Wikipedia:Reference desk/Archives/Mathematics/2009 August 23

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August 23

Pi

In the same vein as my question above on significant figures, how can we say that pi extends indefinitely if it is the answer to a division problem, named 22/7, in which the LCD of significant figures is 1? Is it because we are not measuring anything with the 22 and the 7, and we really therefor mean 22.000000000000000000000000000000000000000000000000000/7.0000000000000000000000000000000000000000000, but of course with the zeros continuing forever? DRosenbach ^{(Talk | Contribs)} 02:18, 23 August 2009 (UTC)

π ≠ 22/7 Staecker (talk) 02:21, 23 August 2009 (UTC)

So how do we know what pi is other than dividing a circumference by a diameter, neither of which we are able to measure to an infinite number of decimal places? DRosenbach ^{(Talk | Contribs)} 02:56, 23 August 2009 (UTC)

Pi also can be derived with a variety of other techniques. The most common ones that come to mind are some easy integrals, which are part of elementary calculus. You can read pi, and Proof that π is irrational. I think you might be stuck on the "repeating decimals" thing - you have to recognize that irrational number is a more subtle mathematical concept than just the number of decimal-places. The fact that the digits never repeat is an artifact of the way we represent numbers with decimal place-values, but in reality, the concept of irrational numbers is much more precisely defined than the non-repeating decimal representation. There is no number system (not binary, not octal, not rational numbers, nothing)... which can express the exact value of an irrational number; it can not be exactly bound by two other numbers greater and less than it (because for any arbitrarily sized distance, there are smaller numbers which make a better bound). There's a lot of subtle advanced mathematics here - the best place to start would be to really read and understand our irrational number and real number articles. That being said, we can both prove that pi is irrational (meaning we can not find an exact value for it); and at the same time, we have lots of techniques and algorithms to approximate its value to any desired level of precision (if we are willing to work out the math to that accuracy). In any case, 22/7 is a terribly inaccurate approximation - they differ by more than 0.001 (which is a lot!) Nimur (talk) 03:02, 23 August 2009 (UTC)

A small nitpick: there are number systems in which pi can be expressed exactly. Base pi is the obvious one. Not that I would ever want to use base pi for anything. Rckrone (talk) 05:53, 23 August 2009 (UTC)

Rckrone, the nitpick wasn't very constructive. I think Nimur gave a very nice reply to the earlier post. I have only ever come across number systems to integer bases. Talking about base π seems a little wishy-washy to me. What's the expansion of 7 in base π? It would be irrational! Infact every number except multiples of π would be irrational. Would it even be well defined? ~~ Dr Dec (Talk) ~~ 17:06, 23 August 2009 (UTC)

Some irrational bases are perfectly sound—Golden ratio base comes to mind. But π is a quite different matter, of course. —JAO • T • C 17:28, 23 August 2009 (UTC)

Looking at the Golden ratio base article, it seems to support what I was saying. The arcticle says that 11_φ = 100_φ. So in this base 11 = 100. So there's no uniqueness without talking a standard form. ~~ Dr Dec (Talk) ~~ 18:47, 23 August 2009 (UTC)

A standard convention for eliminating these duplicates is to insist that there are no consecutive 1's. -- Meni Rosenfeld (talk) 19:29, 23 August 2009 (UTC)

That is no worse than base 10 where, for example, 0.999...=1. This kind of notation is usually non-unique, regardless of the base. --Tango (talk) 19:39, 23 August 2009 (UTC)

But the article also says that golden base ratio has this problem too, i.e. equalities along the lines of 0.999... = 1. It seems that every choice of base gives this same problem with infinite decimal expansions. But what base 10 doesn't say is that 11 = 100. ~~ Dr Dec (Talk) ~~ 12:30, 24 August 2009 (UTC)

Yes, this is topological. Infinite strings from a finite alphabet are homeomorphic to Cantor space and therefore totally disconnected (indeed, zero-dimensional), whereas the reals are connected and one-dimensional.

On another note, there is, of course, an exact decimal representation of pi. It happens to contain infinitely many digits. But that's nothing special. Even the decimal representation of the real number zero has infinitely many digits. It just so happens that they're all 0. --Trovatore (talk) 20:57, 24 August 2009 (UTC)

Dr. Dec, I'm sure you know this but "irrational" means not a ratio of integers, which is a property of the number rather than its representation. 7 is rational whether you express it in base 10, 2 or π - "having no repeating expansion in base π" is the correct term.

Now, whether Rckrone's comment is central to the discussion is irrelevant - Nimur has made a factually incorrect statement, and correcting it is appropriate. -- Meni Rosenfeld (talk) 19:29, 23 August 2009 (UTC)

Meni Rosenfeld, number theory isn't really my thing so you're going to have to help me out. I'm getting a bit confussed. What is π in base-π? Well one representation would be that π_π = 1. Now π₁₀ has a non-repeating decimal expansion where as π_π has a repeating decimal expansion. Now π is irrational base 10, but π is rational base π. ~~ Dr Dec (Talk) ~~ 12:39, 24 August 2009 (UTC)

There is no such thing as "rational in base 10" or "rational in base π". A real number is either rational or not. From Rational number:

"a rational number is any number that can be expressed as the quotient a/b of two integers, with the denominator b not equal to zero".

The number 7 can be expressed as the quotient 7/1, where 7 and 1 are integers. Therefore, 7 is a rational. It doesn't matter at all how you choose to represent the number. You can represent it as "7", "3+4" or "$111_{2}$" - it is the same number, and it is rational.

