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11:00 AM
I have a request for all
 
To cram it where the sun don't shine???
 
is there someone who is willing to take me as a minor collaborator for solving some problem that is not too advanced and preferably belongs to number theory?
 
How advanced are you? I need someone to check my equations 8-) 67 pages full of equations.
 
Do you want to write my thesis for me? I have the results, just can't be arsed to write them properly :-D
 
You guys talk as if you're professors already =)
 
11:04 AM
@Jonas: whoa not that kind of advanced! I have a reasonable grounding in elementary number theory, statistics and a very rudimentary smattering of analysis.
 
@Srivatsan No the equations are quite basic 8-).
Just many of them.
The optics professor actually delegated that to me.
 
@Srivatsan You talk like you're an Indian computer scientist in the USA.
 
@Asaf hehe :D
 
How can I hit 20k if no one asks about the axiom of choice?!
 
@Asaf By answering analysis questions.
 
11:13 AM
@JonasTeuwen oh, the horror...
 
Sure I already have 19 answers in ... I can try to get one really good and hit the badge.
 
do we have some nice results regarding distribution of quadratic and higher power residues modulo a given n? I did some google search, but mostly they turned out to be analytic results giving some lower/upper limit on least quadratic residue/nonresidue. Terence Tao has recently asked a similar question in MO I guess. My question is whether we have some exact results? are we in some sense able to determine the precise location of higher power residues?
 
@NikhilBellarykar What do you mean precise location? What kind of information about these residues are you looking for?
Given $a \in \mathbb Z /n\mathbb Z$, do you want to know if $a$ is a residue?
 
@Srivatsan : amongst the integers 1 to (n-1) , exactly which integers are residues for a given power a>1, yes.
 
@Srivatsan I don't want to know.
 
11:20 AM
@AsafKaragila =)
@NikhilBellarykar That's very imprecise, no?
 
how so? I did not get you I am afraid.
 
The resides are scattered everywhere (and in some sense, they are random or are thought to be random, I think). Even to point out which ones are (quadratic) residues, I will need to give you a list of around $n/2$ numbers.
 
Ok, fine. What about higher powers? I guess the situation should improve (not monotonically, of course) with increasing power since the no. of distinct residues modulo n for a given power a are phi(n)/gcd(n,a); i.e. no. of residues decreases; so the pattern should somehow be not random?
 
@NikhilBellarykar Perhaps. I am not sure.
 
k thanks.
 
11:27 AM
But I would imagine that higher powers will be even more complicated than quadratic case even if their number is smaller.
 
@Srivatsan Higher powers such as your various gods? :-P
 
quite possible, I guess I will stack up some numerical evidence regarding this and then get abck.
 
@NikhilBellarykar Oh, btw. Huge disclaimer: I have no expertise on these subjects. Please do not assume my word counts for anything. =)
@AsafKaragila Yes, they are more complicated than the quadratic gods. =)
 
@Srivatsan Fine, although I guess I know better :)
i.e. better than to trust this statement of yours
 
@NikhilBellarykar His statement about quadratic gods?
 
11:34 AM
@AsafKaragila, no, the one before that.
 
About even higher gods?
 
oh, I guess the Hierarchy of Gods is ok
 
11:49 AM
Anyone willing to only suggest a problem for me to work on? it should be preferably related to number theory, and not very advanced-i.e. algebra and analysis requirements should not be much if at all while some statistics is ok?
 
Pick up a book about number theory. Study and solve the exercises.
Also, stop asking the same question repeatedly. It's an annoying habit.
 
Indeed. thanks for the suggestion.
 
12:17 PM
A number theory problem using statistics? Well, that is odd.
 
@JonasTeuwen Some results on normal numbers can be shown using Strong law of large numbers. I guess probability results might be useful in study of number theoretic densities. Probabilistic number theory. I would call it using probability, but perhaps the line between probability theory and statistics is not that clear...
 
Sure, but statistics isn't probability theory.
My stuff can be done using probability. Well. Almost anything can.
But right, the distinction might not be that clear.
 
