search AMC AIME LaTeX inequalities function geometry

[quote="JBL"]I don't know if you realize it, but when someone asks for help on a problem, the answer "this is a well-know result" is the second-most unhelpful thing you could say, right after "Oh, that's trivial!"[/quote]Don't worry about it.. I already read the solution. I usually don't even attempt the problems in this book before I read the solutions.

That's probably not a good strategy for developing problem-solving skills.  One thing you might do is go back to problems that you've looked at the solutions to several days or a week ago and see if you can solve them then -- in other words, check to make sure that the solutions you read are sticking.

[quote="JBL"][quote="pbornsztein"]This is a well-known result from Beatty (and the two involved sequences are so-called Beatty's sequences).
Note that this kind of partition is no longer possible with three (or more) sequences.

Pierre.[/quote]

I don't know if you realize it, but when someone asks for help on a problem, the answer "this is a well-know result" is the second-most unhelpful thing you could say, right after "Oh, that's trivial!"[/quote]I think Pierre was reffering to the fact that this problem has appeared at least 5 times on the forum before, the latest topic being in http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Beatty&t=22013

[quote="JBL"]I don't really want to turn this into an argument, but to me it appears that he was making a reference to the literature without providing any way of reaching it or any idea of the proof.  If he wished to indicate that this problem has appeared before on this site, it would be logical for him to make a reference to this problem on this site, and possibly even provide a link to one of the previous times it was discussed, but he did not.  Thus, I do not think your conclusion is logical.[/quote]

Don't worry, JBL. This is one of the most common problems I always encounter in this site. I have emphasized about this situation before but no one seemed to care about it. It's like Andrew Wiles saying that "The proof of FLT is trivial, and QED; OK kiddos, you guys get it?" If there is something to say, please post links/references/hints on how to do it. If not, please save a post. Thank you all for finally [b]comprehend[/b].

I think some of us must start using the search function. There are more than 5 topics on the same subject (Beatty sequences) here on the forum.

There were instances when people were asking for say shortlisted problems, and some members, moderators even, redirected them to other websites, instead of showing the [b]tons[/b] of problems presented here.

At this point, with 120,000 posts, I think that more than 75% of the problems being posted were already discussed on the forums, so a search for them appears natural.

Ok so Pierre didn't give a link, but he usually does, and he didn't do it this time probably because he was either in a hurry, or tired of the same and same problems showing up again and again :) Let's not jump to conclusions here.

In addition to what Valentin said, I'm sure that a fair number of us have the aforementioned USSR book, and as such know that it contains complete solutions.

Ooops... :blush:  I didn't look at this thread for a while, and it seems that I've made something wrong (I do not say 'bad', but... :P ).
I agree that it is more useful to give an appropriate link, but as Valentin said, this problem appeared many times on the forum, the latest before this one being less than one week ago :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=beatty&t=22013
for which I gave a link after only writing 'Beatty' on Google, which seems to be the keyword here, and that's what I meant when I said that this problem is very well-known (so that any personal search about it should be very easy).

Pierre.

[quote="3X.lich"]It's like Andrew Wiles saying that "The proof of FLT is trivial, and QED; OK kiddos, you guys get it?" If there is something to say, please post links/references/hints on how to do it. If not, please save a post. Thank you all for finally [b]comprehend[/b].[/quote]

3X.lich is right.  You may find it bothersome that the same problems appear multiple times: fine, but that's an immediate consequence of having a large and growing community, something I assume you wanted when you decided to join your forum.  Not everyone will have seen everything on this website.  Perhaps people should do a search before every single post, but to me that seems like a major imposition on people, especially newcomers to the site.  And I find that a lot of "old-timers" simply don't do anything to make this community more friendly and inviting and are frequently counter-productive, especially in the intermediate and pre-olympiad forums, Pierre's post being just a single example.  (Other popular methods include the belittling of the difficulties of problems and posting solutions below the level at which the poster really is, especially without hiding the solution.)  I note that the link Pierre gives is to the advanced section of this site -- I would hardly expect someone with 26 posts to have thoroughly surfed through the large forum structure of that section.

Finally, I'm not jumping to any conclusions whatsoever about Pierre -- I don't think he's a bad person, or even a bad member of this community.  I do, however, think that this post is representative of a class of posts that makes this a (sometimes) unattractive community.

[quote="JBL"]Finally, I'm not jumping to any conclusions whatsoever about Pierre -- I don't think he's a bad person, or even a bad member of this community.  I do, however, think that this post is representative of a class of posts that makes this a (sometimes) unattractive community.[/quote]I personally do not think that there are more "bad members" than my fingers on my hands in this community.

What we have to agree upon is the never-solved problem of difficulty. In my opinion this is an advanced subject, and should be treated there.

Futhermore, all new students, especially those thinking about olympiad-like (that is more difficult than the average multiple-choice aime-type question) should search the forums because almost every problem they have has been previously solved before (especially if given in a ussr olympiad, or if it is a nice problem like this one). I must concur that when I join a very large board for the first time (happens every now and then) the first thing I usually do before posting anything is [b]search[/b] :)

As at this point we don't have the merge topics ability (will come in the future) it's hard to have tons of topics on the same subject, knowing that a simple search on the site (or google for that matter) will reaveal numerous  solutions (also the book itself has a solution).

With all due respect the original author of the thread left the impression that he doesn't poses the solution, which in fact he did (either solved by himself or found in the book) so I woudln't call Pierre's answer as he didn't wanted to help him.

To make long things short, this topic was posted in the wrong place. :)

:mad:  :mad:  :mad:
I am very angry.
People on this site have different aims. Who granted rights to JBL to judge them?

Valentin, you didn't even respond to my points.  But, if we're getting into this fundamentally unrelated issue, I shall defend myself: I didn't call anyone on this forum a bad member, nor did I even suggest that there are bad members -- only that there are bad posts, something that I think anyone can agree upon.  (Myth apparently thinks there is at least one bad member, but that is beside the point.)  I was very careful not to judge any individuals because I feel that the majority of the input of the majority of the people on this forum, especially those of you who were in mathlinks before the two forums joined, is wonderful.  Nonetheless, I think that even some of the generally good contributors make the type of posts I complained about.  Rather than addressing my complaint, you implied that I was attacking individuals and made the obvious statement that, if people did a search before they posted, it might reduce a little bit of clutter, without even mentioning my central complaint.  You have yet to address, or even mention, it in 3 posts on this topic.

[quote="JBL"]Valentin, you didn't even respond to my points.  You have yet to address, or even mention, it in 3 posts on this topic.[/quote]I've addressed your points. In this case Pierre's posts as I said above, is not out of line, or improper, considering the background and difficulty level of the problem at hand.

For the "making a reference to the literature without providing any way of reaching" point, I've stated that Pierre usually (as other people do too) post links where such references can be obtained. Again he did not post any in this  case because the simple mentioning of Beatty sequences give you [b]tons[/b] of ways to get the info (which was already avaialable in that book) as stated.

For the "Other popular methods include the belittling of the difficulties of problems and posting solutions below the level at which the poster really is" point, I think that if people whoose level is higher than the one of the problems posted in intermediate and pre-olympiad would stop posting then the people asking help would have no place to get it (as themselves could not solve the problems, and higher people would not be able to post).

As a final notice, this is in fact a forum of discussions, and discussions are welcomed in all forums. So what if Pierre didn't supply a solution? Is he by any means bound to supply one? I don't think so. Each member of this community contributes to it as he sees fit, and there are moderators (and admins :P) that must make sure that things are posted where they are supposed to, no vulgar language is used, and that's about it.

If he just wanted to give a hint, without actually saying the "this is a hint: look for Beatty sequences" and others would have posted a solution (which they did, btw) I do not see where the problem is. Again, no-one is forced to have his posts necesarily under the [hide ] tag or all of one's posts in a forum to represent solutions.

PS Due to the fact that this discussion has really no connection with the actual problem (in a mathematical sense) I will split it.

PPS [absolutely personal opinion - off the record] Besides the fact that the topic should have been posted in the Advanced Section >> Algebra >> Algebra Proposed Problems forum (which moderators should have moved it too), the author was not looking for a solution because he didn't solve it himself (as he proved by posting another solution) but rather for a discussion on the problem. In this case Pierre's post (which gives more information about the problem) is much helpfull than a solution.

I guess this topic should be sticky.  :)

I think an important point is [b]STRUCTURE[/b]. And I guess JBL's argument is somehow misleading. It is obvious that you do not fully understand something new which is complex, i.e. theories, forum structures etc. Structure is also important though most people just will tend to stick to a small subset of the available forums. Some important points to make the best use of the search engine are as follows:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21455]"please explicitely define all notations used instead of hoping that the reader will be able to guess their meaning" (Darij)[/url]

- proper naming of thread subjects/titles: not just "easy and nice"

- [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15585]if from a certain source several problems are posted collect them in a thread, for example the IMO ShortList 1998 here[/url]

-  [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]please learn $\LaTeX$ and post your problem in this style, this makes reading the problems much easier and people will be easier attracted by your problem[/url]

- classification of problems as suggested by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=285]Harazi[/url], for instance:

Discrete Mathematics, Graph Theory, Labeled Graphs, Labeled Tree

[quote="JBL"]So we disagree about whether or not this single post of Pierre's was appropriate: fine.

[quote]I think that if people whoose level is higher than the one of the problems posted in intermediate and pre-olympiad would stop posting then the people asking help would have no place to get it (as themselves could not solve the problems, and higher people would not be able to post). [/quote]

Somehow, in the many months I spent on this forum prior to the union between AoPS and Mathlinks, I managed to make many posts in the Getting Started and Intermediate levels without either solving all the questions there or demeaning anyone else's questions as being "too easy" to appear on the forum.  Both activities now seem quite popular.  I can give you a large number of links to threads where such activities occur, if you wish.  As someone who is most comfortable in the pre-olympiad forum, I note an extraordinary number of solutions that are written by people who probably should be solving in the advanced forums, frequently unhidden.  This makes the forum, for me, a much less attractive place to turn to look for problems.

I don't think Pierre should have supplied a solution, if for no other reason than he had seen the question solved within the past week.  There were a number of things he could have done that I would have found more reasonable than what he did do, each of which would have been more helpful to someone (such as myself) who has not seen the problem before.[/quote]I think that Pierre's post could not have possibly shinder your chances of solving a problem. He mentioned that it is well-known (which it is!) and it's called Betty sequence because the [b]thread's name[/b] was "I don't know how to name this thread" (or something of the sort). In my opinion it's the best reply anyone could have given (aside from the fact that the a link was omitted).

I am curious what exactly would have you expected him to post? Exactly, what do you think that was appropriate to post (that is post the "post" as it should have been posted :)  - too many damn "post" words here).

On the other hand, I also do not agree with people saying that users' questions are "too easy", but it's exactly as when people say they are too hard (which also happen). These should be solely regarded as a person's personal opinion, and it is true that each of us defines his own "simple" problem and "difficult" problems, so one's mentioning them is really not that of a "offence" I think. It rather might demonstrate someone's *modesty* at most.

I also think that you might be deliberately giving the Advanced Sections the etiquette "too hard" without even taking a closer look. Many of the Advanced Section problems seem to be that need to be placed in the intermediate or pre-olympiad and some of the intermediate or pre-olympiad need to be placed in the advanced sections.

I have to go to have some dinner right now, but I'll be back to continue this conversation :)

I'm not sure we're communicating fully, probably my fault: I am expressing 2 different things.  1 of them is about Pierre's post, which I did not think incredibly greivous and which I have not been highly concerned with in my previous 2 posts.  The other is a whole class of posts, and we could argue about whether or not Pierre's post belongs in that class and for what reasons.  It seems to me that you address my more general comments as if they were all about Pierre's post -- this is why I said that I thought you were not addressing my major complaints.  I do not think Pierre's post would have hindered anything -- I did think it was unhelpful, something you disagree about.  I claimed it was a member of a class of "counter-productive" posts on this forum, particularly in the lower-level sections.  As far as I can tell, your response was that people should do a search before they post something, which neglects to address my point entirely.  I can imagine a number of things that Pierre could have said that would have been more helpful than what he did: among them, a link to one of the previous discussions of this problem on this site or to any other website discussing them or a suggestion for solving the problem (presumably coupled with the comment that this problem is difficult and has been tackled on the advanced fora).

