foundation in algebra and trig?

user174558

@OverlyExcessive It is not possible to tell.

Overly Excessive

Why?

user174558

3:34 PM

Everyone is different. No stranger on the internet can give you a good answer when they don't even know you.

user174558

It's like asking am I pretty when they have not seen you.

Overly Excessive

Okay let me rephrase

Would proficiency in algebra and trig help with learning Calculus?

user174558

I also suggest using proper books rather than online material like Khan Academy.

Overly Excessive

Why is that?

user174558

@OverlyExcessive Yes, it would help.

user174558

3:37 PM

@OverlyExcessive Because online materials are very sketchy and not as polished.

user174558

It's alright if you want a quick overview though, but real books are always the best.

Overly Excessive

Here's the thing about Khan Academy that I like though

It gives me a very guided kind of learning, it provides the learning material, and exercises and it tells me when I have "mastered" a certain concept and what concepts that I need to put extra time into.

Books don't do that.

user174558

If you want free materials online for learning calculus, I suggest you visit Paul Dawkins's page where you can download the books.

user174558

@OverlyExcessive Calculus should not be capitalised in that sentence as it is a subject and not a title of a book.

user174558

Hello @Jake1234! How is your OCD?

3:41 PM

Hey man, it's been ok.

user174558

@OverlyExcessive Well, use what you like, I guess. Good luck!

My sleeping schedule keeps going forward (now going to bed at 7am... ), how about you?

Overly Excessive

Okay -- but I've found that many math books are a bit technical and hard to understand, though I haven't read many.

user174558

@Jake1234 I have nothing new to add. I sleep irregularly, and I don't know what will happen to me in future, if I have a future at all.

A day at a time then.

user174558

3:42 PM

@OverlyExcessive Are you planning to be a mathematician? Why do you need calculus?

Overly Excessive

@JasperLoy I'm a programmer and I'm going into machine learning, calculus and linear algebra are prereqs so I need to get a decent understanding but I'm not pursuing pure mathematics.

user174558

@OverlyExcessive I see. Then I have no further comments.

@OverlyExcessive Khan Academy is great, it's a good way to get some clear uses of some concepts and their applications. Retrospectively, if I were to learn a some subject from the beginning, I would probably watch all the videos on the subject on khan academy, spend a little time trying to understand all that is said in the video, and then move to learning the subject from a book/lectures + books.

Overly Excessive

@Jake1234 It's a bit embarrassing to say I guess but I started from the beginning, with the K-2 mission planning to go K-2 through K-8 and then into calculus and linear alg. But I suppose books would be very good too, I picked up Sergey Lang's basic mathematics but I'm finding it a bit.. Technical I suppose... It may just be that I'm still unused to reading mathematical notation.

user174558

@OverlyExcessive Serge Lang's Basic Mathematics is very basic. You should get used to it!

user174558

3:50 PM

@OverlyExcessive Serge Lang's calculus books are great too.

Overly Excessive

@JasperLoy Thanks now I feel stupid :D .. No but I suppose you're right -- it's not that I don't understand it but I guess it's just that I'm unused to reading more "complex" expressions and I know that they are not complex but I haven't touched math in like 8 years prior to this so ..

user174558

@OverlyExcessive Aha, you can ask in this chat if you have any minor problems reading texts.

user174558

@OverlyExcessive Did you know there is the word overexaggerate? LOL

Overly Excessive

@JasperLoy It's on purpose, I'm being oxymoronic ;) .

user174558

@OverlyExcessive Are you sure that is the meaning of oxymoronic?

3:55 PM

"overly excessive" is not quite an oxymoron.

Overly Excessive

Some people are saying it's redundat but in my interpretation it would be an oxymoron

Something which is more than excessive would be by definiton, moderate, would it not?

user174558

You need to look up the definition of oxymoron.

Is there an uncountable set of $\R^2$ that contains countably many points along each vertical and horizontal line?

Overly Excessive

Well I don't know what to call it then.

It's contradictory in my book.

Guys, could help me find some info on rigorous proof that $X^X$ is a set, and it is non empty? (contains a constant function)

4:01 PM

you say "contains a constant function." so... you already know it's nonempty? and isn't it obviously a set? or is there some kind of set-theoretic super-pedantry I don't know about going on here?

