Wikipedia:Reference desk/Archives/Mathematics/2008 March 30

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March 30

x + 2 = x + 2

What is the solution to this equation? Does it depend on what x represents? Certainly the solution is broader than "x is any number." It could also be any abstract quantity, any vector space - the space has to be closed under addition of a scalar 2. Maybe I'm just rambling--but what is the most precise and pedantic solution to such an equation? What is the solution to the equation x + yi = x + yi, where i is the imaginary unit? Is it sufficient to just say that x and y are elements of the real field? —Preceding unsigned comment added by 124.191.113.222 (talk) 07:13, 30 March 2008 (UTC)

While you could certainly find something more general if you wanted to, a relatively general setting for an equation like this is a unital ring. Essentially, you have a set with operations that behave like addition and multiplication, and there is a multiplicative identity (which we happen to call 1). You can then interpret 2 as 1 + 1. Since rings contain additive inverses (i.e. there is an element called -1 such that 1 + (-1) = 0, the additive identity), any element of the ring would satisfy x + 2 = x + 2. A more general setting would be any abelian group, but in this case there would be no element that would necessarily be called 2. The short story is, it depends on how you define "+" and "2". If you don't have a (right) additive inverse for 2, the equation might be non-trivial. 134.173.93.127 (talk) 07:28, 30 March 2008 (UTC)

Whenever x+2 is defined, x is a solution to x+2=x+2. Besides that, the equation contains no information about x. The equation x + yi = x + yi is satisfied whenever x + yi is defined. x and y can be complex numbers or quaternions or elements in any complex vector space. Bo Jacoby (talk) 08:26, 30 March 2008 (UTC).

I think I messed up in my question. The fact is we can say x is a real number for the first one; someone can object and say "it can also be a complex number;" someone can object and say, "more generally, it can be any quarternion." I really like the first answer. I thought the "unital ring" pretty much covers all bases - neat. But as for the second equation, I messed it up: if x + yi = a + bi and a = x and y = b, then I'm wondering what the solution for x and y is (probably just real numbers). —Preceding unsigned comment added by 124.191.113.222 (talk) 09:07, 30 March 2008 (UTC)

In common mathematical usage, we can take an equation like $x+2=x+2$ and assume it is clear from context that x should be a real number (or whatever), and then the set of solutions will be the set of all real numbers. But to deal with questions such as yours, we have to be more precise with what we do. We can't just take an equality in vacuum and assume that it itself says something about the realm we are discussing. A specification of where x comes from is required. Thus, you cannot treat a formula like $x=x+2$ as a valid equation. You need to say something like "$x=x+2$, where $x\in \mathbb {R}$", or "$x=x+2$, where $x\in \mathbb {Z} _{2}$" or "$x=x+2$, where x is a cardinal number". For the first case, there are no solution; For the other, there are. Implicit in my last example is that unital rings don't cover all settings where equations like those might make sense. Indeed, wherever you have addition and you have 2 you can discuss them.

The same goes for equations like $x+yi=2+3i$. You need to specify over which structure you are working; It needs to have addition, multiplication, 2, 3 and i. You can also require the unknowns to belong to some subset. Thus you can talk about "Solve $x+yi=2+3i$, where the operations are taken over the field of complex numbers, and x and y are real". -- Meni Rosenfeld (talk) 09:52, 30 March 2008 (UTC)

Indeed. Another way to say much the same thing is to ask what you mean by "+" and "2". We use the symbol "+" for lots of different things (addition of natural numbers, addition of integers, addition of rational number, addition of real numbers, addition of complex numbers, addition of quaternions, addition of integers modulo 3, addition of vectors, addition of matrices, etc, etc, etc), we also use "2" to mean lots of things (the natural number, the integer, the rational number, the real number, the complex number, the quaternions, the integer modulo 3, the integer modulo 4, etc, etc, etc). Once you narrow down what you mean by those symbols, it should be obvious what values x can meaningfully take. (I think there may a small element of confusion over whether it's meaningful to add an integer and a real number - it depends on how you define and construct them. The integers may be a subset of the reals, or they may just be isomorphic to a subset of the reals. In the former case, the addition makes sense, in the latter, I'm not sure it does. I may be being overly pedantic here to the point of actually being wrong...) --Tango (talk) 11:22, 30 March 2008 (UTC)

