Wikipedia:Reference desk/Archives/Mathematics/2010 January 9

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January 9

Arrow's impossibility theorem and real elections

Arrow's impossibility theorem says that given three or more candidates, we can't convert a list of individual rankings into an aggregate ranking that satisfies certain criteria. But in real elections, we don't need a complete aggregate ranking, just a choice of the most preferred candidate (we don't have to rank the losers against each other). How much of the theorem still applies? NeonMerlin 06:31, 9 January 2010 (UTC)

Well, basically the same difficulties are present; you may want to study then the Gibbard-Satterthwaite theorem. What G-S says is (essentially) that no voting procedure can satisfy simultaneously strategy-proofness and non-dictatorship.

Interestingly enough, Gibbard's version linked voting schemes with social preferences, and applied Arrow's Impossibility theorem to derive his result. Pallida  Mors 19:09, 9 January 2010 (UTC)

It is also noteworthy the fact that normal voting procedures you are mentioning are just one set of possible voting schemes: the outcome they produce is formed using as informational input just the tops of individual preferences; since more general schemes are manipulable, that is of course true especially of traditional majority voting, under which many individuals are tempted to vote (i.e. declare the top of their preferences) for a second choice whenever their first choice is believed to have no real electoral chance. Pallida  Mors 02:08, 10 January 2010 (UTC)

Actually, I'm also talking about alternative systems that take a complete ranking of preferences from each voter but output only a first choice. NeonMerlin 08:24, 10 January 2010 (UTC)

Well. Then, that is the general concept explored under the G-S setting. Pallida  Mors 14:55, 11 January 2010 (UTC)

1001 (number)

Is there a known counterexample to the conjecture that 1001 is the only product of consecutive primes that is one more than a perfect power?

If not, is there a heuristic estimate of the probability that the conjecture is true? (I assume it has not been proven, though someone may set me straight on this.)Julzes (talk) 06:48, 9 January 2010 (UTC)

There's the trivial counterexample of 1, which is the empty product. —Bkell (talk) 12:45, 9 January 2010 (UTC)

Also silly counterexamples such as 5 and 17, which are products of a single prime. —Bkell (talk) 12:50, 9 January 2010 (UTC)

I have noticed a worrying trend of this sort of behaviour on Wikipedia reference desks. There are often many ways in which one may "misinterpret" a question in order to respond, but it is also important to note that which the OP desires. In this case, you have answered the question to some extent which is appreciated, but you should bear in mind that such answers will have little effect on the questioner unless he/she has not even thought about his/her own question! Of course, your contributions are much appreciated, but you should let others answer a question if you cannot respond with a non-trivial answer and know that your answer is trivial (if you truly thought that your answer shed light on the question, then there is not a problem; if you knew that your answer did not, but included it anyway, then you should have let others respond instead of frustrate the OP with trivial answers). This is merely my opinion, however, and you may ignore it if you wish. --PST 14:23, 9 January 2010 (UTC)

It's not always perfectly crystal clear what exactly it is that an OP intends to ask, or how much effort he has already put into the question. Sometimes an OP himself is confused about what he is after. Pointing out trivial cases can help an OP focus the question, which benefits other responders as well.

An OP has an obligation to specify unambiguously what he wants more than responders have an obligation to guess it. -- Meni Rosenfeld (talk) 17:10, 9 January 2010 (UTC)

I agree with that myself, but in my case isn't 'consecutive' implying something pretty clearly. Is anybody going to say what's known? Let me Google it to see if I've made a booboo.Julzes (talk) 17:37, 9 January 2010 (UTC)

But the question, in this case, was crystal clear! The phrase "product of consecutive primes" is pretty clear. Consecutive means to follow, to be in succession^[1], so the empty product and the "silly counterexamples" given by Bkell above clearly don't match the conditions given by the OP. The only problem I have is as to how many consecutive primes we allow. Is it the product of two, three, four, …, or arbitrarily many consecutive prime numbers? ~~ Dr Dec (Talk) ~~ 18:25, 9 January 2010 (UTC)

