Mathematics - 2012-06-13 (page 2 of 3)

Martin Sleziak

13:00

regular matrix = invertible matrix

Jaydon Zhao

not necessarily

Martin Sleziak

Wiki says: Regular matrix may refer to invertible matrix (this usage is rare).

I did not know that it is an unusual usage.

Jaydon Zhao

haha

Martin Sleziak

@JaydonZhao Well, then $Q=0$ is a counterexample.

Jaydon Zhao

yeah, I've always used invertible

well, if we know for sure

that $Q \neq 0$

Henning Makholm

13:01

@MartinSleziak Thanks!

Jaydon Zhao

but nothing suggests that $Q$ is invertible

Martin Sleziak

You're welcome.

Eugene

oh boy do i hate mornings.

Martin Sleziak

What about $Q=\begin{pmatrix} 1&0&0\\0&0&0\\0&0&0\end{pmatrix}$ and $A=\begin{pmatrix} 1&0&0\\0&0&1\\0&0&0\end{pmatrix}$?

Henning Makholm

@JaydonZhao If all you know is that Q is nonzero, then it tells you nothing. For example if Q has two columns of zeroes, then $QA$ does not tell you anything about the corresponding rows of $A$.

Martin Sleziak

13:04

Which is based on what Henning just said.

Jaydon Zhao

@Henning @Martin: thanks, makes sense. :)

however

I assume that if $Q$ is invertible

the $A = A^T$ follows?

Henning Makholm

Sure, then you can just multiply by $Q^{-1}$ from the left.

Jaydon Zhao

okay

Matt N.

@ZhenLin Are you there?

If yes: is it true that a functor $F$ is exact (i.e. maps short exact sequences to short exact sequences) if and only if it maps exact sequences $M \to N \to P$ to exact sequences $F(M) \to F(N) \to F(P)$?

I think Paul told me this but I'm not sure anymore.

Zhen Lin

I think you need it to be additive.

Preserving monos + epis + biproducts is enough to be exact.

Matt N.

13:17

Ok, so the proof is slightly more involved from the point of view of someone who doesn't know much about category theory.

I tried to prove it because I thought it might be a one-liner.

Then failed and thought that maybe it's not true. So I need the additional assumption that $F$ is additive... good to know : )

Zhen Lin

I guess it must be true though, because every exact sequence breaks up into little ones like that.

Matt N.

That's what I tried but then after applying $F$ I couldn't stick them back together to get the sequence I wanted.

Zhen Lin

Ha?

Matt N.

I tried something like this:

If $0 \to M \to N \to P \to 0$ is exact then $0 \to ker(f) \to M \to N$ is exact. Since $F$ is exact, $0 \to F(ker(f)) \to F(M) \to F(N) $ is exact.

Then I want to do the same for $g: N \to P$.

Zhen Lin

No, don't bother with breaking it up into kernels and cokernels. Just cut it up directly.

Matt N.

13:20

And then stick it back together into $0 \to F(M) \to F(N) \to F(P) \to 0$.

I see.

Zhen Lin

To show that preservation of short exact sequences implies preservation of long exact sequences, you need to cut it up using kernels and cokernels, yes.

Matt N.

And do I need that $F$ is additive to do this direction?

Or is cutting up with kernels and cokernels enough?

Ah let me try and see.

Zhen Lin

Hmmm... I don't think you need it to be additive either way. Exactness implies additive, after all.

Matt N.

If $M \xrightarrow{f} N \xrightarrow{g} P$ is exact I get $0 \to ker(f) \to M \xrightarrow{f} im(f) \to 0$ is exact and hence
$0 \to F(ker(f)) \to F(M) \xrightarrow{F(f)} F(im(f)) \to 0$

Zhen Lin

I'm afraid I have to go off now, but I'm sure @DylanMoreland can help. :p

Dylan Moreland

13:34

I just think about modules!

Matt N.

@ZhenLin Sure, np : )

Thanks for your help!

I'll ask it on SE.

Meh. First I'll try for a few more minutes though.

Dylan Moreland

@MattN You're working on this, right?

Matt N.

@DylanMoreland Yes and I have one direction.

Now I want to show the other, i.e. that if F maps ses to ses then $F$ maps an exact sequence $M \to N \to P$ to an exact sequence.

Dylan Moreland

What's your definition of exact?

I also thought the other direction would be harder. I'm not even sure that the other direction is true.

Matt N.

