Wikipedia:Reference desk/Archives/Mathematics/2007 October 12

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

October 12

Need help: grazing cow

a cow is tied to a circular silo. The rope connecting the cow to the silo is half the circumference of the silo. Upon how much area can the cow graze?

How do I start? We just finished parametric equations in class.--Mostargue 02:15, 12 October 2007 (UTC)

It depends on how a cow is tied to a circular silo. if it is tied to a fixed point or if the rope is free to move around the silo. 202.168.50.40 04:24, 12 October 2007 (UTC)

I think it's safe to assume that the point is completely fixed and is on the circle.

Be careful, the region created by the rope is not actually going to be circular, the reason why is as it gets close to the silo, the silo will get in the way of the rope and prevent the cow from having his rope in a straight line. There will be a semi-circular region the cow may graze, plus the region where the rope is effectively shortened by the silo. A math-wiki 06:27, 12 October 2007 (UTC)

It should be fairly easy to write a set of parametric equations for that curve and then use integration to find the area, I am not familiar enough with parametrics, and specifically parametric calculus to be of any real help, but Algebraicly, I may be able to solve this A math-wiki 06:41, 12 October 2007 (UTC)

EDIT: On thinking about it, disregard some of that. I just noticed that, since the rope is half the circumference, your algebraic equation will be a lot simpler. The basic equation your trying to right create a tangent line from a point (x,y) on the silo circle such that the distance from endpoint to silo PLUS the distance still against the silo is always equal to $\pi r$. A math-wiki 06:53, 12 October 2007 (UTC)

I should note that if you think about the shape, it's a well known polar equation (that I can't remember the name of!!) It looks like a circle with one side partially flattened. Hope that helps. A math-wiki 06:55, 12 October 2007 (UTC)

I believe it's a form of the rose curve, specifically one that looks like $r=\alpha sin(k\theta )$ A math-wiki 07:04, 12 October 2007 (UTC)

It seems to me that you can integrate to get the answer without knowing the shape of the curve. Consider the angle θ between the anchor-point and the point where the rope leaves the silo; you know that the length of free rope is π-θ, and it infinitesimally sweeps a triangular area ... I may come back to this later. —Tamfang 02:15, 13 October 2007 (UTC)

That's the idea, but I haven't learned much about parametrics and nothing about (parametric calculus) so the problem is I can't set up the equation. A math-wiki 05:05, 13 October 2007 (UTC)

It is the involute of a circle, which is treated as an example in our Involute article. More precisely, the perimeter of the area where the cow can roam and graze freely is formed by (1) the perimeter of the silo; (2) a half disk (whose radius is the full rope length) in front of the spot where the cow is tethered; (3) a section of the involute; and (4) the mirror image of that section. You don't need to know anything about parametrics to solve the problem: as suggested above, just integrate the infinitesimal area swept out by an infinitesimal change in angle dφ. If the free rope length is given as a function r(φ) of φ, then – by the formula for the area of a triangle – the infinitesimal triangular area swept out is one half times height r(φ) times base dφ = ¹/₂r(φ)dφ.  --Lambiam 11:41, 14 October 2007 (UTC)

First, assume a spherical cow... --Carnildo 19:30, 15 October 2007 (UTC)

Rational Number

Consider the following:

${\frac {0}{1}}={\frac {0}{2}}={\frac {0}{3}}$

Now if we have a set of

$\left\{{\frac {0}{1}},{\frac {0}{2}},{\frac {0}{3}}\right\rbrace$.

Do we have a set of 3 rational numbers or a set of 1 rational number? 202.168.50.40 05:11, 12 October 2007 (UTC)

I would say set of 1 rational number, namely, $\{0\}$ --Spoon! 05:51, 12 October 2007 (UTC)

Are you sure? I don't know much set theory, but I think that'd be a pretty sticky technical question. The rational numbers are ordered pairs of integers, right? So that set contains three elements, each an ordered pair. (Each ordered pair (a,b) is then a set of two elements, the set {a} and the set {a,b}.) I'm not getting much out of the intuitive definition, though, that an element of a set is entirely determined by whether the set contains it. Asking the question, "What rational numbers does this set contain", the answer would be "0/1". That's the only yes-answer to the question, something like 5/4 would be a no-answer. But if you asked what ordered pairs the set contains, you'd get three things that claimed they were in it. Black Carrot 06:27, 12 October 2007 (UTC)

I believe if I read the set theory section here correctly, it would actually be three. This is because (at least in naive set theory) it doesn't matter whether or not the elements are equal. A math-wiki 06:30, 12 October 2007 (UTC)

The answer is one rational number. The numbers 0 ⁄ 1, 0 ⁄ 2, 0 ⁄ 3 are all equal, so the set contains only one element (namely the number zero). With respect to the other posters, there is absolutely no legitimate argument that the set contains three elements, and this should not be considered a difficult or technical question. Trust me on this. Jim 07:44, 12 October 2007 (UTC)