Overly Excessive

Anyhow, I have an easy question if anyone cares to answer. I asked earlier but I feel I need a more thorough explanation, how is it that $\sqrt{n} = n^{\frac{1}{2}}$ ?

I don't know it, I know that's the approach in all answers.

@OverlyExcessive by definition

But I don't understand why that thing is a set, and I don't know why there's a set $X^X$ that contains it.

@Jake1234 while reading this sentence, at first I that the phrase "that thing" was referring to X^X, but then the second part makes me second guess that. So: what?

4:02 PM

I mean if I were to be really exact about this, I know how from the basic ZF axioms, you can make a cartesian product of a finite amount of sets.

Overly Excessive

@anon Sorry what? I am trying to understand what $n^{\frac{1}{2}}$ would be.

@OverlyExcessive n^(1/2) is by definition sqrt(n)

Overly Excessive

Right but what is $n$ raised to the power of $\frac{1}{2}$ ?

Anon, I want proof that for $X$ set, $X^X$ is a set. I also want proof that it is non empty.

@OverlyExcessive it is defined to be sqrt(n)

4:05 PM

@OverlyExcessive $n^{1/2}$ is defined to be $\sqrt{n}$, as anon said. The rationale behind this definition is that $(\sqrt{n})^2 = n$. $(n^{1/2})^2$ would be $n^{2/2}$ by the exponent laws, which is in turn $n$.

$n^{1/2}$ of course does not make sense literally. We just define $n^{1/2} : = \sqrt{n}$.

@Jake1234 to me that sounds like asking to prove $5^5$ is a number. this is some kind of set-theoretic super-pedantry I have no familiarity with.

like, if you define a subset of a given set, what does it mean to prove that it's a set?

That you can create it from the axioms.

Overly Excessive

@BalarkaSen Hang on.. Let me try and make sense of what you just said.. So for example $(\sqrt{36})^2 = 36(36^{1/2})^2 = 36$ ?

Eh.

Do you agree that $(\sqrt{n})^2 = n$?

I mean, that's how square roots are defined.

Overly Excessive

Yes

4:09 PM

@OverlyExcessive there was a period and a space between those two equations that you seemed to have missed

Overly Excessive

Oh ..

haha

Sorry I suck at this haha

@OverlyExcessive So, assume as a pipe-dream that $n^{1/2}$ makes sense. What would $(n^{1/2})^2$ be?

Recall the laws of exponents.

we may first define positive integer powers by repeated multiplication. it obeys the nice laws $a^na^m=a^{n+m}$ and $a^{n-m}$ and $(a^n)^m=a^{nm}$. if we define $a^0$ to be the multiplicative identity $1$, and $a^{-1}$ to be the multiplicative inverse of $a$ (so $a^1a^{-1}=a^0$), and $a^{-n}$ to be $(a^{-1})^n$ or $(a^n)^{-1}$ (which are the same), then all of these exponent laws remain true. notice the pattern then: $\cdots,a^{-2},a^{-1},1,a,a^2,a^3,\cdots$.

I see what you're saying anon, but what I'm really trying to say is... $X$ is a set, from ZF axioms, construct a set, that would conventionally be called $X^X$ ?

Overly Excessive

@BalarkaSen It would be $n^{2.5}$ wouldn't it?

4:11 PM

IOW the multiplication of powers of $a$ function like addition of integers, since $a^na^m=a^{n+m}$ just as $n+m$ is, well, $n+m$, and also exponentiation of powers of $a$ function like multiplication of integers by integers, since $(a^n)^m=a^{nm}$ just as $nm$ is, well, $nm$

@OverlyExcessive Why so?

$(a^n)^m = a^{nm}$, not $a^{n+m}$.

@OverlyExcessive $(n^{1/2})^2=n^{1/2}n^{1/2}$, can you simplify that?

Overly Excessive

But if the exponents share the same base don't you add the exponents?

@OverlyExcessive if you're multiplying two powers of the same thing yes... which is why $n^{1/2}n^{1/2}=n^{1/2+1/2}$.

You're confused. $(a^2)^3 = a^2 a^2 a^2 = a^{2 \cdot 3} = a^6$.