As it happens, there is a theorem that says that a rational number has a repeating expansion in any fixed integer base numeral system. However, the word "rational" does not mean "has a repeating expansion", and the distinction becomes important when we discuss non-integer bases.

So the number 7 has an expansion in base π which starts with "20.202112..." and happens to be non-repeating. But the number 7 is still rational.

Similarly π has a base-π expansion "10", which is repeating (terminating, even). But that does not make it any more rational, since it is still not a ratio of integers.

Anyway, you have also mixed up your notation - digits_base means "the number whose expansion in the given base consists of the given digits". So 10_π = π rather than π_π = 10. -- Meni Rosenfeld (talk) 13:22, 24 August 2009 (UTC)

Meni Rosenfeld, I've already confessed that number theory isn't my thing, so I would ask you to kindly tone down your rhetoric a couple of notches, and to show patience and good humour. We have that π₁₀ = 10_π. Now 10 is an integer and so a rational number. Now 10 is a rational number, but π₁₀ = 10_π, so why isn't π₁₀ writen to the base π a rational number? ~~ Dr Dec (Talk) ~~ 13:46, 24 August 2009 (UTC)

I guess you need to say that a number is rational when it can be writen as the quotient of two integer, base-10. Like we have seen: the rational number 7₁₀ written in base-π gives a number that would be irrational if it were to base-10. Similarily the irrational number π₁₀ written in base-π gives a number that would be rational if it were to base-10. ~~ Dr Dec (Talk) ~~ 13:53, 24 August 2009 (UTC)

I'm not sure what you mean by "tone down my rhetoric". I'm using whatever rhetoric I think would be effective, while remaining civil.

Please read User:Meni Rosenfeld/Numbers (work in progress), and then the rest of my post.

"π₁₀" doesn't make much sense. The subscript "10" means that we use the decimal system to represent a number with a sequence of decimal digits. "π" is not a sequence of decimal digits, it is a Greek letter. This letter denotes the number π in a way which does not involve any radix system.

Now, 10 is an integer, not "10". 10 is the number of slashes in "//////////", and is an integer with any conventional way of defining integers. "10" is a string consisting of the symbol "1" followed by the symbol "0", which can mean anything at all. Usually this string is interpreted using the decimal system, in which case it means the number 10. But the string itself is not a number and certainly not an integer (unless we choose to embed all finite ASCII strings as integers, but that is beside the point). 10 is of course also a rational number, while "10" is not.

As you have noted, the representation of the number π in base π is the string "10". But the fact that the string "10" denotes a rational number in some other system doesn't say anything about the rationality of the number it denotes in base π. And indeed, we know that the latter is π, which is an irrational number.

A rational number is the ratio of two integers. I can represent these integers in decimal, in Roman numerals, in Gematria, in base π or whatever - they are the same numbers in any case.

"the rational number 7₁₀ written in base-π" does not give a number, it gives a string of symbols (an infinite string). In theory we can take this string and interpret it in base 10, resulting in a number which is indeed irrational. But doing this doesn't make any sense. -- Meni Rosenfeld (talk) 20:50, 24 August 2009 (UTC)

I didn't see anything offensive in Meni's language. —Tamfang (talk) 17:05, 27 August 2009 (UTC)

I apologize; I should not have said "there is no number system..." which can represent pi. What I meant to say is, there is no number system which can represent all irrational numbers exactly. If you switch to a number system based on an irrational number, you will be mapping a different subset of real numbers into rationals and irrationals; but there will still not be an exact representation for all real numbers. These are moot points though - I shouldn't have made such a strong statement about pi. In any case, the OP is clearly unaware of the concept of rational and irrational numbers - so let's try to phrase our responses to help him/her understand that, before diving into such subtleties. Nimur (talk) 21:55, 23 August 2009 (UTC)

We can't measure a diameter and a circumference with infinite precision, true; we also can't know π with infinite precision. To put it another way, we can (in principle) measure a circle to as many digits as we want, and compute π to as many digits as we want. —Tamfang (talk) 17:05, 27 August 2009 (UTC)

DrDec: "Rational" does not mean rational relative to some base. "Rational" means a ratio of two integers. Thus 22/7 is a rational number. What base you express numbers is has absolutely nothing to do with it. Michael Hardy (talk) 23:26, 24 August 2009 (UTC)

Michael you're not being very clear. There's no point being curt and just barking a definition at me without giving an explanation. The integer article does not specify a base, and says that 10 is an integer. When you talk about integers, you seem to mean those numbers with integer values when expressed to base 10. So the integers base 10 seem to hold a special place in our arithmetic. I asked this question earlier, but didn't get a reply. If not, well then isn't 10_π an integer? If so then 10_π/1_π would be a rational number. But hang on: π₁₀ = 10_π/1_π. Any real number x can be expressed as an integer, since 10 is a (baseless) "integer" and x₁₀ = 10_x. Once again, I would appreciate patience and good humour in any following posts. ~~ Dr Dec (Talk) ~~ 11:09, 25 August 2009 (UTC)

The integers are the closure under addition and negation of {1}. Just as with rationals, no mention of bases. --COVIZAPIBETEFOKY (talk) 12:34, 25 August 2009 (UTC)