@MartinSleziak Results on normal numbers!?! Are there any? =)
[I am kidding...]
 
Ergodic Theory <3.
 
@JonasTeuwen <3333 Ergodic theory. =)
[I hope you remember the <3333 question though.]
 
12:34 PM
The solution given in this problem could be called statistics, even though the problem is finding sum of a series: sosmath.com/CBB/viewtopic.php?t=28258
The problem is to find $\lim\limits_{n\to\infty}e^{-n}\sum_{k=0}^n\frac{n^k}{k!}$.
 
@Srivatsan I don't 8-).
@MartinSleziak Oh, I call that a probabilistic argument.
 
12:55 PM
@Srivatsan And it looks like someone very famous has seen your answer :)
 
@tb Who is that very famous person?
 
@tb Should I go like: Yay, he commented on the other answer in that page! =)
 
Do you mean R?
 
I was thinking PS, but R also works =)
 
PS: Not R.
 
12:59 PM
I understood. I was just kidding...
 
Oh right, that guy.
I had a discussion with him about ad hominem 8-).
No ad populum.
 
@JonasTeuwen Interesting. When did you meet PS?
 
Just on the internets.
 
Done with Mechanical Physics exam
WOH!
 
@N3buchadnezzar How went the exam?
 
1:03 PM
I think I will pass <3
 
@JonasTeuwen Quite cool. :-)
 
@Srivatsan they know that $dim U^\perp$ doesn't give the desired result but don't know that \dim does the trick...
Looking at how people type their stuff here it is incredible how many are completely oblivious to typographical things...
 
@tb Nice point; I agree with that. But why pull the n-k outside the $$? =)
 
@Srivatsan That's a mystery I can't explain :)
 
Not just that. I also see things like $f$ $:$ $R$ -> $R$ =)
 
1:13 PM
Well: I know how to write n-k. The only thing I do not know how to do is that upside down T. I'll try ^\perp - it does not work, since in needs something in front of it. So I'll write $U^\perp$.
 
@MartinSleziak Um, that might be it, after all.
Nice point as well.
 
@MartinSleziak Better to use \top and \bot anyway.
 
BTW is there a difference between \perp and \bot?
Test: $\perp$ and $\bot$.
 
test: $\perp$ $\bot$ $\top$
 
We wrote that almost simultaneously :-)
 
1:15 PM
@MartinSleziak I imitated your action actually. =)
 
@MartinSleziak Compare $A \bot B$ and $A \perp B$: \bot is a math symbol while \perp is a relation
 
So they have different catcodes....?
 
Different spacing, anyway. I don't know what a catcode is...
 
Just like a \vert b, a|b and a \mid b.
test: $(a \vert b) = (a \mid b) + (a | b)$ gives $(a \vert b) = (a \mid b) + (a | b)$
 
I thought that catcode is that thing that says whether it is relation, operation etc. (and thus it defines spacing in math mode), but I remembered it wrong: en.wikibooks.org/wiki/TeX/catcode
I think that \makeatletter is equivalent to changing catcode...
 
1:20 PM
@MartinSleziak Does the right hand side of your comment (that I am replying to) have a border-like thing on the right? (Half gray and half white)
 
@Srivatsan I don't see there anything like that.
 
@MartinSleziak Thanks. Strange; I see such a thing in some of the comments.
What's the policy on Project Euler problems? I know about Jonas M.'s meta post (and his exchange with Project Euler on posting their problems here), but was some course of action agreed upon?
 
This is what I see: http://i.stack.imgur.com/X63wS.png
and http://i.stack.imgur.com/VdeAE.png
 
@Srivatsan That was Will J.
 
@tb I see. =/
 
1:27 PM
 
@MartinSleziak This is what I was talking about: i.stack.imgur.com/yVcji.png
 
@Srivatsan That's a mini-scrollbar because Martin's comment was multiline.
 
@Srivatsan I don't remember seeing something like that. What browser are you using? (I have FF.)
 