I posted my first comment after just a little bit of thinking about the problem, thinking perhaps that it was the only kind of trivial counterexample, with the idea that often trivial counterexamples are easy to overlook but can shed light on the original problem. Then shortly after I realized that there were lots of silly counterexamples in which only one prime is "multiplied"; since I had already posted one silly counterexample, I figured I should mention the others as well. I was and am fully aware that this is not what Julzes was asking. I was attempting to post my thoughts, the idea being that other contributors could join in and we could make some progress on the question by working together (as I was also doing in my answer to the letter grid problem). I was not aware that the mathematics reference desk is only to be used for complete, final answers that spring fully formed from the minds of great mathematicians. I apologize for my problem-solving technique, in which sometimes I investigate silly cases in an attempt to understand how the problem behaves. It won't happen again. I'm sure one of you would have posted a complete "non-trivial answer" to this question by now if I hadn't screwed everything up by posting a couple of innocuous, true, but trivial facts. —Bkell (talk) 20:05, 9 January 2010 (UTC)

Bkell: I sincerely apologize if I have upset you with my comment; it was a suggestion that people could consider, rather than a criticizm of your contributions. I firmly believe that you are an excellent contributor to the reference desks; it was not my intention to represent you otherwise. In fact, I will be the first to admit that I posted nothing in the direction of the question, so of course I should be criticized if you are. However, please do note that my comment was merely a suggestion that you need not follow, although it does have a purpose. In some cases, when people observe that a question has already been answered, they feel that there is no furthur need to respond (some merely glance over questions and answers, therefore, without carefully examining the content). Therefore, it may well be that no one responds furthur to Julzes' question, thinking that your answer is complete without examining it thoroughly. Secondly, I just feel that, given the sorts of questions Julzes has asked in the past, it is unlikely that he had not even thought about his question (in fact, the wording of his question suggests that he has thought about the trivial cases already). However, I did not intend to fuel a heated debate regarding your post, and for that I shall apologize. --PST 03:08, 10 January 2010 (UTC)

Arbitrarily many. That makes for less likelihood that the conjecture is correct, but I doubt it makes that much difference except for the estimate I was seeking. There is nothing in Unsolved Problems in Number Theory (2nd ed.), and I couldn't find anything on Google. I'm deletingI've deleted what's below this as out of sequence and irrelevant.Julzes (talk) 19:27, 9 January 2010 (UTC)

Out of interest, I searched for small (less than 10²⁰) instances of power + 1 that are the product of distinct primes close together. Apart from 1001, closest I could find are:

$11543117=226^{3}+1=211\times 227\times 241$

$86350889=442^{3}+1=421\times 443\times 463$

Those factorisations don't even contain a pair of consecutive primes, so this suggests that there aren't any small counterexamples. Gandalf61 (talk) 10:06, 10 January 2010 (UTC)

Interesting. Less than 10²⁰? And these 8-digit numbers have the closest factors? That's acording to their relative orderings, I assume. Anyway, that's what I would choose. My general intuition is and has been that this is an essentially unprovable fact, but now you've actually told me something that strikes me as peculiar. It seems certain to me that whatever measure of closeness you chose, this is a little strange. Hmmm. Thanks for filling me in. Now I'm more curious about this basically useless subject.Julzes (talk) 10:36, 10 January 2010 (UTC)

Oops, missed out an even smaller example with close but non-consecutive prime factors:

$1729=12^{3}+1=7\times 13\times 19$

Also found some examples of power+1 with close (but again non-consecutive) pairs of prime factors, although the separate pairs are not close to each other:

$551369=82^{3}+1=(7\times 13)\times (73\times 83)$

$11259713=108^{3}+1=(7\times 13)\times (109\times 127)$

I haven't exhaustively tested power+1 up to 10²⁰ - I only tested first 10⁵ squares and cubes, for example. Gandalf61 (talk) 11:15, 10 January 2010 (UTC)

Of course, I knew about 1729. So your measure is then absolute ordering. I'm going to have to look at this question of close-but-no-cigar myself some time. There is one rather strange thing about your whole collection, as you've related it. I can't say off the top of my head about the factors of 86350889, but the rest all look like pairs of almost-consecutive primes. Well, enjoy the strange task I set you on. I'm going to sleep.Julzes (talk) 11:41, 10 January 2010 (UTC)

I have a related very strong conjecture now: Among all n>1, the ratio of the largest and smallest prime factors of n^p+1 for a fixed prime p>3 is smallest for n<5 except for an isolated exception at p=43 and n=7.Julzes (talk) 13:29, 11 January 2010 (UTC)

Conjecture false: $157518^{5}+1=102461*134081*157519*170341*263071\,$. Are there other exceptions?Julzes (talk) 11:47, 17 January 2010 (UTC)

I'll also make the conjecture that the record-setters for the ratio between largest and smallest prime factors (excluding the cases where they are equal) of squares plus one and cubes plus one always have two and three distinct prime factors, respectively, and that there is not a last record-setter in either case.Julzes (talk) 15:53, 11 January 2010 (UTC)