Assuming that at least the symbol "=" has its usual meaning, it is safe to say that any equation whose two sides are identical expressions – whether x + 2 = x + 2 or ξ∬η = ξ∬η – is either meaningless, or vacuously true for all unknowns in the domain of the operations. Which of the two, and what the domain of the operations is, depends on the context. The normal rule is that whoever produces the equation should also supply the context in which it is to be interpreted – with a lot of latitude if that context can reasonably be assumed because of standard notational conventions.  --Lambiam 20:15, 30 March 2008 (UTC)

You can do that, or you can take the domain to be absolutely everything and define meaningless statements to be false. That's the approach I was taking. --Tango (talk) 21:38, 30 March 2008 (UTC)

The domain of 'absolutely everything' is a moving target, leading to Russell's paradox. Bo Jacoby (talk) 13:23, 31 March 2008 (UTC).

Unless you use New Foundations, which avoids the paradox of having a universal set.  --Lambiam 00:01, 1 April 2008 (UTC)

Probability of selection - both with and without replacement

I took probability and statistics in college, but I can't find my book. I take one-half a tablet of a prescription medicine every day. If I shake out a whole tablet, I take 1/2 and put the other half back. If I shake out a 1/2 tablet, I take it. If the container starts out with n tablets, how many selections must be made before the probability of shaking out a 1/2 tablet reach 0.5? Or, in other words, Starting with a container of "n" green balls, select a ball from the container. If the ball is green, remove it and replace it with a red ball. If the ball is red, remove it without replacement. How many selections must be made until the probability of selecting a red ball reaches 0.5? —Preceding unsigned comment added by Johnerbes (talk • contribs) 20:04, 30 March 2008 (UTC)

This is a lightly disguised version of the birthday problem. Algebraist 20:50, 30 March 2008 (UTC)

If you say so... --Tango (talk) 21:39, 30 March 2008 (UTC)

Apparently I need to be clearer (alcohol always dulls my clarity-detectors). The n tablets in the container correspond to n days in a (modified) year. Taking a tablet out and breaking it in half corresponds to putting a new person in the room with a random birthday. Getting a tablet you've already broken corresponds to adding a new person with a birthday you've seen before. Thus all the calculations in the article I linked apply, replacing 365 with n. Algebraist 22:00, 30 March 2008 (UTC)

But doesn't removing the half-tablet after you take it out again mess that up? --Tango (talk) 22:22, 30 March 2008 (UTC)

Eg. For n=3, I've worked out (by just drawing a probability tree) this problem gives probabilities of 1/3, 4/9 and 11/18 after 1, 2 and 3 picks respectively. The birthday problem with a 3 day year gives probabilities of 0, 1/3, 7/9 with 1, 2 and 3 people in the room respectively. Where's the correspondence? (It is possible I just can't add up, of course.) --Tango (talk) 22:42, 30 March 2008 (UTC)

At least one of us is very confused about what the problem is. For n=3 and 1 pick, we're taking one tablet out of our container of three whole tablets, and want to know the probability that the tablet we've taken is a half-tablet. Surely this is zero? In general, removing the half-tablets doesn't matter since everything's decided once we've picked a half-tablet anyway. Algebraist 23:04, 30 March 2008 (UTC)

I think that's just an off-by-one error - I mean after one pick, the probability for the next pick is 1/3. A rather odd way to phrase it, now I think about it - just shift all my numbers by 1! However, more importantly, I think we've interpreted the problem differently. I think you interpreted it as how many times do you have to pick a tablet before the chance of having had at least one half tablet is greater than 50%. I interpreted it as how many times do you have to pick a tablet before the probability of the next pick being a half-tablet is greater than 50%. From the phrasing of the question, either interpretation seems possible, although yours does seem slightly more sensible (mine requires considering lots of different branches, each with a different probability after k picks, and taking a weighted average, which is a pretty strange thing to do). Yours is also an easier calculation (it is, indeed, just the birthday problem). --Tango (talk) 23:19, 30 March 2008 (UTC)