I am using "$F$ is exact if and only if it maps ses to ses".

I think there is more than one way to define exactness of a functor.

Dylan Moreland

13:46

Yeah.

Matt N.

I have seen something about preserving limits somewhere.

Dylan Moreland

How'd you prove the other direction?

Matt N.

Let $0 \to M \to N \to P \to 0$ be exact. Then $0 \to M \to N$ and $N \to P \to 0$ are exact. Hence $0 \to F(M) \to F(N)$ and $F(N) \to F(P) \to 0$ are.

Hence $0 \to F(M) \to F(N) \to F(P) \to 0$ is exact.

I'll ask the other direction on SE. I shouldn't be doing this really, the exam will be about commutative algebra mostly, I suppose. So I should focus on that.

: S

bbl

Dylan Moreland

That looks good (maybe say something about exactness in the middle, but that's even more obvious). I used to get confused about this and I got confused again :)

You seem to be on the right track for the other direction.

Ilya

14:08

@robjohn: and what about now?

Ilya

14:24

@Jonas: hi!

Jonas Teuwen

@Ilya Hi!

Ilya

@Jonas: how are you in Spain?

Jonas Teuwen

Great. Doing many theorems!

About square functions. They are so cool.

Ilya

Do you actually know a good reference for Banach fixpoint theorem (aka contraction mapping theorem)?

@JonasTeuwen which square functions?

Jonas Teuwen

Conical ones.

Why do you want a reference for such an easy theorem? :-).

Just iterate like a monkey!

2

Ilya

14:30

@JonasTeuwen to put it in a paper. don't you think I'm looking for the proof because it's too hard for a guy who is working in the systems and control department?

Jonas Teuwen

Haha, no, you're a mathematician.

Well, maybe check out wikipedia and see what references they have? I don't know from the top of my head :-).

Ilya

sometimes looking around and talking to my supervisor I can forget about that

@JonasTeuwen I put Real&Complex Analysis by Rudin, there is a more general theorem

if one really does not know that (strong) contraction operator has a unique fixpoint, well - it certainly worth reading the whole Rudin's book!

Jonas Teuwen

Yes.

I need to go down! There is a coffee break, don't want to miss the cookies...

Ilya

@JonasTeuwen eat well! :)

was nice to talk to you

@Jonas: it appeared to be Rudin, Principles of Mathematical Analysis

robjohn

@Ilya how about now?

:-)

Ilya

14:41

@robjohn already found it :) was lurking for contraction mapping theorem

a reference on

robjohn

@JonasTeuwen I never knew monkeys iterated...

Ilya

@robjohn even if they do random iterations, they write "Hamlet"

robjohn

@Ilya Aha... I think it was Eugene that was talking about shrinking maps (slightly different)

@Ilya Why don't they ever write Macbeth?

Eugene

@DylanMoreland i answered the 3-torsion question yesterday

Ilya

@robjohn they even write the story of your life

robjohn

14:51

Geez... the answers to this question have gotten almost no votes. There seem to be some questions that get unwarranted numbers of votes, and then some that almost eschew readers.

@Ilya but not Macbeth? ;-)

Ilya

and her story also

Eugene

@robjohn it's like peer review. the more difficult the solution, the less people who can review it i guess.

robjohn

@Ilya I have the card catalog to that library :-D

Ilya

@robjohn you store it in your Hilbert cube?

I usually use Hilbert cube to put there my socks - to be sure that I always have a black pair in the morning

Mariano Suárez-Alvarez

socks are stored in Klein bottles

that's why they get lost

robjohn

14:58

@MarianoSuárezAlvarez Kleinmore driers are where all those socks go :-)

Mariano Suárez-Alvarez

sinc I don't have a drier, I have to lose them in my washing machine

Eugene

@robjohn my posted solution to this question didn't receive any votes but strangely enough the question got an upvote!

robjohn

@Eugene I'm sorry to say that I don't understand any of it :-(

Not that your answer is poorly written, I just don't know algebraic geometry.

Eugene

@robjohn haha. yeah, and that's why the harder answers just get no votes.

Rajesh D

I have been out of PhD program and i have been trying for a job for the past six months, I have not got a single interview call!

Eugene

15:03

huh. curiously it just got a downvote. hopefully the person will explain what is wrong with it.

robjohn

@Eugene I upvoted it because I could barely make sense. So that offsets the downvote.

@Eugene I also commented to the downvoter.

Eugene

@robjohn oh thanks!