Here's more base-ten stuff: The record-setters for the square case appear to always be round numbers with the exception of one case after a short start in which it's not true. The single biggest example I have is $32000000^{2}+1=31992001*32008001.\,$Julzes (talk) 12:39, 12 January 2010 (UTC)

PrimeHunter has given an explanation that n=50m^2 is the form that produces a lot of the record-setters (at his talk page). Getting an approximate probability that this is always true will be a task for me at some point in the next couple of weeks.Julzes (talk) 16:04, 12 January 2010 (UTC)

I pointed out that (50m²)²+1 = (50m²-10m+1) × (50m²+10m+1) gives a low ratio (50m²+10m+1)/(50m²-10m+1) when both factors are prime, and that probably happens infinitely many times. It happens for m = 152129×10⁵⁰⁰ with two 1013-digit primes found and proved by PrimeForm/GW. The ratio is approximately 1 + 2.6×10^-506. PrimeHunter (talk) 17:23, 12 January 2010 (UTC)

m was chosen so n ends in more than 1000 zeros: n = 115716163205×10¹⁰⁰¹. PrimeHunter (talk) 17:31, 12 January 2010 (UTC)

The record-setters and the cases where the two factors are prime coincide after a short run, unless there is a semiprime with very close factors by chance for a different type of n. It has only been remarked on in an edit summary, so I'll repeat it here the last case is likely to be one giving the factorization 1973*1993^2.Julzes (talk) 17:50, 12 January 2010 (UTC)

I did a little work on this on paper. What I came up with is:

If the largest prime factor of 2s²+2s+1, p, and its cofactor, k, satisfy 2s+k²+1<p, and if p-2*(2s-k+1) is prime; then n=p-(2s+1) outperforms all smaller n of the form 50m².

Whether such large s other than those give n in the form 50m² exist is then the question.Julzes (talk) 23:53, 12 January 2010 (UTC)

Note that it's easy to check that the k=1 case doesn't give any possibility that n is not of that form.Julzes (talk) 00:21, 13 January 2010 (UTC)

No, the above recent algebraic result is false.What I get doing it more carefully is much more complex and more limiting:

Assuming the ratio is better with the factorization $4n^{2}+1=[2n+(2s+1)]*[2n-(2s+1)+2k]$ yields the requirements that

$(2s^{2}+2s+1)/k$ and $(2s^{2}+2s+1)/k-2*(2s-k+1)$ are prime

and that the inequality

(2s-k+1)*{[(2s^2+2s+1)/k-(2s+1)]^1/2+1+[(2s^2+2s+1)/k-(2s+1)]^-1/2]}<(2s²+2s+1)/k holds or almost does.

I'll be back with more on what this actually means later.Julzes (talk) 08:02, 13 January 2010 (UTC)

Pairwise coprime integers

What is the probability that three randomly chosen integers are pairwise coprime? --84.61.151.145 (talk) 09:26, 9 January 2010 (UTC)

The question's formulation seems to me to require improvement (randomness from a countable set), but it is clear what is meant. I doubt that the result I got can be simplified (unlike the question with just two numbers), but it can be computed. What one should do to get what I have is consider divisibility by each prime as independent from divisibility by each other prime. Then you should be able to get what I have. Unless someone beats me to it, I'll get back to you with a numerical value (which I'm certain is already in print somewhere and is easy to generate).Julzes (talk) 11:20, 9 January 2010 (UTC)

Assuming I have interpreted the question correctly, the probability equals the reciprocal of Apéry's constant. In general, the probability of k randomly chosen integers being coprime is ${\frac {1}{\zeta (k)}}$ where $\zeta$ is the Riemann zeta function. I quote a relevant discussion from one of our articles below:

Given two randomly chosen integers $A$ and $B$, it is reasonable to ask how likely it is that $A$ and $B$ are coprime. In this determination, it is convenient to use the characterization that $A$ and $B$ are coprime if and only if no prime number divides both of them (see Fundamental theorem of arithmetic).

