Wikipedia:Reference desk/Archives/Mathematics/2011 April 25

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April 25

Directed-complete posets

"By a directed set in a poset (P, <=), we mean a nonempty subset D such that any pair {x, y} of elements of D has an upper bound in D. We say (P, <=) is directed-complete if each directed subset D ⊆ P has a least upper bound in P . Show that a poset is complete if and only if it is directed-complete and has joins for all its finite subsets."

I believe to the best of my knowledge all my definitions are standarad: complete means every subset of P has a supremum. I was under the impression however that a 'join' was a least upper bound, but if so then I'm somewhat confused as to why both terminologies are used separately here. Could anyone perhaps explain to me what I'm meant to be proving? I couldn't find any similar sounding theorems or lemmas, but perhaps I overlooked something. Staytime101 (talk) 00:50, 25 April 2011 (UTC)

I think quite possibly the author is simply using two different terms for the same thing. Maybe he wants to avoid the word "join" in cases where the set not known to be finite. Michael Hardy (talk) 19:28, 25 April 2011 (UTC)

Graph Theory doubt

Hi. I have a doubt in a theorem from a paper on graph factorizations which I am reading. The complete result is too long to state so I'll just state the sentence where I have the doubt. Hopefully I'll not be overlooking some extra hypothesis. To begin let $n_{1}.\cdots n_{k}$ be positive integers of which at least 1 is even and $N=\sum _{i=1}^{k}(n_{i}-1)$ and that G is a N-regular graph of even order. I have deduced due to certain other statements in the proof of the theorem that N is odd(dont know whether it matters or not). My doubt is this: The proof contains the statement: If G has an $n_{i}-1$ factor for each even $n_{i}$, then the graph obtained from G by removing each of these factors is regular of even degree, and hence 2-factorable, and so has $n_{i}-1$ factors for odd $n_{i}$ as well. I have been struggling with this statement for days, in particular why will the graph so obtained be "even" regular, when the number of even n_i is even. If someone can help me I will be most obliged. If you want further clarification about the theorem I will be happy to provide it.-Shahab (talk) 06:33, 25 April 2011 (UTC)

Lie algebra

Hi. I am trying to teach myself Lie algebra from Erdmann and Wildon. On page 20, it seems to imply that $[x,y]=x$ describes a Lie bracket. But this doesn't satisfy the Lie algebra axioms! What gives? (also, would the refdeskers recommend this book? It seems light on motivation. What book would be a good one for someone like me, interested in computational mathematics?) Robinh (talk) 10:31, 25 April 2011 (UTC)

You've taken that formula completely out of context. They're discussing a non-abelian 2-dimensional Lie algebra, and have concluded that the derived algebra must be 1-dimensional. Then they let {x} be a basis of the derived algebra, and {x,y} be a basis of the whole algebra with y scaled such that [x,y]=x. Since the Lie bracket is bilinear and alternating, this formula determines the value of the Lie bracket on every pair (since [x,x]=0, [y,y]=0, [y,x]=-x and {x,y} is a basis). Furthermore, it's not hard to show that for any two-dimensional space with basis {x,y}, there's a unique Lie algebra structure such that [x,y]=x. Algebraist 13:49, 25 April 2011 (UTC)

Thanks for this, algebraist. What you have said matches what the book says. But I'm still not getting it. In what sense is there a "unique Lie algebra structure such that [x,y]=x", when [x,y]=x does not satisfy the axioms? Robinh (talk) 20:50, 25 April 2011 (UTC)

It does satisfy the axioms. Explicitly, the bracket in question is given by [ax+by,cx+dy]=(ad-bc)x. You should be able to check that all the axioms are satisfied. Algebraist 21:25, 25 April 2011 (UTC)

Got it! Thanks: My mistake was to misinterpret 'x' in the formula.

Resolved

Robinh (talk) 21:52, 25 April 2011 (UTC)

The Reality of negative numbers.