NOT $a^{2+3} = a^5$.

Overly Excessive

4:14 PM

@anon $n^2 times 1/2$ ?

?

Overly Excessive

@BalarkaSen Okay but that is true because those numbers all share the same prime factors right?

Getting help by two persons at once is unhelpful, so I leave this to anon.

I'm sure there's something weird about what I'm writing, but I mean... not everything you "define" has to be a set. A "set of all sets" is not a set, or more exactly, something that contais all sets is not a set. So what I'm asking for, is how would we show, that $X^X$ is a set, when $X$ is a set. Maybe it would be precise to say that $X^X$ is the class of all functions $X \rightarrow X$.

@OverlyExcessive prime factors have nothing to do with anything here

4:15 PM

@OverlyExcessive Where does prime factors jump in now?

@Jake1234 can you explicitly write down a definition of the set of all sets as a subset of a given set in symbols? I can write down a definition of $X^X$ as a subset of ${\cal P}(X^2)$.

Overly Excessive

@BalarkaSen Is that not why we can write $a^2a^2a^2 = a^6$?

anon ... finally, that';s what I was asking for... :D

$a^2a^2a^2=(aa)(aa)(aa)=aaaaaa=a^6$

$a^2a^2a^2 = aaaaaa = a^6$. That's it.

No prime factors whatsoever.

Overly Excessive

4:17 PM

ahh. Okay that was actually an easier answer than I thought

@Jake1234 $X^X$ is all subsets $A$ of $X\times X$ such that for all $x\in X$ there exists a unique $y\in X$ for which $(x,y)\in A$

So what is $(n^{1/2})^2$?

Yup, I get it now, great, thanks.

@OverlyExcessive it's important to understand what things mean. like, $a^2$ means $a\cdot a$.

Overly Excessive

$n^1$ ?

4:18 PM

Yep.

And how is it with the existence of some function in $X^X$ ?

So it makes sense to define $n^{1/2} :=\sqrt{n}$, right?

Overly Excessive

@anon Right, I know that, but I am not confident in the properties of all operators, in terms of what operators that are associative, commutative, distributive, etc.

@Jake1234 well, you said yourself, pick an $x\in X$ and consider the function $X\times\{x\}$ (that's the constant function $x$)

@OverlyExcessive I am not talking about properties, I am talking about meaning.

Overly Excessive

@BalarkaSen This is because $\sqrt{n} = n^ 2 $ ?

4:21 PM

huh?

that's like saying x/2 = 2x

Overly Excessive

It's uh

What I was thinking about was the thing you were talking about earlier about finding the square roots of n.

and again, that's $X \times X$, so it is indeed a function, right?

I meant, that's a subset of $X \times X$

$(\sqrt{n})^2 = n$, not $\sqrt{n} = n^2$.

Overly Excessive

Ah that's what I meant to say, but it came out wrong

4:24 PM

The latter identity is blatantly false, as $\sqrt{4} = 2$ and $4^2 = 16 \neq 2$

@Jake1234 well, it's a subset satisfying the appropriate property, yes

Overly Excessive

But this, $(\sqrt{n})^2 = n$ this is related to $n^{1/2}$ ?

yes, right, I forgot.. subset that has one y for each x.

Ok, great. Finally I get it, thanks.

@OverlyExcessive you wrote a wrong thing again

@JasperLoy Hi, are you around?

Overly Excessive

4:26 PM

@anon Yes, I spotted it haha.

But okay I understand that first thing, the square root of n squared is the same as n. That's what you're saying right.

@OverlyExcessive: I think you're pushing around symbols you're not entirely comfortable with. Every time you write down a formula, you should try to plug in a couple numbers to check that it makes sense.

Overly Excessive

@MikeMiller Yes that is good advice, I do it sometimes but I should do it more ..

But I still don't understand this completely.. I have to admit, to me $n^{\frac{1}{2}}$ seems like a nonsense thing.. I understand that this is out of ignorance, but I am trying to understand it..

I'm not going to comment to avoid stepping on others' pedagogical toes. Just thought the previous statement was worth making.

Overly Excessive

@BalarkaSen How is this $(\sqrt{n})^2 = n$ related to $n^{1/2}$ ?