Ah, I see. Another downside of your version is that it corresponds to a variant that is not treated in the birthday problem article, and I am going to bed. Unless you or someone else feels like solving this version, I suggest awaiting clarification from the OP. Algebraist 23:30, 30 March 2008 (UTC)

And I read it as looking for the expected number of picks before the number of remaining halves exactly equals the number of remaining wholes. (If this is actually what Johnerbes meant, he may not have realized that this condition is in some cases not reached until the container is empty.) From a starting position of w wholes and h halves, my model is f(w,h)=0 if w=h, otherwise f(w,h)=1 + [w/(w+h)]*f(w-1,h+1) + [h/(w+h)]*f(w,h-1) and my answer is f(n,0).

For n=1 through 6 I get 2, 1, 11/3, 35/12, 40487/7500, 83612/16875, and then the numbers start getting ugly. We obviously need some clarification on the original question, but I notice that from a pretty simple definition I've generated a sequence that doesn't appear in OEIS. Is it interesting enough? --tcsetattr (talk / contribs) 00:49, 31 March 2008 (UTC)

While your interpretation is certainly an interesting question and it makes a little more sense than mine, the fact that the word "expected" (or any synonym, eg. average) does not appear in the statement of the problem suggests to me that it's not what the OP had in mind. I can only guess, though. We've managed to come up with 3 interpretations, all getting completely different answers - I think I'll go with Algebraist's plan and go to bed and hope the OP clarifies things. --Tango (talk) 01:15, 31 March 2008 (UTC)

OP here. Although the problem procedure is well defined, I'm not sure that the question is well defined or if I am even asking the right question. The problem is best defined as starting with a container of "n" green balls and selecting from the container until it is empty. If a green ball is selected, it is replaced with a red ball. If a red ball is selected, it is removed. As an example, if you start with 100 green balls, the container will be empty after exactly 200 selections, as 100 green will have been removed (and replaced by 100 red), and the 100 red will also have been removed. Starting with 100 green, the probability of selecting red on the second selection (following the first red replacement) is 1/100. Likewise, on the 200th selection, the probability of selecting red is 1/1. At one extreme, if the first 50 random selections were all green, the container would then contain 50 green and 50 red, so, the probability of selecting red must be less than 1/2 for the first 50 (or n/2) selections. I think that tcsetattr came closest to interpreting my question when he stated "And I read it as looking for the expected number of picks before the number of remaining halves exactly equals the number of remaining wholes." This may be a question that can't be answered. Let me redefine the question to ask: Starting with "n" green balls, what is the probability of selecting a red ball on selection "p". Example: Starting with "n" green balls, the probability of selecting red is 0 on the first pick, 1/n on the second pick and 1/1 on pick 2n. —Preceding unsigned comment added by Johnerbes (talk • contribs) 15:25, 31 March 2008 (UTC)

All right, I'm taking another shot at it. Given an initial state of w wholes and h halves, we want to know the probability that the pth pick will be a half. If we knew the state immediately before the p'th pick, it would be easy. But there are several possible states, depending on how many of the preceding picks were halves. We need to know the probability of each of those states before we can answer the main question. The general form of that question is: from a certain initial state, what is the probability that exactly r of the next p picks will be halves? Let that be the definition of a new function: f(w,h,p,r). A few base cases, and then the recursive rule:

f(w,h,p,r) = 0 if r<0 (you can't pick a negative number of half-pills)
f(w,h,p,r) = 0 if r>p (2 picks will never yield 3 halves)
f(w,h,0,0) = 1 (0 picks always yields 0 halves)
f(w,h,p,r) = [w/(w+h)]*f(w-1,h+1,p-1,r) + [h/(w+h)]*f(w,h-1,p-1,r-1)