People keep saying the same thing without ever actually addressing my question. I think we do need a base. Why do we say that 10₁₀ is an integer, but 10_π is not an integer? Base 10 holds a special place. ~~ Dr Dec (Talk) ~~ 12:48, 25 August 2009 (UTC)

I think you have a very good point here: even the introduction of our Integer article seems to say that the concept "integer" can be defined as a number that can be written without a period. As you have discovered, this definition only works in base 10, 2, 16, etc—or in other words, only in bases which are, themselves, integers. A much better definition is "an integer is an element of Z", after Z has been constructed as in Integer#Construction (with the natural numbers given by the Peano axioms). Or more loosely but more intuitively, n is an integer iff at least one of the numbers n and −n is a natural number/a finite cardinal/a finite ordinal. Regardless of whether we write π as 3.14159...₁₀, as 10_π, or just as the symbol π (with no need for a base), we cannot create a set with that many elements, and that's why it's not an integer. (By the way, I apologize if my earlier answers have offended you, and I hope you'll find this one better.) —JAO • T • C 13:07, 25 August 2009 (UTC)

Are you familiar with the constructions of natural numbers, integers, rational numbers and real numbers? Very few of these constructions make any mention of base representations. A base-ten representation of a number may look pretty, but, really, it's a rather clunky representation to go by for the purposes of mathematics (partly because it contains multiple representations for the same number, and partly because the definitions of addition and multiplication aren't exactly trivial). The integers are not defined by their decimal representation.

If you have the complete set of real numbers without having taken any of the intermediate steps in defining natural numbers, integers and rationals, you can still produce the set of all integers, as I said above, as the smallest possible closure of the set {1} (where 1 is the multiplicative identity) under addition and negation. --COVIZAPIBETEFOKY (talk) 13:11, 25 August 2009 (UTC)

Jao, thanks for your reply. The construction of the integers starts of by saying that they are equivalence classes of pairs of natural numbers. And this is okay because I understand what COVIZAPIBETEFOKY is saying about the integers being generated by {1}. Since 1_x = 1_y for all real numbers x and y. And indeed the closure of {1_n} under addition and negation, where n is an integer, gives the set of integers; but there's the problem - I'm going around in circles. We need to know what an integer is, before we can understand what an integer is. I know I sound insane, and I'm getting quite worried. But, for example, what does 10_π actually mean?

$10_{\pi _{10_{\ddots }}}\ ,$

What base is the base in? And what base is the base of the base in? We have to take base 10 as a starting point and move on. Otherwise I find myself going round and around in circles. I guess that's the problem with expressing the integers in any base. It reminds me of relativity, and there being no idea of absolute space or absolute time; only inertial frames. I'm going for a lie down...~~ Dr Dec (Talk) ~~ 13:29, 25 August 2009 (UTC)

As for that specific example, remember that "π" is not the base-10 representation of the number π; the base-10 representation of π is "3.14159...". π is just a symbol whose value does not depend on base—just like the symbols 1, 2 and 5 denote the same numbers in octal, decimal or hexadecimal. Base dependence only comes into play when two or more such symbols are juxtaposed. But with another example, your remark is quite apt. When we write "30A6₁₆", for instance, we assume the subscript to be in decimal, and trying to explicitly state this by subscripting another "10" leads nowhere. You can go around this, of course, by writing the subscript as "ten", "dec" or even "5+5", all of which always mean the number ten, which "10" does not necessarily do. (The convention that numbers are in decimal unless otherwise stated is pretty useful.) —JAO • T • C 17:07, 25 August 2009 (UTC)

You make a good point about π! As for the point about bases, it shows that base 10 does indeed play a special role. We always assume that numbers are base 10, and then within this framework we can talk about other bases (base 10). To define integers and the like without any mention of a base requires more sophisticated machinery. It's just like coordinate geometry and coordinate-free geometry. Jao, thanks for your patience :o) ~~ Dr Dec (Talk) ~~ 18:24, 25 August 2009 (UTC)

No, base 10 is not "special", its just customary. You confuse numbers with their representation in a certain writing system. 10100₂ is the same number as 4₁₀, and is (e.g.) the cardinality of the set S={{{{{}}}}, {{{}}}, {{}}, {}}. We can specify or construct (many) numbers and all integers without resorting to a certain base X notation. --Stephan Schulz (talk) 18:36, 25 August 2009 (UTC)

Of course, you meant "100₂". -- Meni Rosenfeld (talk) 19:11, 25 August 2009 (UTC)

Well, of course I meant "10₄" ;-). --Stephan Schulz (talk) 19:47, 25 August 2009 (UTC)

Dr Dec - I have already addressed all of your concerns. Please read User:Meni Rosenfeld/Numbers like I asked you. If you missed the post in which I asked this, please go over this thread to locate it. I'm sure your confusion will be dispelled if you do, or at least we will be able to intelligently discuss this. -- Meni Rosenfeld (talk) 18:02, 25 August 2009 (UTC)

Meni, your rhetoric was quite strong. I felt like you were speaking down to me and trying to belittle me. Just stop and read what you've written. Phrases like "As it happens, there is a theorem...", "Anyway, you have also mixed up your notation" and "I have already addressed all of your concerns" and quite strong, and could be written, IMHO, in a much more friendly way. (Although I can be quite sensative.) Many mathematicians – and I have sometimes been guilty of this myself – seem to suffer from a superiority complex. There is no need for it. I'm sure that there are areas of mathematics where our levels of knowledge are reversed, and I would take great pleasure in fostering your interest in those areas by being patient, warm, helpful, and respectful. ~~ Dr Dec (Talk) ~~ 18:24, 25 August 2009 (UTC)

Frankly, I think you really are too sensitive. I completely agree that we should be friendly and respectful, but I still don't see where I failed that.

