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December 31

Ryan's hypo-whatsisis?

I've read milenium problems and wow! The math's mindboggling! My math teacher's a math degree and he sais he couldn't solve Ryan's hypo-whatsisis (I don't know what a hypothises is?). He's tried many calculators to solve it failed (he sais he'll solve it though since he's a math degree, but Wiki sais its to hard to solve). Now I know why math's there after 12th grade. Many thanx! Though there's another milenium problem Hodge conjecture? My math teacher sais it's fake math because there isn't numbers and formulas. Is Hodge real math? Please explain Hodge conjecture though! I know many of youy think I'm not serious but I'm serious. It's my English (i'm not a native speaker), so may be it sounds I'm not serious. But I now understand math's more than calculators, that's math is more than 12th grade (my high school teacher's wrong then). Sorry and many thanx. —Preceding unsigned comment added by 122.109.239.199 (talk) 04:09, 31 December 2009 (UTC)[reply]

Based on previous questions from the same IP address, this is not a good-faith, serious question. Expand to see off-topic discussion (which really belongs on WT:RD)

The Hodge Conjecture is real math. Eric. 131.215.159.171 (talk) 04:55, 31 December 2009 (UTC)[reply]

Thanx but what's Hodge conjecture about? My math's teacher's a math degree and he sais it's fake math since there's no number/formula but there's weird cross symbol. He sais capital letters can't be crossed since math's about x's & y's crossed he's probably wrong since you all have dozen math degress (meni rose&field sais) and sais he's wrong. Please explain Hodge conjecture to me though?? What's it about? Many thanx. —Preceding unsigned comment added by 122.109.239.199 (talk) 06:47, 31 December 2009 (UTC)[reply]

There are many unsolved problems in mathematics. You can read about the Hodge Conjecture on the relevant article. Perhaps you would be more interested in simple to understand unsolved problems like Goldbach conjecture and whether odd perfect numbers exist? These problems, although simply stated, have withstood attacks by highly talented people for centuries.-Shahab (talk) 06:53, 31 December 2009 (UTC)[reply]

Thanx but please explain Hodge though??? I want to understand Hodge because my math teacher doesn't sais it's fake math so I can be better than my math teacher (who's a math degree and's a math nerd). Please explain. I've solved Godbach. is it a math problem? You sais that math isn't about calculating? 4=2+2 6=3+3 8=5+3 10=7+3 ... does that solve goldbach math problem? Many thanx. Math hurted my brain but I wanted to challenge myself like it. Now I like it thanx! I wnat to understand Hodge though. —Preceding unsigned comment added by 122.109.239.199 (talk) 07:05, 31 December 2009 (UTC)[reply]

<stuffing the troll> So much to choose from! I'll go with this - Your first post didn't have any grammatical or spelling errors. This one has plenty of them. So it seems your control of the English language is deteriorating. Perhaps you should spend less time worrying about mathematical problems and more on honing your language skills? </stuffing the troll> -- Meni Rosenfeld (talk) 07:07, 31 December 2009 (UTC)[reply]

Another good point is that he appears to respect his "omniscient mathematics teacher", while concurrently referring to him as a "math nerd" (or expanding on your point, he appears to alternate between spelling "says" as "says", or "says" as "sais"). --PS T 09:38, 31 December 2009 (UTC)[reply]

I can't speak english good sorry. English isn't my native language. I write my question quickly I make many mistakes. Please don't insult me though. —Preceding unsigned comment added by 122.109.239.199 (talk) 07:26, 31 December 2009 (UTC)[reply]

But your IP address tells me that you're near Perth, Australia. If you live, and are educated, in Australia then... ~~ Dr Dec (Talk) ~~ 12:25, 31 December 2009 (UTC)[reply]

While I do appreciate the layman's enthusiasm in mathematics, some things like the Hodge conjecture are a bit overboard to explain to the non-technical audience. I'm a beginner in algebraic topology and I have no idea what that article is talking about. Maybe someone more experienced can explain it right, but I've seen lots of layman asking what a Calabi-Yau manifold is (from string theory) without even knowing what a manifold is. However, the two conjectures Riemann zeta hypothesis and Poincare's conjecture (which has been solved) are fairly intuitive. Money is tight (talk) 07:12, 31 December 2009 (UTC)[reply]

No you haven't solved Goldbach's conjecture what you did is copy and paste 4=2+2 6=3+3 8=5+3 10=7+3 from the article. Money is tight (talk) 07:17, 31 December 2009 (UTC)[reply]