[quote="JBL"]  As far as I can tell, your response was that people should do a search before they post something, which neglects to address my point entirely.  I can imagine a number of things that Pierre could have said that would have been more helpful than what he did: among them, a link to one of the previous discussions of this problem on this site or to any other website discussing them or a suggestion for solving the problem (presumably coupled with the comment that this problem is difficult and has been tackled on the advanced fora).[/quote]

JBL, you must understand that people like Pierre are here for their joy. They are NOT payed teachers. I think it is already very kind to say what he said and posting a link. Rewriting all the hints again each time is a very tiring job. I think it is not their job to do that just because people are too lazy to use the search engine. They are no service and convenience staff employed on ML/AoPS. If all the details are covered in other threads why should one do that ? Just that a certain user does not need to follow the links ? Personally I appreciate the efforts of Pierre and others. If users think hints are not sufficient they still may ask for more.

Yes that's exactly what I wanted to say orlando. JBL you still haven't made me the pleasure of seeing:

1) what [b]exactly[/b] you would have wanted Pierre to post (literally I mean - put it in a quote)
2) the links to the topics in which people are *offensed* by other people by calling their problems *not worth to be on the forum*.

As for the "other whole class of posts", I must repeat that this is an open forum and people are invited to discuss mathematics, especially problem solving, but that doesn't mean restricting one's expression about certain problems.

Of course it would have absolutely no sense for me to do in the getting started and post in each of the problems there "this is very easy for me", and for any of us for that matter.

What the issue is in fact is what some people consider difficult and what some people easy. Clearly there is no such thing as a general easy problem or a hard problem. It might be hard for you, easy for me, or vice-versa. Now if the problem that caused this topic would have been posted where it stands now in the first place, then this discussion would have not taken place.

Similarly there are many problems posted in the advanced topics which are too easy to be there (or so think some people): for example
http://www.mathlinks.ro/Forum/viewtopic.php?p=146807&highlight=#p146807
http://www.mathlinks.ro/Forum/viewtopic.php?t=22839
and many more. This is more a matter of structure.

We are discussing, and we will at some point in the future implement post ratings, which will help in separating general hard from general easy problems.

Untill then I wonder why you have posted in exactly 8 topics in the advanced sections (all of which ended up there after being moved from different parts of the site). Do you find it [b]that[/b] hard? Or is it more of a personal reason?

I unstickied this - while an interesting discussion among members of this community I highly respect, it doesn't need to be stuck on top of the forum as it isn't really a discussion about Other Problem Solving topics (or really anything besides 'How I Would Like People to Use the Forums').

I would like people to try to post in a constructive, clear, and educational way, but you can't force people to do what you want them to - people will use the board in different ways.  Perhaps when we institute post rankings this will somewhat solve itself - maybe we can even implement a filter a la slashdot so you don't have to be bothered with low-rated posts.  We'll see what we can do when we have time.

I don't see why there's all this ruckus going in this topic. If it is your style that "Hey that's trivial!!!" or "Hey that's super-easy!!!" or "Hey that's well-known!!!", maybe you should move on because these phrases [b]do not[/b] help at all to the readers. For example (and this is true), I saw a couple of posts about "Popovicius Inequality" and I do see many other posts where they say "that's well known, just use Popovicius". I remember the first time I saw that I was like, "what is going on? Am I stupid to miss what Popovicius is?" I think fairness has been worn out when such phrases like "well-known" or "triviality" comes up. Try writing "Popovicius is well-known" in an IMO paper, I'm sure you will get full marks (you will?).
If anyone disagrees to what I say, OK fine, I apologize to have stirred up your inner fire, and perhaps you can ignore me in the future.

[quote] Try writing "Popovicius is well-known" in an IMO paper, I'm sure you will get full marks (you will?).  [/quote]

If you were to write something like - 'this inequality follows immediately from Popoviciu's inequality applied to the convex function $f(x)=x^2$, then yes, you would receive full marks.

And I think that there's a few things that should be noted.  First of all, if a problem comes up that is a named theorem, e.g, Nesbitt's Inequality, I think that it's helpful to say that.  That means that you can use 1) the forum search, and 2) google, to look for a solution to the problem.

Second of all, if someone says, 'this inequality is an immediate consequence of $X$ theorem', I also think that that's helpful.  I mean, again on inequalities, you can prove a lot of them using AM-GM.  However, once you learn something like Muirhead, it TRIVIALIZES (oops) a lot of the traditional AM-GM inequalities.  And there are similar results in the combinatorics, geometry - heck, all of the sections.  So this should also be noted; people shouldn't be expected to manufacture an artificial solution using only the knowledge known to the proposer when there is a simple, clean solution using a more advanced theorem.

[quote="Valentin Vornicu"][quote="3X.lich"]If it is your style that "Hey that's trivial!!!" or "Hey that's super-easy!!!" or "Hey that's well-known!!!", maybe you should move on because these phrases [b]do not[/b] help at all to the readers.[/quote]As me and Richard pointed out, helping other readers is [b]not[/b] mandatory. It is for Richard and me up to a level, and to the moderator of the respective forum, again up to a certain level. But the rest of the users are [b]not obligated[/b] to make each of their posts as helpfull as they can get. We cannot ban users from certain forum for these reasons (especially when the affirmations they make can also be interpreted by some people as a good helping post!). We introduced age ranges to create a certain feeling that "hey, maybe you should post here if this is a forum where stuff that is usually discussed by 13 years olds" but some of the visitators of those forums [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16644]weren't too happy[/url] about this (although it was somewhat in their winning, at least from JBL's point of view).

And as I said, one cannot expect every post in a topic to be a "here's how to solve this problem" post. This is a forum and if someone expresses his opinion, and his opinion is "this is too easy" then you just skip to the next post. There are people on the site for which almost all problems are "easy", and some of them state that. If you disagree, just post your opinion .. "it's hard" :)

We understand that this must be for some users an annoying thing, but look on the bright side: if you further increase your problem solving skills, then you will like having also advanced people out there to help you, and the problems will become easy to you also (the ones that you at first said that are not easy).

We have plans to upgrade the forum in the future, which will further help you with these issues (introducing posting ranking, ability of ignoring users  (so if you ignore a user, you will not see his posts, reveal only the first post some topics, etc.) while some of them are effective now (we have two accessing urls, collapsing forums to hide the unwanted stuff, favorite topics, local communities, descriptions systems) but you must try to fit everyone's shoes in this story :)[/quote]

OK, thx for clarifying me that no one is obliged to help others. I see your point, understood, and I apologize again to everyone who felt offended by my thoughts. My sole purpose was to help people in this forum but unfortunately the forum is not compromised to do that. I apologize to everyone for stirring up all these troubles reading through my posts and I will posts whenever I can help. If someone requests me to provide an answer, I am not obliged to provide any sort of information. It is up to me. Thank you for giving me the freedom of speech which I enjoy the most so I can express my opinions the way I want over the internet as long as it does not concern anything else besides it is related to the subject Mathematics. Thank you and I will not post in this topic anymore. Bye.

[quote="Valentin Vornicu"]I think some of us must start using the search function...[/quote]
[quote="Valentin Vornicu"]But the rest of the users are not obligated to make each of their posts as helpfull as they can get. [/quote]

I said, right at the beginning, that I didn't want to turn this into an argument, but it's happened anyway.  Take your pick, Valentin: I just took 2 things you've said in this discussion and put them at the top of this post.  In the first, you argue, more or less, that the blame for certain types of comments should be put upon those who post topics without first searching through the website.  In the second, you say that you can't possibly force people to post in a certain way.  That's logically inconsistent, and worse than that, it's a completely arbitrary distinction of demanding certain things of certain people and not of others for some unknowable personal reason.

Aside from that, which I think is just outrageous, are other smaller flaws in your argument that any mathematician should be apalled about.  For example, you sent me a link to a post about people comlaining about the age guidelines added to the forum.  (2 people in fact -- the same number complaining here.  Apparently, you have again decided that certain people's complaints have more merit than others.)  However, their complaints line up far more with mine than against them -- they are complaining on behalf of people who are older, but less skilled -- not those likely to be swooping down into lower-level forums, but those who the guidelines might push, unfairly, into higher-level forums.  So that little sentence, which you seem to have thought was very clever, is completely bunk.

Blahblahblah -- if I posted a question in the Getting Started forum, and you proved it with Muirhead and acted as if that was it, don't you think I would have a right to be annoyed?  The post, by nature of where it lies, was obviously designed for solvers of a certain level who wouldn't have heard of Muirhead.  Your using it isn't helpful at all to the hypothetical me, a middle schooler from Indiana.  (Although, if you commented that it could be solved by a relatively advanced inequality named Muirhead and pointed me to some posts on the topic, I could perhaps learn something from it.)

Valentin, (in reference to your penultimate post), I don't see why you need me to give you specific example of bad things in order to admit that bad things exist -- you should recognize, again as a mathematician, that my refusal to produce an example does not prove that none exist.  Instead, I feel that I am causing enough trouble between individual members of this forum and myself through my posts without needing to add the extra strain of labeling certain individuals posts negatively, since I already seem to be doing a number with you and Pierre. [color=red][censured by Admin - please do not mock other individuals][/color]

Taking two statements of mine, and putting them out of their context, and thus conlcuding that they are illogical is not a correct assumption. Please read carefully the context in which they were made.

I am not discussing your ideas of my arguments being "just outrageous, smaller flaws that any mathematician should be apalled about", etc. and not the ones that I deleted in which you mock other people.

Anyway, even if supposedly my arguments would be as you described them above, the problem which is discussed here is not a mathematical problem, and one should learn that mathematical arguments are not always the best and the most correct ones.

As I've said before, and you have reiterated it here, it seems that you have a personal problem with some members, including me probably, so henceforth this discussion will be taken on an e-mail or private message, to avoid further personal conflicts.


Let $ \langle F_{n}\rangle$ be a sequence of nonempty closed sets of real numbers with $ F_{n\plus{}1}\subset F_{n}$. Show that if one of the sets $ F_{n}$ is bounded, then $ \cap^{\infty}_{i\equal{}1}F_{i}\neq\emptyset$. Give an example to show that this conclusion may be false if we do not require one of the sets to be bouneded.

Try to come up with an unbounded closed subset of $ (\minus{}\infty,\infty)$ that involves $ n$.  :maybe:

There are tons of ways to prove this depending on what you are allowed to use. One way would be to use the Theorem: $ X$ is compact iff for every collection $ \mathbb{C}$ of closed sets in $ X$ having the finite intersection property, the intersection $ \cap_{C\in\mathbb{C}}C$ all of the elements of $ \mathbb{C}$ is nonempty.

Since these are sets of real numbers: compact iff closed and bounded. You have one of the sets $ F_{i}$ is bounded and all the sets are sitting inside of it after $ i$ (by nested property), so since $ \cap_{n \equal{} 1}^{\infty}F_{n}\subset\cap_{n \equal{} i}^{\infty}F_{n}$ you can just look at your compact space $ X$ being $ F_{i}$ and apply the theorem in the forward direction.

There are ways to do it without theorems. You look at complements and use DeMorgan and things.


abstract algebra
vector
superior algebra
superior algebra unsolved

I need examples of groups , rings , and in fact for bands (boolean semigroups )?

I know some standard examples :

1) $ (\mathsf(P(X),\Delta,\cap)$.

2) $ \mathbb Z_2 \times \mathbb Z_2 \cdots \times \mathbb Z_2$

3)$ \mathbb Z_2 \times \mathbb Z_2 \cdots\cdots$

Please put down any example you know , may be it will helpful  :)

I don't know what you mean by a "Boolean group."  Do you mean a group in which every element has order $ 2$?  Any such group must be abelian, hence a vector space over $ \mathbb{F}_2$.  Vector spaces over a field are uniquely determined by their dimension.

For Boolean rings construct the corresponding Boolean algebra and use [url=http://en.wikipedia.org/wiki/Stone%27s_representation_theorem_for_Boolean_algebras]Stone's representation theorem[/url].