Now we're ready for the main question, what is the probability that the p'th pick will be a half? We need to take the probability of each preceding state (given by f) and multiply it by the probability of picking a half-pill from that state, then add those together. The general description goes like this: We had w wholes and h halves. We made p-1 picks, and removed r halves. We must have also removed p-1-r wholes, replacing them with halves. We now have w-(p-1-r) wholes and h+(p-1-r)-r halves. The probability of reaching the current position was f(w,h,p-1,r). The probability of the next pick being a half is [h+(p-1-r)-r]/[w-(p-1-r)+h+(p-1-r)-r] = (h+p-1-2r)/(w+h-r). Put that all together and we get your requested probability function:

$g(w,h,p)=\sum _{r=0}^{p-1}{\frac {h+p-1-2r}{w+h-r}}f(w,h,p-1,r)$

Starting with 100 wholes and 0 halves, the probability that the p'th pick will be a half-pill is g(100,0,p). Actually computing these numbers takes a long time because the recursion in f is so heavy. After spending a few CPU-minutes on it, I see that the probability crosses 50% between p=90 (0.49917...) and p=91 (0.50257...). The 91st pick is the first one to have a greater than 50% chance to be a half-pill. Something more interesting happens at the end of the list though. g(100,0,199)=0.85782... meaning there's about a 14.22% probability that a single whole pill survives to the end, being broken only on the next to last day, counter-intuitive to me but confirmed by simulation.

I do not see a way to simplify the formula, and I don't see the resemblance to the birthday paradox either. Maybe I'm just not clever enough. --tcsetattr (talk / contribs) 08:10, 1 April 2008 (UTC)

The nature of mathematics

I just wanted to share my (probably quite ignorant) view on mathematics to establish what other people think about the matter.

It is my current belief that pure mathematics is a completely human construct that has no "reality" outside of our minds (on the assumption we are alone in the universe). My reasoning for this is that the basic concept of number, on which all mathematics is based, is surely a human construct. We regard, for instance, a tree on its own in the desert "one tree" even though this is a designation based on our unique perspective - its perhaps more accurately a collection of cells, or configuration of atoms and on all scales we can consider it to have a range of quantities aside from unity. The universe doesn't "count" trees.

The system of mathematics that arises from the numbers and operations we use to analyse our surroundings is, I consider, a "flawed" construct that has particular nuances and structure grown out of our requirements for the science. If we had no computing technology, for instance, would many areas of discrete mathematics be as rich as they currently are? Because we create rules that are suitable to our needs, it helps to explain the patterns we see so regularily in mathematics - perhaps there are no ODD perfect numbers because this result stems from the way we have constructed our traditional algebras?

It would be quite an interesting thing to consider that something as deep and beautiful as complex numbers are "encoded" in the set of rules we have created in the "game" of mathematics, waiting to be revealed at some point in time- suggesting that whilst mathematics may be a completely artifical construct, we can still investigate and "discover" parts of it that follow in direct conclusion. Any thoughts?

Damien Karras (talk) 20:36, 30 March 2008 (UTC)

While some would share your view that mathematics doesn't exist outside of our minds, I strongly disagree. I find it absurd that we would be able to invent anything that didn't exist already.

It is not correct that all of mathematics is based on the concept of number. Numbers are just one example of an abstract entity which mathematics can deal with. Regardless, I also don't agree that numbers don't exist in nature - take for example photons, to which there is no substructure as far as I know. Do you not agree that there is in nature the concept of "one photon, two photons, three photons..."? Even taking trees as an example, while there are definitely many ways and many scales in which you can investigate them - surely one of these scales is treating the tree as a whole unit and counting "one tree, two trees...". You don't need humans for that.

I am platonistic in the sense that I believe there is an abstract "world of ideas", exhibiting incomprehensible richness and containing any abstract idea there could possibly be (the word "be" here is tricky, since it can be associated with physical existence in our natural universe, but the world I describe is at a much higher level than it). When we set up axioms and definitions, we basically construct a window into this world. What we see through that window is the theorems that follow from them. These theorems exist without us - all we have done is decided to look at them.