Intuitively, the probability that any number is divisible by a prime (or any integer), $p$ is $1/p$ (for example, every 7th integer is divisible by 7.) Hence the probability that two numbers are both divisible by this prime is $1/p^{2}$, and the probability that at least one of them is not is $1-1/p^{2}$. Now, for distinct primes, these divisibility events are mutually independent. (This would not, in general, be true if they were not prime.) (For the case of two events: A number is divisible by p and q if and only if it is divisible by pq; the latter has probability 1/pq.) Thus the probability that two numbers are coprime is given by a product over all primes,

$\prod _{p}^{\infty }\left(1-{\frac {1}{p^{2}}}\right)=\left(\prod _{p}^{\infty }{\frac {1}{1-p^{-2}}}\right)^{-1}={\frac {1}{\zeta (2)}}={\frac {6}{\pi ^{2}}}$ ≈ 0.607927102 ≈ 61%.

Here ζ refers to the Riemann zeta function, the identity relating the product over primes to ζ(2) is an example of an Euler product, and the evaluation of ζ(2) as π²/6 is the Basel problem, solved by Leonhard Euler in 1735. In general, the probability of $k$ randomly chosen integers being coprime is $1/\zeta (k)$.

Unfortunately, the exact value of the probability of which you inquire is not known; an approximate numerical value is ${\frac {1}{1.20205690315959428539973816151144999076498629234049888179227155534183820578631309018645587360933525814619915}}\approx 0.831907373$ which is approximately 83.2%. --PST 11:39, 9 January 2010 (UTC)

What you've given looks like the probability that there will be no prime dividing all three, while the questioner asked about pairwise coprimality (no prime divides more than one). I can't TeXify it yet, but that's the product of $1-3/p^{2}+2/p^{3}$.Julzes (talk) 11:58, 9 January 2010 (UTC)

You are right; I did not read the OP's question carefully enough (I missed the "pairwise" bit). Thanks. --PST 12:04, 9 January 2010 (UTC)

The probability of picking three integers between 1 and N at random being pairwise coprime goes to 0 as N increases toward infinity. Proof?Julzes (talk) 12:20, 9 January 2010 (UTC)

I'd have thought it would be more like the cube of the chance of two numbers being coprime considering the 3 possibilities of being coprime, of course that leaves a lot out! Dmcq (talk) 12:57, 9 January 2010 (UTC)

If the density (=limit probability) of the set of triples of pairwise coprime natural numbers is what you suggested above, that is $\scriptstyle \prod _{p}(1-3p^{-2}+2p^{-3}),$ that I trust very much, then it can't be 0. An infinite product of non-vanishing terms of the form $\scriptstyle \prod _{k}(1+a_{k})$ with $\scriptstyle \sum _{k}|a_{k}|<\infty$ always converges to a non-zero limit. --pma 15:33, 9 January 2010 (UTC)

And, since for all prime $p$ it is true that $\scriptstyle 1-3p^{-2}+2p^{-3}\geq (1-p^{-2})^{3}$, Julzes' formula is also consistent with Dmcq's bound, for $\scriptstyle \prod _{p}(1-3p^{-2}+2p^{-3})\geq \prod _{p}(1-p^{-2})^{3}=6^{3}/\pi ^{6}=0.22...$ --pma 16:20, 9 January 2010 (UTC)

The argument is sound, my program was not (apparently), and I've been busy. Let me try again with a calculation.Julzes (talk) 16:44, 9 January 2010 (UTC)

Okay, I figured out what I was expecting from PARI/GP wasn't working and, more or less why. The result I got from the first 500000 primes using rounding at twenty digits is $0.28674743539888982872\,$.Julzes (talk) 17:31, 9 January 2010 (UTC)

I'd call that 0.2867474354 and guess it's accurate that far.Julzes (talk) 17:33, 9 January 2010 (UTC)

I agree with your product formula (the finite probabilities indeed do converge to the infinite product by a dominated convergence argument). The Plouffe's inverter [2] seems to have nothing special about the value 0.286747.., but Google gives [3] and this [4] (pag.11). But there should be for sure some more classical result. --pma 18:27, 9 January 2010 (UTC)

So, there is (almost certainly) a closed form given at the bottom of that first Google result. I imagine the other--later--source says more (but I can't get it right now).Julzes (talk) 18:55, 9 January 2010 (UTC)

But the value of that expression is 0.286747419846(..), possibly found by an inverse calculator; it's close but seems different from the value you computed. An interesting remark there, is that the product formula easily generalizes to the case of n numbers. The other paper has no relevant information.--pma 20:37, 9 January 2010 (UTC)

I'm a little skeptical of both my precision and the idea that a nice coincidence of something looking reasonable times Apery's constant was found by inverse calculator in this instance. I'm going to explore the question more precisely.Julzes (talk) 03:54, 10 January 2010 (UTC)