You haven't proved the Goldbach conjecture. You need to show that every even number greater then 2 is a sum of 2 primes. Mathematical proofs are based on deductive and not on inductive reasoning.-Shahab (talk) 07:20, 31 December 2009 (UTC)[reply]

My friend help's me write my question. He sais he researches algebra and that algebra isn't calculators I didn't beleive him so he sais ask Wiki (he sais Wiki is very reliable). He helps me wrote my first question. Sorry for grammer and spelling error. He didn't help me anymore with questions. I written my questions quickly. Please don't insult me though. Thanx shahab. Why does meni rose&field insult me though??? —Preceding unsigned comment added by 122.109.239.199 (talk) 07:24, 31 December 2009 (UTC)[reply]

Nobody has insulted you. Your explanations for certain strange things about your posts are unconvincing.Julzes (talk) 07:42, 31 December 2009 (UTC)[reply]

Why does the OP misspell my name though? -- Meni Rosenfeld (talk) 07:51, 31 December 2009 (UTC)[reply]

He or she is aching to be banned from wikipedia. I'd certainly treat it as an attack on the site. I don't know what the procedure is for dealing with that, of course, but someone here does.Julzes (talk) 07:55, 31 December 2009 (UTC)[reply]

I think WP:ANI is the place to report these things, but also that we're not nearly at the point where this is necessary. -- Meni Rosenfeld (talk) 08:07, 31 December 2009 (UTC)[reply]

Personally, I'm a little more trigger happy than you. But I'm also a little bit of a novice, so someone else will have to report it if that turns out to be more clearly necessary. I'll look into these incident procedures for future reference.Julzes (talk) 08:34, 31 December 2009 (UTC)[reply]

Usually we do not encounter "mathematics trolls" at this reference desk (in my experience). In fact, even OP's with good intentions often leave after asking their questions (and thus while they may see our responses, they usually do not add furthur comments to the responses). Consequently, I agree it is somewhat surprising for this particular IP to be periodic in his antics. However, hopefully this is the last time we see him (I am holding my breath...); even if we do see him again, I think that it is best not to give him much attention. Ignoring his question, if truly not genuine, will give him indication that he needs to improve his conduct to obtain furthur responses. Reporting him at WP:AN/I may fuel a heated discussion on a trivial matter that could be resolved without the intervention of administrators (at least in my view). Unless he starts to vandalize/post multiple times in quick succession at this reference desk, no serious disciplinary action is really necessary. --PS T 09:29, 31 December 2009 (UTC)[reply]

I cannot actually comprehend how one could mistake "Rosenfeld" as "Rose&field" (or how one could mistake "Riemann" as "Ryan", if you were referring to "Riemann's hypothesis"); it appears too awkward to be genuine. Please make an attempt to respect us, and we will make a similar attempt to respect your questions. Your (inconsistent) grammar and spelling are not really the problem were it not for your copious references to your "math teacher", how he possesses a "math degree" and how this excuses him from calling true mathematics "fake math" (I doubt your mathematics teacher actually exists, to be honest; more likely is that he (and your "algebraist friend") are hypothetical). If you actually ask us genuine questions (rather than seek a predictable disagreement with your "omniscient mathematics teacher's" words, and use this as a motivation to claim that he possesses a mathematics degree), I am sure that your posts will be treated with greater seriousness. However, at present, I would not recommend posting furthur questions; if you are not serious, it is likely that someone will report you (and even in the unlikely case that you are serious, we are quite convinced otherwise and may report you nonetheless). Consider posting furthur questions when you have thoroughly reflected on the comments given to you, and the time spent in answering your questions. --PS T 09:19, 31 December 2009 (UTC)[reply]

I just looked up the IP, they come from Australia. So yes I would have thought that learning English should be a top priority, or at least Strine. Dmcq (talk) 09:55, 31 December 2009 (UTC)[reply]

It is possible, of course, that the OP was an immigrant. But your evidence, in addition to the fact that the OP persistently refers to "high school", suggests that he demonstrates a more competent display of English elsewhere (but then again, even in English speaking countries, high school students cannot be assumed to be competent in writing English...). --PS T 10:31, 31 December 2009 (UTC)[reply]

I suggest that we do not feed this troll by ignoring all of their future posts on this matter. ~~ Dr Dec (Talk) ~~ 12:35, 31 December 2009 (UTC)[reply]

I have recommended (to the OP) that he post questions after thoroughly reflecting on the responses given here, and that posting such questions in quick succession could result in a potential block. Hopefully his/her conduct will improve, in which case we would be happy to respond to his/her questions. However, I agree with you that, in the case of furthur trollish behavior, we should ignore his/her posts rather than reply or defer to WP:AN/I. --PS T 12:54, 31 December 2009 (UTC)[reply]