I do agree that we have a choice of where to look and how to interpret what we see. The axioms we work with often are just carefully constructed windows through which we can see those ideas that we consider useful due to the specifics of our physical universe. If we chose to look elsewhere, we would see different things. But all those ideas exists in a way which transcend humanity, time, and our physical universe. -- Meni Rosenfeld (talk) 21:20, 30 March 2008 (UTC)

Well put. --Tango (talk) 21:42, 30 March 2008 (UTC)

I generally agree with Meni here, but I do think you touch on an interesting (and rather philosophical) point. Mathematics is a completely abstract construct that does not necessary have any actual impact on reality, but our observations have shown a great many times how mathematics has accurately predicted things in our universe. And that adds lots of credibility to the claim that mathematics does have a direct and significant relationship to our reality. The only real assumption that I find in mathematics (as a science) is the assumption that the human brain computes logic correctly, this is not necessarily true, but the amount of evidence supporting that assumption is enormous. A math-wiki (talk) 22:44, 30 March 2008 (UTC)

Maths on its own doesn't predict anything in our universe - maths combined with a model does. The fact that it works so often is a sign that our models are good, it's not a sign that maths is good. Modelling trees as natural numbers when you're planting them and chopping them down turns out to work rather well. It's not natural numbers that are good - they just are what they are - it's the model that's good. Had we modelled trees are integers modulo 3, we would have found out that it doesn't work well at all. That wouldn't mean that integers modulo 3 are bad, it's just a bad model. --Tango (talk) 22:49, 30 March 2008 (UTC)

On a tangent (I am far away from being a mathematician): There are some experiments with the cognitive capacities of animals which seem to imply that abstracting numbers is not solely a human achievement. I recently read about an avian (Alex, An African grey parrot researched by Irene Pepperberg, Ph.D. / Brandeis) who "invented" a sort of Zero concept. Alex, by the way was an acronym for Avian Leaning EXperiment. --Cookatoo.ergo.ZooM (talk) 23:40, 30 March 2008 (UTC)

I think I would tread a middle (or more nuanced) ground on this. I do think it is potentially fair to say that numbers don't exist outside of our minds. Let's take your example of photons. There may be photons, but to give them a number, to count them and say "three photons", that takes us (or some other like minded observer). I don't think nature in itself goes around counting photons. Once could argue that some event happens after a certain number of photons hit a target (or some such, physics isn't my field) and that nature was thus counting the photons, or energy levels, or somesuch. I don't see that as a natural conclusion though. Nature does what nature does, and the counting is our doing. I would suggest that the counting is something that occurs in our model of nature because that is how we understand nature; nature doesn't need to do any counting, but we do to keep track of things within our model. Let's be honest, counting is setting up a bijection, and I suspect nature doesn't bother with that, it just runs along as it is; the bijection is something we set up to hopefully elucidate those workings.

On the "world of ideas" and "what could possibly be", there is the difficult question (as you note) of deciding what can "be". Does logic proscribe what is possible? If so, what logic exactly? Do we accept things as possible even if they contradict the law of excluded middle? Constructivist mathematicians would, and a case can certainly be made that classical logic is rather too strict. So what else are we allowed to drop and still admit as possible? Can we accept something as able to "possibly be" if it contradicts the law of identity? Such a thing seems rather hard to conceive, but then so are many other things (like completed infinities); maybe we just need some examples and some practice. Given such freedom the "world of ideas" begins to seem rather too unconstrained and not just a little paradoxical (though what that means when we're willing to let go of logic is a question in itself). I tend to fall into the non-platonist camp here; it seems better to forgo the "existence" of this universal world of ideas, and simply accept those "local" worlds of ideas that we care to delineate/define ourselves -- thus local existence only within our minds, and no platonist absolute universal eternal existence.