What about the set {0, 0, 0, 1}? Does this contain 4 elements or 2? -- JackofOz 07:53, 12 October 2007 (UTC)

from the reply i received below, it would seem {0, 0, 0, 1} is not a set, since a set can only have each element once. (Likewise, a set could contain {} as an element 0 or 1 times, but not more times than that. {0, 1, {} } is not the same as as {0, 1}. Though I suppose you could have a set like this: {0, 1, {}, { {} }, crabapple } but not like this: {0, 1, {}, { {} }, { {} }, crabapple } since it would contain an element twice. Then again you COULD have a set like this {0, 1, {}, { {} }, { {}, {{}} }, crabapple } Folks, am I reading this right????

a related question. is the set { {}, {}, {} } a subset of {42}? Why or why not? What about simply { {} } (a set containing only the null set), as opposed to just {}? —Preceding unsigned comment added by 81.182.100.107 (talk) 08:26, 12 October 2007 (UTC)

Okay, let's take this minefield one careful step at a time:

• Our article on naive set theory says "Repetition (multiplicity) of elements is irrelevant; for example, {1,2,2} = {1,1,1,2} = {1,2}". So {0,0,0,1}={0,1} and contains just two elements as long as the three 0s are identical. If there is some way of distinguishing between the three 0s (if we can call them "blue 0", "red 0" and "green 0", for example, and keep track of which one is which if the order of elements is changed) then {<blue> 0, <red> 0,<green> 0,1} contains four elements.

• { {} } is the set containing the empty set - it contains one element, which is {}, so { {} } is not the empty set itself. The empty set {} is a subset of {4,2} but it is not a member of {4,2}. So { {} } is not a subset of {4,2} because it contains an element {} that is not a member of {4,2}.

• The set S=$\left\{{\frac {0}{1}},{\frac {0}{2}},{\frac {0}{3}}\right\rbrace$ contains three distinct fractions or ratios. Like "blue 0", "red 0" and "green 0", we can clearly distinguish between them. The fact that they are equivalent ratios and are all equal to 0 does not make them identical.

• However, if we take the image of S under the usual mapping from fractions to rational numbers, we get a set T={0,0,0}={0} with only one member, which is the rational number 0. Note, however, that T is not the same as S - it is the image of S.

So the answer to the original question "Do we have a set <meaning S> of 3 rational numbers or a set of 1 rational number?" is neither - S is a set of 3 ratios, and contain no rational numbers. Gandalf61 09:21, 12 October 2007 (UTC)

But surely the OP's set is a set of three ordered pairs $\left\{(0,1),(0,2),(0,3)\right\rbrace$ because it contains the operation of division which defines it as such? I beleive there is a case for which set operations on ordered pairs produce different results from operations on single elements, although I cannot find an example because I'm drunk. —Preceding unsigned comment added by 91.84.143.82 (talk) 10:02, 12 October 2007 (UTC)

It depends on context. See Rational number#Formal construction. In most contexts the definitions would say that { 0/1, 0/2, 0/3 } contains one rational number which could also be written as 0. If your context has definitions where 0/1 = 0/2 = 0/3 with equality and not just equivalence, then I would certainly say one rational number unless a special context is given. { (0,1), (0,2), (0,3) } contains three ordered pairs but the notation 0/1 usually implies that it's no longer an ordered pair but is either the result of a division or an equivalence class of ordered pairs. If it's defined as the result of division then 0/1, 0/2, 0/3 have the same result, just like 0+2, 1+1, 2+0 have the same result. If it's defined as an equivalence class then 0/1, 0/2, 0/3 are three ways to represent the same equivalence class which is the set of all (0, n) where n ≠ 0, and can be represented by 0/n for any n ≠ 0. PrimeHunter 10:54, 12 October 2007 (UTC)

Folks, this is trivial. Sets collapse equal elements. The first thing stated in the post is that the three elements are equal.

An anonymous poster asked about a situation involving the empty (null) set. Please note that the empty set is an object. A set containing the empty set is not empty. See Set-theoretic definition of natural numbers. --KSmrq^T 11:40, 12 October 2007 (UTC)

That's right, I asked that. (I'm not the OP). So let me get this straight: {} is a susbset of every set, but exactly once. { {} } is not usually a subset of a set, unless {} had been in the set explicitly for whatever reason (For example {3, 4, 7, {}, apple}. So, if I asked about how many subsets {42} had, the answer would be 2, namely: {} and {42} (but not { {} }, since it wasn't { 42, {} } but merely {42} which doesn't contain the empty set as a member.). Similarly, if I asked how many subsets the set {apple, banana} had the answer would be 3 ( 1: {apple} 2: {banana} 3: {apple, banana} 4: {} ). Is that right????? Thank you!

No, I would say that sets collapse identical elements and that identity is a stronger relationship than equality. So 0/1, 0/2 and 0/3 are all equal to 0 but they are obviously not identical to 0, or to one another. Gandalf61 12:28, 12 October 2007 (UTC)

Gandalf, I think this is the first time I have ever heard the claim that identity is different from equality (mathematics) (other than the word identity (mathematics) being sometimes used to refer to a certain type of theorem). Two objects are equal iff they are the same object, i.e., identical.