I have no doubt that there are areas of mathematics where our levels of knowledge are reversed - but here, in this discussion, you have made several statements which reflect very fundamental misunderstandings about the subject matter. These misunderstandings don't resolve themselves, and the more deeply they are rooted, the stronger the rhetoric required.

The "I have already addressed all of your concerns" comment was inevitable after I explained that numbers are distinct from their representations, that you do not need a base to represent numbers, that "π₁₀" doesn't make sense, and what the "10" in the subscript of "7₁₀" means -- and yet you continued to repeat the same mistakes. -- Meni Rosenfeld (talk) 19:09, 25 August 2009 (UTC)

But that oversight on my part exemplifies my point: I took π as a given, i.e. 3.14159265... = π. Most people always assume that the numbers are given to base 10. I agree that numbers exist independently of their representations. My problem seems to lie with the representations themselves. I have read you user page section on numbers and it does help a little. I am begining to think that my problem is not (thankfully) with the idea of integers and natural numbers, but with the notation used to give representations of these numbers. I have learned something over the last few days; but it has been quite painful. I'm sure people could have been more gentle. Anyway... ~~ Dr Dec (Talk) ~~ 19:20, 25 August 2009 (UTC)

Perhaps you will be interested in the last post here, where I try to define what a decimal expansion really is. If you're interested in more general radix representations, you can read pages such as Numeral system, Non-standard positional numeral systems and those linked by them. -- Meni Rosenfeld (talk) 19:46, 25 August 2009 (UTC)

DRosenbach: See:

• Computing π
• Bailey–Borwein–Plouffe formula
• Numerical approximations of π
• Chronology of computation of π
• Leibniz formula for pi
• Liu Hui's π algorithm
• Proof that π is irrational
• Proof that π is transcendental
• Gauss–Legendre algorithm
• Software for calculating π
• Viète's formula
• Wallis product
• Borwein's algorithm
• List of topics related to π

Michael Hardy (talk) 23:39, 24 August 2009 (UTC)

Lagrangian is invariant

What does that mean? Let's for simplicity take a system where the Lagrangian L is invariant under U(1) (which I understand contains the complex numbers with absolute value 1). But invariance can not mean -L = L. 93.132.179.71 (talk) 07:58, 23 August 2009 (UTC)

Just to fix the vocaboulary: if $f:X\to Y$ is any function (usually with $Y=\mathbb {R}$) and if G is a group with an action on $X$, one says that $f$ is invariant for the action of G, or G-invariant, iff for all $x\in X$ and $\gamma \in G$ there holds $f(\gamma \cdot x)=f(x)$. One also says that G is a group of symmetries of f. So in any case there's no "-L=L". Maybe you have in mind a situation where there is a G action on Y too, (usually, but not necessarily, the same action, with X=Y), and $f$ satisfies $f(\gamma \cdot x)=\gamma \cdot f(x)$ for all $\textstyle x$ and $\textstyle \gamma$; in this case one says that $f$ is equivariant wrto the two G actions, or G-equivariant. --pma (talk) 10:19, 23 August 2009 (UTC)

Thank you, that excludes some of the possibility for misinterpretation. So f is constant on the equivalence classes of X induced by G. Now with the Lagrangian itself operating on functions I'm still not clear if in this case G acts on these functions (as the domain of L) or on the domain of each of these functions. 93.132.179.71 (talk) 13:26, 23 August 2009 (UTC)

(By the way, the equivalence classes induced by a group action have a special term, "orbits", for those days when you get tired of writing "equivalence class induced by the group action". Eric. 98.207.86.2 (talk) 21:01, 23 August 2009 (UTC))

Proof?

Hi, indeed I mostly reply questions at Arabic Wikipedia, and one question stopped me where I couldn't think where to classify (regardless whether or no being a homework). I'll try to translate it in English symbols as follows: For any natural numbers n, k; Prove that

$3n^{2}+3n+7\neq k^{3}$

always holds.

I guessed it can be done by mathematical induction but failed. This is the original question in Arabic. Thanks in advance, --Email4mobile (talk) 11:11, 23 August 2009 (UTC)

These problems are usually best done with modular arithmetic. Take the equation mod 6 ( $\scriptstyle n(n+1)$ is even), what does this tell you about $\scriptstyle k$ mod 6? If $\scriptstyle k$ is $\scriptstyle m$ mod 6 then you have $\scriptstyle 3n(n+1)+6+1=(m+6r)^{3}=m^{3}+18m^{2}r+108mr^{2}+216r^{3}$ for some $\scriptstyle r$... 129.67.186.200 (talk) 11:40, 23 August 2009 (UTC)

Taking the equation mod 6 and probing for the possible values of x^3 mod 6 shows k = 1 mod 6. But I can't see any use of this fact. 93.132.179.71 (talk) 13:31, 23 August 2009 (UTC)

Me neither. Maybe there is an elementary reason, maybe not so elementary; for instance the classical diophantine equation $\scriptstyle x^{2}+2=y^{3}$ is studied factorizing the LHS in $\scriptstyle \mathbb {Z} [{\sqrt {-2}}]$ (so here one would work in $\scriptstyle \mathbb {Z} [{\sqrt {-3}}]$ instead). --pma (talk) 13:51, 23 August 2009 (UTC)