@tb No, I do not see it now =)
 
I don't see it there either....
 
1:29 PM
@MartinSleziak I use Chrome.
 
Test2: Let us try something multiline containing TeX. $x^2+y^2=z^2$.
$\sum k=\frac{n(n+1)}2$
$\int x \mathrm{d}x$
Now I have scrollbar there. It looks differently in FF.
 
Yes, that's what I meant.
Okay Srivatsan, you've got some teaching to do :)
 
When I view transcipt without applying bookmarklet, it's without scrollbar, it gets there only after I use it. So probably in the main chat people who do not have bookmarklet don't get that scrollbar either.
 
I need some help in Q, May i ask the Q?
 
1:34 PM
Ok, we should stop fooling around and listen to @FreakEnum's question.
So what's the question?
 
@: A(guy) can do certain job in 12 days. B is 60% more efficient than A , then how much time does B takes to do the job?
 
@FreakEnum Would you know the answer if B would be twice as efficient as A?
 
@MartinSleziak yes
 
And the answer would be?
 
A:B :: 2:1
 
1:37 PM
I.e., 6 days.
Here we divided 12/2, because B does 2 unit of works in the same time A does 1 unit.
Now, 60% more efficient means that B does 1.6 units in the same time A does 1 unit.
 
@FreakEnum The ratio 2:1 is not the answer. We are asked to find the number of days B would take.
 
@MartinSleziak aah , right
sec I do the solution
15/2
 
@FreakEnum Well, do you expect us to check that your answer is correct? [Does the textbook also mention the answers at the back?]
 
@Srivatsan no , my answer matches the book answer
Thanks @MartinSleziak @Srivatsan
 
@FreakEnum Ok, that's great. // I didn't help you this time =)
 
1:42 PM
@Srivatsan but mostly you do ( from previous stats :D )
 
(Some silliness.) I once had a Thermodynamics textbook. It had all solutions at the back, except for every third problem, for which it mentioned only the answer and not the full solution. All the problems for which the solutions were available -- they were quite routine. But it turns out that each of the skipped problems was a monster. I used to struggle for hours to find some approach to the problem. And at the end of it, my answer will invariably be wrong. =)
 
All of my books just include the answers
Even my physics books =(
 
QED
physics
Where can I find a completely rigorous complex analysis book or notes?
 
It's strange thing that in Russian you write Физика (fizika), and other languages have the letter Y there. But maybe it's because it would change the pronunciation.
 
@QED I'm don't think any book out there will meet your standards of rigor =)
 
QED
1:56 PM
I'm having trouble connecting things like homotopy and simple connectedness with the complex integrals. There seems to be a gap between them
it's obvious what the pictures of these things are but that's no proof
 
@QED It might be helpful to mention which books you tried to follow...
 
I have only skimmed through it but I liked the book "Visul Complex Analysis"
Althoguh, I doubt it is very rigorous.
 
QED
For example, this has an interesting remark
 
@QED What about the remark?
 
QED
You would just think "homotopy equivalent" paths etc. and that is true, but to prove it is much harder than giving a restriction for a smooth homotopy
So we are likely using theorems which we have not proved.
 
2:07 PM
I don't follow you at all. I do not know the proof off the top of my head. Without seeing the proof, I cannot of course tell if it uses other theorems that aren't proved.
 
@MartinSleziak Well, the Greek root is φύσις, clearly a y. Isn't Russian mostly phonetic?
 
@QED Are you worried that the book is not proving theorems in full generality? Or are you worried about the rigor?
 
By the way Italian and Spanish write "fisica"
 
QED
Sorry, All I meant was that in this complex analysis stuff it seems like I am meant to take a lot of things for granted and the proofs of most important are often omitted because they are 'out of scope'
 
@tb I would say so. But I don't hear spoken Russian very often. On the other hand, there are plenty of good math books available in Russian here.
 
2:10 PM
@QED The book gives the proof for the smooth case, right? Perhaps this author does not require the non-smooth case for her purpose?
 