Boolean group means the square of any element gives itself.

In fact I tried to find a band (boolean semigroup ) of odd order ($ \geq 3$)

Thx  :)

The only such group is the trivial group.

Could you define more thoroughly what a Boolean semigroup is?  I can't find a good definition online.

[quote]
Could you define more thoroughly what a Boolean semigroup is?  I can't find a good definition online.[/quote]

The terminology "Boolean" means in algebra (as I know) if you take the element with itself in the operation it will gives itself .In other words, Boolean means every element is idempotent.

Thus, Boolean semigroup $ (S,\bullet)$ , it means if $ s\in S$ , then $ s^2 \equal{} s$.

 :)

The semigroup $ \{ e, a, b \}$ with identity $ e$ and relations $ a^2 \equal{} ab \equal{} a, b^2 \equal{} ba \equal{} b$ is "Boolean" (I am not convinced this terminology is universal) of order $ 3$.  Every word evaluates to its leftmost element.

Ok. that's nice one.

but , what about commutative Boolean semigroup of odd order?

(I believe it should be universal terminology)

Take $ \{ a, b, c \}$ with multiplication as follows: any word evaluates to the letter in it which is lexicographically first.  Hence $ a^2 \equal{} a, b^2 \equal{} b, c^2 \equal{} c, ab \equal{} ba \equal{} ac \equal{} ca \equal{} a, bc \equal{} cb \equal{} b$.  This works for any totally ordered set.

That's really very nice  :)

Now, could we find infinite Boolean semigroup ?infinite Commutative Boolean semigroup?

Thank you so much  :)

The above works for [b]any[/b] totally ordered set.  For example, take the integers with binary operation $ \text{min}(x, y)$.  The power set of any infinite set with either union or intersection is also still a commutative Boolean semigroup.

Can you please just give the entire question at once?

Sorry, I just like to collect examples of several properties .

Last question :

I need two commutative Boolean semigroups , the first one is finite (odd order) and the other one is infinite . Such that both of them have a zero divisors (like the definition we know in the ring theory for zero divisors) and they have at least two elements are not a zero divisors .

Thank you in advanced  :)

[quote="t0rajir0u"]This is starting to sound like homework.  At this point I've given you enough ideas that you should be able to do this yourself.[/quote]

[b]This is not homework[/b]  :!: .In fact,I'm trying to study the properties of zero graph of commutative bands , and I could not find a finite (odd order ) and infinite bands with some elements are zero divisors and some element are not zero divisors (at least two elements )

I can see we can add some elements in your finite commutative example and trying to make an example as I asked for , but I need to collect more of commutative bands  :)

Thank you  :wink:


We get :

$ \sum_{cyc} \frac{1}{1\plus{}\frac{b}{a}\plus{}(\frac{b}{a})^2} \ge 1$

We put $ \frac{b}{a}\equal{}x$ and so on.so we have $ xyz\equal{}1$

Now we have two solutions;we can either expand it and then we are done!
Or we can put $ x\equal{}\frac{st}{r^2}$ and so on to get :

$ \sum \frac{r^4}{r^4\plus{}r^2st\plus{}s^2t^2} \ge 1$

Now we can use Cauchy :

$ \sum \frac{r^4}{r^4\plus{}r^2st\plus{}s^2t^2} \ge \frac{(\sum r^2)^2}{\sum r^4\plus{} \sum r^2st \plus{} \sum s^2t^2}$

So we have only to prove that $ \sum s^2t^2 \ge \sum r^2st$
Which can be done by using AM-GM.

please learn to use TeX to write down your problem here...

$ \sum{\frac{a^2}{a^2\plus{}ab\plus{}b^2}}\ge1$

$ \iff a^4b^2\plus{}b^4c^2\plus{}c^4a^2 \ge a^3bc^2\plus{}a^2b^3c\plus{}ab^2c^3,$

by AM-GM, we have

$ a^4b^2\plus{}b^4c^2 \ge 2a^2b^3c,$

$ b^4c^2\plus{}c^4a^2 \ge 2ab^2c^3,$

$ c^4a^2\plus{}a^4b^2 \ge 2a^3bc^2,$

adding them together, done.

the solution of shoki is similar to me.I think it is an interesting way

[quote="duc 95"]Can you find another way?[/quote]
it is very old and well known inequality.
Another sulution:
let:
$ {a}^{2} \plus{} ab \plus{} {b}^{2} \equal{} x$
$ {b}^{2} \plus{} bc \plus{} {c}^{2} \equal{} y$
$ {c}^{2} \plus{} ca \plus{} {a}^{2} \equal{} z$
we have:
$ \left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1} {z} \right)\left(\frac {{a}^{2}}{x} \plus{} \frac {{b}^{2}}{y} \plus{} \frac {{c}^{2}} {z} \minus{} 1 \right) \equal{} \sum \frac {{a}^{2}}{{x}^{2}} \plus{} \sum \frac {{b}^{2} \plus{} {c}^{2}}{yz} \minus{} \sum \frac {1}{x}$
$ \equal{} \sum \left(\frac {{a}^{2}}{{x}^{2}} \minus{} \frac {bc}{yz} \right) \plus{} \sum \frac {{b}^{2} \plus{} bc \plus{} {c}^{2}}{yz} \minus{} \sum \frac {1}{x} \equal{} \sum \left(\frac {{a}^{2}}{{x}^{2} }\minus{} \frac {bc}{yz} \right)$
$ \equal{} \frac {1}{2}\sum {\left(\frac {b}{y} \minus{} \frac {c}{z} \right)}^{2}\geq 0$
So $ \sum{\frac {a^2}{a^2 \plus{} ab \plus{} b^2}}\ge1$
Equality occurs for $ a \equal{} b \equal{} c$ :)


You just post ways of making pictures using only your keyboard. You can put a word describing it.

(-(-(-(-(-(-(-(-(-(- -)-)-)-)-)-)-)-)-)-) ninjas

(\__/)
(+'.'+) bunny
(")_(")

The Joker
  (.)(.)[color=red]
 \____/[/color]

Whoa nice one 1=2!:coolspeak:

(=^_^=) Kitty

        m          m
        | |(o o)| |
          ||(~)||                  Monkey

~~~{~~@          Rose

d-_-b     Headphones

<D  Ice cream

[quote="winner2009"]to ytrewq: what is that??? it looks sorta like a snakeskin if u scroll fast... is that it?[/quote]

[quote="ytrewq"]MWAHAHAHAHA You're computer has been invaded by an army of 1337 fishes MWAHAHAHAHAHA[/quote]

That's what it is.

O-|--( STICK MAN!!!! (look from side and turn head to the left)

O-|--< Stick man

___
|||||-)
|||||-) Four ninjas looking
|||||-) around a corner.  :ninja:
|||||-)
=-=


Το επόμενο θέμα το αφιερώνω σε όλους εσάς ... :)

Δίνονται οι αριθμοί  $ x_i$,   i = 1,2, ... n,  που ανήκουν στο σύνολο [0,1].

Αν υπάρχουν $ i, j \in$ { 1, 2, ... n }, τέτοια ώστε $ |x_i \minus{} x_j|> \frac{1}{2}$

να αποδείξετε ότι για έναν τουλάχιστον από τους αριθμούς $ x_i$,   έστω $ X$, ισχύει

$ ............$  $ \frac{3}{4}\le 4X^2\minus{}6X\plus{}3 \le 1$

Η τελευταία συνθήκη δίνει ισοδύναμα

$ \frac{3}{4} \leq \left(2x\minus{}\frac{3}{2}\right)^2\plus{}\frac{3}{4}\leq 1 \Leftrightarrow 0\leq \left(2x\minus{}\frac{3}{2}\right)^2 \leq \frac{1}{4}$

Αρκεί λοιπόν (από την τελευταία) να δείξουμε ότι υπάρχει κάποιο $ x_i$ για το οποίο

$ \minus{}\frac{1}{2} \leq 2x_i\minus{}\frac{3}{2}\leq \frac{1}{2} \Leftrightarrow \frac{1}{2} \leq x_i \leq 1$ το οποίο ισχύει λόγω της υπόθεσης.

(Εαν για όλα τα $ j \in \{1,\ldots,n\}$ είχαμε $ 0\leq x_j < \frac{1}{2}$ τότε $ |x_k\minus{}x_l|<\frac{1}{2}$, για όλα τα $ k,l \in \{1,\ldots,n\}$, άτοπο).

Αλέξανδρος

$ 3/4 < \equal{} 4x^2 \minus{} 6x \plus{} 3 < \equal{} 1$
Για καθε $ x$ ισχυει οτι $ 4x^2 \minus{} 6x \plus{} 3 > \equal{} 3/4$
Για να ισχυει οτι $ 4x^2 \minus{} 6x \plus{} 3 < \equal{} 1$ θα πρεπει $ 1/2 < \equal{} x < \equal{} 1$  που ισχυει γιατι αν $ x_i < \equal{} 1/2$ και $ x_j < 1/2$ τοτε $ |x_i \minus{} x_j| < 1/2$ ατοπο αφου υπαρχουν $ i,j$ ωστε $ |x_i \minus{} x_j| > 1/2$ Αν εχω κανει λαθος πεστο μου. :)

ΟΚ  παίδες ... :)

Έχω υπόψη μου και γεωμετρική λύση αν ενδιαφέρεστε πείτε μου

[i]Μακρής Βαγγέλης[/i]

Φυσικα ενδιαφερομαι  Βαγγελη...Σε παρακαλω οποτε μπορεις γραψτην. :)

Παιδιά να ο γεωμετρικός τρόπος ... :)   ( σε attachment )

Πείτε μου τη γνώμη σας ...

Βαγγελη πολυ ενδιαφερουσα λυση....και σιγουρα η καλυτερη. :wink: Απο περιεργια ποσο χρονων εισαι? :)


Solve in real numbers:

$ x \plus{} y \plus{} z \equal{} 2$

$ (x \plus{} y)(y \plus{} z) \plus{} (y \plus{} z)(z \plus{} x) \plus{} (z \plus{} x)(x \plus{} y) \equal{} 1$

$ x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y) \equal{} \minus{} 6$

[hide]Set $ \sigma_1 = x + y + z,\sigma_2 = xy + yz + zx, \sigma_3 = xyz$ and get from the equations
\begin{align*} \sigma_1 & = 2 & {} &{}\\
\sigma_1^2 + \sigma_2 & = 1 & \Rightarrow \sigma_2 &= - 3 \\
\sigma_1 \sigma_2 - 3\sigma_3 & = - 6 & \Rightarrow \sigma_3 &= 0
\end{align*}
So $ (x,y,z)$ are (distinct) solutions of $ 0 = x^3 - 2x^2 - 3x = x(x + 1)(x - 3)$ (Vietà). This gives the solutions $ (0, - 1,3),(0,3, - 1),( - 1,0,3),( - 1,3,0),(3, - 1,0),(3,0, - 1)$ and we easily check that these are indeed solutions of the original system.[/hide]


geometry
quadratics
trigonometry
Pythagorean Theorem
algebra
quadratic formula

Circle a is tangent to both the x and y axis. Circle b is tangent to circle a, and is tangent both the x and y axis also. Both circles rest completely in quadrant I and their point of tangency is the only point that they share. If the radius of circle a is 1, then what is the radius of circle b?

plz help me solve this!

Smaller circle has radius $ 1$ and larger cicle has radius $ r$ (or vice versa).

Pythagoras gives $ (r\plus{}2)^{2} \equal{} 2r^{2}$

$ r^{2} \minus{}4r \minus{}4\equal{}0$

r= 2 plus or minus the square root of 2

where did you get $ (r\plus{}2)^{2}\equal{} 2r^{2}$ from?

The equation should be $ (\sqrt{2}\plus{}1\plus{}r)^2\equal{}2r^2$ since the smaller circle doesn't intersect the origin. [geogebra]b423b85e232eb58d5ef25e02c645e4c926bbda9b[/geogebra]

Let me know if you don't see where the equation comes from.

You're right, fishythe fish. I was a bit hasty.

It's okay. When I did the explanation, I almost didn't catch it myself until I drew it out.

No problem. Using the Pythagorean Theorem on the larger triangle seems obvious, but the smaller right triangle you also have to solve is not.

How would you solve the equation $ (\sqrt{2}\plus{}1\plus{}r)^{2}\equal{}2r^{2}$ for r?
I can't get r by itself!!! sry, I'm kind of a beginner...

Expand first:

$ (\sqrt{2}\plus{}1\plus{}r)^2\equal{}2r^2$

$ r^2\plus{}2r(1\plus{}\sqrt{2})\plus{}1\plus{}2\sqrt{2}\plus{}2\equal{}2r^2$

$ 0\equal{}r^2\plus{}(2\plus{}2\sqrt{2})r\plus{}(3\plus{}2\sqrt{2})$

This is kinda ugly, especially with the quadratic formula:

$ x\equal{}\frac{\minus{}(2\plus{}2\sqrt{2})\pm\sqrt{(2\plus{}2\sqrt{2})^2\minus{}4(1)(3\plus{}2\sqrt{2})}}{2(1)}$

$ x\equal{}\frac{\minus{}2\minus{}2\sqrt{2}\pm\sqrt{4\plus{}8\sqrt{2}\plus{}8\minus{}12\minus{}8\sqrt{2}}}{2}$

$ x\equal{}\frac{\minus{}2\minus{}2\sqrt{2}\pm\sqrt{0}}{2}$

$ x\equal{}\minus{}1\minus{}\sqrt{2}$

So it all works out in the end. Maybe it's not that ugly.

It's actually easier if you take the square root of both sides. We obtain

$ \sqrt{2} \plus{} 1 \plus{} r \equal{} \sqrt{2} r \implies (\sqrt{2} \minus{} 1)r \equal{} \sqrt{2} \plus{} 1\implies r \equal{} (\sqrt{2} \plus{} 1)^2 \equal{} 3 \plus{} 2\sqrt{2}
\\
\sqrt{2} \plus{} 1 \plus{} r \equal{} \minus{}\sqrt{2} r \implies \minus{}(\sqrt{2} \plus{} 1)r \equal{} \sqrt{2} \plus{} 1\implies r \equal{} \minus{}1.$

Since both our answers are negative, I think we need to take absolute values. (Like in trigonometry, you can reverse the sign of the radius if you add $ \pi$ radians to the angle, right?)

And TachyonPulse, you can't go from

$ (\sqrt{2}\minus{}1)r\equal{}\sqrt{2}\plus{}1$

to

$ r\equal{}(\sqrt{2}\plus{}1)^2$

If you divide both sides out, you get

$ r\equal{}\frac{\sqrt{2}\plus{}1}{\sqrt{2}\minus{}1}\equal{}\frac{(\sqrt{2}\plus{}1)(\sqrt{2}\minus{}1)}{(\sqrt{2}\minus{}1)^2}\equal{}\frac{2\minus{}1}{2\minus{}2\sqrt{2}\plus{}1}\equal{}\frac{1}{3\minus{}\sqrt{2}}\equal{}\frac{3\plus{}\sqrt{2}}{7}$

And something seems intuitively wrong with the second solution.

I messed up in my earlier post. The equation should be

$ 0\equal{}r^2\minus{}(2\plus{}2\sqrt{2})r\minus{}(3\plus{}2\sqrt{2})$

$ r\equal{}\frac{2\plus{}2\sqrt{2}\pm\sqrt{(2\plus{}2\sqrt{2})^2\plus{}4(1)(3\plus{}2\sqrt{2})}}{2(1)}$

$ r\equal{}\frac{2\plus{}2\sqrt{2}\pm\sqrt{4\plus{}8\sqrt{2}\plus{}8\plus{}12\plus{}8\sqrt{2}}}{2}$

$ r\equal{}\frac{2\plus{}2\sqrt{2}\pm\sqrt{24\plus{}16\sqrt{2}}}{2}$

$ r\equal{}1\plus{}\sqrt{2}\pm\sqrt{6\plus{}4\sqrt{2}}$

but if you multiply TachyonPulse's answer in a different way...

$ r\equal{}\frac{\sqrt{2}\plus{}1}{\sqrt{2}\minus{}1}\equal{}\frac{(\sqrt{2}\plus{}1)^{2}}{(\sqrt{2}\plus{}1)(\sqrt{2}\minus{}1)}\equal{}\frac{2\plus{}2\sqrt{2}\plus{}1}{(\sqrt{2})^{2}\minus{}(1)^{2}}\equal{}\frac{3\plus{}2\sqrt{2}}{2\minus{}1}\equal{}3\plus{}2\sqrt{2}$

Yes and I am right because fishythefish dropped a $ 2$ in front of the $ \sqrt {2}$ in the following step:

$ \frac {2 \minus{} 1}{2 \minus{} 2\sqrt {2} \plus{} 1} \equal{} \frac {1}{3 \minus{} \sqrt {2}}.$

I also checked both the solutions I obtained through Wolfram Alpha and both worked.

Wow. I feel like an idiot.  :blush:

$ \frac{1}{3\minus{}2\sqrt{2}}\equal{}\frac{3\plus{}2\sqrt{2}}{1}\equal{}3\plus{}2\sqrt{2}$

So it turns out that if I do the math right, you don't need to multiply it a different way... :blush:


[quote="Karth"]I have a question regarding WOOT: Is it possible for a company to sponsor you/pay part of the money for you to take WOOT? There are 4 or so people in my school who would really enjoy taking the class, except that we don't have that much money to pay for the class. Any help would be appreciated  :)[/quote]
Yes, but I think you'll have to figure out how to do that on your own. There was another topic similar to this one here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100121


geometry
circumcircle
angle bisector
geometry unsolved

Let $ ABCDEF$ be a convex hexagon, such that:

$ \angle CAB\equal{}\angle FBA, \angle ECD\equal{}\angle BDC$

$ \angle AEF\equal{}\angle DFE, \angle EAC\equal{}\angle FBD$

$ \angle ACE\equal{}\angle BDF, \angle CEA\equal{}\angle DFB$

Prove that the "central" hexagon formed by the intersection of the lines $ AC,BD,CE,DF,EA$ and $ FB$ is circumscribable.

let $ A': AC \cap FB$, $ B': BD \cap AC$ and cyclicaly define C',D',E',F'. By angle the quads FF'DE', DD'B'C, BB'F'A are cyclic, so $ \angle A'B'F' \equal{} \angle A'F'B' \equal{} a$, $ \angle C'B'D' \equal{} \angle C'D'B' \equal{} b$, $ \angle E'D'F' \equal{} \angle E'F'D' \equal{} c$ and $ \angle FF'E\equal{} \angle FD'E \equal{} \angle DD'C \equal{} \angle DB'C \equal{} \angle BB'A \equal{} \angle BF'A\equal{}k$. Then the angle bisector of $ \angle F'A'B'$, $ \angle B'C'D'$, $ \angle D'E'F'$ concur on the circumcenter O of $ \triangle B'D'F'$. So we have to prove that the angle bisector of $ \angle A'B'C'$ pass trought O and so that $ 2 \left ( 90\minus{}\angle F'D'B' \plus{} a \right ) \equal{} a\plus{} b\plus{} \angle F'B'D' $$ \Longleftrightarrow \ (180 \minus{} \angle F'D'B' \minus{} \angle F'B'D') \plus{}a \equal{}\angle F'D'B' \plus{} b$ that is true because $ (180 \minus{} \angle F'D'B' \minus{} \angle F'B'D') \plus{}a \equal{} B'F'D' \plus{} a \equal{}$ $ 180 \minus{}\angle FF'D' \equal{} 180 \minus{} \angle F'D'E \equal{}  \angle F'D'B' \plus{} b$.


Two concentric circles are drawn such that the inner circle covers 25% of the area of the outer circle. Given that the radius of the outer circle is 10 units, how many units are in the radius of the inner circle?

[hide]Since the small circle covers $25\% = \frac{1}{4}$, the ratio of the radii is $\sqrt{\frac{1}{4}}= \frac{1}{2}$. So the radius of the small circle is $10\cdot \frac{1}{2}= 5$.[/hide]

EDIT: thanks

[hide]The ratio of the areas of the two circles is $4: 1$.  Thus the ratio of the radii is $sqrt{1}: sqrt{4}=1: 2$.  Thus the radius of the smaller circle is $10/2=5$[/hide]

[quote="vishalarul"][hide]Since the small circle covers $25\% = \frac{1}{4}$, the ratio of the radii is $\sqrt{\frac{1}{4}}= \frac{1}{2}$. So the radius of the small circle is $50\cdot \frac{1}{2}= 25$.[/hide][/quote]
Where'd you get 50?


Let a,b,c>0 such that $ ab\plus{}bc\plus{}ca\equal{}1$ and $ \alpha$ be real number
Prove that
 $ \frac{a^\alpha}{b\plus{}c}\plus{}\frac{b^\alpha}{c\plus{}a}\plus{}\frac{c^\alpha}{a\plus{}b} \geq \frac{3\sqrt{3abc}}{2}(\sqrt{a^{2\alpha\plus{}1}}\plus{}\sqrt{b^{2\alpha\plus{}1}}\plus{}\sqrt{c^{2\alpha\plus{}1}})$


Suppose that lists are formed in the following manner. Start with the list
$1,1$
Then, insert in between two consecutivey entries of the list $x$ and $y$, the number $x+y$. So the next list would be
$1,2,1$
and then
$1,3,2,3,1$
$1,4,3,5,2,5,3,4,1$
and so on.

Prove that $a$ and $b$ can be consecutive in at most one list $L_{n}$.

A consecutive pair (a,b) can only come from exactly 1 of the pairs (a-b,b) or (a,b-a), more precisely the only in which both entries are positive(that is the smallest number of the 2 is preserved). Same story for that pair, and so on. This gives a unique path leading to (a,b) that must clearly terminate in a finite number of steps. Remark that (1,1) is the only possible pair of equal consecutive numbers. It must end with one of the entries being 1, and the issue is now settled because there is only one pair of the form (1,x) or (x,1)


superior algebra
superior algebra solved

Let $ S$ be any commutative ring, $ x\in S$, $ I\subseteq S$ an ideal and $ P_1,\ldots,P_k$ prime ideals of $ S$. Suppose that the coset $ x\plus{}I$ is contained in $ \cup_{i\equal{}1}^kP_i$. Then there exists $ j$ such that $ xS\plus{}I\subseteq P_j$.

[quote="hjbrasch"][url]http://www.math.lsa.umich.edu/~hochster/615W07/L03.19.pdf[/url][/quote]

Well, that is the course that I am following. I gave this problem for you to think at, not for you to show me where I can find it.
All the results that I have posted so far that have the subject description "Tight closure course" can be found there or close to there.

well, it is a slight modification of the normal prime avoidance lemma


Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of the larger sphere?

the answer is : 3+root 69/3

anyone can explain?[/img][/code]

find the height of the tetrahedron formed by the centers of the 4 spheres, then add 3 (=1+2).


function
inequalities proposed
inequalities

Let $a,b\ge0$ and $n \in \mathcal N$:

$\sqrt[n+1]{a^{n+1}+b^{n+1}} \le \sqrt[n]{a^n+b^n}$

if ab=0 then equal.

ab>0

WLOG b=1

and define a function f(x)=(1+a^x)

g(x)=f(x)^(1/x)

g'(x)*x^2 *(1+a^x)(a^x(lna-ln1+a^x) -ln(1+a^x)))

g'(x)<0 (for all x>=1)

therefore g(n)>g(n+1)

Q.E.D.


inequalities
search
inequalities proposed

Iura, do not drop the atomic bomb! And the mongolian inequality was not refering to some order between the $x_i$?  Anyway, even if it is a consequence of it(though I do not realize now), I want a solution in 3 lines.

Very strange...
Here is my idea..
I consider
\[
x_i  + x_{i + 1}  + k\frac{1}{k}x_{i + 2}  \ge (k + 1)\left( {\frac{1}{k}x_{i + 2} } \right)^{\frac{k}{{k + 1}}} \left( {x_i  + x_{i + 2} } \right)^{\frac{1}{{k + 1}}
}\]
Multiply all of them together,
\[
3.1^n  \ge (k + 1)^n \left( {\frac{1}{k}} \right)^{\frac{{nk}}{{k + 1}}} \left( {\prod {\left( {x_i  + x_{i + 2} } \right)} } \right)^{\frac{1}{k}}
\]
Suppose k satisfies
\[
3.1^n  \ge (k + 1)^n \left( {\frac{1}{k}} \right)^{\frac{{nk}}{{k + 1}}} 2.1^{\frac{n}{k+1}}
\]
then we are done.
I used mathematica to solve the equation,
\[
\frac{{3.1k^{k + 1} }}{{k(k + 1)^{k + 1} }} = 2.1
\]
it seems to have a root but i doubt if that's the mistake of mathematica. (the root is right between 0.4761904 and 0.4761905), because the graph is very weird...
can anyone verify if there is a root for this equation?

[quote="Fedor Petrov"]Maximum over $k$ is about $2.109362$. Computed by Maple.[/quote]
do you mean this value of k satisfies my equation?
is my idea valid?

Sorry. Why did I search for a maximum? We are interested in minimum. Put $k=1/(2.1)$ and get desired result.

After all, I do not consider this problem too easy. At least, it seems more interesting than usual inequalities which I see on olympiads. Or is there much easier solution?

I will try to remember wy idea, I know it was nice. I agree with Fedor about the inequalities in contest, they are all the same!

May I help you? We have $\root n\of{a_1a_2\dots a_n}+\root n\of{b_1b_2\dots b_n}\le \root n\of{(a_1+b_1)(a_2+b_2)\dots (a_n+b_n)}$. Proof: wlog, $a_i+b_i=1$ (else multiple $a_i$ and $b_i$ by the same factor $t$), then we have $\root n\of{a_1a_2\dots a_n}+\root n\of{b_1b_2\dots b_n}\le \sum a_i/n+\sum b_i/n=1$.

Then just put $a_i=x_i$ and $b_i=x_{i+1}+x_{i+2}$.

Oh, yes! Thanks, Fedor. Oh, so this is the reason for which you said that after all you do not consider the problem that easy: you always had this extremely easy solution?  :)

I said it before I found this solution. But nevertheless, I consider a problem tricky enough after more than all. It looks stupid when you have a solution, but not before (at least for me).

[quote="iura"]Sprry if I make a mistake, Harazi,  but isn't this just consequence of Mongolian Inequality?[/quote]

What is the Mongolian inequality?

If $x_1\le x_2\le x3\le x_n$, then $\prod_{i=1}^n \frac{x_i+x_{i+1}+x_{i+2}}3\ge
\prod_{i=1}^n \frac{x_i+x_{i+1}}2$. (products are cyclic). I am sure it was posted on the forum.


geometry
analytic geometry
graphing lines
slope
trigonometry
articles
vector

Hi,

This might sound like a stupid idea but here it is (I didn't know where to put it so I'll just place it here.)

I just had a thought that occurred to me when it was raining:

Depending on how heavy or how light the rainfall is, is there a certain speed at which you can run so that you can avoid a majority of the rainfalls that would have otherwise hit you? Can you avoid them all?

Again, this question might not make sense, but I'm just curious to see you guys' opinions/views on it.

Thank you very much :)

Unless you have a lightning cloud over you that you are trying to outrun, I don't really see any way to do it. The rain in front of the spot you just were at would hit you instead. In addition, running would provide a greater bird's eye view area than walking, so I think you would be hit by more, not less raindrops.

The rain that falls on your head varies with the speed you move. The rain you walk into does not. If you think of the rain as a continuous space of water that doesn't move, you move diagonally upward through it. The faster you move, the shallower the slope, and since you're going the same horizontal distance, you want to have as shallow a slope as possible.

If raindrops fell in a predictable array of some sort (for example, a falling 3D lattice array), then theoretically, you could probably get through without getting hit, if you were small enough compared to the distance between rain drops. However, that's not the case.

I've thought about this, since I may have to decide between biking or walking in the rain.

There's two ways you get hit by rain -- rain that falls on you and drips down, and rain that you walk into. As it happens, if you're going faster, you tend to lean forward, so the extra surface area of your body as viewed from above will initially cancel out the reduced amount of travel time, and roughly the same amount of rain falls on you. But, as you speed up more and more, the reduced time has a larger impact than the increase in your surface area viewed from above, and it starts to become effective. In the limit case, your body will be horizontal and you'll be moving at velocity infinity (forget Einstein for a moment), and you won't be hit from above with any water at all.

As for the rain you walk into, it's about the same. The only difference is, if you're going really slowly, then the rain you walk into is more likely to hit your shoulders, and when you're, say, biking, it's more likely to get your whole body. The fact that you lean forward while going more quickly has little effect. Think about it -- at 90 degrees, a small change in angle will cause a negligible change in sine but a large change in cosine.

Ultimately, the only useful solution is to cover yourself by a means of your choice.

For some reason my English teacher of a few years ago was talking about his really smart physics buddy from Caltech.

And apparently this buddy wrote a paper on how fast you would have to go through a rain storm to avoid getting hit.

So a solution exists, somewhere, I think.

Anyways, if you go above mach one or something I'm sure the air pressure around you would cause any raindrops to slide off without technically hitting you, which is what I imagined this buddy's solution entailing.

I remember someone writing an article on this at least 20 years ago - probably the [i]American Mathematical Monthly,[/i] if someone wants to look for a reference. The author was using a continuous model - treating rain as having a certain (vector) mass flux. I vaguely remember that the conclusions were that in a headwind or crosswind, you intersect the least water by going as fast as possible, but in a fairly strong tailwind, there was another optimum where you roughly match the wind speed.

I think the author he was treating the pedestrian as an essentially vertical rectangular box - the actual geometry may matter, and in particular, I would think that leaning forward while moving forward rapidly would help quite a bit. That's certainly an argument for riding the bicycle - at least as long as it's only the falling rain that matters; wheel-splash would be another issue.

[quote]If raindrops fell in a predictable array of some sort (for example, a falling 3D lattice array),[/quote]
Not a chance. I would assume that a good model for how many raindrops hit a specified area in a specified time would be a Poisson probability distribution.

Many references on the topic...

http://www.flyingcircusofphysics.com/pdf/Chapter1_Ref_Com.pdf

Also, the rain isn't falling at the same rate all day long. Clearly if the rain is getting bigger you would do well to go fast, while if it's lightening up (arrgh... pun) you may as well take your time.

Interesting views. I was curious to know what you guys think about this. Thank you for your time :)

Yes, you can avoid them by bringing an umbrella!  :lol:

[quote="Kent Merryfield"]I vaguely remember that the conclusions were that in a headwind or crosswind, you intersect the least water by going as fast as possible, but in a fairly strong tailwind, there was another optimum where you roughly match the wind speed.[/quote]
Yes, that sounds about right.  Roughly, if you match the wind speed no drops hit you horizontally, and the vertical drops are unavoidable.


function
algebra
functional equation
algebra proposed

I'd like to see your solutions to this functional equation:

Find all functions $f: \mathbb{Q}\rightarrow \mathbb{Q}$ such that:

$f(x+y)=f(x)+f(y)+xy$

Denote $g(x)=f(x)-\frac{x^{2}}{2}$, we have $g(x+y)=g(x)+g(y)\forall x,y\in\mathbb{Q}$ and $g(x)=ax(a\in\mathbb{Q})$. Final $f(x)=\frac{x^{2}}{2}+ax$.


limit
arithmetic sequence
number theory unsolved
number theory

Questions:

 (1)  Is there 4 CONSECUTIVE primes such that they below an  aritmetica progression ??

 (2) Is there 3 primes in aritmetica progression with arit razon: r=8??

Thanks in Advanced

Faby  :rotfl:

Look at the paper [url=http://www.dms.umontreal.ca/~andrew/PDF/PrimePatterns.pdf]Prime number patterns[/url] by Andrew Granville.

Ok.. thankz maxal...

bt ..
what about in these especific cases ..?

Faby.  ;)

[quote="fabycv"] (2) Is there 3 primes in aritmetica progression with arit razon: r=8??
[/quote]
Let the primes be $p, p+8, p+16$. Since the three numbers are pairwise incongruent modulo 3, there is exacly one of them being a multiple of 3. Thus the only possibility is $p=3$. Hence the only sequence satisfying your condition is $3,11,19$.
P.S. I hope "arit razon" means "common difference". :P

And of course for any $n$, there are $n$ primes in arithmetic progression.

[quote="maxal"][quote="fabycv"]
 (1)  Is there 4 CONSECUTIVE primes such that they below an  aritmetica progression ??
[/quote]
For example:  251, 257, 263, 269[/quote]

Maxal ....how could you find this arit progression??...
i 'd like to know the way ... to try  the t0rajir0u's QUESTION -"Generalization"....

Thank you for your answer!!!

Carlos Bravo ;)
Lima - Peru

[quote="maxal"][quote="carlosbr"]Maxal ....how could you find this arit progression??...[/quote]Simple [url=http://pari.math.u-bordeaux.fr/]PARI/GP[/url] program:

[code]forprime(p=2,10^5,if(isprime(p+6)&&isprime(p+12)&&isprime(p+18)&&primepi(p+18)-primepi(p)==3,forprime(q=p,p+18,print1(" ",q));print()))[/code][/quote]

well....
i hoped ... a mathematical process....

Thanks...

Carlos Bravo ;)
Lima Peru

[quote="seamusoboyle"][quote="t0rajir0u"]Now, let us ask the more interesting question:  for $n \in \mathbb{N}$ let $d(n)$ denote the smallest arithmetic difference that a sequence of $n$ primes in arithmetic progression can have.

What is $\lim_{n \to \infty} d(n)$?  :)[/quote]

I think it tends to $\infty$

For an IrMO, we were asked to show that if we had 15 primes in aritmetic progression, then the common difference must be divisable by $2,3,5,7,11,13$ So continuing the pattern the result follows. Of couse, we'll need to prove it  :lol:[/quote]

Well, it makes sense - if the common difference wasn't divisible by those primes then the remainder $\bmod 2, 3, 5, 7, 11$ or $13$ from the first prime to the last would reach zero in $14$ steps or less... or something along those lines :)

For example, the progression maxal found had a common difference divisible by $2, 3$ for a $4$-length sequence.  Yeah... that makes sense.  So $d(n)$ would just be $\prod_{p_k < n} p_k$?

No, $d(n)$ could be bigger, at least there are a lot of really hard problems about that.
Also take a look at http://www.mathlinks.ro/Forum/viewtopic.php?t=6075 .

[quote="seamusoboyle"]When it says 'primes in amithmetic order', does it mean consecutively?[/quote]
Yes, everything else is senseless.

As i realised just after i posted! Just thinking of $4a+1$  !!!


Έχω μία απορία για τη γενίκευση μιας βασικής ανισότητας. Έχουμε ότι $ 2(a^{2}\plus{}b^{2})\geq (a\plus{}b)^{2}$ και $ 3(a^{2}\plus{}b^{2}\plus{}c^{2})\geq (a\plus{}b\plus{}c)^{2}$. Άρα ισχύει αυτή η μορφή ανισότητας και για άπειρους αριθμόυς, δηλαδή $ n\cdot \sum^{n}_{i\equal{}1}a^{2}_{i}\geq (\sum^{n}_{i\equal{}1}a_{i})^{2}$???