Once you know that $\scriptstyle k=1+6r$, you get $\scriptstyle 3n(n+1)+6+1=1+18r+108r^{2}+216r^{3}$. This gives $\scriptstyle n(n+1)+2=6r(1+6r+12r^{2})$, but $\scriptstyle n(n+1)+2$ has no roots mod 6. 129.67.186.200 (talk) 15:38, 23 August 2009 (UTC)

I'm not very good with number theory: it's not my thing. I've given this a once over and here are my thoughts. The LHS $3n^{2}+3n+7$ is always congruent to 1 modulo 6. That means that $3n^{2}+3n+7$ can always of the form $6m+1$ for some integer m. This has reduced the problem from a quadratic to a linear problem. Why not try more modular arithmetic on this linear expression. ~~ Dr Dec (Talk) ~~ 17:19, 23 August 2009 (UTC)

p.s. I've just tested it for $0\leq n\leq 1,000,000$ and it holds. ~~ Dr Dec (Talk) ~~ 17:30, 23 August 2009 (UTC)

Yes, seeing that it's 6m + 1 is certainly the first step. Seeing why m can't be 57 (or any other number making 6m + 1 a cube) is the second. (This is what 129.67.186.200 did two hours ago.) —JAO • T • C 17:49, 23 August 2009 (UTC)

Really? If he did then it's very convoluted. I didn't see that he said that. If I didn't see that then I'm sure the person making the original post didn't see it either. Jao, please do me a favour and calm down. From this post, and earlier ones, you seem to be quite confrontational. So what if someone said something earlier? Does it really matter? Is there any need to make a sly comment? No! ~~ Dr Dec (Talk) ~~ 18:09, 23 August 2009 (UTC)

Hi, 129 here, I don't think my solution is convoluted. All I did was to observe 1) the LHS is 1 mod 6 so k must be 1 mod 6 (I left this as a hint, 93 spelled it out) 2) knowing this you are lead to the equation n(n+1)+2 = 0 mod 6 which is easily seen to have no solutions (in my 15:38 post). Jao is only saying (I think!) that you should read the thread carefully before you reply, it gets confusing otherwise. Anyway, the problem is done. 87.194.213.98 (talk) 19:33, 23 August 2009 (UTC)

I did read the thread carefully, it wasn't very long. You gave very little explanation and just wrote a lot of maths. I found that convoluted. Clearly, you wouldn't think that your solution is convoluted: you wrote it! People ask questions because they don't understand the topic. There's no point writing a reply that assumes they have the required knowledge to solve the problem. If they did then they would have solved it, and if they'd have solved it then they wouldn't be asking the question. And whatever Jao was trying to say, he didn't say it in a very nice way. Scroll up and read more of his posts, he's not the most civil or polite. ~~ Dr Dec (Talk) ~~ 20:54, 23 August 2009 (UTC)

Injecting myself into this conversation, I'd like to say that Jao has always been civil in the lengthy period of time I've seen him contributing to the RD. I can sort of see how Jao's comment in the unit normal discussion appears sharp, but that can't possibly be a big deal, can it? Eric. 98.207.86.2 (talk) 21:14, 23 August 2009 (UTC)

I found Jao's remark a gentleman post. Very opportune, too. --pma (talk) 06:57, 24 August 2009 (UTC)

Added emphasis to "appears". Eric. 98.207.86.2 (talk) 07:53, 25 August 2009 (UTC)

PS: I just meant to agree with your remark, it was no objection --pma (talk) 15:37, 26 August 2009 (UTC)

Ah, cool. Eric. 216.27.191.178 (talk) 19:36, 26 August 2009 (UTC)

I'd like to thank you all for you wonderful cooperation and interaction. To tell you the truth, I feel such proofs are totally new to me because I didn't study about "modular arithmetic"; although I think the one who asked this in Arabic Wikipedia must have studied it unless there's another way to treat such a problem.--Email4mobile (talk) 18:27, 23 August 2009 (UTC)

Well, 129's proof is clever and clear. It reduces to an elementary fact (i.e., n(n+1)+2 is never divisible by 6) that one can easily check. Most likely you are able to understand it perfectly, but of course, if any point is still not clear to you, I guess you are welcome to ask for further details. --pma (talk) 06:57, 24 August 2009 (UTC)

I started reading about modular arithmetic in order to learn more, and I hope that you give me some more recommendations to study this subject. --Email4mobile (talk) 10:14, 24 August 2009 (UTC)

Average

In a philosophically mathematical manner, is it ridiculous to ask if "the average person is average"? I suppose there could be some term switching in that both averages would not refer to the same method of obtaining an average (e.g. mean vs. median), but that would sort of be a fallacy of four terms. My sister-in-law asked me this question, and at first I thought it was insightful, but shortly thereafter figured that it's probably not that insightful at all. Any comments? DRosenbach ^{(Talk | Contribs)} 13:38, 23 August 2009 (UTC)

I'd say not; I guess it can be translated in a more mathematical language as asking whether the variance is small. An average gives only a first description of a distribution; then you can ask how close the individuals are to the average value. For instance, in a society of clones, everybody is very close to the average person (and I'm trying to recall a sentence in a novel by Oscar Wilde, asserting that the typical Englishman is not typical...)--pma (talk) 14:16, 23 August 2009 (UTC)

If "the average person" refers to someone chosen at random, there is a 1 in 2 chance that he is not average because only 50 of the population placed on a bell curve would be considered average. So perhaps it's not such a bad statement after all. DRosenbach ^{(Talk | Contribs)} 14:36, 23 August 2009 (UTC)

I've never heard of "average" meaning "random" or "in the middle 50%". --Tango (talk) 19:43, 23 August 2009 (UTC)

Another interpretation would be to relate this to ergodicity. (My understanding of ergodic theory isn't great, so please correct me if this makes no sense). For instance, if you were interested in which parks in a city were well-liked by people, you could can do two things. You can look at a lot of people in a single instant and see which parks are busy and which are empty. You can also follow a single person for a year, and see which parks the person frequents.