In Polish it's completely different: fizyka. Weird. en.wiktionary.org/wiki/physics
 
QED
basically would like to find a development of complex analysis that proves every result and doesn't skip anything. I don't need exercises or examples or anything like that (the sort of things which would go well in an undergrad class).
I think the books and notes I find are targeted at the wrong audience
 
What book is this, by the way?
 
QED
pretty much all the ones I looked at
 
Did you try Conway's book?
 
QED
2:13 PM
no
thanks
 
Alternatively (and better in my opinion) Remmert's book.
 
Springer! =)
 
I read Howie's Complex Analysis a couple years ago. It seemed reasonably rigorous.
 
@Srivatsan If ever you wondered why the Springer symbol looks the way it looks: Springer (literally: "jumper") is the German name for the chess piece knight.
5
 
It seems like everyone here took complex analysis a few years ago...
feels young and immature
 
2:19 PM
If few years includes more than 10 then yes :)
 
QED
Well don't worry, even though I took it I didn't really learn it
 
@tb I have found it curious, but didn't expect it to have a reasonable explanation =) . Thanks.
 
Gowers once commented on his blog that he didn't really learn complex analysis until he had to lecture it...!
 
@tb That's not too surprising, I guess. I myself took it 4.5 years back. =)
 
@Srivatsan Is there anything special that happens if an answer is accepted half a year later? :)
 
2:26 PM
@tb Apart from your satisfaction, that is...
 
@Srivatsan Didn't expect that to happen, honestly.
 
@tb Congrats! Today is your day. Sadly there isn't too much of it left though... =)
 
Yes, apparently. One more vote and I'm capped...
 
Maybe that someone just have a really slow connection
 
@Srivatsan why not, it's still eight and a half hours.
 
2:30 PM
"The problem is "nonsingular" means smooth for some people. This is regrettable." — Too many subtly different concepts in algebraic geometry...
 
@Srivatsan someone else
 
@tb But why s.o.e. I mean, is that an abbreviation?
I got who it referred to, of course.
 
Hello there.
Term is over. Yay : )
 
Congrats!
 
2:32 PM
@tb Thanks! phew : )
"phew" is an understatement.
 
@Matt Hey, congrats!
 
Thanks Srivatsan : )
I'm feeling bonged out.
Now one more to do item until my world is whole again: successfully flog the puppy tomorrow.
 
@Srivatsan Yes, it's an abbreviation. No it's not referring to anyone specific. The story is this: A friend of mine invited a guy who is somewhat nuts to a dinner. That guy asked can I bring someone? My friend answered, yes, sure. The guy wrote on a piece of paper "ask s.o.". They talked a few minutes more and then the guy crossed out "ask s.o." and wrote "ask s.o.e." instead...
 
@ZhenLin Sure, it's just $(\Omega,\Sigma) \to (\mathbb{R},\mathcal{B})$-measurability.
 
2:38 PM
Or perhaps the OP wants an explanation of why we don't ask for preimage-of-Lebesgue-is-Lebesgue...
 
@tb I feel so clueless that it's amusing to me. =) Never mind though. (Thanks!)
 
@ZhenLin It is a good exercise to figure out why not.
 
@ZhenLin Just tell him that the Borel sigma algebra on R is generated by the open sets. So it's the same?
 
@tb It was never explained to me. But then again, I don't think I ever grasped that subtlety in the definition of Lebesgue-measurable when I was studying measure theory...
 
2:43 PM
Now I can finally stop worrying about homework deadlines and start worrying about (oral) exams. Yay.
 
Oral. Yay!
 
@tb measurability of $\mu$-negligible sets?
 
@Matt It doesn't make sense to speak of the completion of a $\sigma$-algebra without a measure (or measure class) present. The completion of the Borel $\sigma$-algebra with respect to a point measure is the power set.
 
@tb Google search for "point measure" is inconclusive but there is the Lebesgue measure present.
 
2:52 PM
@Matt It wasn't mentioned by yoyo and that's what I'm complaining about. Point measure = Dirac measure.
 