Rajesh D

@Eugene : You suck! youre upset for a single downvote!

Eugene

@robjohn you can remove your upvote if you want. the downvoter retracted it.

robjohn

@Eugene Nah, I didn't upvote to counteract the downvote. I was going to upvote anyway.

Eugene

15:06

@robjohn oh thanks then! i hope i didn't make a genuine mistake though. i'm pretty sure i checked the math.

oh wait the downvoter is back.

this is so confusing!

Rajesh D

Hi @robjohn

Eugene

is it right or is it wrong???

robjohn

@RajeshD hey there.

@RajeshD If it were a downvote on an answer that I think is correct, I would be miffed as well.

@RajeshD It warrants at least a comment as to why the downvote

Eugene

it's gone again??? i'm so confused!!!

somebody is trolling me for sure now!!

robjohn

@Eugene don't let it get to you.. soon they will settle down.

@Eugene perhaps they just can't figure out whether it's right or wrong. Hopefully they will comment.

Eugene

15:10

@robjohn like you said though. i'm mostly worried if it's correct. the easy ones i don't really care much about but these ones i put some effort into them!

christ they're back again.

bah. whatever

@robjohn oh ok it doesn't matter now. i'm 100% sure the math is right.

thanks again.

robjohn

@Eugene That's what counts. I have a couple of answers that I know are right but have been downvoted without comment. You just have to let it slide.

Eugene

@robjohn yeah. i'm trying to get used to MSE voting. i get worried if it's right, but once i'm sure it is i don't really get bothered anymore.

Rajesh D

@robjohn I am getting downvoted in real life and career everyday! I am not even able to find anyone whoom i can tell this

robjohn

@RajeshD are you with the same adviser who is abusing you (they really are)

Rajesh D

I am out of it, i am out of the phd program, but technically i am on academic leave from university but my ties with the current advisor is over. He is happy to get rid off me

me being rejected by my advisor, no one (in this university) will be willing to take me, any never mind @robjohn : these problems can't be solved here. Sorry for being off tiopic. @Eugene Sorry about my previous comment

Eugene

15:22

@robjohn anyway i have to get to work now. thanks again!

@RajeshD it's not a problem. i really didn't take any issue with it at all.

Rajesh D

thanks @Eugene. Bye

robjohn

@Eugene later

@RajeshD people change advisers here without problems. It's too bad they don't let you do that. :-(

Nirakar Neo

15:40

@Ra

@RajeshD who was your advisor?

Rajesh D

Hi @Nirakar

I don't think its good to post it on the internet.

experimentX

Hi ...

if some sequence is less than (-1)^n

does it converge?

Rajesh D

$0.5 (-1)^n$ ??

does it converge?

sorry it not actually less than it

@ExperimentX : You can define a sequence any way you want. So i guess there are sequences less than that but do not converge

experimentX

16:00

Oh ... thanks

Rajesh D

for example $a_n = 0.5(-1)^n$ when $n$ is even, $a_n = 5(-1)^n$ when $n$ is odd

experimentX

@RajeshD I'm posting an answer ...please review ... if error i'll delete before it gets downvoted

1

Q: determine whether this series converges for this value of z

Does $$f(z)=\displaystyle \sum_{n=0}^{\infty} \frac{2^n+n^2}{3^n+n^3}z^n$$ converge for $z=\frac{-3}{2}$?

are there any errors??

experimentX

16:42

0

Q: Is it possible to evaluate this limit without series expansion and l'Hôpital's rule?

I have a limit which is to be evaluated without using l'Hôpital's rule or series expansion. $$ \lim_{x \rightarrow 0}\frac{\frac{\sin x}{x} - \cos x}{2x \left ( \frac{e^{2x} - 1}{2x} - 1 \right ) }$$ Is it possible to evaluate it without l'Hôpital's rule or series expansion only using trigonometr...

limit

Gigili

17:17

Well, is there anything else you want to tell us?

Like, umm, (removed)?

experimentX

Hi ...

Is there any specific set of rule while changing the order of integration in double integral like in this question math.stackexchange.com/questions/157494/…

robjohn

@AméricoTavares I looked into your question and I couldn't come up with more than what WimC did. I agree with Generic Human's comment that the series may or may not diverge (as with the Euler-Maclaurin Sum Series), but the point is that in general it is hard to show convergence/divergence, and we are only interested in a finite number of terms anyway to get the asymptotic behavior of $f$.