It seems you are of the opinion that behind the ridiculous stories lie genuine questions. To me it looks like the OP doesn't have any real confusion at all, or at least no interest in its resolution. My working assumption is that he is trying to mock mathematicians in retaliation to arrogance he has encountered among some of them. -- Meni Rosenfeld (talk) 13:18, 31 December 2009 (UTC)[reply]

You are probably right. However, my ultimate aim is to communicate with him politely, in which case he may change his behaviour. The chances of this are, of course, slim given that many have already done this with no success. But it is not completely impossible to change a person, even though it is highly improbable in this case. --PS T 13:31, 31 December 2009 (UTC)[reply]

Just to clarify, I do believe that he is a troll (the biggest indicator being his repeated reference to his mathematics teacher (not to mention that he insults him by calling him a math nerd)). But he is not the worst of the trolls I have seen. --PS T 13:34, 31 December 2009 (UTC)[reply]

I'm not sure if our questioner fulfills the definition of troll, but it seems quite sure that the reason why he put his posts here is not to get an effective answer to his questions. I think he just wishes to be funny and make us laugh of his performance of the "stupid student". In fact he can't ignore that there are millions of maths teachers in the word, therefore he may not seriously believe that being a teacher is a title for having any hope to get a millennium prize. He also plays the stupid when he quotes the "Ryan hypo-what", like somebody who has just heard about the Riemann hypothesis wihout knowing the spelling, while he said he has just read about it. So, if he's still around, I wish to tell him: OK funny, but understand that this is not the right place for your performance; it will not be understood and will turn out to be annoying noise. Go and try playing the stupid at a Police Station instead, they will laugh and offer you a beer. --pm a (talk) 21:11, 1 January 2010 (UTC)[reply]

you guys are cocks. I am sure there is a contradiction between 'assuming good faith' and stating that something is bad faith. You are the ones who have broken wikipedia rules. —Preceding unsigned comment added by 92.22.51.77 (talk) 19:29, 3 January 2010 (UTC)[reply]

Yes; I include your remarks in the collapsed part, since they belong to it.--pm a (talk) 21:10, 3 January 2010 (UTC)[reply]

Gauss–Jordan elimination = reduced row echelon form?

Is Gauss–Jordan elimination the same as Reduced Row Echelon Form? --33rogers (talk) 07:46, 31 December 2009 (UTC)[reply]

Gauss Jordan elimination involves taking the coefficient matrix and reducing it to reduced row echelon form, with corresponding changes on the right hand side. Then the system is solved by back substitution. This whole procedure is called Gauss Jordan elimination.[1] Reduced row echelon form is the type which the coefficent matrix ultimately becomes.-Shahab (talk) 07:51, 31 December 2009 (UTC)[reply]

Some basic topology of the torus

I want to show that the torus, described as "the surface of revolution obtained by rotating a circle about an axis in its plane and disjoint from it", is homeomorphic to $S^{1}\times S^{1}$ (part of Problem 4 here). I suppose this is meant to be easy but my analytic geometry is sufficiently weak that I'm having a devil of a time turning this description into something concrete enough to write down an explicit map and formally verify all the conditions. Here's my attempt so far:

It's not too hard to see that $S^{1}\approx \mathbf {R} /\mathbf {Z}$ . Define $f:\mathbf {R} /\mathbf {Z} \to \mathbf {R} ^{3}$ by $f(s)=(\cos 2\pi s,0,\sin 2\pi s)+(2,0,0)$ so that the image of f is the circle of radius 1 in the xz-plane centred at (2,0,0). Let R(t) be rotation around the z-axis by the angle 2πt. Finally define $h:\mathbf {R} /\mathbf {Z} \times \mathbf {R} /\mathbf {Z} \to \mathbf {R} ^{3}$ by $h(s,t)=R(t)f(s)$ . It is clear that h is injective, and it's continuous because each component of $R(t)f(s)$ is a linear polynomial in sines and cosines of s and t (as R(t) is just a rotation matrix). Since the domain of h is compact, this implies that h is a homeomorphism onto its image.

