Wikipedia:Reference desk/Archives/Mathematics/2010 April 26

= April 26 =

A number that is itself
Hello,

I'm dealing with what I believe is an intractable problem. Here it is stated as succinctly as possible: A positive integer in its letter form can equal itself (given that a is 1, b is 2, and so on) when the sum of the letters are taken. What is the second one? Naturally, I want to be exhaustive in the answers I can give: does one write the number in traditional English as "TWENTY FIVE", "ONE HUNDRED AND THIRTY SIX", does one leave out "and", or does one resort to calling out each number within the greater string of numbers, e.g., "111" would be "ONE ONE ONE"? Besides that linguistic dimension of the problem, there is also the immense difficulty of finding the damn numbers! I'm trying to find these answers quickly and efficiently, but seeing how open-ended the problem is (but don't think there isn't an answer, because there definitely is), I'm wondering if anyone can provide a suitable approach to tackle this problem, even if I have to learn to write a program to do it. Thanks. I know this isn't your everyday sort of problem, and it probably will require a lot of thought. Another possibility that I don't know about is whether such a number would have a certain, known property, like an "honest number" or perfect number. If anyone knows of such numbers without having to go through the bricks and mortar of finding the answer, then please let me know that as well.Coitoergosum (talk) 02:02, 26 April 2010 (UTC)
 * The number of letters in the spelled-out form will increase linearly with the number of digits, while the actual number will increase exponentially. So if the two graphs ever cross, they will do it for fairly small n.  You have to decide for yourself what spelling scheme to use, and then either use cleverness to solve for an answer, or use a program to exhaustively search for one.  I get "two hundred and fifty one" and "two hundred and fifty nine" as solutions.  Without the "ands", I get that there are no solutions.  Maybe there are more solutions under other schemes, or in languages other than English. 69.228.170.24 (talk) 04:50, 26 April 2010 (UTC)
 * Thanks for that. I'm aware of the divergence that arises when one compares the additive sequence and the number sequence; however, neither of the answers you have given are the answer to the problem. I'm guessing that this problem is a little trickier than I had anticipated. There may be a solution involving the ordinal numbers (first, second, etc.) that I didn't think of before, so I'll have to try that. Again, thanks. This isn't a resolved problem by any means. Another aspect to the problem is whether this is a known property of numbers, cf. "honest number"; it probably wouldn't be one found in formal mathematics, but it certainly can be construed as a linguistic property of numbers. If anyone knows about it, I'd like to know what this property is called.Coitoergosum (talk) 16:14, 26 April 2010 (UTC)
 * Well, this isn't a property of the number itself - it depends on which base you represent the number in. In binary you would be calculating the values of "one zero zero one" intead of "nine" etc. And it is language dependent. And it depends on your convention for valuing alphabetic strings (Why ignore spaces ? Why add letter values rather than multiplying or using a place value calculation ? Why not use ASCII character values instead of 1-26 ? etc.) There are so many arbitrary conventions here that it would't surprise me if this property has never been named or studied. Gandalf61 (talk) 16:31, 26 April 2010 (UTC)
 * While in principle I agree with your evaluation of the situation in general, the problem very explicitly and unambiguously states that one has to use letters when giving/finding the answer (a "space" isn't a "letter", either). The problem is also stated in English, so the answer must be one found in English. Different number bases came to mind, but I don't think the problem is a test of one's knowledge on that level (if it is, then it is extremely devious and nasty).Coitoergosum (talk) 20:41, 26 April 2010 (UTC)
 * Just assume that no number above 1000000 will have this property and use a brute force algorithm. You must have a specific encoding scheme decided before starting, but the question is ambiguous if you don't so that is not really something we can help with.  Once it's clear how you are writing your numbers just start with 1 and go up, a reasonable program should reach 1 million in less than a minute. It's possible that there are tricks you can use to speed the computation up, but the problem is small enough that it's not really worth trying to find such tricks.  The hard part might be proving the upper bound, ie. that no number above 1000000 has this property.  Taemyr (talk) 17:03, 26 April 2010 (UTC)
 * Thanks. I've already thought this myself, but I have practically zero programming skills, so this will be an uphill battle. You guys have been great; thanks for your help. I won't give up.Coitoergosum (talk) 20:41, 26 April 2010 (UTC)
 * It's actually a fairly decent task for introductory programming. Parsing a number into text is slightly tricky if you haven't done any programming, but will teach you the basic of string manipulation.  Taemyr (talk) 20:29, 27 April 2010 (UTC)