QED
3:10 PM
It's too hard to get back into study, especially with low motivation because I doubt I can actually find out anything I wanted to know
 
@ZhenLin Nice answer. Although I'm not surprised that I see simplicial sets... :)
 
It was an exercise I had a couple weeks ago...
The 5-adjoint characterisation of Set is quite interesting. I had seen it mentioned on the Categories list recently, but I didn't bother looking it up.
 
@ZhenLin I looked it up a few minutes ago but I decided that I would read it when I'm more in a categorical mood.
The morphisms in an abelian category example reminded me of my favorite homological algebra exercise: Compute the right derived functor of $\ker: \mathcal{A}^\to \to \mathcal{A}$.
 
Having been told that the third adjoint is the (co)kernel, I'm at a loss as to what the intermediate ones might be...
Also, a remark about using $\mathscr{P}\mathbf{C}$ to denote the presheaf category $[\mathbf{C}^\textrm{op}, \textbf{Set}]$: the resemblance power sets goes deeper than notation!
Marcelo Fiore pointed out to me that the presheaf category is a generalisation of the power set.
(Hint: Recall that a 0-category is a set, and a (-1)-category is a truth value...)
 
3:26 PM
Oh :)
 
QED
sounds like you're have a lot of fun with this categories stuff
 
@QED I wouldn't be applying to do a Ph.D. in it if I weren't. ;)
 
@tb Now that it's clear to me what you meant I feel stupid for answering a rhetorical question : (
 
@Matt Don't worry, I do that all the time :)
 
3:51 PM
[I flagged it, btw.]
 
QED
why didn't you downvote it?
what I mean is, is it wrong for me to downvote it?
 
@QED Not wrong. But it will be more useful if you leave a comment.
The user might simply be inexperienced and lost.
 
This is not really a new trend. This happens every other day.
 
It's dawning on me that I have absolutely no idea what hocolim is...
 
@tb Really? I haven't seen this before.
 
3:56 PM
@Srivatsan I would estimate that about a quarter of my flags is triggered by such "follow-ups"
@ZhenLin Have you ever read Freyd's Splitting homotopy idempotents in the La Jolla proceedings?
I found that pretty helpful for understanding at least why it is useful.
 
@tb No. I still know very little about homotopy theory...
 
It's not really about homotopy theory. He basically proves that an additive category with countable coproducts is idempotent complete.
(of course he phrases that as "Idempotents split in a +'ive category with countable copowers")
 
4:50 PM
@tb I recognize those as English words dealing with mathematics... :-)
 
Does that page work for y'all?
 
@Srivatsan I have seen some reasonable answers there
 
@robjohn What do you mean by answers?
 
@Srivatsan It works for me...
 
@Srivatsan It works for me, but with great lags, even for planetmath's standards.
 
4:53 PM
@Srivatsan articles.
 
It does have some good pages, I agree -- which is why I want to visit it. =) But it is invariably slow/does not load/keeps reloading etc.
Thanks.
Who maintains it, btw? Any idea?
 
@Srivatsan I assumed it was a Wiki-type set up.
Interesting: `No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.`
I thought I had installed the jsMath fonts.
That could explain the slowness.
 
I'm impressed. Compare this to this.
 
@tb Who would think such comments had any effect? :)
 
5:09 PM
@tb much nicer.
@Srivatsan evidently, they do.
 
I can never remember the conditions of L'Hospital. Wonder if there's a catch to my answer.
 
I respond a lot better to comments than to downvotes. :-)
 
@robjohn But I don't think you would post a question that would require such comments. =)
 
I'm just curious on the culture in this chatroom-- if I have a general question asking for pointers, is it okay to ask here?
 
Just ask. :)
 
5:11 PM
I think it's time I jot down a meta thread introducing the chat's general feel.
 
@Incognito You need to go through some initiation rituals. =)
 
@tb Thanks :). I don't have specific questions at the moment, but I've decided my math skills are deplorable for where they could be. I understand a lot of it, I just can't put it into practice easily.
@Srivatsan Is pi involved?
 