I'm not sure how satisfactory a solution to the original problem this is. Any advice appreciated! — merge 13:37, 31 December 2009 (UTC)[reply]

This is not much of an answer, but can you intuitively feel why the result is true? Can you imagine the torus as a cartesian product in the manner described? In my eyes, the most important aspect of topology is attaining an intuitive feel for the topology of a space; although many textbooks encourage defining explicit homeomorphisms between topological spaces, it is not necessary as long as you can intuitively visualize the homeomorphism. However, when confronted with a first example, it is sometimes instructive to construct an explicit map. In this case, consider using your intuitive feel for the torus as a cartesian product when constructing a homeomorphism. Does this help? --PS T 14:59, 31 December 2009 (UTC)[reply]

Thanks for your response! Yes, the basic intuition is clear enough to me for all the spaces mentioned in the problem. What I'm specifically concerned about at this stage is grounding the intuition in correct proof. The above was exactly my attempt to do the explicit construction you suggested (modulo writing out all the rotation matrix details, which don't seem all that relevant). — merge 15:28, 31 December 2009 (UTC)[reply]

How about this:

(x,y,z)=(\cos(2\pi t),\sin(2\pi t),0)(2+\cos(2\pi s))+(0,0,\sin(2\pi s)).

Michael Hardy (talk) 01:48, 1 January 2010 (UTC)[reply]

Thanks! I'm less concerned about the actual formula than whether the argument is sound, though (and whether there's a better way to do it). — merge 11:20, 1 January 2010 (UTC)[reply]

What you did seems fine. Demonstrating a homeomorphism seems like a good way to go about proving that they're homeomorphic. 67.100.146.151 (talk) 05:01, 2 January 2010 (UTC)[reply]

Yes, after looking at it for a while it seems all right to me. I think I've also managed to show the other spaces in the problem are equivalent. Thank you! — merge 12:21, 2 January 2010 (UTC)[reply]

Looking for random "shopping" algorithm

I'm looking to see if an algorithm for a certain problem (described below) exists. I've run into this problem a number of times in different contexts, so suspect it might be solved, but don't know what it might be named. For the time being, I've coined the name "random shopping algorithm".

The problem begins with presented with a collection of "goods" with different costs and a given sum of money. The algorithm must then choose a random selection of goods to purchase whose total is equal to the given sum. It does not work to simply choose items at random until the sum is reached as that would select one high cost item more often than a lot of small cost items.

--jwanders^Talk 22:25, 31 December 2009 (UTC)[reply]

I'm not sure what your intended criteria for selection is but see subset sum problem. If you want all subsets with an equal sum to have the same probability then you have to know the total number of solutions and then pick one of the solutions. Or do you want each individual item which is both inside and outside at least one solution to have the same probability or as close to the same as possible? Then we would also have to define a closeness measure when there are items with different probabilities. PrimeHunter (talk) 23:33, 31 December 2009 (UTC)[reply]

Please: "This criterion is..." or "These criteria are....". Michael Hardy (talk) 01:15, 1 January 2010 (UTC)[reply]

Hmm, no, I don't think the subset sum problem quite matches. As an example, say you have $20 and the items available are A for $10 or B for $1. The three possible outcomes here are (2As), (1A + 10Bs), or (20Bs). The algorithm I'm trying to find would chose one of these three with an equal probability. You right that I need some way of A) counting the number of possibilities N, randomly selecting n<N, and B) creating possibility n directly. As I said above, I'm hoping this is a solved problem that someone will recognise and know the name of. --jwanders^Talk 00:16, 1 January 2010 (UTC)[reply]

So your algorithm is:

generate the N possibilities by the subset sum algorithm, and remember them in a list
choose a random number between 1 and N, call it x
output the x-th element of the list in step 1

Doesn't that do it? Staecker (talk) 01:19, 1 January 2010 (UTC)[reply]

In fact, if you have a method of enumerating all the possible combinations (perhaps using a recursive algorithm), then you don't even need to store them all in a list; the following algorithm will work to pick a random combination uniformly:

Let k equal 1 and let X be the first combination.
If there are no more combinations to generate, stop and return X.
Otherwise, let Y be the next combination and increment k by 1.
With probability 1/k, set X ← Y.
Repeat from step 2.

In fact, with a slight modification this algorithm can be used to uniformly pick n random combinations:

Let k equal n and let X₁, ..., X_n be the first n combinations.
If there are no more combinations to generate, stop and return X₁, ..., X_n.
Otherwise, let Y be the next combination and increment k by 1.
Generate a uniformly distributed random integer j between 1 and k inclusive.
If j ≤ n, set X_j ← Y.
Repeat from step 2.

If you shuffle the first n combinations in step 1, the output should also be uniformly shuffled. However, it's just as easy to shuffle the output afterwards if you want. (Of course, there are cases where Staecker's method is better, especially if you expect to need more combinations later and if storing the list is cheaper than regenerating it.) −Ilmari Karonen (talk) 08:49, 4 January 2010 (UTC)[reply]