As for the original question: That depends on whether by ${\frac {0}{1}}$ we mean the literal string of symbols (rare) or the number represented by the string of symbols (common). Under the unusual interpretation, $\left\{{\frac {0}{1}},{\frac {0}{2}},{\frac {0}{3}}\right\}$ is a set of three distinct strings of symbols. Under the usual interpretation, $\left\{{\frac {0}{1}},{\frac {0}{2}},{\frac {0}{3}}\right\}$ is the set $\{0,0,0\}$ which has 1 element, 0.

As for the representation of rational numbers: Rational numbers are constructed as equivalence classes of ordered pairs, not just ordered pairs (where $(a,b)\equiv (c,d)$ iff $ad=bc$). So in this construction, both ${\frac {0}{1}}$ and ${\frac {0}{2}}$ are the rational number 0, which is the set $\{(0,a)|a\in \mathbb {Z} ^{*}\}$ (0 here is the integer 0).

As for the new questions of anon: I think you have a few typos but it seems more or less right. If a set has $n$ elements then it has exactly $2^{n}$ subsets. So, under the assumpion $\mathrm {apple} \neq \mathrm {banana}$, $\{\mathrm {apple} ,\mathrm {banana} \}$ has exactly 4 subsets. -- Meni Rosenfeld (talk) 13:01, 12 October 2007 (UTC)

Why can't apple == banana? For example, if x * y = 0 and x + y = 0 then there IS no set {x, y, {} } ? (A set containing the variable x, the variable y, and the empty set? ) —Preceding unsigned comment added by 84.0.126.201 (talk) 13:21, 12 October 2007 (UTC)

There is a set $\{x,y,\{\}\}$, but it has 2 elements - 0 and $\{\}$ (even worse, 0 is sometimes defined as $0=\{\}$, so with this definition, the set has only 1 element). -- Meni Rosenfeld (talk) 13:30, 12 October 2007 (UTC)

Gandalf: You may want to take a look at the article on extensionality. Mathematics doesn't make the distinction you insist upon. For example, the functions $x\mapsto (x-1)^{2}$ and $x\mapsto x^{2}-2x+1$ are considered to be identical even though their presentations (intensions is the technical term) differ. Morana 13:19, 12 October 2007 (UTC)

So how many members does the set $\left\{{\frac {x}{1}},{\frac {y}{2}},{\frac {z}{3}}\right\}$ contain ? I say it always contains three members, which are three different and distinct expressions (strings, if you like). Depending on the values of x, y and z etc. these expressions may or may not be equal to the same rational number - but the set always contains three members. To make the size of the set $\left\{{\frac {x}{1}},{\frac {y}{2}},{\frac {z}{3}}\right\}$ depend on the values of x, y and z seems illogical to me.

Responding to Morana's example, why do you say the functions $x\mapsto (x-1)^{2}$ and $x\mapsto x^{2}-2x+1$ are identical - is it because they are equal everywhere if x is restricted to some domain such as the real numbers ? The functions $x\mapsto (x-1)^{2}$ and $x\mapsto x^{2}+1$ are equal everywhere in a field of characteristic 2 - does this make them identical ? I would say not. Are the functions $x\mapsto cx$ and $x\mapsto xc$ identical ? They are equal everywehre in an Abelian group - but not in a non-Abelian group. Is 10-3 identical to 7 ? They are equal in decimal, but not in octal. Is 1000 identical to M ? Bottom line is equality depends on domain, context and interpretation; identity does not. Gandalf61 14:39, 12 October 2007 (UTC)

But if you don't restrict the domain, you don't have a function at all. In the ZFC framework, a function is a certain type of set. If by "$x\mapsto x+1$" you mean "the collection of all ordered pairs where the second element is the first object plus 1, where "plus" and "1" are as appropriate within the context", not only do you get something ambiguous, but you obtain a collection larger than any set (i.e., a proper class), so it is certainly not a function.

Morana probably meant the function $f:\mathbb {R} \to \mathbb {R} ,x\mapsto (x-1)^{2}$, which is definitely the same function (that is, the same set) as the function $f:\mathbb {R} \to \mathbb {R} ,x\mapsto x^{2}-2x+1$.