David Wheeler

17:33

@experimentX google Fubini's theorem.

experimentX

thanks for the clue @David

David Wheeler

HAI @Gigili

Gigili

High @David.

David Wheeler

ah...that ship has sailed for me, i'm afraid.

experimentX

lol ... instead of Google i youtubed it ... lol

robjohn

17:44

@AméricoTavares however, I have added my 2 cents as an answer.

Gigili

18:04

@anon OMG, you cannot be serious.

David Wheeler

serious how? i've always thought anon is perfectly capable of being serious, but perhaps we've been played.

i suspect Arturo keeps entire paragraphs pre-composed and ready to copy-and-paste into his answers.

Gigili

That must be the case.

As it is, there's no other possibility.

David Wheeler

why, gigili, a reducio ad absurdum. i'm impressed.

Gigili

He has some servants to type the answer for him.

David Wheeler

i was reading a question that had been posted 3 minutes before i read it. as i finished reading it, his answer appeared. fully grown, and with a driver's license to boot. sigh

Américo Tavares

18:15

@robjohn Thanks! I think I understand everything but "the remainder (not the remaining terms) can be bounded by something smaller than the preceding terms", because I do not see what's the (subtle?) difference between "remainder" and "remaining terms". (It may be some trouble with my English).

experimentX

Hi ... is there a bijection relation between R and R^2 ... R being set of real number??

Gigili

The question is, What is the motivation behind this?

David Wheeler

@experimentX yes, they have the same cardinality.

Gigili

Dixi.

David Wheeler

@Gigili huh?

experimentX

18:17

there's isn't relation between Z and R right?? ... but how could there by for R and for R^2

Gigili

@DavidWheeler Umm.

David Wheeler

Z is countable, R is not.

Américo Tavares

@robjohn I've found your explanation easier apart from that last sentence.

David Wheeler

suppose we call the cardinality of R, k. then the cardinality of R^2 is k^2. assuming the axiom of choice, k^2 = max{k,k} = k.

experimentX

still would there be one to one correspondence between that are not countable??

David Wheeler

18:24

for an explicit idea of how this can be done, consider a peano space-filling curve, which is a surjective map from R to R^2. using the axiom of choice, we can construct an injection from R^2 to R ("choosing" one pre-image point in R for every point in R^2).

we then have an injection from R^2 to R, and the obvious one from R to R^2, so we can apply cantor-schroeder-bernstein to get a bijection between the two.

experimentX

Hmm ... lol, i m encountering terms i've never before ... thank you @DavidWheeler ... i have searched for these terms!! for now i'll keep in mind what you told me :)

David Wheeler

keep in mind this bijection will NOT be differentiable. space-filling curves are necessarily "kinky" (in the geometric sense, not the...well, you know)

experimentX

are these space filling curves??

David Wheeler

infinite sets are...trixy. to get from countable, to uncountable, it's not enough to have "infinite copies of an infinite set". for example, the set of all polynomials with integer coefficients is countable, even though we have "infinite choices for the coefficients", and "infinite possible powers of x"

yes, i believe that is David Hilbert's variation of Peano's original construction.

which shows you how a 1-dimensional line can "cover" a plane.

experimentX

Hmm ...

mathematics is sure interesting :)

David Wheeler

18:34

yes...surprising things happen, sometimes.

experimentX

David, I checked Fubini's theorem ... it was not i was looking for :( ... I tried to evaluate $ \int_0^1 \int_0^{\sqrt y} dx dy $ like in Fubini's way ... but here $ \sqrt y $ involved ..

robjohn

@experimentX Those look like the first several levels of the Hilbert Curve

David Wheeler

yes, @robjohn they are

robjohn

@DavidWheeler Thanks ;-)

experimentX

Hm ... i need to do some deep mathematics study ... i guess after exam :)

robjohn

18:39

I have to go proctor a Logic exam. BBL

David Wheeler

@robjohn @experimentX <-- was asking if R^2 has the same "number" of elements as R

robjohn

@DavidWheeler it does

David Wheeler

which is why i mentioned Hilbert's curve...seeing is believing

robjohn

ttfn

Gigili

Talk To, umm, talk to Fredrick now?

David Wheeler

18:41

ta-ta for now

or as we netizens say: l8terz

experimentX

@DavidWheeler about my earlier question ... can you do something ... is there some set of standard rules ... I could get the same result by $ 1 - \int_0^1 \int_0^x^2 dy dx$

Gigili

But I'd talk to Fredrick now.