How can negative numbers be integers in the set of Real numbers? I don't believe they are. I have long finished school, and this isn't a homework question. It's just something which doesn't make sense to me. It keeps me awake at night. I honestly think that people just go along with what they're taught, not questioning the absurdity of it. It's like the emperor has no clothes. Sure, you can be overdrawn at the bank by let's say 25 bucks, but that's not actually less than zero. That's the bank temporarily loaning you positive $25. Compare that to the thug who tries to steel $25 off the guy who has only a picture of his ex and two fives in his wallet. The poor bloke isn't gonna limp home with -$15 bucks! Or take the frequent example of temperature below zero. If we talk about minus 18C, that's not actually negative. There's still plenty of heat, it's just 18 degrees less than the point at which ice liquifies. Absolute zero, the real zero temperature, cannot be reduced by any quantity. We could also talk about the Cartesian Coordinate system in similar terms. We can use the negative symbols in -4X, -3Y to plot a point in reference to 0,0. But these are just to describe how far, in absolute value units, the point is to the left, and down from the named point zero. The negative sign is a useful symbol, and it "works" in that we can apply operations to it, and even some of the solutions will bear negative symbols (names of points to the left and or below the named point zero), however, none of the solutions will actually be less than nothing. They will still be some distance, measured in positive numbers, away from point Zero and all other points (unless of course the distance is zero). You might say that 1234 Lincoln Avenue East is no more "negative" than 1234 Lincoln Avenue West. Furthermore, the Origin zero in that case doesn't represent "nothing." The zero names something, one point in space. I think all negative numbers should be reclassed as imaginary numbers. I think it's the negative component of a negative square root which mathematically casts it into i. Can someone out there please tell me when there is ever actually less than nothing of anything? Anything at all. Why are negative numbers Real numbers?175.32.24.111 (talk) 12:50, 25 April 2011 (UTC)

Why do you believe that any number is real anyway. They are all concepts in our head.-14.97.55.144 (talk) 13:15, 25 April 2011 (UTC)

"Real" number is not meant to mean that they are a perfect reflection of reality; the name merely refers to the fact that the real numbers have a much stronger, more intuitive geometric interpretation than the imaginary numbers do (note: I'm not saying there is no geometric interpretation for the imaginary numbers. I'm just saying that it's not as intuitive).

The negative numbers do serve a pretty good model of the situations that you describe, even if, in those situations, the people involved don't actually have less than no money, and the temperature isn't actually less than no heat. Just because the actual quantities involved aren't truly less than nothing, doesn't mean that we should refuse to use the negative numbers to model them.

In answer to your question, you probably won't find such an example, because anytime you are measuring something, it is arguably a substantial thing which physically cannot be less than nothing. However, using negative numbers does simplify many models, including the ones you described.

Another prime example is electric charge: you know you have a neutral charge when the sum of all charges is 0; a negative charge is not actually less than 0 charge (it merely indicates the presence of more electrons than protons), but just talking about adding charges up, after arbitrarily assigning negative and positive to opposite types of charge, is a simpler model than what you would have to come up with without the use of negative numbers.

Certainly, you could just say to add up all the charges of one type, add up all the charges of the other type, and then subtract the larger from the the smaller. But then you've sacrificed the freedom to do a whole lot of algebra to solve various problems, because the operation you have described is less susceptible to those techniques.

Before mathematicians accepted the use of negative numbers, problems that we would consider to be different cases of the same form were handled differently. We have a single formula for solving $ax^{2}+bx+c=0$. Historically, there were many different forms of this equation, such as $ax^{2}=bx+c$, $ax^{2}+c=bx$, etc, where all the coefficients are positive. We have found, through the use of negative numbers, a way to handle all these equations identically. --COVIZAPIBETEFOKY (talk) 13:35, 25 April 2011 (UTC)

Thanks. I appreciate your thorough and well reasoned explanation. Yes, I was trying to match numbers up with physical reality. That's how I learned to count, by one to one correspondence. It is quite confusing because you can't do that with values less than zero. Now I see negation as a mathematical instrument for manipulating quantities, but not as descriptive for quantities. Right? Wrong? — Preceding unsigned comment added by 175.33.141.37 (talk • contribs) 18:11, 25 April 2011

Yeah, I'd say that's about right, although the negative numbers certainly do describe the quantities we've mentioned before. It's just not the "less than nothing" aspect of the numbers that we choose to focus on, but rather, the existence of quantities of two types which will cancel when added to each other.

Also, the reality is that we don't know exactly how all quantities behave in reality. There is no fundamental reason to believe that certain quantities in everyday life could theoretically hold every single value on the real number line; in many cases, we actually know this to be patently false. Heat is one example, because there is no temperature smaller than absolute value, and money is another example, because it always comes in integer multiples of a unit amount. Truly, the real numbers are just an abstraction, with a purely mathematical construction, that happens to be extremely useful in describing certain things in our universe, as far as we know. --COVIZAPIBETEFOKY (talk) 17:19, 25 April 2011 (UTC)

If you wish to be able to deal with the numbers without all this angst how about just renaming them to 'credit numbers' and 'debit numbers' since you seem happy enough with those? Dmcq (talk) 14:01, 25 April 2011 (UTC)