You are saying that the size of $\left\{{\frac {x}{1}},{\frac {y}{2}},{\frac {z}{3}}\right\}$ should not depend on the values of x, y and z. Would you also say that the size of the set $\{x,y\}$ does not depend on the values of x and y? -- Meni Rosenfeld (talk) 15:21, 12 October 2007 (UTC)

Anyway, the statement "+1 and -1 are not identical", in the sense you mean it here, is entirely nonmathematical; at best, it is a typsetting statement, which specifies that the collections of pixels + and - do not represent the same mathematical symbol. However, the statements "the real numbers +1 and -1 are nonequal" and "the GF(2) numbers +1 and -1 are equal" are mathematical statements. -- Meni Rosenfeld (talk) 15:28, 12 October 2007 (UTC)

You asked "Would you also say that the size of the set $\{x,y\}$ does not depend on the values of x and y?". Yes, I would. The set $\{x,y\}$ always contains two members, regardless of what values we assign to x and y. If we set x=y=0 then we transform $\{x,y\}$ to the set $\{0,0\}$, which only has one member, 0, but this is no longer the set $\{x,y\}$. Would you say that you don't know how many members the set $\{apple,banana\}$ contains because maybe in some world apple=banana ? Does the set $\{1000,{\mbox{one thousand}}\}$ contain one member for English speakers but two members for non-English speakers ? If tomorrow is my 30th birthday, does the set $\{29,{\mbox{my age in years}}\}$ contain one member today but two members tomorrow ? I say that each of the sets $\{x,y\}$, $\{apple,banana\}$, $\{1000,{\mbox{one thousand}}\}$ and $\{29,{\mbox{my age in years}}\}$ always contains two members, regardless of context. These two members may sometimes be equal, but they are never identical. Gandalf61 15:45, 12 October 2007 (UTC)

I would rather not discuss the set $\{\mathrm {apple} ,\mathrm {banana} \}$, as neither are mathematical objects. A slightly better example is $\{10^{9},\mathrm {billion} \}$. This is a string of Ascii symbols. When written by a Frenchman, it is meant to represent the set which has $10^{9}$ and $10^{12}$ as elements. This set has 2 elements. When written by an American, this string of symbols is meant to represent the set which only has $10^{9}$ as an element.

Likewise, the string $\{x,y\}$ can mean the singleton $\{1\}$ if $x=y=1$ or the set $\{1,2\}$ if $x=1,y=2$.

So a string of symbols can refer to different sets depending on the context. But it is not a set if it is not given the required context, and being not a set, you cannot discuss its number of elements. What you can do is discuss the number of elements of the sets which may be represented by it under different circumstances. -- Meni Rosenfeld (talk) 16:25, 12 October 2007 (UTC)

<rmv indent>Meni, you seem to be identifying the set $\{x\}$ with $\{{\mbox{the number represented by }}x\}$ - but, as you have realised, $\{{\mbox{the number represented by }}x\}$ is not a set. You have no way to determine whether a given number is a member of $\{{\mbox{the number represented by }}x\}$. You can't tell how many members $\{{\mbox{the number represented by }}x\}$ contains - what if x is "the second smallest even prime number". And you can't even define a rule for telling how many members $\{{\mbox{the number represented by }}x\}$ contains - what if x is "the smallest odd perfect number". If we want $\{x\}$ to be a set (which seems natural to me) then the simplest way to avoid this confusion is to say that $\{x\}$ is indeed a set with a single member, the expression "x", which is entirely independent of whatever value (if any) we might assign to x. Similarly, the set $\left\{{\frac {0}{1}},{\frac {0}{2}},{\frac {0}{3}}\right\}$ is a set containing three expressions, and it is distinct from the set of the rational numbers represented by ${\frac {0}{1}},{\frac {0}{2}},{\frac {0}{3}}$, which is the singleton set $\left\{0\right\}$. Gandalf61 18:53, 12 October 2007 (UTC)

Yes, I am taking the common intepretation that when a string of Ascii characters is written in a mathemacal statement in a natural language, it means the object represented by the string rather than the string itself. Otherwise I wouldn't be able to say that 1+1 is equal to 2, as the former is a string of 3 symbols and the latter of 1 (and no, I still don't agree that equality is distinct from identity). "1+1" is different from "2" (the strings are different) but 1+1 is the same as 2 (the numbers are the same. And I do mean the integers 1 and 2 and the ordinary addition on integers).

And yes, again in the framework of ZFC, where all objects are sets, when I write $\{x\}$ I mean that the letter "x" is a symbol representing some (currently unspecified) set, that when I speak about x I am speaking about that set, and that $\{x\}$ is the set containing a single element, that element being x.

x cannot be the second smallest even prime positive integer, as there is no such integer. No matter what x is, $\{x\}$ is always a singleton (contains exactly one element). -- Meni Rosenfeld (talk) 19:09, 12 October 2007 (UTC)

So, Meni, under your interpretation of $\{x\}$, is $\{x\}$ still a singleton set when x is "the smallest odd primeperfect number" ? How about when x is "the counterexample to the Riemann hypothesis with the smallest absolute value" ? Gandalf61 19:21, 12 October 2007 (UTC)

I'm assuming you meant "perfect"... And you can't say "x is the smallest odd perfect number" unless you have already proven that this number exists and is unique. It's just not a valid mathematical definition. Compare {x|x is a perfect odd number less than all other perfect odd numbers}, which is a well-defined set but I just don't currently know if it has 0 or 1 elements (and I guess it may not even be decidable in ZFC). -- Meni Rosenfeld (talk) 19:50, 12 October 2007 (UTC)

You seem to be putting rather a lot of restrictions around how I can and cannot define x in order to make the set $\{x\}$ fit your rules. Can you tell me which definitions of x you will allow ? Is there a rule that I can follow so that I can be certain I only produce valid definitions of x ? Gandalf61 20:17, 12 October 2007 (UTC)

???