These method will likely give very different results, whatever person you choose. This means that even if you choose the "most average" person in the city (by whatever definition) his path (the parks he visits) will still diverge from the average. In that sense human beings have a tendency to be non-average (which, I think is more subtle than just having a high variance, although that may be an effect). risk (talk)

A game with positions expressable as vectors

I'm looking for a reasonably complex game of perfect information, where each position (ie. the state of the game after a player has made her move) can be fully expressed as an n-dimensional vector in a straightforward manner. I'm sure that the board-positions of Chess and Go can be encoded in this way, but any method I can think of is rather convoluted, 'hiding' the relevant information, or leads to a massive, very empty space.

Can anybody think of a game that fits this description? I'm sure such a game would be very useful in Game Theory (relating game theory to geometry), so I figured someone might have invented one just for that purpose. risk (talk) 16:02, 23 August 2009 (UTC)

A very simple example would be naughts and crosses. There are nine positions (represent this by $\{0,\ldots ,8\}=\mathbb {Z} _{9}$), and each place can have a naught, a cross, or be empty (represent this by $\{0,1,2\}=\mathbb {Z} _{3}$). So any game would be a vector in $\mathbb {Z} _{9}\times \mathbb {Z} _{3}.$ Alternatively, a game could be given by a mapping $f:\mathbb {Z} _{9}\to \mathbb {Z} _{3}.$ Of course not all of these vectors and maps would represent a valid game: for example, you couldn't have a cross in all nine positions. As to your problem about big empty spaces. The layout of a game has myriad posibilities, and so the space of all game states would be huge – if not infinite. So you would only ever have a single vector for a given game state sat in a massive space of game states. I've thought about something similar with snooker. The pocket is on a two-dimensional table and the trajectories of the cue ball which lead to a pot are limited. But when you add side, top, bottom and power and other such factors you can think of your pockets being multi-dimensional pockets on a multi-dimensional table. ~~ Dr Dec (Talk) ~~ 18:34, 23 August 2009 (UTC)

I was thinking along similar lines for Chess or Go, but I figure that even if the space of all games is near infinite, the space between 0 and 1 in even one dimension is actually infinite, so it could easily accommodate all those positions. Consider, for instance a representation of a chess position as a binary string. You can interpret each binary string as a number in (0, 1), which gives you the required densely packed representation (informally; it wouldn't actually be a dense set, I guess).

The problem with this is that the transformation removes too much information and the geometric interpretation of this set is fairly meaningless. I was trying to think of a game where there is a more natural transformation from positions to euclidean points, so that things like the euclidean distance are more meaningful. risk (talk) 19:37, 23 August 2009 (UTC)

Wouldn't $\mathbb {Z} _{3}^{9}$ be a more obvious vector space for describing games states in noughts and crosses? Each position is represented by a dimension and what is filling that position is given by the coordinate in that dimension. I'm not sure how your suggestion would work - you would need 9 points in that space to determine a game state and the first coordinate is completely redundant as long as those points are ordered (and the coordinates in a vector are always ordered). --Tango (talk) 20:03, 23 August 2009 (UTC)

Yeah, I'm not sure why I wrote $\mathbb {Z} _{9}\times \mathbb {Z} _{3}.$ I was thinking one thing and writing another. $\mathbb {Z} _{3}^{9}$ would indeed be the vector space to use. An empty board would be prepresented by the zero vector. A board filled with naughts (although not a valid game state) would be (1,…,1) and a board filled with crosses (although not a valid game state) would be (2,…,2). Tango, thanks for the correction. ~~ Dr Dec (Talk) ~~ 20:26, 23 August 2009 (UTC)

Risk, I see what you're saying about the interval. Is there really a difference? Is a point any more or less isolated in the interval [0,1] as a vector is in $\mathbb {Z} _{3}^{9}?$ I would say that a point is more isolated in the interval than a vector in $\mathbb {Z} _{3}^{9}.$ The vector $(v_{1},\ldots ,v_{9})$ where $v_{i}\in \{0,1,2\}$ could be set to correspond to the number $0.v_{1}v_{2}\ldots v_{9}.$ What about irrational numbers like π/4 ≈ 0.7854? They don't correspond to any element of $\mathbb {Z} _{3}^{9}.$ In fact the numbers in [0,1] which don't correspond to elements of $\mathbb {Z} _{3}^{9}$ form an open and dense subset of [0,1]. ~~ Dr Dec (Talk) ~~ 20:42, 23 August 2009 (UTC)

That's a good point. In $\mathbb {Z} _{3}^{9}$ every point is a legal board position. In that sense, you're absolutely right, the space is fully packed with positions (though not all legal). But the structure that those points make when you view the set as a subset of $\mathbb {R} ^{9}$, isn't very interesting (essentially a 3x3x3x... grid), and in that sense it does create more 'empty space'. Of course, a carefully designed game could get around this by making all the real or rational vectors possible/legal moves, which is more or less what I'm looking for.