My question isn't about substituting words. I'm not happy enough with debits and credits. Wherever did you get that idea? Not from me. My question is about the possibility of true values less than zero. Sarcasm isn't helpful. — Preceding unsigned comment added by 110.21.50.67 (talk • contribs) 18:23, 25 April 2011

I was not being sarcastic. What do you mean by you are not happy enough with the idea of debits and credits? Neither has a minus sign so exactly what causes you problems with them? In mathematics the concept of negative umbers is normally handled by considering pairs of natural numbers (c,d) and saying a pair (c,d) is equivalent to (a,b) if c+b=a+d. Just letting c stand for credit and d for debit gets a mapping directly, see Integer#Construction for details. It is defined using the straightforward numbers you are happy with in modern mathematics. Dmcq (talk) 19:25, 25 April 2011 (UTC)

Doron Zeilberger would say that there only exists a finite number of numbers, adding 1 to the largest number (N) yields zero. So, if he is right, -n should be N+1-n, so everything is actually positive. Count Iblis (talk) 15:56, 25 April 2011 (UTC)

He looks like a fun guy to take classes from.Naraht (talk) 16:12, 25 April 2011 (UTC)

Indeed :) .Count Iblis (talk) 16:24, 25 April 2011 (UTC)

175.32.24.111 - it is certainly possible to describe the physical world using only positive quantities. Indeed, most European mathematicians and natural philosophers continued to do this until as recently as the seventeenth century, regarding the concept of netaive numbers (an import from China, India and the Arab world) with suspicion - see negative number#History. By way of analogy, it is also possible to go through life with one hand tied behind your back; it is, however, somewhat inconvenient. The physical world does not force the concept of negative numbers on us; it is, however, much easier to describe certain aspects of reality in mathematical terms if use them. The same can be said of imaginary and complex numbers, which is why the term "real number" is an unfortunate misnomer, which we are stuck with for historical reasons. Gandalf61 (talk) 16:48, 25 April 2011 (UTC)

What about things like the electron? They carry negative electrical charges, and they existed a long time before humans. Don't they "force the concept of negative numbers on us"? — Fly by Night (talk) 01:01, 26 April 2011 (UTC)

No. --COVIZAPIBETEFOKY (talk) 04:15, 26 April 2011 (UTC)

No, because electrons used to carry positive charges until humans messed up. – b_jonas 22:27, 26 April 2011 (UTC)

But as a whole, the nucleus has to have almost zero charge to be stable. If you relabeled the electrons so as to have positive charge the then protons would also need to be relabeled to have negative charge. There's no escaping it. — Fly by Night (talk) 00:28, 28 April 2011 (UTC)

Fly by Night, I addressed this exact point above. You can easily define what the effective charge of a collection of protons and electrons will be without making any reference to negative numbers. The negative numbers simplify both the calculations and the definitions, but they are not a necessity. --COVIZAPIBETEFOKY (talk) 03:18, 28 April 2011 (UTC)

Yes, you can certainly define things this way, and say that there are two kinds of charges, and that when you add one kind of charge to another, the two things subtract (in one way or another, depending on which kind of charge is the greater). This is something like Dmcq's "debit numbers" and "credit numbers". However, it seems to me that this is just a physical model for one axiomatization of negative integers (up to an isomorphism interchanging "positive" and "negative" depending on whether the electron has positive or negative charge). The difference is purely semantic. Sławomir Biały (talk) 22:12, 28 April 2011 (UTC)

It's certainly true that most people just go along with what they're taught. And if you suggest they think about it and understand it instead, they ask whether that will be on the final exam, and if not, why should they bother?

One small point: Most negative numbers, like most positive numbers, are not integers.

I suspect no really good account of this matter for the curious layman has ever been written. Michael Hardy (talk) 19:24, 25 April 2011 (UTC)

The OP requested: please tell me when there is ever actually less than nothing of anything . The inequality a<b means that a is less than b only when a and b are positive numbers. Generally it means that a comes before b in the ordering. So there is nothing less than zero, but there may be something before zero in the ordering. Bo Jacoby (talk) 10:18, 27 April 2011 (UTC).