Ναι. Είναι η Cauchy-Schwarz για $ x_i \equal{} a_i$, $ y_i \equal{} 1$.

Cheerio,

Durandal 1707

Συμφωνώ πως η ανισότητα που λέει ο Γιώργος ισχύει λόγω της bcs  για n όρους  όπου n φυσικός αριθμός, δηλ.ισχύει για πεπερασμένο πλήθος όρων.Για [u]άπειρους[/u] όρους τι γίνεται?

$ s_n\geq t_n \implies \lim s_n \geq \lim t_n$

Η LHS της ανισότητας πάντα υπάρχει στο όριο $ n\to\infty$. Αν και η RHS υπάρχει, θα ισχύει η ανισότητα.

Cheerio,

Durandal 1707

Ευχαριστώ Durandal ,όμως δεν έχει διαφορά το $ n\rightarrow\infty$ από το $ n\equal{}\infty$?
Bέβαια αυτό σχετίζεται με μια άλλη απορία μου:Τί διαφορά έχει όταν λέμε το πλήθος των όρων μιας ακολουυίας είναι άπειρο από το να τείνει στο άπειρο :?:

Δεν υπάρχει καμία διαφορά νομίζω ... στην στην πρόταση που γράφεις ...
Είναι σίγουρα λάθος όμως το  $ n \equal{} \infty$

[i]*Βέβαίως όμως μπορεί να σου πεί περισσότερα ο Panagiotis ![/i]

Heh, Βαγγέλη με πρόλαβες :)

Guys, μην μπερδεύεστε...

$ n \equal{} \infty$ δε θα το δεις ποτέ γιατί δεν έχει νόημα (εκτός κατά σύμβαση), μόνο το $ n\to\infty$ έχει νόημα. Γι'αυτό απάντησα έτσι, προσπαθώντας να δείξω με ποιον τρόπο είναι σωστά διατυπωμένη η ερώτηση.

Το $ \infty$ είναι το μοναδικό σημείο συσσώρευσης των φυσικών αριθμών, και αν ποτέ γράψουμε πχ $ a_\infty \equal{} \ldots$ αυτό που στην πραγματικότητα εννοούμε είναι ότι $ \lim_{n\to\infty}a_n \equal{} \ldots$. Γι'αυτό όταν λέμε $ \sum_{n \equal{} 1}^\infty$ είναι απλώς συντομογραφία του $ \lim_{n\to\infty}\sum_{k \equal{} 1}^n$. Αν θέλεις, αυτό το "άπειρο" είναι ένα αφηρημένο [i]σημείο[/i] εκτός του $ \mathbb{N}$ το οποίο μπορούμε να το προσαρτήσουμε τοπολογικά ώστε να έχει νόημα η έκφραση $ n\to\infty$.

Το ότι το πλήθος των όρων μίας ακολουθίας είναι "άπειρο" έχει να κάνει με τον πληθυκό αριθμό $ \aleph_0$ του συνόλου $ \mathbb{N}$ των φυσικών αριθμών... εντελώς άλλη ιστορία! Δεν είναι πεπερασμένος, άρα είναι "άπειρος", όπως και τα στοιχεία των πραγματικών είναι "άπειρα" αλλά είναι διαφορετικό "άπειρο" αυτό επειδή είναι της τάξης του συνεχούς $ \mathfrak{c}$ και.........

Όπως καταλαβαίνεις, τα δύο "άπειρα" στα οποία αναφέρεσαι είναι δύο εντελώς διαφορετικές έννοιες (το ένα είναι σημείο συσσώρευσης, το άλλο είναι πληθυκός αριθμός) και αν δεν είσαι προσεκτικός στη χρήση τους οδηγείσαι σε πονοκεφάλους και παρερμηνείες ;)

Cheerio,

Durandal 1707


HOw many times does the digit "4" appear in the integer part of the quotient 10^100/7

calculators allowed

would that be (10^100)/7 or 10^(100/7) because one's a heck of a lot easier than the other...

Ok. SOlution:

[hide]10^0/7=0.142857142857....

Now, since we have powers of 10, the decimal is shifted.

So in every 10^(6n), there are n 4's.

So 10^96 has 16 4's.

But notice that the integer part of 10^96 ends in a 7. This means that 10^98 has a 4 as its last digit.

SO 10^100 has 16+1=17 fours[/hide]

I believe that is correct. Let's try something else. What is the sum of the digits in the integral part of that expression?


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $K$ a finite extension of $Q$ and $A$ the set of algebraic integers in $K$. Prove that $A$ is a free abelian group of rank $[K:Q]$.

I know one proof of this is to show that the ring of integers of the Galois extension of K is finitely generated, which means that the ring of integers of K, which is a subset of that, is also finitely generated, and therefore forms a  finitely generated torsion free z-module, meaning that it's additivie group is a free abelian group.  Not sure if there is anything more basic.

it is classical because i don't want to copy the proof and being too lazy to type solution you can find it in this pdf (theorem 41 p.35)
http://www.math.sc.edu/~filaseta/gradcourses/TheMath784Notes.pdf


inequalities
quadratics
analytic geometry
inequalities proposed

[quote="nickolas"]The real numbers  $a,b,c$  with  $bc\neq0$  satisfy $\frac{1-c^2}{bc}\geq0.$ Prove  that  $10(a^2+b^2+c^2-bc^3)\geq2ab+5ac.$[/quote]
If $bc<0$ then the result is obvious,else
$c^2\leq 1\Rightarrow 10(a^2+b^2+c^2-bc^3)\ge 10a^2+10b^2+10c^2-10bc\ge 2ab+9a^2+9b^2+10c^2-10bc\ge 2ab+9a^2+(10-\frac{25}{9})c^2> 2ab+9a^2+7c^2> 2ab+5ac$

hehe,I think use $bc(1-c^2) \geq 0$ will be a little easier. :)

Can it be made stronger?
Subtract $ 10a^2$ from both sides, then the right side is a quadratic, since no restriction is plaed on a, we can assume a is the x-coordinate of the vertex, which is $ \frac{2b\plus{}5c}{20}$. Substituting in, it remains to show that
$ 10(b^2\plus{}c^2\minus{}bc^3) \ge \frac{(2b\plus{}5c)^2}{40}$, or
$ 396b^2\plus{}375c^2 \ge 400 bc^3\plus{}20bc$
But as others have shown, the given implies $ bc(1\minus{}c^2) \ge 0$, or $ bc \ge bc^3$. So we have to show
$ 396b^2\plus{}375c^2 \ge 420 bc$
But by AM-GM, $ 396b^2\plus{}375c^2 \ge 770.713954 bc$. Is this correct?

[quote="nickolas"]The real numbers  $ a,b,c$  with  $ bc\neq0$  satisfy $ \frac {1 - c^2}{bc}\geq0.$ Prove  that  $ 10(a^2 + b^2 + c^2 - bc^3)\geq2ab + 5ac.$[/quote]

$ \frac{1-c^2}{bc}\geq 0\Longleftrightarrow -bc^3\geq -bc$

$ 10( \sum a^2-bc^3)\geq 10( \sum a^2-bc)$ So now we need to prove that $ 10( \sum a^2-bc)\geq 2ab+5ac\Longleftrightarrow 10\sum a^2\geq 2ab+10bc+5ac\Longleftrightarrow 30\sum a^2\geq 3( 2ab+10bc+5ac)$.

But from Cauchy we have that $ 30\sum a^{2}\geq ( a+2b+5c)^{2}$. Thus we need to prove that $ ( a+2b+5c)^{2}\geq 6ab+30bc+15ac$ which is obviously true...


Is it true that lefties are better at math and logic and stuff than righties?
I am a right hander I've happened to notice that each time someone beats me at a competition or a math exam or something they're always lefties...is there some sort of connection?
If yes, then what about ambidextrous people?? :lol:

Ambidextrous people are extremely lucky in the fact that if one hand is broken, they can use the other in a competition. (However, most ambidextrous people pretend they're not during a huge final exam  :D  when they can't write).

I am left-handed and I believe it started when my father was teaching me how to write when I was young and he sat across from me. Naturally when he wrote with his right hand, my brain interpreted that I am supposed to write with my left. I guess it caught on over the years  :wink: . My right hand-writing capabilities are okay but I need to work on speed and accuracy.

[quote="Xevarion"]Handedness is usually determined by birth, I thought. (I don't know if it's genetic or part of early brain development.) It's unlikely that you "chose" it by observing your father. [/quote]

Yeah I know this isn't true but for me its the earliest memory I have of myself and I remember observing my father writing with his "left" hand.

I'm ambidextrous but mainly use my right hand...and I know loadz of math whizzes who are lefties...more than the number whom I know are righties...I don't think there's any connection between doing well and hand...it doesn't matter which hand you use, but what you write with it  :)
I feel it's all in the brain...  But if there's any bio-fan here then maybe you could just confirm...this is one subject I suck at so... :P

Now that you mention it, the topper in our class (an IMO bronzie) is a leftie, and he in the first two sems he has amassed around 980 out of 1000.

But no, if you seriously want to know it, I can assure you there is no connection. My intuition as a statistician tells me so!

Oh and that reminds me:
Q: What do you call somebody who puts his right hadn into a monster?
A: A leftie.

[quote="Xevarion"][quote="7h3.D3m0n.117"]I am left-handed and I believe it started when my father was teaching me how to write when I was young and he sat across from me. Naturally when he wrote with his right hand, my brain interpreted that I am supposed to write with my left. I guess it caught on over the years  :wink: . My right hand-writing capabilities are okay but I need to work on speed and accuracy.[/quote]
Handedness is usually determined by birth, I thought. (I don't know if it's genetic or part of early brain development.) It's unlikely that you "chose" it by observing your father. It is possible to force someone to make the "wrong" hand dominant, but this will cause them to be relatively clumsy. It was not unusual for this to happen even as recently as within the past century. My grandfather was left-handed, but that was seen as bad so he was forced to do everything with his right hand.[/quote]

Centuries ago, they used to think left-handed people were witches. Hence the Latin "sinistra" for left, from which we derive "sinister".

I am left-handed when it comes to writing, eating, etc. but I am right-handed when it comes to playing sports, etc.  However, I am not ambidextrous...

I have a book about left-handed people... it does say that there are more neural connections between the two hemispheres of the brain.

I am right-handed. I do everything with it, writing, cooking, eating. I can do math with my left hand, but you can barely read it.

I heard that which hand you are dominant with is on the same side as your dominant leg.

This year, there are three lefites in B&H IMO team  :lol:

The theories of "lefties' intelectual supremacy :D " don't have scientific arguments on their side... But on the other hand, it is scientificaly proven that in a number of sports (tennis and ping-pong, for instance) lefties have an advantage (correct me if I'm wrong)... :)

BTW, I'm a leftie.

I seem to recall an abnormally large number of lefties at MOP last year......not sure if it matters, but statistically....

[quote="hsiljak"]But on the other hand, it is scientificaly proven that in a number of sports (tennis and ping-pong, for instance) lefties have an advantage (correct me if I'm wrong)... :) [/quote]

I'm not sure what "scientically proven" means, but there's an obvious simple explanation: most people, and thus most players, are right-handed.  Thus, most games any given player plays will be against right-handed people.  In a game where handedness matters (of which tennis and ping-pong are surely examples, due to the distinction between forehands and backhands), a left-handed player playing a right-handed player will be used to that configuration while the right-handed player will be unused to it.

Baseball (pitcher vs. hitter) is another famous example of this phenomenon, at least in America.  However, left-handed players have other problems -- picking off players at first base requires a more difficult motion, and lefties can't play third base.

One would expect that other sports (e.g. soccer and American football) should not exhibit much leftie-rightie separation (although perhaps left-footed players are more likely to play on the left wing?).

Familiarity isn't that much of the difference between left-handers and right-handers in baseball; the "platoon" advantage is mostly about how typical pitches curve. Most right-handed pitchers throw pitches which tend to break toward the first base side, which is toward a left-handed hitter and away from a right-handed hitter. Left-handed pitchers throw pitches which break in the opposite direction. It is generally easier to hit pitches that are curving toward you than pitchers that curve the other way.
The size of the platoon advantage is very consistent across hitters. It's not for pitchers- some pitchers (such as Randy Johnson) have very large advantages against their own handedness, while some others have essentially none or even get the opposite handedness out better.
This also leads to the phenomenon of batters switch hitting, or even batting left-handed full time while throwing right-handed.

As for handedness on defense: being left-handed is a tremendous disadvantage at second base, shortstop, and third base, to the point where no left-handers play those positions at the major league level. At first base, being left-handed is generally an advantage- some of the hardest plays for a first baseman involve throws to second base, which are hard for a right-hander. Most teams choose a first baseman for his hitting, so there are still plenty of right-handers there. Being left-handed probably hurts a pitcher's fielding, but helps with holding runners on and picking them off. This is of course mostly ignored, since pitching left-handed is such a useful specialty. Handedness doesn't matter much in the outfield, so there are a lot of lefties chosen for their hitting.