That restriction is maths 101 stuff, at least for the kind of math I know. If we can't agree on that then I really don't know what else to say. -- Meni Rosenfeld (talk) 20:25, 12 October 2007 (UTC)

Meni, I am happy to agree with whatever rules you put forward. I simply want to be sure that I understand your rules so that I can explore their logical consequences with confidence. So, I think you will allow x to be "n|n is a perfect odd number less than all other perfect odd numbers", but in this case you can't say whether $\{x\}$ has 0 or 1 elements. Yet a little while ago you said "No matter what x is, $\{x\}$ is always a singleton (contains exactly one element)". Isn't there an inconsistency there ? Gandalf61 20:38, 12 October 2007 (UTC)

But I am not putting forward any rules. I am stating as obvious rules and notations which I have learned during my first year of mathematical education and that are consistent, both explicitly and implicitly, with everything I have learned since then. If your perspective on mathematics is so different that you have to actively agree to agree with them, then we are clearly each talking about different kinds of mathematics, making this futile.

And "I" certainly do not allow x to be "n|n is a perfect odd number less than all other perfect odd numbers". I am using a standard notation where $\{x\}$ is the singleton containing the element x, and {n|n is a perfect odd number less than all other perfect odd numbers} is not a singleton containing the element n|n is a perfect odd number less than all other perfect odd numbers (indeed, that doesn't even make any sense) but rather the set of every object which is an odd perfect number less than all other perfect odd numbers. -- Meni Rosenfeld (talk) 20:51, 12 October 2007 (UTC)

Gandalf: There's a fairly fundamental point here that you're missing. Sets, as contemporarily conceived, are determined by their elements, not by the rule (if any) that determines whether an element is included. They should be thought of as just grab bags full of objects. Sometimes one of these bags happens to contain exactly those objects that satisfy some rule, and in that case you can use the rule notation for defining the sets. But that isn't what makes a set. To use technical language, sets are extensional, not intensional.

So if there aren't any odd perfect numbers, then {the least odd perfect number} is a symbol that simply fails to denote anything. But you can't actually know whether it denotes anything, without knowing whether there's an odd perfect number. --Trovatore 20:55, 12 October 2007 (UTC)

The "grab bag" analogy sounds nice and simple, but it doesn't work. If a set were like a grab bag then you could reach into the set {n|n is the smallest prime factor of RSA-2048), pull out a member (it definitely has exactly one member, so how hard can it be ?) and you are $200,000 richer. Or you could reach into each set in an uncountable set of sets, pull out a member, and show that the axiom of choice does not need to be an axiom. Of course, you can't do either of those things, because sets are not just grab bags. And my problem with Meni's position (if I understand it correctly) is that he seems to accept that $\{x\}$ is always a set containing the single member x, regardless of how x may be defined, but he doesn't seem to accept the consequence that $\{x,y\}$ is always a set containing exactly two members x and y, regardless of how x and y may be defined. And that is, I think, inconsistent. Gandalf61 21:54, 12 October 2007 (UTC)

Hm? It isn't a physical grab bag; you can't put your physical hand in it. And even if you could put your hand in it, if someone would give it to you, how do you know anyone will?

I think you have a basic issue here with objects versus names for objects. What you seem to want to do is identify objects with their names. But you can't do that; you have to keep them separate and straight. RSA-2048 has a least prime factor, which is an abstract object existing in the Platonic heaven. It has various names. One of them is "the least prime factor of RSA-2048", which won't win you the prize. Another is some decimal representation, though we don't know which; that one will win you the prize. But the object denoted is the same.

With the name (or more properly description) "the least odd perfect number", it's unclear whether it names anything at all (the smart money, I think, is that it doesn't). So if the phrase "the least odd perfect number" fails to denote, then clearly "the set whose unique element is the least odd perfect number" also fails to denote.

When Meni was talking about {x}, though, x was not a description, but rather a variable, ranging over all objects (or perhaps all mathematical objects). For any value of x, the set {x} has exactly one element; that's a true statement. But you can't replace x by a description here, because that's a different level of discourse. --Trovatore 22:09, 12 October 2007 (UTC)

Oh, not a physical grab bag ? Some sort of non-physical grab bag then ? Some sort of grab bag that you can't always pull things out of, and you don't always know what it contains, or how many objects it contains, or even whether it contains any objects at all ? And when you do manage to pull something out, you're not always sure what you have got, because you might find you are holding a name, not an object ? So it's not really very like a grab bag at all, then.

But, just to make sure I correctly understand the distinction between objects and names, perhaps you can answer a simple question for me. If x and y are variables, then does {x, y} stand for the set containing the variables x and y ? Or does {x, y} stand for {the object represented by the variable x, the object represented by the variable y} ? Gandalf61 22:41, 12 October 2007 (UTC)

To your first questions, yes, a non-physical grab bag, but there is never any ambiguity about how many objects it contains or what they are. Your state of knowledge about these things is another matter, of course; just because you don't personally know truth doesn't mean there is no truth.