apparent misprint:S
Hi all, My problem is little but very frustrating. I am reading from a book that has very little explanations and is not very rigorous too, and am stuck at a little point. I belive there is a misprint but then I can't make out what to substitute for the correct values: Consider the equation (in Z) $$x_1+\cdots+x_{k-1}-x_k=-b$$ (k,b are even). The book says that $$x_i=a_t;(1\le i\le \frac{k}{2}), x_i=a_{t+1}+1;(\frac{k}{2}+1\le i\le k-1), x_k = \frac{k}{2}a_t+(\frac{k}{2}-1)(a_{t+1}+1)+b-1$$ is a solution, which I fail to see. I believe the -1 at the end of the xk term is an error. But reading onward it is evident that the -1 is very important in the subsequent discussion and in fact the value of xk must be exactly what is proposed. So I feel the error must be in defining the other xi terms. Subject to the additional contraints that xi's don't belong to [at+1,at+1], at's are increasing and $$a_{t+1}\ne a_t+1$$ how should I modify the solution. i.e. the xi's. Also is some modification actually needed or I am just plain missing something. Thanks-Shahab (talk) 05:35, 26 April 2010 (UTC)
 * I also get that the two sides are off by 1. However, without the context to know why these values were chosen it's impossible to say what the "correct" fix is.  You could change any term you wanted by 1 to make the sum work, or even change lots of terms. 67.100.146.151 (talk) 05:58, 26 April 2010 (UTC)
 * Thanks, at least now I am assured that this is a misprint. The proper context will take too long to describe. Hence I gave the required constraints.


 * Let me rephrase my question now. Is there a fix where xi (i is not k) is chosen as either at or at+1+1 and the only change that I have to do is to decide from where to where xi takes either of these two values?-Shahab (talk) 06:17, 26 April 2010 (UTC)

Factorisation
Hello. I'm looking for information on the best way to factorise $$ab(a^2-b^2) + bc(b^2-c^2) + ca(c^2-a^2)$$. I notice that swapping a and b negates the result, which suggests that the expression is divisible by (a-b), but I'm not really sure how to proceed from there. The people on IRC advised that considering a, b, and c as points in three-dimensional space and employing Lagrange multipliers could work, but I don't know much about that. Thanks for the help. — Anonymous Dissident  Talk 12:18, 26 April 2010 (UTC)


 * Well by symmetry (b-c) and (c-a) must then also be factors which since the original had fouth powers means you only have a linear expression in a, b and c left which must also be symmetric in them. Now I wonder what kind of expression is symmetric and linear sum of a, b and c? Dmcq (talk) 12:38, 26 April 2010 (UTC)
 * a + b + c. So inspection works best. Thanks. — Anonymous Dissident  Talk 13:12, 26 April 2010 (UTC)
 * Yes I think it is kind of funny that for two variables you've got two options a+b and a-b whereas for three you've only got one. In other circumstances one can have a+bω+cω2 where ω is the cube root of 1 but then you'd need other terms like this to cancel out the ω's. Dmcq (talk) 13:55, 26 April 2010 (UTC)

Period (number)
The article says that "irrational algebraic numbers and functions are themselves expressible as integrals of rational functions over rational domains". Could someone handwave how algebraic numbers can be expressed that way? —Preceding unsigned comment added by 212.87.13.69 (talk) 16:54, 26 April 2010 (UTC)
 * The reference cited in the article looks very good. It gives the example: $$\sqrt 2=\int_{2x^2\le 1} dx$$.  69.228.170.24 (talk) 07:49, 27 April 2010 (UTC)

Generation of batting sequences
The game of cricket has a team, usually 11 in number, who in an innings bat in pairs, initially person 1 and person 2 together. Whoever is dismissed in a partnership is replaced by the next person in numerical order, this continuing until the 10th partnership is broken, when the innings is complete. If there were only 3 players there would be 2 possible partnership sequences, (1,2), (1,3) or (1,2), (2,3). A 4th player would double the possible number of sequences to these: (1,2), (1,3), (1,4) or (1,2), (1,3), (3,4) or (1,2), (2,3), (2,4) or (1,2), (2,3), (3,4). It's apparent that for n players there will be 2^(n-2) different sequences of length n-1. My problem - to get an algorithm to generate them. I feel that this should be fairly standard, but can't get one to work.→86.166.205.252 (talk) 20:32, 26 April 2010 (UTC)


 * I wrote it and tested it in Fortran:


 * I didn't include the BINARY subroutine, which converts an integer to a 16 character binary string. Hopefully it's all self-explanatory.  If not, let me know. StuRat (talk) 04:19, 27 April 2010 (UTC)