Other sports have more subtle effects. In basketball, being opposite-handed to your counterpart is good on defense and equally bad on offense. This leaves left-handers at about the same proportion as the general population, but likely to be defensive specialists.

[quote="mlok"]You did not mention catchers: they are also right-handed, right?

Lefties should have an option of playing in a league where they run bases clockwise.  :lol:[/quote]

Yes, catchers are right-handed, i play catcher, and have been for the bast 8 years or so.  :P
i mean, you can catch and be left handed, but at the competitive level (high school and beyond...nearly all catchers are right handed, not all), along with first basemen being lefties.

the reason for this though, is not like because you can catch better if your a rightie, but it is because the majority of batters are right handed, and when someone steals or whatever, you want your throwing arm away from the batter; the same reason its hard to throw to third, because the batter is in the way.

back on topic though--
i only know one lefty who is not ambidexterous, and that is ragnarok23. so hes pretty smart.

on my baseball team though, there are many ambidexterous people, maybe 3, and all are...dumb.  but i play on the southside/ghetto of tallahassee, so their lack of intellegence might be more of a cultural/childhood/upbringing thing, than a natural scientific thing.

EDIT: someone wrote this a couple posts ago

"Baseball (pitcher vs. hitter) is another famous example of this phenomenon, at least in America. However, left-handed players have other problems -- picking off players at first base requires a more difficult motion, and lefties can't play third base. "

Are you crazy? lefties have the easiest pick-off move to first.  its righties who cant do it for anything.  but its all fair...righties have that same advantage when picking off to third, but no one does that much.

Running bases clockwise would change the patterns of defensive plays, making lefties preferable at third base, etc. This was the reason I brought it up -- not the baserunning part itself.

Actually, I don't think handedness makes much difference to a catcher's defensive responsibilities. Catchers are right-handed for other reasons; most importantly, lefties with the necessary arm strength and accuracy are pushed into pitching.

Hm about the math, one study showed that mathematically intellegent people were more likely to be male, left handed, and prone to allergies.

[quote="Kent Merryfield"]Baseball players have two types of handedness: throwing hand and batting hand. Throwing hand is right or left, batting hand can be right, left, or both. That makes for 6 possible combinations, and there have been historically great players in each of the 6 types. But here's a look at a few of those of the throws left, bats left type: Ty Cobb, Tris Speaker, Oscar Charleston, Babe Ruth, Lou Gehrig, Ted Williams, Stan Musial, Barry Bonds. The representation of that list among the greatest hitters of all time says something about the intrinsic advantages of left-handedness. (Note also that most of that list were outfielders, with the exceptions being Gehrig who played first base and Musial who divided time between first base and outfield.)[/quote]

that could be misleading though, because it is easiest to hit your opposite handed pitcher, and since there are more right handed pitchers, it is an advantage to be a left-handed hitter in that matter.

but then again, a lefty-lefty matchup is much tougher than righty-righty.

When proctoring calculus exams, I take notice of who writes with which hand (there is not much else to do anyway). So far, I have not seen any clear correlation between handedness and test scores, except for the following: the students who hold their head with both hands for long periods of time (making it difficult to determine their handedness) usually do poorly.

[quote="Kent Merryfield"][quote]because it is easiest to hit your opposite handed pitcher, and since there are more right handed pitchers, it is an advantage to be a left-handed hitter in that matter. [/quote]
That's exactly what I meant by "intrinsic advantage of left-handedness."[/quote]

Lets remember this is a math forum, i get D's in English  :D

i thought you meant that they were perhaps better just because they were left handed, i.e. left handed people have better hand-eye or something of that nature.

sorry about that... :oops:

I'm a leftie.
I know people who have still lived under the ruthless oppressive regime by the righties. :mad:  Back in the days when my grandparents grew up, a kid would be hit on his left hand if he dared to use it to write.

But the shameless discrimination still continues.  One-person-tables are often only designed for righties.  Classrooms always have the blackboards on that side such that the window is on the left hand side for the pupils.

Lefties around the world, unite! :!:

Does not it make sense to have blackboard on the side that is convenient for most pupils? You can't have both.
For better or for worse, I teach in windowless rooms. It would be nice to have some chairs designed for lefties, though.

Hmm... as I inventory in my head the classrooms I often teach in, it seems there are about as many with windows on the right as there are with windows on the left. And there are a couple of lefty desks in each room, although the left-handed students may or may not wind up sitting in them. Also, if you're in a room with left-side windows and stuck with a righty desk, don't you partially turn in your chair so that your back is to the window?

[quote="Treething"]Considering that Daniel Li is a leftie and probably the smartest red MOPper this year, and I'm a rightie and perhaps the dumbest red MOPper this year, I'd say there's some correlation.[/quote]

Just as a sidenote, I know Jacob Steinhardt is a lefty, but I don't remember much about the other MOPpers...

On the other hand, David Xiao was a lefty, and I'd rank him around my level at MOP.

[quote="Klebian"][b]On the other hand[/b], David Xiao was a lefty, and I'd rank him around my level at MOP.[/quote]

Worst pun ever.

Right-handed and I am useless at everything I do. I know the top student at my school is a leftie.

[quote="sonny"]is ambidexterousness something one teaches themselves to do? or something you happen to have by birth?[/quote]
I don't think there's complete agreement on this in the scientific community, but I'd say it's leaning more towards birth than learning. At least I know that if you start too late, you can force yourself to learn everything with the off hand but it will still remain your off hand. So if it is learned, that happens very early (first couple of years probably).

I'm sort of ambidextrous, as I can do a lot of things with both hands, but my left hand is slightly weaker than my right.

I have a friend does everything with his left hand, but he writes with his right.

I'm left-handed, but I've adapted to using right-handed mice. I think my grandma was born left-handed, but then was forced to turn into a right-handed person. Its really funny how my Chinese teachers (I don't take Chinese anymore, it was SO boring!) from a Chinese school called Yellow River Chinese school used to complain how I wrote the strokes in the wrong direction because I'm left handed. Some people think of being left-handed as having a disability. One parent was actually amazed that I could use chopsticks  :furious: .

[quote="Nerd_of_the_Ages"]One parent was actually amazed that I could use chopsticks  :furious: .[/quote]

Chopsticks are an interesting challenge for lefties...I avoided them for a while because it was difficult to use (not to mention your family members staring at you when you mess up).

Hmph another thing against lefties: when writing gigantic essays, normally your left hand would smear across the paper as you write from left to right. Every time I take Cornell notes for AP bio it always smears  :mad: and I can  barely read a thing.

I'm a leftie, and I absolutely HATE having "half-desks" made in favor of right-handed people.  I wish they'd just make "full-size" desks.  When I took my SAT Subject Tests, there were 3 lefties, but only 2 leftie desks.  I was the odd one out.  :furious:

But, I've adapted somewhat to a right-handed world.  Although, it is annoying when I have sit next to someone and eat at a restaurant...  Oh, well.

I read somewhere that before long-lasting tools were made (around the Bronze Age, when tools were made more durable), about half of the people were righties and about half were lefties.  However, once tools could be made that would last several generations, people had to choose a handedness; apparently, right-handedness won out.  I wonder why?

My dad is a leftie, but he has adjusted to a right-handed world, like me.  We are the only two lefties in my immediate family.  However, my younger brother, who is a rightie for most things, does many things in sports left-handed.  Go figure.  I wonder why that is?  It's curious, to say the least.

[quote="Zakary"]However, my younger brother, who is a rightie for most things, does many things in sports left-handed.  Go figure.  I wonder why that is? [/quote]

Well, Rafael Nadal (number 2 seed tennis player) is right handed but plays a leftie because his forehand was ironically stronger on his left hand.

i dont think there is a very big relation between IQ and handedness.
people who are left handed arent automatically geniuses, and people who are right handed arent automatically dumb.


http://www.clickmazes.com its a great site and has a bunch of good games my favorite is 3-some maze in in the 2nd from the top on the right. btw to get more mazes for each one look at the top of the screen.

the tilt mazes are fun. also there is another site with similar puzzles
http://www.logicmazes.com/theseus.html


logarithms
function
number theory unsolved
number theory

lets say i have the following simple maths division

a = b / c

now is there a formula to test the if the result in "a" has

1) a fix number of digits

e.g.  a = 2, 2 digit answer
    or a = 123, 3 digit answer
    or a = 12345, 5 digit answer

2) a unlimited number of digits

e.g. a = 1.12312312312312312312402389403289402349803985034985034589.......
   or a = 6.23409823409823409823049823048920398420398402398402384023894.....

now, if I have to write a C procedure called Checknum()

b = Checknum( a );

if a is of a fix digit answer, eg in example 1) above, Checknum() will return into b the number of digits in a

if a is of an infinite number of digits, eg in example 2) above, Checknum() will return into b the number '0'

now the problem is how do i write the procedure Checknum()

int Checknum( int number )
{
    ???
}

Thank you for your time.

[quote="Fiachra"]One could do it as follows:

(i) Loop until $2 \not| c$, replace $c$ by $\frac{c}{2}$ at each step.
(ii) Loop until $5 \not| c$, replace $c$ by $\frac{c}{5}$ at each step.
(iii) Now if $c \not| b$, return $0$, otherwise return the floor of $\log_{10} \frac{b}{c} +1$.[/quote]

can someone please kindly explain this in plain simple english ???

my maths is very poor, but i am learning ...

thank you so much for your time...

it is very simple

The formula is this >>>   a = b / c

Example (1)

a = 9999 / 13

means 'a' = 769.153846153846840343495345345........   to infinity

a has an infinite digit answer

Example (2)

a = 6 / 2

means 'a' = 3

a has a 1 digit answer

Now how do I test if a number after division has an infinite number of digits as in Example (1) and how do i test if a number after division has a set number of digits as in Example (2)

Thank you so much for your time

Andy

[quote="Fiachra"][quote="vendetta"]
can someone please kindly explain this in plain simple english ???

my maths is very poor, but i am learning ...

thank you so much for your time...[/quote]

The loop is an ordinary programming loop (e.g. a for loop). $5|c$ means that 5 divides c, and $5 \not| c$ means that it does not. Basically we eliminate the factors 2 and 5, and then check for divisibility. If $c$ divides in evenly, $a$ has a finite number of digits.