Oh, I don't know. It looks to be moving solidly toward a higher degree of precision with more factors included, but the source already said it was close and it doesn't look likely to be true on paper. I'm inclined to reverse any skepticism I had that they were merely close. It probably was done by reverse calculator. It's simple, but doesn't look simple enough.Julzes (talk) 06:27, 10 January 2010 (UTC)

What is the probability that four, five, or more randomly chosen integers are pairwise coprime? --84.61.151.145 (talk) 10:06, 10 January 2010 (UTC)

The same counting argument that gives Julzes' product formula applies; as the first link above remarks, for (n+1) pairwise coprime numbers it's $\scriptstyle \prod _{p}(1-1/p)^{n}(1+n/p).$ (note: We should speak of natural density rather than probability, for there is no uniform probability on $\scriptstyle \mathbb {N}$).--pma 13:04, 10 January 2010 (UTC)

$0.286747428437\,$ is accurate based upon 500000000 primes. The equality proposed in the link does not hold.Julzes (talk) 10:01, 14 January 2010 (UTC)

calculating pi

I would like to know whether value of pi could be calculated in this way.

First let us assume, there are pi for all shapes like, regular triangle (3_pi), square (4_pi), pentagon (5_pi), hexagon (6_pi), .. n-gon (n_pi) and, these values can be calculated from their perimeter,

 2 * 3_pi * r = 3 * a
 2 * 4_pi * r = 4 * a
 2 * 5_pi * r = 5 * a
 2 * 6_pi * r = 6 * a
 2 * 7_pi * r = 7 * a
 2 * 8_pi * r = 8 * a
 2 * 9_pi * r = 9 * a
 ....................
 ....................
 ....................
 2 * n_pi * r = n * a

Let,

 r be the radius of circle you draw through all the corners of the regular polygons mentioned above
 a be one side of a regular polygon (this can be ignored, it is equal to 1, for simplicity)

As we know values of a, and r, we can easily calculate corresponding 'n_pi' values.

Now by considering a circle as a regular polygon, with infinitly big number of sides, Is it possible to extrapolate the data what we got from first few polygons (3_pi, 4_pi, etc.), and calculate n_pi ?!

Anything wrong with this process ? Anyone already tried like this before ? --V4vijayakumar (talk) 14:14, 9 January 2010 (UTC)

I think the problem is you'll just end up proving π == π. I.e. the formula for the boundary is N tan (π / N) for a regular polygon with N sides around a circle radius 1. As N -> ∞, π / N -> 0 and tan (π / N) -> π / N. So the length of the boundary tends to π. --JohnBlackburne^words_deeds 16:03, 9 January 2010 (UTC)

"I think the problem is you'll just end up proving π == π." No, we don't know 3_pi, 4_pi, ... n_pi etc. values yet. —Preceding unsigned comment added by V4vijayakumar (talk • contribs) 06:58, 10 January 2010 (UTC)

Archimedes did something very similar to this over 2000 years ago (see Method of exhaustion), with a few differences:

• Rather than using perimeters, he used areas.
• Rather than a limiting process, he found bounds by inscribing and circumscribing a polygon.
• Rather than considering a general n (which, as JohnBlackburne points out, leads nowhere), he considered $n=3\cdot 2^{k}$, starting with known values and applying the ancient equivalent of the half-angle trigonometric formulae. Once the geometric argument is extracted into pure arithmetic, we end up with a simple calculation involving square roots.

-- Meni Rosenfeld (talk) 16:22, 9 January 2010 (UTC)

Also, François Viète considered the 2ⁿ-gone to produce an infinite product for π (possibly the first infinite product in the history of maths). --pma 16:37, 9 January 2010 (UTC)

Archimedes died around year 212 BC, which is now 2221 years ago! Bo Jacoby (talk) 07:51, 10 January 2010 (UTC).