All of that said, I don't think that means we "have a choice of how we see things" in the free sense that all choices are equal. We are, ultimately, products of our biology, and that has a great deal to say about how we see and categorise (which is ultimately the start of mathematics in many ways) things. We, and most anything else like us (e.g. aliens across the galaxy that have even roughly the same scale-wise view of things), are necessarily predisposed toward certain abstractions due to their relative efficacy. We see a tree not a collection of atoms because that's productive for us. We count because it is a fairly canonical abstraction from such discrete categorisation, and efficacious in our internal model of the world. This gives some sense of universality, but doesn't provide for some external eternal platonic truth; instead we merely have efficacy for our situation. -- Leland McInnes (talk) 13:58, 31 March 2008 (UTC)

Heisenberg said: If a question cannot be asked experimentally, you do not need to answer it theoretically. Can the question, whether pure mathematics is a completely human construct that has no "reality" outside of our minds, be asked experimentally? Bo Jacoby (talk) 00:45, 31 March 2008 (UTC).

That's science. We're talking about either maths or philosophy (I'm not sure which!), so falsifiability doesn't really apply. --Tango (talk) 01:08, 31 March 2008 (UTC)

Bo Jacoby: You may be interested in indispensability arguments in the philosophy of mathematics.--droptone (talk) 11:43, 31 March 2008 (UTC)

The question is definitely not one of mathematics. Only if it is a scientific question of the real world do we have a hope of finding out. Which observation or experiment will tell us? If no observation will tell us, then the question is pure philosophy. Mathematics, like physics and economy and biology, is historically inspired by the reality outside of our minds. Bo Jacoby (talk) 15:25, 31 March 2008 (UTC).

You can take ZF as an axiomatization of set theory and consider the Axiom of determinacy (AD). As is well known, AD is independent of ZF: it does not follow from the axioms of ZF. It is a quite plausible statement, though, so we may want to add it as a new axiom to ZF, giving a new system ZFD = ZF+AD. Likewise, the Axiom of choice is a plausible statement that is independent of ZF, so we may want to add it, giving a new system ZFC = ZF+AC. Why not add both and obtain an even more powerful system ZFCD = ZF+AC+AD?

Well, ZFCD is a bit too powerful. In ZFCD you can also prove that 2 + 2 = 5. So unless ZF itself is terribly wrong, AC and AD cannot both be true. In any mathematical universe at least one of the two is false. Does it make sense to ask which one is false? Can we hope to conceive of some argument that finally settles the issue, like "AD definitely holds, and therefore AC is false" (or the other way around)? Any such argument has to be based on a new axiom or deduction rule not in ZF, but then how can we argue that that axiom or rule is sound? We know we can only prove it if we posit yet another axiom, and so on, ad infinitum.

There are consistent geometries in which the parallel postulate holds, and others in which it does not hold, and by now it has been generally accepted that it is meaningless to ask which of these possibilities is "true" and which is "false". Both Euclidean and non-Euclidean geometries can serve as useful models of parts of observable reality, but that does not make them in some sense more true. We should likewise accept that there are different consistent set theories, none of which is more true in any meaningful sense than any of the others. Unlike for the divergent geometries, there are no useful applications of set theory as a model of part of observable reality where the difference is testable by an experiment.

If a conclusion is necessarily true in mathematics, it is because we have gamed the rules such that the conclusion is inevitable. Let me count: 1+1+1+1+1+1+1+1+1+1+1 = 11. Now I know how many trees I have in my garden. But do the rules of arithmetic also apply if you want to count the number of waves in the Atlantic ocean on April 1, 2008? These rules hold for discrete systems whose individuals have a permanent identity. To the best of our understanding, nothing in physical reality is fundamentally discrete and permanent.

Although we can communicate mathematical thought – an ability that has to be innate – to other human beings, and apply this successfully to models of reality, the mathematical content itself is a mental construction, and the assignment of discrete persistent identities to, say, "trees", is a mental act reflecting an ultimately subjective (even if intersubjective) way of imposing a never fully fitting order on reality.  --Lambiam 23:21, 31 March 2008 (UTC)