Well as I said above one can define negative numbers as a pair of natural numbers so it is always possible to argue negative numbers do not exist by dealing with two different numbers like credit and debit. If you hae 100 quatloos in your pocket and owe 200 then it is using the equivalence relation that defines negative numbers that one can say one has −100 quatloos. Dmcq (talk) 12:51, 27 April 2011 (UTC)

One thing that most people don't really learn in school is that, from a theoretical (i.e. neither practical nor ontological/philosophical) point-of-view, mathematics is all symbol-pushing within an axiomatic system. You can define anything however you want, regardless of whether it has any connection with physical reality or even conceptual reality. Or, to put it more (?) formally:

(That needs to be turned into a song.) Anyway, see philosophy of mathematics - you're hardly the first person to wonder about such things. « Aaron Rotenberg « Talk « 22:15, 27 April 2011 (UTC)

Hi, um, I'm the OP back again. I've got 3 cookies to share with you. Please help yourself to 5 of them. If you look up Wikipediea's article on real numbers you'll see that the first part of the first sentence is "In mathematics, a real number is a value that represents a quantity along a continuum ..." I see how negative numbers are useful operationally, but I don't agree that they are real, because as far as I can tell they never do or could represent any quantity. Please read on.

From this forum I have learned how helpful it can be to pretend that there are numbers less than zero. Doing this allows us to model things more simplistically to help us quantify real things in the real world that have quantities greater than 0 or equivalent to it. Let's call negative numbers a method, or tool in the toolbox to manipulate numbers. Operators such as addition and multiplication are some of the other tools in this toolbox. Subtraction is in the tool box too, but is limited, I think to results of zero or above. The problem is that negative numbers could never actually match up to anything in the real universe. In some ways it's like trying to divide by zero I think. The answer to this kind of problem is never zero or any other number, or if I remember from school all numbers would satisfy the equation or something like that. My point is that it's the null set. You just can't divide by zero mathematically. That's similar to what I'm trying to say about numbers less than zero. You just can't truthfully have any real quantity, as mathematically defined, equal to less than zero I think.

I'm just a taxi driver, and I haven't had much formal education past high school. I don't know what a mathematical proof actually is in an academic definition, but I would really like to know if negative numbers have ever been proven or disproven, or if the whole knowledge and educational system is running on a false axiom. I know there are a lot if really smart mathematically minded people reading this, so maybe some of you can give me a definite answer on this. Should negative numbers have even been placed in the set of real numbers? Again, can you ever actually have less than nothing of anything? Can a<b be true where b=0? —Preceding unsigned comment added by 49.180.24.185 (talk) 08:14, 28 April 2011 (UTC)

Short simplified answer: yes it has been proven that negative numbers can exist in a consistent system.

Longer answer: It's a good idea to compare subtraction beyond zero (typically allowed) to division by zero (not allowed). The important difference is that if you were to allow division by zero, then the resulting system would be inconsistent. You would in fact be able to prove that 0=1, which would contradict the nature of the numbers that you started with. Therefore division by zero cannot be allowed if you are to maintain any integrity at all in your number system.

Similarly we could ask if subtraction beyond zero can be allowed. The question is not whether any quantities less than zero "really exist", but can they be defined in a consistent way without destroying the whole system of numbers? And it turns out that the answer is yes, the system including negative numbers is consistent, and this can be proved (if you assume that the natural numbers are consistent to begin with). For this reason, mathematicians are comfortable saying that negative numbers exist, regardless of whether or not they represent any "real quantity" like you said. This is exactly the same sort of reasoning applied to justify the use of irrational numbers and imaginary numbers, both of which can seem from certain viewpoints to "not really exist" in nature. Staecker (talk) 12:04, 28 April 2011 (UTC)

Thanks for restating what I said about division by zero, it was much clearer than the way I phrased it.  Please tell me what the negative number proof it is. I'll try to understand it. What I have really been meaning this whole time is that I think negative numbers are not real numbers but imaginary or maybe complex. I think they're in the wrong set. Isn't the difference between the square root of 1 and the square root of negative 1 the negative part?  I do assume that the number system is consistent.  Does the negative number proof rely on an axiom?175.34.183.34 (talk) 21:42, 28 April 2011 (UTC)

OK I'll give it a try: Assuming that the natural numbers are consistent, then define the integers (including positive and negative numbers) according to integers#construction, which defines an integer as a pair of natural numbers. Now if you assume that the natural numbers are consistent, then the set of pairs of natural numbers are consistent, and thus the constructed set of integers are consistent. There are several technical details that would need to be checked to fully formalize this proof, but this would be the general idea. (I'm sure there are other strategies.)

As for whether or not negative numbers are real or imaginary or complex, I think you're just debating terminology at this point. The phrase "real number" refers to a specific abstract mathematical construction, and the phrase "imaginary number" refers to a different abstract mathematical construction. But neither of these is really any more or less "real" than the other- they are both abstract mathematical constructions. It sounds like you're saying that the word "real" isn't a good word to use when describing the real numbers, and actually I think many mathematicians would agree with you. Mathematicians in general especially don't like the term "imaginary number" because it implies that this concept is somehow fictitious, which it isn't. But the term's been in use for so long that we're stuck with it.