@OverlyExcessive We want $n^{1/2}$ to fit into the pattern of exponent rules. That means whatever $n^{1/2}$ is to be, we should have $(n^{1/2})^2=n$. In other words, in order for $n^{1/2}$ to fit the pattern of exponent rules, we would need to define it to be a thing that squares to $n$, which is precisely $\sqrt{n}$.

4:39 PM

Is it true that $P(|X| + |Y| > a + b) \leq P(|X| > a) + P(|Y| > b)$? I see it used in a proof but it somehow feels... wrong.

@OverlyExcessive If $n^{1/2}$ makes sense, you have just proved above $(n^{1/2})^2 = n$. $(\sqrt{n})^2 = n$ too. So...

Overly Excessive

@anon I think.. I understand it now.

@BalarkaSen, @anon, Thank you both very much, I think I understand it now, thanks for your patience.

To reiterate, since it's been a while since it was said, the exponent rules are $(a^b)^c=a^{bc}$ for positive $a$.

The proof looks something like: $P(|X - Xn| > 2k) \leq P(|X| + |Xn| > 2k) \leq P(|X| > k) + P(|Xn| > k)$. But it feels like |X| > a OR |Y| > b should not imply |X| + |Y| > a + b.

Overly Excessive

Yes yes, I understand now

I'm not good with maths but I think I know why it is true now.

4:44 PM

not "why it is true," rather "why we choose to define it that way"

@peteykun: You're misinterpreting the inequality. I'm lazy so won't write absolute values. What $P(X+Y>a+b) \leq P(X>a)+P(X>b)$ means is that if $X+Y>a+b$, then either $X>a$ or $Y>b$. This is true and you should convince yourself of it.

You're correct that one of those being true does not imply that $X+Y>a+b$. That's why it's an inequality!

Overly Excessive

@anon Could we have defined it in another way you mean?

@OverlyExcessive sure, we could have defined it to be anything you want. anything that does not already have a definition you can define to be anything you want - think of words and the dictionary. if xuiosdgkl is not in the dictionary already, then there's no issue making up a definition for it - that's not going to conflict with anything! we wanted to define $n^{1/2}$ to be something that fit with the pattern of exponent rules, and it's that desire that forced our choice.

@MikeMiller Ah, thanks! I was looking in the wrong direction. :)

Overly Excessive

@anon And specifically in this case, it must be $\sqrt{n}$ because $(n^{1/2})^2 = n^1 = n$ ?

4:49 PM

mmhmm

Not "must", but "should".

Overly Excessive

Could we also leave it as undefined?

Like $n/0$ ?

yes

but then you're missing out on all the math beyond basic arithmetic

Overly Excessive

In what way?

are there multiple ways to miss out on something?

like, is there more than one way to not go to a party?

4:53 PM

@MikeMiller I have been introduced to some basic differential topology (assuming the inverse function theorem) and have done some exercises from Guillemin-Pollack chapter 1 during my stay on the topology & physics school. While that makes me feel like Neo ("I know kung fu"), I am promptly going back to calc because I don't kno how to prove IFT.

Overly Excessive

@anon What I mean is. How does it miss out on anything beyond basic arihmetics by being undefined?

Ok.

Thought I should let you know.

@OverlyExcessive how much of society can you participate in and appreciate if you can't read, write, talk or comprehend speech or text?

Overly Excessive

@anon You can't participate much except for being present I suppose.

4:55 PM

it's a bit of an overexaggeration, but to the extent that math beyond basic arithmetic requires familiarity with exponent laws and these definitions to understand...

Overly Excessive

So you're saying I can't understand it, I see ~

I'm saying any math that to be understood requires you understand exponent laws and the meaning of fractional and negative (and even real) exponents... will require you understand these things for you to participate in them

which is why choosing to leave them undefined is not an option if you want to learn more math

I know what transeversality means now and can prove that transeversality is stable (if $f : X \to Y$ is transeversal to $Z \in Y$, $X$ compact, then for every htpy $f_t$ of $f$ there is a $\varepsilon$ such that for all $t < \varepsilon$, $f_t$ are all transeverse to $Z$). :)

Overly Excessive

@anon Okay, that makes sense, I thought it was some form of veiled insult.

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