 * I think that problem may intended as an exercise in recursion. That's probably simpler than the explicit binary expansion.  Is there some reason you want to use fortran?  It's mostly for numerics, not so much for these combinatorial things. 69.228.170.24 (talk) 10:57, 27 April 2010 (UTC)


 * The OP didn't ask for Fortran, but I'm a Fortran programmer, so that's what I know best. StuRat (talk) 13:48, 27 April 2010 (UTC)


 * (OP) Thanks, StuRat, I checked that it worked. The problem was an exercise only in the sense that it occurred to me while watching cricket, bur afterwards I was unable to generate a complete list. If anyone feels inclined to derive a recursive method I'd be interested to see it.→86.155.208.113 (talk) 13:08, 27 April 2010 (UTC)


 * That's not too hard. For 2 players there is only one batting sequence: (1,2). For n > 2 players, you take each batting sequence for n-1 players, and use it to generate two new n player batting sequences as follows. If the final pair in the n-1 player sequence is (a,b), then one new sequence is created by appending (a,n) to that sequence and the other is created by appending (b,n). So from (1,2) you get the two 3 player sequences (1,2)(1,3) and (1,2)(2,3), then the four 4 players sequences (1,2)(1,3)(1,4), (1,2)(1,3)(3,4), (1,2)(2,3)(2,4), (1,2)(2,3)(3,4) and so on. Gandalf61 (talk) 13:20, 27 April 2010 (UTC)


 * OK, here's a recursive version, also in Fortran:


 * It also prints out each sequence on one line instead of multiples, but I could have done that on the non-recursive version, too. StuRat (talk) 15:36, 27 April 2010 (UTC)


 * Here are the runs for 2-6:

Enter N (>2): 2 ( 1, 2)

Enter N (>2): 3 ( 1, 2)  ( 1, 3)  ( 1, 2)  ( 2, 3)

Enter N (>2): 4 ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 1, 2)  ( 2, 3)  ( 3, 4)

Enter N (>2): 5 ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 1, 5)  ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 4, 5)  ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 3, 5)  ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 4, 5)  ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 2, 5)  ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 4, 5)  ( 1, 2)  ( 2, 3)  ( 3, 4)  ( 3, 5)  ( 1, 2)  ( 2, 3)  ( 3, 4)  ( 4, 5)

Enter N (>2): 6 ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 1, 5)  ( 1, 6)  ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 1, 5)  ( 5, 6)  ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 4, 5)  ( 4, 6)  ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 4, 5)  ( 5, 6)  ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 3, 5)  ( 3, 6)  ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 3, 5)  ( 5, 6)  ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 4, 5)  ( 4, 6)  ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 4, 5)  ( 5, 6)  ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 2, 5)  ( 2, 6)  ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 2, 5)  ( 5, 6)  ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 4, 5)  ( 4, 6)  ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 4, 5)  ( 5, 6)  ( 1, 2)  ( 2, 3)  ( 3, 4)  ( 3, 5)  ( 3, 6)  ( 1, 2)  ( 2, 3)  ( 3, 4)  ( 3, 5)  ( 5, 6)  ( 1, 2)  ( 2, 3)  ( 3, 4)  ( 4, 5)  ( 4, 6)


 * StuRat (talk) 20:44, 27 April 2010 (UTC)


 * Here's a C++ metaprogram that generates the sequences at compile-time :)


 * End example.  (talk) 16:31, 30 April 2010 (UTC)

Ouch, decltype, that was brilliant but cruel ;-). I like Haskell for problems like this: This is recursive: the "cricket [a,b]" equation handles the 2-element case, and "cricket (a:b:c:ds)" handles the recursive case (3 elements a,b,c followed by a possibly empty remainder of the list).  The bracketed expression in the recursive case is a list comprehensions.  The mapM_ in the "main" function runs "print" on each of the sublists, so they are shown one per line.  [1..4] is shorthand for [1,2,3,4]. 69.228.170.24 (talk) 09:07, 2 May 2010 (UTC)