I should perhaps elaborate a little on what I'm trying to do. One of the things I'm toying with is the idea that in such a representation you can talk about the (fractal) dimension of the total set of board positions, or the dimension of the sets of winning/losing positions, or the dimension of the boundary between these points. (You could do that anyway by just defining a metric on the positions, but that would introduce the possibility that the choice of metric is more important to the dimension than the game). Of course any game with a finite number of positions has dimension zero, but there are ways around that.

Another option would be to treat a move as a map. This would open up the possibility of treating the game like a discrete dynamical system. risk (talk) 21:40, 23 August 2009 (UTC)

Are you restricting yourself to vector spaces or is any manifold acceptable? There is an additional requirement that the subsets you are interested in actually be subspaces/submanifolds, otherwise the dimensions you talk about aren't well defined. I'm sure we can invent a game that works, but finding an existing one (other than the kind of strange games mathematicians invent for various purposes) seems unlikely. Most games have a finite number of possible game states since there is usually a finite number of possible moves at any given point and a finite amount of time to play the game (although I suppose it is possible that time could be unbounded, leading to countably infinitely many game states, but that isn't much help anyway). If we want to have interesting geometry over the real numbers we obviously need uncountably many game states, that means uncountably many decisions at a given point. The only such games I can think of are sports where you have things moving around in space, but they aren't usefully described in the way you want. For football (soccer) you can have the position of the ball represented by a 2D vector, the possible game states are a rectangle and the winning game positions are a line segment at each end of the pitch. Not very interesting, but it fits your requirements technically. You could try snooker with 2 dimensions per ball, the possible game states would be those where the balls are in a rectangle and there is a minimum distance between them (the diameter of the balls). The end game states would be a discrete set where all the balls are in the pockets (the minimum distance rule doesn't apply there), and you need 2 extra discrete dimensions for the scores. So, you can do the kind of things you are talking about, but they aren't useful. I think it is only going to work nicely if you invent a game for that purpose. --Tango (talk) 22:29, 23 August 2009 (UTC)

I guess you're right. Specifically, the difference between a countable and an uncountable number of positions may not be so easy to gloss over as I had thought. I gues I have some more thinking to do. Thank you both for the insights. risk (talk) 11:11, 24 August 2009 (UTC)

If it helps, most variations of Nim, and most take-away games in general, can be viewed as motion in a discrete subset of a vector space in an obvious way. I've also seen an analysis of Peg Solitaire that viewed playing the game as vector addition in a high power of Z₂. And of course, there's the classic probabilistic games in economic Game Theory, which are explicitly viewed as points in Rⁿ. Black Carrot (talk) 06:31, 26 August 2009 (UTC)

Those are very helpful suggestions, thank you. risk (talk) 15:08, 27 August 2009 (UTC)

BN-pairs: conjugate generators weyl group give double cosets of same size

Hello,

I read (B, N) pair, but I still have a question about the correspondence described there between elements of the Weyl group and double cosets.

Suppose $w_{1}$ and $w_{2}$ are both generators of the Weyl group, and suppose that they are conjugate (within the Weyl group). Is it correct, and trivial to see(?), that the double cosets $Bw_{1}B$ and $Bw_{2}B$ have the same cardinality? (I'm rather sure it should be, because in geometric language, this is the number of chambers through one panel in a thick building).

However, I'm afraid it's not like we can prove that by proving that both double cosets are conjugate as well (because I think they aren't). Do you have any ideas? Thanks!

Evilbu (talk) 21:25, 23 August 2009 (UTC)

So by w_1 and w_2 being generators, I'm assuming you mean they are both part of a single (Coxeter) generating set. The Weyl group of type An is a little weird because individually all of its generators are conjugate (they are just two cycles in the symmetric group), but within the particular Coxeter system they are quite different. For instance, take the group to be SL(4,p^f) of type A3. B is a Borel subgroup, the normalizer of a Sylow p-subgroup. B is a semidirect product of T and the Sylow p-subgroup, and N is the normalizer of T in G. N/(BnN) is isomorphic to the symmetric group on 4 points, a Coxeter group of type A3 with Coxeter generating set w1=(1,2), w2=(2,3), and w3=(3,4). Letting n1, n2, n3 be preimages of w1, w2, and w3, one has that the double cosets Bn1B and Bn2B have different sizes. For instance, when p^f=3, Bn1B has size 157464 and Bn2B has size 17496. Now, if w1 and w2 are conjugate under a graph automorphism of the Weyl group, then I think the group itself should have a graph automorphism so that Bn1B and Bn2B are conjugate. The graph automorphism is unlikely to be an inner automorphism, so probably they are not conjugate within the group itself, but they would be in the holomorph. JackSchmidt (talk) 00:34, 24 August 2009 (UTC)

Hello, thanks for your reply. Indeed, I mean that they are both part of a single coxeter generating set (sorry if that was unclear). I must admit that I do not know what a "graph automorphism of a group" is. I do know "automorphisms of a graph" (and there is plenty on that on the internet) but that's something else (it seems). I was looking at the numbers you gave me so I could try it myself. Do you also happen to know the size of B? If $p^{f}=3$, I find that it should be 2^3 3^6=5832. In general, if I let q denote p^f, I find that: $B=(q-1)^{3}q^{6},Bn_{1}B=Bn_{2}B=Bn_{3}B=q^{7}(q-1)^{3}$ (which is 17496 if q=3). So I don't understand what is going on. Note that the cardinality of each double coset must be a multiple of the cardinality of a coset of B. (And to be honest : it is in fact the quotient that I am interested in). Many thanks.ıı