To the second question: In this context, {x,y} means the set that contains the value of the variable x, the value of the variable y, and nothing else; it may have cardinality either 1 or 2 (depending on whether x and y have the same value). If you wanted a set containing the literal variables x and y, that would also be a legitimate set, but you'd need some other notation for it. --Trovatore 22:52, 12 October 2007 (UTC)

Then, as we do not have definitions for x and y, or any way of deciding what their values are or whether they are equal, I believe your {x,y} is not well enough defined to be called a set. In other words, your {x,y} has no extensional definition, so it is not a set. It is not just that neither you nor I know what is in your grab bag - there is no conceivable way of ever telling what is in your grab bag or what is not in your grab bag. See my reply to Meni below. Gandalf61 23:03, 12 October 2007 (UTC)

This objection of yours has nothing to do with sets. It's just about variables. Take the symbol x for the moment to be a variable restricted to range over the real numbers. Analogously to your objection, we might as well say that x+1 is not a real number, because we have no way of telling whether it's greater, less than, or equal to zero (simply because we don't know what x is). --Trovatore 23:19, 12 October 2007 (UTC)

rmv indentYes, Trovatore, you have reached the truth. x+1 is not a real number - it is a variable that represents a real number. Although we may loosely and informally say that x+1 is a real number, this is not strictly correct, because every real number is greater than, less than or equal to 0, whereas x+1 is at the same time greater than, less than and equal to 0, depending on how x is defined. It is not just that we do not know whether x+1 is greater than, less than or equal to 0 - it is fundamentally unknowable. Even if we had instant divine access to all of Platonic heaven, we would not know whether x+1 was greater than, less than or equal to 0, until and unless we defined x. Therefore x+1 cannot be a real number. In the same way, your {x,y} is not a set - it is a variable representing a set. If it were a set then it would have either one member or two members in Platonic heaven. Because it has both one and two members, depending on how x and y are defined, then it cannot be a set. QED. Gandalf61 06:43, 13 October 2007 (UTC)

Well, you're not really going to be able to communicate with most mathematicians if you can't agree with the proposition "if x is a real number then so is x+1". If you want to pursue this line of thinking, though, you might be interested in the work of Kit Fine on what he calls "variable objects". It never made a lot of sense to me, but maybe it would resonate with you. --Trovatore 06:54, 13 October 2007 (UTC)

Trovatore, you are setting up a straw man. I completely agree with the proposition "if x is a real number then so is x+1". My whole point is that an undefined x is not a real number - it is a variable representing a real number. If x were a real number then in Platonic heaven x would be greater than, less than or equal to 0. Undefined x is not greater than, less than or equal to 0, even in Platonic heaven (or at best it is greater than, less than and equal to 0, all at the same time), therefore undefined x cannot be a real number. Gandalf61 07:30, 13 October 2007 (UTC)

To the extent your comments are not just a silly quibble, which is frankly what I think they mostly are, they seem to be an attack on the standard treatment of variables. Variables are a bit difficult to deal with philosophically; I grant that. However this difficulty never winds up causing me any trouble, so I mostly just don't bother with it. If you want to bother with it, go ahead, knock yourself out. But there's nothing different about sets in this connection, and I'd appreciate it if you'd avoid muddying the issue as it relates to sets, which confuse people enough as it is. --Trovatore 07:41, 13 October 2007 (UTC)

Oh yes, I completely agree that for almost all pratical purposes the distinction between the variable x and the value of the variable x is not significant. It is just a little piece of mathemtical precision that I happen to find interesting. But one man's precision is the next man's "silly quibble". If you don't wish to "bother with it" then there is no reason why you should have to - don't worry about it, and have a good day. Gandalf61 08:09, 13 October 2007 (UTC)

Gandalf, these variable problems will disappear as soon as you use a perfectly formal system. Mathematicians are very lazy. Whilst they are, in principle, working under the very strict rules of a formal system; to write out proofs of theorems from the axioms, following the rigid symbol-shunting rules of logic, would be so tedious that nobody would ever get anything useful accomplished because they'd be so busy writing out tiresome and astonishingly long proofs of basic, obvious facts. So we relax the rigour requirement. Even a proof that is considered highly rigourous and formal by mathematicians is really just a sketch of a proof. To get from the premises of some theorem to the difficult-to-accept conclusion, a mathematical "proof" simply breaks the idea down into a lot of small, easy-to-swallow steps. It is (hopefully) obvious to anybody reading the proof that the small steps could be broken into even smaller, easier-to-swallow steps if it was so desired; and so on and so on until, if you so desired you could (in principle) do it from the axioms of your logical framework. Note: for any even moderately interesting result, a completely rigourous proof from the axioms would be absolutely horrifyingly awful.