Work Word problem
hi the problem i have says "50 people build 10 bridges in 500 hours. If 15 people build 40 bridges, how many hours will they take to make them?" how do i solve this?? —Preceding unsigned comment added by Needlotsofhelp (talk • contribs) 21:38, 26 April 2010 (UTC)
 * Hint: If 50 people can build 10 bridges in 500 hours, then 200 people can build 40 bridges in 500 hours (assuming they do not work more efficiently, the more there are). Now it's a matter of simple division. —Preceding unsigned comment added by 76.230.149.101 (talk) 22:02, 26 April 2010 (UTC)

well my teacher told me about a way to find the answer using a fomrula but i dont remember it was it like multiply people by hours and then divide by stuff they built??? i'm not sure.... —Preceding unsigned comment added by Needlotsofhelp (talk • contribs) 22:20, 26 April 2010 (UTC)
 * Forget about the formula, you can do it yourself. To make it concrete, let's say you pay each man 1 dollar per hour. Building 10 bridges costs 500 hours of 50 men. How much is it? And 1 bridge? And 40 bridges? How much will you pay each man if they're 15 of them? --84.221.198.37 (talk) 22:57, 26 April 2010 (UTC)


 * Your teacher's method will have used the following formula to calculate the man-hours required to build each bridge:


 * $$ \frac{\text{people} \times \text{hours}}{\text{bridges}} = \text{man hours per bridge}$$


 * Then we assume that the man hours per bridge is the same in both parts of the problem. So we have two ways of calculating the man hours per bridge figure, which must both give the same value. In other words, if it takes 15 people x hours to build 40 bridges then


 * $$\frac{50 \times 500}{10} = \frac{15 \times x}{40}$$


 * Now you can rearrange this equation to find the value of x. Gandalf61 (talk) 12:11, 27 April 2010 (UTC)

Uncountably Large Subfields of the Real Numbers?
Well? --128.62.32.123 (talk) 23:00, 26 April 2010 (UTC)


 * Well what? Dmcq (talk) 23:07, 26 April 2010 (UTC)
 * Do they exist? --128.62.32.123 (talk) 23:10, 26 April 2010 (UTC)
 * Yes, of course. R is one, for a start. If you want a proper one, then you could pick a transcendence basis of R over Q and take the field generated by some uncountable proper subset of it. Algebraist 23:21, 26 April 2010 (UTC)
 * Thank you. --128.62.32.123 (talk) 23:23, 26 April 2010 (UTC)
 * Wait, doesn't that mean you have to take a transcendence basis of infinite degree? Is it obvious that you can do that without getting all of R? 69.228.170.24 (talk) 06:14, 27 April 2010 (UTC)
 * It is, because of the algebraic independence of the bases. Different subsets of it generate different fields. --131.114.72.230 (talk) 09:08, 27 April 2010 (UTC)

Complicated Integration based on a Cylinder Cross-section
Is it possible to solve... ((ay^2/(b+c))-cy)*(1-(y^2/b^2))^0.5 dy from y=b to -b Many thanks for any help. Here is a link to it on wolfram alpha if that helps to visualise the problem.

From Andrew McArthur --137.222.114.238 (talk) 23:00, 26 April 2010 (UTC)
 * Hint: Substitute in cos2 θ for 1-(y^2/b^2). Do some arithmetic to figure out what θ is and don't forget to reset the bounds! 76.230.149.101 (talk) 02:25, 27 April 2010 (UTC)

If I'm reading the question right, the problem is to evaluate
 * $$ \int_{-b}^b \left(\frac{ay^2}{b+c} - cy \right)\left( 1 - \frac{y^2}{b^2} \right)^{1/2}\,dy. $$

I'd separate it thus:
 * $$ \int_{-b}^b \frac{ay^2}{b+c}\left( 1 - \frac{y^2}{b^2} \right)^{1/2}\,dy - \int_{-b}^b cy \left( 1 - \frac{y^2}{b^2} \right)^{1/2}\,dy.$$

The second integral is that of an odd function over an interval symmetric about the origin; therefore the second integral is zero. The first integral is that of an even function over an interval symmetric about the origin; therefore its value is twice that of the integral over the half-interval:
 * $$ 2\int_0^b \frac{ay^2}{b+c}\left( 1 - \frac{y^2}{b^2} \right)^{1/2}\,dy. $$

Now let

\begin{align} \sin\theta & = \frac{y}{b} \\ b\cos\theta\,d\theta & = dy \\ \end{align} $$ etc. (And of course &theta; will go from 0 to &pi;/2.) Michael Hardy (talk) 17:28, 27 April 2010 (UTC)

Redefining The Comparison of Cardinal Numbers
The standard definition for two sets X, Y of | X | ≤ | Y | is that there is an injection from X to Y. An alternative definition may be that there is a surjection from Y to X; under the assumption of the Axiom of Choice, this latter definition is equivalent. How does this definition behave if the axiom of choice isn't assumed? --128.62.32.123 (talk) 23:26, 26 April 2010 (UTC)