Oops, sorry, my explicit calculation was wrong (it took a B and an N, but not form the same (B,N) pair, so my n1 was actually of length 3 in the true Coxeter generators). The general calculation made it clear:

For SL(n+1,q) of type An, B has order (q-1)^n*q^(n(n+1)/2) and it has a very important subgroup U of order q^(n(n+1)/2). U is made up of subgroups X_i, each of order q, called root subgroups, that are in one to one correspondence with the positive roots. For a given element w of the Weyl group, there is a subgroup U_w of w which is generated by all X_i such that w(i) is a negative root. So U_1 = 1, and U=U_{w0} where w0 is the longest element of the Weyl group. The double coset BwB is also equal to BwU_w and has cardinality |B|*|U_w|. In particular, for w=w1, U_w1 = X_1 has order q, and for w=w2, U_w2 = X_2 has order 3, so the quotient |BwB|/|B| = |U_w| is equal to q whenever w has length 1 (that is, whenever it is one of the generators from the Coxeter generating set). Indeed, if w has length l (it is a product of l generators, but not a product of fewer than l generators), then |U_w|=q^l.

For general (B,N) pairs I don't know a formula for the size, but for so called "split (B,N) pairs of characteristic p satisfying the commutator relations" this same thing occurs, BwB = BwU_w and |BwB|/|B| = |U_w|. However, even for algebraic groups, the size of U_{w1} can depend on which root you take, though this only happens in the twisted groups. Most of this can be found in Carter's Simple Groups of Lie Type. Hopefully I'll have time later today to write out the SU(n,q) case. JackSchmidt (talk) 13:06, 24 August 2009 (UTC)

Hello, I certainly enjoy "debating" with you (and I think we have also crossed paths elsewhere). If it suits you better, this discussion can also be continued on my talk page. I somewhat recognise the stuff you wrote about U_w. I read that in general, there is a q_i associated with each generator w_i of the coxeter generating set. The correct formula for the size of U_w would then be $q_{1}^{a_{1}}\ldots q_{n}^{a_{n}}$ where a_i is the number of positive roots, conjugate to $w_{i}$, turned negative by w. (so we use weighted lengths instead. So for SU(6,q) there would be 6 positive roots associated with $q^{2}$, and 3 positive roots associated with q. So the size of $U=U_{w_{0}}$ would be $(q^{2})^{6}(q)^{3}=q^{1}5$.

It is exactly those q_i that I am after. They should also be the cardinalities |B w_i B|, divided by |B| itself. And they should be equal if the corresponding generators are conjugate. (so that is why there is only one q involved in the projective case, the A_n case). A lot of people have told me that Carter's book is a good reference, but I couldn't find an answer in there. That is in generaly my problem with theory like this, too much stuff I picked up "from the street". I was very curious if there is a really quick way to see that those q_i must be equal for conjugate generators Evilbu (talk)

It's nice to have the question asked this way. I'll try to check the SU case today (I was thinking that this would be an example where the wi could be conjugate but have different $q_{i}$, but perhaps they happen to be). Unfortunately the Weyl group is type B_n in the SU case, and so there does seem to be a good chance the q_i will respect conjugacy. The size of the q_i though is supposed to depend on the size of the orbit of the root under the graph automorphism that defined the twisted group, and I don't see why that should be particularly related to conjugacy in the Weyl group. If it is true (in some generality), then Carter's Finite Groups of Lie Type (the thick brown one, not the thin black/gray one) might have the appropriate setup in its early chapters. I prefer the thin "Simple" book for most things, but it treats twisted and untwisted separately. JackSchmidt (talk) 15:52, 24 August 2009 (UTC)

Perhaps you know all this, but in $A_{n}$ all generators (and hence all roots) are conjugate, in $B_{n}$ they all are except that one on the end. In F_4 the two on the left are conjugate, and so are the two on the left. The $q_{i}$ can be equal- as far as I understand it- even if the generators are not conjugate, but conjugacy does imply that they are equal. I must have a look at that other book by Carter then. But in the mean time, could you tell me what exactly you mean by "graph automorphism"?Evilbu (talk)

Cool, I checked all the twisted groups and it is always true that if w1 and w2 are conjugate in the Weyl group then q1=q2 for the root subgroups. I still don't quite understand why this is true other than as coincidence, but it seems to work. By graph automorphism I mean five related things: an automorphism of the Coxeter diagram, an automorphism of the Weyl group described by permuting the generators according to the diagram, an automorphism of the algebraic group (over the algebraically closed field) inducing that automorphism on the Weyl group, an automorphism of the finite untwisted group that is just the restriction of the one on the algebraic group, and, when talking of a twisted type group, the automorphism of the algebraic group that gets combined with the frobenius endomorphism to define the twisted group. I think the fourth and the fifth don't conflict, since I think a twisted group doesn't have any graph automorphisms in the fourth sense, though 2E6 as F4 worries me a little.

Is there a simple reason why conjugate generators have equal q_i? JackSchmidt (talk) 21:33, 24 August 2009 (UTC)