...which is 'over 2000 years ago', as Meni Rosenfeld points out. Simple mathematics! --KageTora - (影虎) (Talk?) 16:19, 10 January 2010 (UTC)

Standard deviation of 4 numbers

For 4 numbers, is it ALWAYS true that the 2nd largest number lies in 1 standard deviation? Mathematically, for 4 numbers a, b, c and d where a>b>c>d, is b≤μ+σ ALWAYS true? —Preceding unsigned comment added by 219.79.198.73 (talk) 15:43, 9 January 2010 (UTC)

I think your inequality wants to be μ − σ ≤ b ≤ μ + σ, where μ is the arithmetic mean and σ is the standard deviation. ~~ Dr Dec (Talk) ~~ 16:52, 9 January 2010 (UTC)

No, I think the OP has it right. Your inequality is obviously false in general, consider the set {1,5,5,9}. --Tango (talk) 17:07, 9 January 2010 (UTC) You changed it while I typing! --Tango (talk) 17:09, 9 January 2010 (UTC)

You must be slow at typing ;oP (I changed it at 16:52^[5] and you made your first post at 17:07^[6]). ~~ Dr Dec (Talk) ~~ 14:10, 10 January 2010 (UTC)

Yes, this is true. Hint: Prove by contradiction. If $b>\mu +\sigma$, then $(c+d)/2<\mu -\sigma$. -- Meni Rosenfeld (talk) 16:41, 9 January 2010 (UTC)

To me the phrase "2nd largest number lies in 1 standard deviation" looks as if it means c ≤ μ + σ. That's a somewhat subtler question. It's obvious that b and c cannot both lie outside of the interval [μ − σ, μ + σ], since then the average of the squares of deviations would be smaller than all four of the squares of deviations—everyone would be above average. So at least one of the middle numbers lies within the interval. The question is whether both of them must. Michael Hardy (talk) 03:10, 13 January 2010 (UTC)

No, b is the second largest number, and it must have $b\leq \mu +\sigma$, by the proof I outlined. c is smaller so obviously $c\leq \mu +\sigma$. By negating the numbers you get $\mu -\sigma \leq c\leq b$. So $b,c\in [\mu -\sigma ,\mu +\sigma ]$. -- Meni Rosenfeld (talk) 09:30, 13 January 2010 (UTC)

Sorry---I hadn't noticed that they were in decreasing order rather than increasing. Michael Hardy (talk) 12:49, 13 January 2010 (UTC)

I think a proof is called for here even if the result is obvious because of the way standard deviations work.

Without loss of generality let the mean be 0

Suppose a,b > σ

Then

a²+b² > 2σ²

a²+b²+c²+d² = 4σ²

c²+d² < 2σ²

c+d = −(a+b)

c+d < −2σ

(c+d)² > 4σ²

c²+d²+2cd > 4σ²

2cd > 2σ²

2cd > c²+d²

0 > c²+d²−2cd

0 > (c−d)²

i.e. the assumption leads to a square being less than zero

Therefore we can't have a,b > σ with mean 0 or in general we can't have a,b > μ+σ Dmcq (talk) 11:15, 13 January 2010 (UTC)

The target audience here is unclear. If you understand what "without loss of generality" means, you don't need to have the proof spelled out this way. -- Meni Rosenfeld (talk) 11:46, 13 January 2010 (UTC)

I just thought if Michael Hardy was asking a question about your hint then it wasn't obvious and deserved a full answer. And it isn't actually immediately obvious how to go about it, I guess I did put in too many steps, it takes more time and effort to make things shorter and simpler and it seemed adequate. Dmcq (talk) 13:04, 13 January 2010 (UTC)

What Michael was confused about (as I originally suspected, and as he now verified) was that the OP wrote $a>b>c>d$ rather than $d<c<b<a$. -- Meni Rosenfeld (talk) 13:46, 13 January 2010 (UTC)

An easier proof using more statistics might be:

The mean of a and b is more than 2σ from the mean of c and d

Therefore the total variance var(a,b)+var(c,d)+4((mean(a,b)−mean(c,d))/2)² is greater than 4σ², which is a contradiction. QED or whatever Dmcq (talk) 13:23, 13 January 2010 (UTC)

I worked this out by a brute-force method involving trivial but opaque algebra and thought there must also be an intelligent way to do it. Now I know what that is and it seems as if I should have seen this right away. Maybe this is what Dmcq had in mind. Split {a,b,c,d} into an upper part {a,b} and a lower part {c,d}. The average of the upper part is

r = (a + b)/2

and the average of the lower part is

s = (c + d)/2.

The mean μ of {r,s} is the same as the mean of {a,b,c,d}. But the standard deviation τ of {r,s} is smaller than the standard deviation σ of {a,b,c,d}, since the variance of {r,s} doesn't take into account the variability within the upper part or within the lower part (see law of total variance). Now observe that the interval whose endpoints are μ ± τ is just the interval [s, r], and that interval contains the two middle values b and c. Therefore the longer interval whose endpoints are μ ± σ must also contain those two middle values. Michael Hardy (talk) 04:38, 15 January 2010 (UTC)