 * Note that even assuming AC the equivalence only holds for X non-empty (for e.g., there always exists an injection form the empty set to any set Y, but there exists a surjection from Y to the empty set only if Y itself is empty). Assuming X is not empty, an injection  j:X→Y easily produces a surjection  s:Y→X (no AC is needed). Indeed, say you have a bijection j':X→j(X), and say c&isin;X. The inverse of j' extends to a surjection s:Y→ X putting s(y)=c for y &isin; Y\j(X). --131.114.72.230 (talk) 08:29, 27 April 2010 (UTC)


 * Right, so I suppose that means the definition would have to be reworked to account for the special case of the empty set.
 * I already knew that the injective definition implies the surjective definition (I hadn't thought about the corner case of the empty set, but again, that's not really terribly important for my question; it's the infinite cardinals that are interesting); of course the interesting question is the reverse, and it seems doubtful that the reverse implication will always hold in non-AC settings, since the most direct way to get an injection from X to Y given a surjection from Y to X would be a right-inverse, and the ability to take a right-inverse for every surjective function is equivalent to AC. There may be another more clever way that I'm not thinking of, but like I say, doubtfully so.
 * The questions I'm more interested in are things such as trichotomy, how various potential definitions of infinitude using this notion of comparison behave (eg we may define infinitude as greater than or equal to every natural number, or greater than or equal to the set of all natural numbers, etc.), and so on. --128.62.32.123 (talk) 13:02, 27 April 2010 (UTC)
 * ZF does not prove (unless it is inconsistent) that a surjection from Y to X implies the existence of an injection from X to Y, and it does not prove that any two sets are comparable (using either definition of comparison). As for definitions of infinity, you can find a lot of information in our article on Dedekind-infinite sets.—Emil J. 15:47, 27 April 2010 (UTC)
 * Another important thing about your relation that ZF does not prove is that it's a partial order (antisymmetry may fail). Algebraist 17:20, 27 April 2010 (UTC)
 * Well, obviously, ZF proves that it is not antisymmetric, as two different sets can have the same cardinality. But this has nothing to do with failure of AC, this is inherent in the notion.—Emil J. 17:33, 27 April 2010 (UTC)
 * I meant as a relation on cardinals, of course. Algebraist 17:45, 27 April 2010 (UTC)
 * I see. However, if the OP is contemplating alternative definitions of comparison of cardinalities, there is nothing stopping them from redefining equality of cardinalities so that it fits. So let us be precise: you are saying that ZF does not prove the surjective variant of the Cantor–Bernstein theorem: if there is a surjection from X to Y and a surjection from Y to X, then there is a bijection from X to Y (whereas it does prove the original, injective version).—Emil J. 18:00, 27 April 2010 (UTC)

Ask user:Trovatore about this one. Michael Hardy (talk) 17:35, 27 April 2010 (UTC)

I notice that mathematicians seem to have difficulty with the notion of an open-ended question. So far, only one point has been brought up that I didn't specifically mention in trying to give some examples of what I'm looking for. --128.62.55.11 (talk) 00:24, 28 April 2010 (UTC)
 * Well, "128.62.55.11", I guess you like congratulating yourself based on the fact that you've figured out that we colored people like fried chicken. Here's a bit of advice: you might seem less like an idiot if you would abstain from shooting your mouth of before putting your brain in gear, as in your posting of 00:24, 28 April 2010. Michael Hardy (talk) 01:42, 28 April 2010 (UTC)


 * EmilJ gave a complete answer: without AC, ZF cannot prove that every pair of sets is comparable, under either definition. &mdash; Carl (CBM · talk) 00:28, 28 April 2010 (UTC)
 * That doesn't seem like a complete answer at all; it says very little. Michael Hardy (talk) 03:17, 28 April 2010 (UTC)

Thanks, people who made an attempt, I guess. I was hoping for more. I'd be very surprised if no one has ever studied this alternative definition in far more depth than the responses here seem to indicate. I'll assume the lack of further answers and insights indicate that no one else has anything substantial to add. I'll see if I can hunt down some real mathematicians who know more about this, since no one here seems to. --128.62.45.2 (talk) 18:00, 28 April 2010 (UTC)
 * Have you asked user:Trovatore this question? He's a mathematician with some expertise in this area. Michael Hardy (talk) 18:57, 28 April 2010 (UTC)
 * I'm sorry that we weren't up to your standards, and that some of us have areas of expertise other than set theory (which is what real mathematicians do). You can have your money back. -- Meni Rosenfeld (talk) 20:06, 28 April 2010 (UTC)