David Wheeler

well, ok, then

Old John

Hi folks

experimentX

$ \int_0^1 \int_{0}^{\sqrt y} dx dy = 1 - \int_0^1 \int_0^{x^2} dy dx$

Hi @OldJohn

How do we do these types of order change of integrals ... ?? is there a standard rule??

David Wheeler

18:45

there are a number of theorems which say you can given certain conditions on f(x,y)

Old John

For examples like the one you give, I always used to sketch a graph to see roughly what the areas look like (and to help working out the limits of the re-ordered integral)

David Wheeler

if f(x,y) is continuous and bounded on the region D, then that is more than enough

Gigili

People are so mean sometimes.

experimentX

Hmm ... really??

David Wheeler

since f(x,y) = 1 (constant function), in this case, it suffices that D is bounded.

experimentX

18:48

for parabola passing through 0,0 -- 1,1 .. (area of rectange - change of order of integral worked) ... I tried with circle too ... though it did not work

David Wheeler

what do you mean "tried with circle"?

experimentX

$$ \int_0^a \int_0^{\sqrt{a^2 - x^2}} dydx \neq a^2 - \int_0^a \int_0^{\sqrt{a^2 - y^2}} dxdy $$

Though it worked with parabola $$ \int_0^1 \int_0^{\sqrt y} dx dy = 1 - \int_0^1 \int_0^{x^2} dydx$$

David Wheeler

that's because you're not computing the area of the same region on both sides.

i will see if i can explain.

experimentX

Ah ... the source of my problem lies in this question math.stackexchange.com/questions/157494/… ... unable do understand what Marvis did

David Wheeler

the region you are computing in the integral $\int_0^1 \int_0^{\sqrt{y}} dxdy$ is the region bounded above by the square root function, the x-axis, and the vertical lines 0 and 1

when you "turn sideways" you compute the area BELOW the parabola, and subtract from the rectangle

but with a circle, it's different...the bounding curve is "above" either way

experimentX

18:58

Oh ... i see

David Wheeler

so get rid of the a^2 and the minus sign, it should work out

as OldJohn said, draw a picture of the region D first. this will help you "get oriented"

experimentX

Hmm ... ah yes, it works out ...

But how could Marvis do it so easily??

David Wheeler

no doubt he has practiced many times

experimentX

Oh ... thanks i can see that!! ... it was like Old John said ... looks like observation is only the key

Thanks @DavidWheeler for all your help ...

David Wheeler

no problem

Gigili

19:10

So, when I +1 an answer, is it plus one or positive one @Skull?

skullpatrol

"plus one" is the way most people say it :) @Gigili

Gigili

Good, good.

Most people could be wrong anyway.

skullpatrol

Yes, it is a soft question...

Gigili

My question wasn't that soft.

Américo Tavares

When we have a series that may or may not converge what is the difference between "the remainder" and "the remaining terms"?

skullpatrol

19:19

@Gigili I meant the question I posted, sorry I didn't mention that :(

Américo Tavares

@robjohn (Continuation of my doubt). Is "remainder" used for convergent series only and "remaining terms" for convergent or divergent series?

anon

19:35

I think "remainder" just means the difference between a partial sum and the function it's intended to approximate.

skullpatrol

I think your question falls under "colloquialisms." @gigili

Américo Tavares

@anon That makes sense in the context of the following sentence "The series in (1) [a series representation of a function] may or may not converge, as with the Euler-Maclaurin Sum Series. As with most asymptotic series, we are only interested in the first several terms; the remainder (not the remaining terms) can be bounded by something smaller than the preceding terms."

Gigili

@skullpatrol May be.

skullpatrol

Maybe not...

Américo Tavares

19:55

@robjohn Based on anon's comment ( " 'remainder' just means the difference between a partial sum and the function it's intended to approximate") I got it.

@robjohn I've just accepted your answer.

skullpatrol

20:18

@Gigili I think, one day, I'm going to leave this site forever, just like you.

Gigili

@skullpatrol Not like me, I hope.

skullpatrol

@Gigili Good answer +1 :D

Américo Tavares

@anon Thanks for your comment.

anon

no prob

@Gigili BTW yes that screenshot was real, someone apparently flagged it as a joke.

Peter Tamaroff

20:42

@anon Thanks for the feedback on the proof.

anon

no prob. (though I don't like the idea of using induction on that proposition)

Hard mode: what are necessary and sufficient conditions to determine if the sum of the reciprocals of a multiset of integers is itself an integer?