You might want to have a look at Negative number#history to see how people over time have wrestled with this. If nothing else you'll find some very intelligent people who share the opinion that negative numbers are absurd. It used to be a mainstream philosophical debate among mathematicians whether or not we should study things that don't seem to "really" exist. After many many centuries of development, we as a discipline (with very few exceptions) have more or less decided that we will study any system which is consistent, regardless of how "real" it is. Staecker (talk) 00:01, 29 April 2011 (UTC)

Just to show how arbitrary it is in computing they are quite happy to have signed zeros, +0 and −0 (and most mathematicians are not happy with that), and say that 1/−0 is minus infinity! Dmcq (talk) 14:30, 28 April 2011 (UTC)

Hello again OP. You've got a point. The explanation a real number represents a quantity is not quite correct. Only the positive real numbers can represent quantities while the negative real numbers can not. Mathematicians have the confusing habit of generalizing or redefining the meaning of everyday words. real nunber does not refer to everyday reality, complex number does not refer to everyday complexity, the empty set is not really a set, an so on. The word number in everyday language means at least two. "I saw a number of ships" means normally that I saw two ships or more, not that I saw only one ship, and not that I saw zero ships, (meaning that I saw no ship). The mathematician however may answer a question like "how many dollars did you win on the casino" by saying "minus 100" meaning "I did not win, but I lost $100". It is a question of language rather than of philosophy. Bo Jacoby (talk) 11:47, 29 April 2011 (UTC).

Hausdorff dimension under the map x -> x^2

Hello everyone,

I'm trying to show that if $f:\mathbb {R} \to \mathbb {R} ,\,x\to x^{2}$ and $M\subseteq \mathbb {R}$ then the Hausdorff dimension ${\text{dim}}_{H}(M)={\text{dim}}_{H}(f(M))$, but I haven't a clue how to go about showing it. I'd prefer, if at all possible, not to use any big theorems, though if it's generally going to be quicker to actually prove and then employ the theorem than to just prove this directly, obviously that's fine. Could anyone offer some help as to how to prove this? Estrenostre (talk) 22:32, 25 April 2011 (UTC)

Prove two inequalities. For ${\text{dim}}_{H}(f(M))\leq {\text{dim}}_{H}(M)$, show that if f is a contraction mapping, then this inequality holds. Then show it's true for a Lipschitz mapping. Next, if ${\text{dim}}_{H}(f(M_{n}))\leq {\text{dim}}_{H}(M_{n})$ for all $M_{n}$ in some nested collection of sets whose union is $M$, then ${\text{dim}}_{H}(f(M))\leq {\text{dim}}_{H}(M)$. To go the other way, apply the same argument to each of $f(x)={\sqrt {x}}$ and $f(x)=-{\sqrt {x}}$ on $[0,\infty )$. Sławomir Biały (talk) 23:03, 25 April 2011 (UTC)

Righty ho, I grasped all that except for the fact that "if ${\text{dim}}_{H}(f(M_{n}))\leq {\text{dim}}_{H}(M_{n})$ for all $M_{n}$ in some nested collection of sets whose union is $M$, then ${\text{dim}}_{H}(f(M))\leq {\text{dim}}_{H}(M)$". I can't see how this is obvious: certainly if we take the 'limits' in n on the outside of the the map '${\text{dim}}_{H}$' then i can see why one limit must $\leq$ the other, but in taking ${\text{dim}}_{H}(f(M))$ and ${\text{dim}}_{H}(M)$, are we not taking the 'limits' of the M_n inside the map ${\text{dim}}_{H}$ and then saying they too must be equal, which implies some sort of continuity of ${\text{dim}}_{H}$? It's not obvious to me that if we take lots of finite intervals and then their images under f, that we can necessarily say that about the infinite union of such intervals and its image: no matter how much of the set we have already 'covered' in any given finite union, there may be infinitely more of it left, if you get my meaning... Unless the sets are nested getting smaller and smaller, but then that wouldn't make any sense. Estrenostre (talk)

Actually, they don't even need to be nested. See Hausdorff dimension#Behaviour under unions and products. Sławomir Biały (talk) 11:02, 26 April 2011 (UTC)

Ah got it, thanks ever so much! :) Estrenostre (talk) 13:38, 26 April 2011 (UTC)