By the way, I assumed above that $b$ and $c$ were integers. If not, but they have a finite number of digits, then we could multiply by a sufficiently large power of ten. If they have infinitely many digits, one cannot store them directly on a computer.[/quote]

Thank you so much for taking the time to explain everything to me.

However, I still do not quite understand you.

I was wondering if you would like to spend a few minutes to kindly write the C code for this function ???

I really will appreciate it very much.

Thank you so much for your time and effort

Your the BEST !

Andy


What is the perimeter of a right-angled triangle if one of its sides is of length $ 47$ and the other two sides have integer lengths?

hello, and what else is given?
Sonnhard.

Sorry Sonnhard (and Tiger100: just seen you post), I should have said that the sides have integer lengths. I'll edit it in a moment or two.

I used $ (m^{2} \plus{} n^{2}), (2mn)$ and $ (m^{2} n^{2})$ to get $ m^{2} \minus{} n^{2} \equal{} (m \plus{} n)(m \minus{} n) \equal{} 1\times 47$.

Then $ m \equal{} 24, n \equal{} 23$ and so the sides are $ 47, 1104$ and $ 1105$ and the perimeter is $ 47 \plus{} 1104 \plus{} 1105 \equal{} 2256$.

[hide] Generally, suppose that $ k$ is an odd positive integer. Then, there is always a Pythagorean triple of the form $ k$, $ \frac{k^2\minus{}1}{2}$, $ \frac{k^2\plus{}1}{2}$. These obviously are all integers. Solving this general case for the perimeter, we have

$ k\plus{}\frac{k^2\minus{}1}{2}\plus{}\frac{k^2\plus{}1}{2}$

$ \equal{}k\plus{}\frac{2k^2}{2}$

$ \equal{}k\plus{}k^2$.

Plugging in the specific case $ k\equal{}47$, we have $ 47\plus{}47^2\equal{}47\plus{}2209\equal{}\boxed{2256}$.[/hide]


Let $ x,y,z$ be pairwise distinct real numbers such that $ k\equal{}\dfrac{1\plus{}xy}{x\minus{}y},$ $ l\equal{}\dfrac{1\plus{}yz}{y\minus{}z}$ and $ m\equal{}\dfrac{1\plus{}zx}{z\minus{}x}$ are integers. Prove that $ k,l$ and $ m$ are pairwise relatively prime.

[quote=cosinator]Nice problem!
[hide]
Letting $ x,y,z\equal{}\tan A,\tan B,\tan C$, respectively gives $ k,l,m\equal{}\cot(A\minus{}B),\cot(B\minus{}C),\cot(C\minus{}A)$.  We have $ A\minus{}B\plus{}B\minus{}C\plus{}C\minus{}A\equal{}0$ obviously.  It is well known that for three angles which add up to a muliple of $ \pi$, the sum of the pairwise products of their cotangents is equal to $ 1$.  That is, we have $ kl\plus{}lm\plus{}mk\equal{}1$.

Now suppose that the claim in the problem was not true.  Without loss of generality, assume $ t|k,t|l$ for $ t>1$.  Then $ t|kl\plus{}lm\plus{}mk$.  Contradiction![/hide][/quote]

Very nice solution


trigonometry calculus integration function algebra polynomial vector

If I raise $a cis b$ to a power, should I reduce b (the theta) to be between 0 and 360? I am confused on this aspect, as if you raise $a cis b$ to a power, and then simplify the theta, wouldn't raising the new complex number to the recipricol power not yield the original theta?

Furthermore, if I have say $\sin\frac{3}{8}\theta = 1$, then I have $\frac{3}{8}\theta = 90^\circ$

So I get $\theta = \frac{8\cdot 90^\circ}{3}$, $\frac{8\cdot 90^\circ}{3} + \frac{8\cdot360^\circ}{3}$, $\frac{8\cdot 90}{3} + \frac{8\cdot2\cdot 360^\circ}{3}$

I don't think it makes sense to reduce the angles to be an angle from between 0 and 360 because then if I multiply the reduced angles by $\frac{3}{8}$, they will not be equal to $90^\circ$?

just because the angles are not equal, does not mean that they have different sines

[quote="breez"]Okay, I gotcha.

But I am confused on this regard:

$(r^{\frac{2}{5}}$ $cis$ $\frac{2\theta}{5})^{\frac{5}{2}}$ is not equal to $r$ $cis$ $\theta$

It appears to be actually $5 \cdot 2 = 10$ different values.

Does raising the multiple values of a power of a complex number to its recipricol power only yield the original complex number if it is an integral power and not a fractional power?[/quote]

The square root function only takes the principal value by definition, so no, it doesn't take on ten values - it isn't $cis$ that takes on multiple values, it's $\theta$, and this is only because every angle takes on the same $\sin, \cos$ angles as all of the angles that are congruent to it $\bmod 2\pi$ (as much as non-integer mods make sense).

$\sin \theta = 1$ has one solution $0 \le \theta < 2\pi$, but it has infinite solutions over the reals, as does $\sin \frac{3\theta}{8} = 1$.  So you get

$\frac{3\theta}{8} = \frac{\pi}{2} + 2 \pi k$
$\theta = \frac{4\pi}{3} + \frac{16 \pi k}{3}$,

Which in fact also only has one solution $0 \le \theta < 2\pi$, given by $k = 0$.

Then it would be equal to 2 values correct? (My equation from above involving the recipricol powers)

[quote="me@home"]I don't know where to post this, but does anyone have a clear explanation of the idea behind cis and the complex plane and powers $e$ etc.?
I can perform all the calculations, but what does it really mean? How did we hatch such concepts?[/quote]

De Moivre came up with the form that didn't involve $e^{i\theta}$, but Euler... I'm pretty sure he came up with it by messing around with Taylor series.  (Short explanation:  You can write $e^x, \sin x, \cos x$ as infinite polynomials, and then Euler's formula is true because the coefficients of the infinite polynomials of both sides are the same.)

Just take for granted that mathematically, $e^{i \theta} = \cos \theta + i \sin \theta$ is consistent, and ends up producing a vector of length $1$ and angle $\theta$ in the complex plane.  And everything else follows.  :)

breez:  $x^{ \frac{1}{2} }$ only returns one value.

i hate using something I don't completely understand... oh well thanks! :(

[quote="breez"]But say your solving an equation, wouldn't you return all possible values? Then it would be 2 right?[/quote]

If you're solving the equation $\sqrt{9} = x$ then the answer is $x = 3$ because $f(x) = \sqrt{x}$ is defined to return only the positive square root.  If you're solving the equation $9 = x^2$ then the answer is $x = 3, -3$ because both are roots of the equation.

In your example, $\left( r^{ \frac{2}{5} } \, cis \, \theta \right)^{ \frac{5}{2} }$ is a definite quantity with a definite value for any given $\theta$ (one definite value), as is $r \, cis \, \theta$.


If A,B,C and D are angles of a quadilateral and sin A/2.sinB/2.sinC/2.sinD/2=1/4,prove that A=B=C=D=/2

A=B=C=D=/2 doesn't exactly make sense.

Do you mean that ${ \sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}\sin{\frac{D}{2}} = \frac{1}{4}}$?

Fix:
If $ A,B,C,D$ are angles of a quadrilateral and $ \sin\frac {A}2\sin\frac {B}2\sin\frac {C}2\sin\frac {D}2 \equal{} \frac14$, prove that $ A \equal{} B \equal{} C \equal{} D \equal{} \frac {\pi}2$.

From Jensen's Inequality,

since $ \frac{A}{2} \plus{} \frac{B}{2} \plus{} \frac{C}{2} \plus{} \frac{D}{2} \equal{} \pi$

$ \frac{\sin \frac{A}{2} \plus{} \sin \frac{B}{2} \plus{} \sin \frac{C}{2} \plus{} \sin \frac{D}{2}}{4} \le \sin \frac{\pi}{4} \equal{} \frac{1}{\sqrt 2}$

Hence, from AM-GM Inequality

$ \left (\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \sin \frac{D}{2} \right )^\frac{1}{4} \le \frac{1}{\sqrt 2}$

or $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \sin \frac{D}{2} \le \frac{1}{4}$

Since, here equality holds, $ A \equal{} B \equal{} C \equal{} D \equal{} \frac{\pi}{2}$


[quote="Beat"]$a, b, c \in R$ and $a+b+c = 1$, prove
that $15(a^{2}+b^{2}+c^{2})+54abc-7$ is non-negative number.[/quote]

I think the idea for this problem is
1) prove that $F(a,b,c)\ge{F(\frac{a+b}{2},\frac{a+b}{2},c)}$
2)then we must prove that $F(x,x,1-2x)\ge0$. that is all.
But maybe it needs some long computations. :)
I will think on it.


function
floor function
ceiling function
combinatorics open
combinatorics

Let $ n>k$ be positive integers.
Let $ S\equal{}\{1,2,3,...,n\}$.
Denote by $ s(A)$ the sum of the elements of the set $ A$.
How many $ k$_element subsets $ A$ of $ S$ satisfying $ s(A) \equiv 0 (mod 2)$

[b]Can we solve this problem by gernerating function?[/b] :maybe: Thanks.

$ A$ should have an even number of odd elements.

there are $ m\equal{}\lfloor \frac{n}{2} \rfloor$ even numbers in $ S$.

there are $ n\equal{}\lceil \frac{n}{2} \rceil$ odd numbers in $ S$.

the number of $ k$ element subsets of $ A$ that have an even sum is $ a_k\equal{}\binom{n}{0} \binom{m}{k}\plus{}\binom{n}{2}\binom{m}{k\minus{}2}\plus{}\binom{n}{4}\binom{m}{k\minus{}4}\plus{}\cdots\plus{}\binom{n}{2\lfloor \frac{k}{2} \rfloor}\binom{m}{k\minus{}2\lfloor \frac{k}{2} \rfloor}$

now note that $ a_k$ is the coefficient of $ x^k$ in $ \left(\binom{n}{0}\plus{}\binom{n}{2}x^2\plus{}\binom{n}{4}x^4\plus{}\cdots \right)\left(\binom{m}{0}\plus{}\binom{m}{1}x\plus{}\binom{m}{2}x^2\plus{}\cdots\right)$

the second factor is simply $ (1\plus{}x)^m$. the first is $ \frac{1}{2}[(1\plus{}x)^n\plus{}(1\minus{}x)^n]$.


Let $ P(n)$ and $ S(n)$ denote the product and the sum, respectively, of the digits of the integer $ n$. For example, $ P(23) \equal{} 6$ and $ S(23) \equal{} 5$. Suppose $ N$ is a two-digit number such that $ N \equal{} P(N) \plus{} S(N)$. What is the units digit of $ N$?

$ \textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

[quote="mysmartmouth"]Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits of the integer $n$. For example, $P(23) = 6$ and $S(23) = 5$. Suppose $N$ is a two-digit number such that $N = P(N)+S(N)$. What is the units digit of $N$?

$\text{(A) }2\qquad\text{(B) }3\qquad\text{(C) }6\qquad\text{(D) }8\qquad\text{(E) }9$[/quote]

[hide]Let $N=10a+b$.  $10a+b=ab+a+b\Rightarrow ab-9a=0$ so $\boxed{b=9}$.[/hide]

[hide="Solution"]Let $a$ and $b$ represent the digits of the number $ab$. So the numerical value of the number $10a+b$. The sum of the digits is $a+b$ and the product is $ab$. Now combining this with the given,

$10a+b = ab+a+b$
$ab = 9a$

$b=9$

So the last digit (units' digit) is $9$. The answer is $E$. [/hide]


linear algebra
matrix
integration
linear algebra unsolved

Let $A,B\in M_n(C)$ having all their eigenvalues with strictly negative real part.Let $C\in M_n(C)$.Show that there exists $M\in M_n(C)$ such that $AM+MB=C$.

Similar stuff has been discussed before many times. The spectrum $\lambda(\mathcal T)$ of the operator $\mathcal T$ on $\mathcal M_n(\mathbb C)$ sending $X$ to $AX+XB$ is contained in $\Lambda(A)+\Lambda(B)$, which does not contain $0$, so the operator is invertible, and, in particular, onto, which is what we wanted to prove.

killer have a look in RMS n°4 (2004-2005) page 77

$M=-\int_{0}^{+\infty}e^{tA}Ce^{tB}dt$


[quote="daermon"]Hi
I am looking for an online, self-paced precalculus course
The two I'm considering so far are EPGY M013 and CTY Honors Precalculus with Trigonometry
Anybody have opinions on those two or other suggestions?
Thanks[/quote]

i've taken M013 (well i never finished it, w/e)
it's okay, i guess

Eh, don't try to go too high in terms of school type classes...

[quote="daermon"]Hi
I am looking for an online, self-paced precalculus course
The two I'm considering so far are EPGY M013 and CTY Honors Precalculus with Trigonometry
Anybody have opinions on those two or other suggestions?
Thanks[/quote]

DO NOT TO CTY! I went there and I hated it.

Do EPGY.

He's not talking about going to CTY, he's talking about taking an online CTY class.

CTY's "ACS" Algebra 1 and 2 classes were helpful, insightful, and impressive. Their precalc class, on the other hand, was not that impressive. However, that was in 2004... it might've changed by now.

IMO, precalculus really isn't worth spending the big bucks on. Get a book off amazon and figure it out yourself. Probably the only thing in precalc that's actually new is trigonometry (possibly conic sections).

From my experience, the best way to get a good feel of trigonometry is to go on wikipedia, get a list of various identities, and try to prove them geometrically and/or algebraically (i.e. not with complex numbers -- that approach is better to save for later).

[b]Note:[/b] [i]I actually recently listed the book I used for CTY Precalc -- which is a good book -- on Amazon. It's in "pristine condition" and it's 32 bucks  :P buy, buy, buy![/i]