Anyway, when dealing with a proper mathematical formal system, there is never any ambiguity about constants, variables, and so forth. We make that kind of distinction primarily to keep mathematical discourse interesting and comprehensible. All variables are properly quantified, the ambiguity disappears. It's more a case of convention than a genuine mathematical issue. If you're interested in knowing more about this kind of stuff, read up on first order logic to get an idea for how variable quantification and formal systems work. Maelin (Talk | Contribs) 12:31, 15 October 2007 (UTC)

Maelin - we can take everything you have said as read; I have no quarrel with any of that, and I am very familiar with formal systems and with both formal and informal proofs. Given all that - have many elements do you think are contained in the set {x,y} ? Remember that x are y are free variables to which we have not yet assigned values. Do you agree with Meni that {x,y} may have either 1 or 2 members, depending on whether or not x=y when (or if) we eventually assign values to x and y ? How can {x,y} then even be a set if it has an indeterminate number of members (not just unknown to us - fundamentally undetermined, even in a Platonic sense) and therefore no extensional definition (not even an unknown Platonic one) ? Is it some "quantum" shadow of a set that somehow collapses to a specific set if and when we assign values to x and y ? Could you formalise such a definition ? Or do you perhaps agree with Trovatore, who thinks that these questions are a "silly quibble" ? Gandalf61 15:00, 15 October 2007 (UTC)

Is it not clear by now that $\{x,y\}$ is a set, but we don't know which set? It's not some mysterious, magical set which changes its number of elements - it is some set, but we don't know which set it is unless we know what x and y are. And the notation can refer to different sets depending on the values of x and y.

I agree with Maelin's point that, in any sentence in a formal system, all variables are quantified so we don't even need to worry what happens with an undefined variable. -- Meni Rosenfeld (talk) 15:06, 15 October 2007 (UTC)

Meni, you say that $\{x,y\}$ is a set, but we just don't know which set until we assign values to x and y. Okay, let's assign values to x and y and see what happens - I will make x=1 and y=2. Now $\{x,y\}=\{1,2\}$. But now I change my mind and decide to make x=1 and y=1. Now $\{x,y\}=\{1\}$. Clearly {1,2} and {1} are not the same set. So what has happened ? Did $\{x,y\}$ suddenly change halfway through the paragraph ? A set that can change its members - and not just its members, but its size as well - that seems like a very strange set to me. Or have I changed the definition of $\{x,y\}$ by assigning values to x and y - in which case we cannot use the properties of $\{x,y\}$ after assignment to infer properties of $\{x,y\}$ before assigment. Or could it be that $\{x,y\}$, with your definition, is not really a set at all ?? Gandalf61 16:07, 15 October 2007 (UTC)

But y can't be equal to both 1 and 2 at the same time. It could be that in some paragraph or part of an argument, y=2 and then "$\{x,y\}$" denotes the set {1, 2}. In some other place, you could have y=1 and then "$\{x,y\}$" denotes the set {1}. So, for the nth time, the string of symbols "$\{x,y\}$" can denote different sets when it appears in different contexts. But in any given context, when x and y have given values (be those explicitly specified values, or unspecified values by way of quantification), "$\{x,y\}$" denotes some particular set and $\{x,y\}$ is that set. In the programming analogy, "$\{x,y\}$" is a pointer that can point to different sets at different times, but it never points to a shapeshifting set (as such a thing certainly does not exist). -- Meni Rosenfeld (talk) 16:21, 15 October 2007 (UTC)

<rmv indent>Exactly !!! Your $\{x,y\}$ points to a set but it is not itself a set. It is not the same as $\{x,y|x=1,y=2\}$, neither is it the same as $\{x,y|x=1,y=1\}$, which are both legitimate sets. In certain contexts your $\{x,y\}$ may point to either $\{x,y|x=1,y=2\}$ or $\{x,y|x=1,y=1\}$ or to any other set with either one or two members - but that does not make your $\{x,y\}$ a set. That is the difference between sense and reference. I think my work here is done. Gandalf61 16:39, 15 October 2007 (UTC)

Um, sorry to burst any bubbles, but I was extra careful to distinguish "$\{x,y\}$" (without the actual quotation mark, which are used to symbolize holding back following pointers), a string of ascii characters understood as pointing to a set, and $\{x,y\}$, a set. This is a distinction I have made uncountably may times, which you have chosen to ignore. I get the feeling that my work here will never be done, and now is a good time to (re-)agree to disagree. -- Meni Rosenfeld (talk) 16:46, 15 October 2007 (UTC)

Oh, I see your quotation marks now. I would agree that "$\{x,y\}$" is a string of 5 characters. However, you say "$\{x,y\}$" is both a string of characters and a pointer to a set. That is as if you said the variable x and the 24th letter of the English alphabet were one and the same thing. Or you confused a painting with its paint. How very strange. Gandalf61 22:47, 15 October 2007 (UTC)