Peter Tamaroff

@anon Multiset?

anon

Like {1,1,2,3,5}, where membership has multiplicity.

Peter Tamaroff

@anon Oh. For starters, the product $x_1 x_2 \cdots x_n$ can't be greater than the numerator.

In that simple case

anon

Maybe. What if the denominator is positive while the numerator is negative? :D

I did say integers (though I should have qualified with "nonzero")

Peter Tamaroff

20:46

@anon $| X|$!!!!!

@anon But do you think the proof is entirely flawed? Do you follow my line of thought?

anon

Sure, it can be salvaged relatively easily.

Peter Tamaroff

@anon Oh, salvage it! Salvage it!

I feel happy the salvatage is easy.

anon

Easy. $x_n\frac{\tau}{x_1\cdots x_{n-1}}$ can only be an integer (given that the fraction is not one) only if $x_n$ shares a factor with $x_1\cdots x_{n-1}$ ....

Gigili

@anon Umm, I didn't laugh. Not even a bit.

anon

You just have to find the right way to formalize and justify this idea.

Well, I laughed :D

Peter Tamaroff

20:52

@anon What screenshot?

anon

this one from yesterday

Peter Tamaroff

@anon Well isn't that justified by the hypothesis?

anon

yes

Peter Tamaroff

@anon So what's the problem?

anon

the problem with what?

Peter Tamaroff

20:53

@anon What's wrong with that?

@anon The formalization.

I mean, what is missing in the proof?

anon

@PeterTamaroff someone flagged a picture of a cute puppy as "spam/offensive"; do you agree?

I was tasked with in/validating the flag.

Peter Tamaroff

@anon Hahah yes, I mean why did he/she?

anon

As a joke.

Peter Tamaroff

@anon Oh. Bah.

Eugene

@anon i posted a solution to the curves question yesterday and arturo or brian didn't beat me to it! =)

Peter Tamaroff

20:55

It is evident that $$\frac{\tau}{x_1\cdots x_{n-1}}$$ is not an integer from the hypothesis.

@Eugene Ai, mate.

anon

@PeterTamaroff The statement in your proof, that $x_1\cdots x_{n-1}|x_n$, is hard to justify and I'm not sure if it's even true. However you don't need this to conclude it can't be an integer.

@Eugene I know, I saw it yesterday and was tempted to comment on it, saying "How's that grading going?" :P

Peter Tamaroff

@anon But I'm saying the opposite.

Eugene

@anon hahaha. looks like i didn't get past the third stage after all! i still have about 30 papers left...

anon

@PeterTamaroff Well you're saying and saying not it, that's where the contradiction comes in.

Peter Tamaroff

I'm saying that if $x_n\dfrac{\tau}{x_1\cdots x_{n-1}}$ was an integer, then $x_1\cdots x_{n-1} \mid x_n$ would have to be true.

But it is impossible.

anon

20:57

I don't see how to prove that implication.

Gigili

Stop editing already. This is just pah and nothing else.

Eugene

@Gigili me?

Gigili

No.

Eugene

ah i see.

@Gigili how's trying to stop using MSE going? =p

Peter Tamaroff

@anon From the hypothesis $$x_1 \cdots x_{n-1} \not \mid \tau$$

anon

20:58

Instead, say if $x_n\frac{\tau}{x_1\cdots x_{n-1}}$ is an integer while $\frac{\tau}{x_1\cdots x_n}$ is not, then $x_n$ shares a factor with $x_1\cdots x_{n-1}$, which contradicts our hypothesis of pairwise coprimality.

Peter Tamaroff

So I use $a \mid bc $ and $a \not \mid c$ then $a \mid b$.

anon

@PeterTamaroff But that's not sufficient. I even gave you an example in my comment to your question.

robjohn

@AméricoTavares When computing asymptotic series, there is usually a remainder term. It is generally around the same size as the next term in the series, so it is usually left out, but it should really be estimated to be sure that everything is copacetic. In most cases, asymptotic expansions are not necessarily convergent (hence assumed divergent), so you don't want to look at the remaining terms.

Gigili

@Eugene Extremely successful, as you see.

anon

@PeterTamaroff $3\frac{10}{6}\in\Bbb Z$ and $6\not\mid 10$ is not sufficient to establish $6\mid 3$.

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$Mathematics$ Mathematics