@Incognito No, but $\pi+e$ does.
Apple, or blueberries please.
 
Apples if it's transcendental, blueberries otherwise.
(I prefer apples to blueberries, of course :)
 
@Incognito And you need to please the owners of the chatroom. Have you taken a set theory course? =)
 
5:14 PM
Perhaps. I have no formal training in math, but I understand a few concepts like real/natural numbers etc.
 
All that's just a joke, of course. You're better off ignoring all that I say. =)
 
I'm taking on some of the project euler problems, right now they're basic sums of arithmetic series, but as I move further down I don't want to rely on me using my programming skills when there's an elegant solution in math.
For instance, not brute-force run through all the numbers 1..999 when I could do (1+999)*999/2. I know about this solution, but as I move on it will be less obvious to me.
 
Hey guys, how do I search for posts by user X? Never mind. I figured it out :)
 
In chat?
 
I've a Q and my method seems to give wrong answer , anyone got some time to help me?
 
5:17 PM
@Srivatsan the easiest way to do it is to go to the user page and do the search from there. The user:xxxx is then automatically added. Otherwise you need to remember the number xxxx of the user
 
May I ask the Q?
 
I am concerned about $$\lim_{u \to \infty}\frac{\int_0^u f(y) \, dy}{u}
$$
Do we know that the numerator $\to\infty$?
Of course, if it doesn't, the limit is simple, but L'Hôpital doesn't apply.
 
@FreakEnum Sure, I'll listen and Srivatsan will answer :)
 
@robjohn Good point. See David's comment. Bill has explained in some posts that the numerator does not matter. [This variation of L'Hospital isn't that well-known apparently.]
 
Q: While covering a distance of 24 Kilometers , a man noticed that after walking for 1 hour and 40 minutes , the distance covered by him was 5/7 of the remaining distance. What was his speed?
please wait a sec I'm typing my solution
 
5:22 PM
@Srivatsan The numerator doesn't matter? I beg to differ, but I should consider in the context of which Bill is speaking.
 
let remaining distance be x , then remaining distance == 24-x == (5/7)*x
which gives x= 14 km
 
@robjohn I think it's this
 
@robjohn Well, as I said earlier, I do not fully remember the constraints of the theorem, but I feel this solution might be correct after all. Let me find the relevant BD's posts and then we'll see if the solution can be redeemed. =)
 
(and as you can see from the comments, I was confused, too :))
 
@tb That's ridiculously fast. I have no clue how you managed it =)
 
5:24 PM
so distance guy covered is 24-x == 24-14 ==10
and time taken to cover the distance is 1 hour 40 min == 5/2
so speed == 4 km/h == 10/9 meter/min
 
Each of Bill D's answers should have a BD number: the number of layers of references to other BD posts.
 
but book's solution is 5/3 meter/min
so what I did wrong?
@tb and @Srivatsan I wrote my solution here which seems wrong , What I did wrong? Thanks
 
@robjohn And the number of equivalents
 
If I have three values {a, b, c}, is there a way I can get {ab, ac, bc, abc} ?
Or more specifically, if I have n values, {n1, n2.. n}.
@FreakEnum Yes, exactly. I'm just not certain how to generate that set.
 
@Incognito c++ program?
 
5:31 PM
@FreakEnum It sounds to me as if he has already covered 24 km, but I can see that it might also be thought that the distance travelled is 5/12 of 24 km
 
"I have three values ..." ... "Or more specifically if I have n values" :D (sorry about that)
 
@FreakEnum I'd like to do this without software if I could, I'm trying to push my understand of math.
 
@robjohn that prefix "while" projects total distance I think
 
@FreakEnum I think your computation of 10km is fine. However, I don't understand this: 1 hour 40 min == 5/2
 
@Incognito I'm very dumb in maths , ask others for help or they're about to help you
 
5:34 PM
@FreakEnum Thanks for the heads up :)
I'm here because I can write the code that does this, but I know better solutions exist here.
Making computers work harder because I know less is a poor way to go about things.
2
 
@FreakEnum okay, so he has travelled 5/12 of 24 km...
 
@tb aah , you're right , that should be 5/3
 
@tb I guess one has to pick one version of L'Hôpital and use it, or learn all the variants.
 
@robjohn or remember how to Google for Bill's posts :)
 
@tb now answer is coming perfect :) , Thanks @tb @robjohn
@robjohn no , 24 is total distance he has to cover
 
5:38 PM
@tb Bill and I never meshed on sci.math. He never liked my answers and would link to lots of his posts as references.
@FreakEnum yes, so he has travelled 5/12 of the 24 km, so the distance he has left to go is 7/12 of the 24 km, that means that the distance he has travelled is (5/12)/(7/12) or 5/7 of the remaining distance.
 
@robjohn he has travelled 5/12 of the 24 km , how?
 
He says that after travelling 1h40m, he realize that he has travelled 5/7 of the distance remaining.
 
@robjohn and what is the remaining distance?
 
well the distance travelled and the distance yet to go must sum to 24 km, and the ratio of those distances is 5/7...
So he has travelled 5/12 and has 7/12 left to go.
of the total 24 km
 
@robjohn you confused me :( I'm unable to understand the logic
 
5:46 PM
Okay... the distance he has travelled is x.
 
@robjohn yes ..
 
The distance he has left to go is 7/5 x (since the distance travelled is 5/7 of the distance left to go)
 
yes
 
so 24 km = x + 7/5 x = 12/5 x
 
@robjohn aah , got it , Thanks a lot :)
 
5:49 PM
and x = 5/12 of 24 km
 
cool
Thanks again :)
 
but if you think about it, there are easier ways to compute.
knowing that the distance travelled / distance yet to go = 5/7
you know that the distance travelled is 5/12 and the distance left to go is 7/12 of the total distance.
because 5/12 and 7/12 sum to 1 and have a ratio of 5/7
 
Ratio thing looks very hard to me , which makes the Q horrible for me
 
practice will make things look easier.
 
@robjohn Can you explain that line in easy to understand way ?
 
5:55 PM
Do you see that 5/12 + 7/12 = 1?
 
@robjohn yes!
 
Do you also see that (5/12)/(7/12) = 5/7?
 
@robjohn yes, but why knowing this ratio thing is important here? what information does it says?
 
we have two parts of the total distance (count the total distance as 1) and they have a ratio of 5/7
thus the two parts must be 5/12 and 7/12 of the total distance.
we are given the 5/7 as part of the question
 
@robjohn woow , that made sense now :), Thanks a lot XXXXXX
 
5:58 PM
@FreakEnum I am glad that it makes sense :-)
it means that I have not failed
 
6:42 PM
Hey guys, I just had my lin alg final, and I absolutely killed it!! Thanks a lot for all your help!!
 
@JohnDoe Congratulations! : )
 
Always happy to help, John D'oh.
 
Hello, I am Ishaan. Can I ask Questions here, by Questions I mean Silly doubts?
 
Hello, you are Ishaan. You were just asking a Question here, by Question I mean a Silly doubt about this chatroom.
 
lol
Sorry, I am new here and I don't want to violate site's policy and get banned.
Hmm I was wondering why can't I take LCM of irrational and rational numbers?
 
6:54 PM
Ask away.
 
@AsafKaragila If I want an injection $$ f: fin (\omega) \hookrightarrow \omega$$ what do you think of $$ \{ a_1 , \dots , a_n \} \mapsto p_1^{a_1} \dots p_n^{a_n}$$ for $p_i$ prime?
 
@Matt looks good to me :-)
 
@robjohn Nice, thank you! Btw: the other day when I asked the question about a sequence that converges to zero and is not in any $\ell^p$ for $0< p < \infty$, do you know a better example? I can't do $\frac{1}{\log n}$ in my head so I was wondering if there was something simpler.
 

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