Surely we must want the English letter "x" to point to something? If it is just a letter and nothing more, what business do we have even mentioning it in a mathematics argument? Perhaps your point in this whole debate was that English letters, being such, must never be used in mathematics? You could, if you really wanted, say that the letter points to a variable, and the variable points to a set, but what's the point (pun unintended)? -- Meni Rosenfeld (talk) 00:38, 16 October 2007 (UTC)

Because that is how things are and how they work and it avoids the confusions that are inherent in the mixing of levels in your scheme. Continuing the analogy, the paint, the painting and the subject of the painting are three different and separate things. See Magritte's The Treachery of Images. Gandalf61

Fine, then "x" is an English letter pointing to the variable x-, and x- is a variable pointing to the set x. Funny how you accuse "my" scheme of mixing levels, when I am the one who was careful about distinguishing levels in this debate. -- Meni Rosenfeld (talk) 15:45, 16 October 2007 (UTC)

[ec] "I" am very consistent about $\{x\}$ being, for every x, the unique set satisfying the following requirements:

• x is an element of the set.
• For every a, if a is not x then a is not an element of the set.

and about $\{x,y\}$ being, for every x and y, the unique set satisfying the following requirements:

• x is an element of the set.
• y is an element of the set.
• For every a, if a is not x and a is not y then a is not an element of the set.

If $x=y$, that is, x and y are the same set (or other object), then $\{x,y\}$ is the unique set satisfying:

• x is an element of the set.
• x is an element of the set.
• For every a, if a is not x and a is not x then a is not an element of the set.

And simple logic (though I am now dumbfounded as to what can be considered "simple") shows that this is the unique set satisfying:

• x is an element of the set.
• For every a, if a is not x then a is not an element of the set.

And this set (namely, $\{x,y\}$) contains exactly one element, x (which is the same as y).

"I" am also consistent about the fact that a set that can be described by braces, within which we have n variables separated by commas and nothing else, can have anywhere from 1 to n elements, depending on how many distinct values exist among these variables. -- Meni Rosenfeld (talk) 22:15, 12 October 2007 (UTC)

I see. Then I believe your $\{x,y\}$ is not a set at all, because without definitions for x and y you cannot say what objects are members of $\{x,y\}$, whether a given object is a member of $\{x,y\}$, or even how many members $\{x,y\}$ has. At best your $\{x,y\}$ is some sort of variable standing for an indeterminate set that has not yet been defined. Gandalf61 22:53, 12 October 2007 (UTC)

A set can be perfectly well-defined even though you have no practical way of knowing its elements. For example, if you have a set $A$ and a predicate $P$ the set $S:=\{x\in A|P(x)\}$ is well defined. The key idea is that in principle, for any $x\in A$, either $P(x)$ or $\neg P(x)$, therefore either $x\in S$ or $x\not \in S$. And that is all it takes to make the set well-defined. The existence of a decision procedure for membership is not a requirement. This make the concept of set very flexible. It is true that some people, like intuitionists for example, do not accept this concept of set but this is a minority position. Morana 23:22, 12 October 2007 (UTC)

That's exactly right, except that I think the term "well-defined" is confusing and unhelpful here. It's quite natural to say, if it's so well-defined, what exactly is the definition? The point is that the existence of the set does not depend on its definability. We should not shrink from using Platonistic language here. You don't have to actually believe that there's a Platonic heaven -- the observed fact is that talking as though there were is the approach that works, the one that avoids descending into all sorts of unproductive confusions such as the one evidenced in this thread. --Trovatore 23:30, 12 October 2007 (UTC)

Actually it appears that I didn't read Morana's contribution carefully enough -- he actually did give a definition in this case. It doesn't change the larger point though. --Trovatore 23:43, 12 October 2007 (UTC)

But in Meni's $\{x,y\}$ there is no predicate. It is not just that we do not have a decision procedure for the predicate - there is absolutely no predicate at all. So if I ask "is 2 a member of $\{x,y\}$", there is no possible way to even begin to answer this question, even if we had instant access to the whole of Platonic heaven. 2 both is and is not a member of $\{x,y\}$, depending on how we define x and y. It is like asking "is the integer that I will write down at 12:00 tomorrow even or odd ?". The question is fundamentally unanswerable, even in Paltonic heaven. This confuses us if we try to treat "the integer that I will write down at 12:00 tomorrow" as a specific, although unknown, integer. The confusion disappears once we realise that "the integer that I will write down at 12:00 tomorrow" is not an integer - it is a variable representing an integer. In exactly the same way, Meni's $\{x,y\}$ is not a set - it is a variable representing a set. Questions such as "is 2 a member of $\{x,y\}$" or "does $\{x,y\}$ have one member or two" are fundamentally unanswerable (or at best have an indeterminate "yes and no" answer), but that is okay because $\{x,y\}$ is not a set. Gandalf61 07:16, 13 October 2007 (UTC)

I believe your confusion comes from the fact that you take on one hand the convention that $x$ is a number (and not a symbol) and on the other the convention that $\{x;y\}$ is a symbol of a set (and not a set). You cannot have it both way. Either you interpret what we write as identified with abstract objects (the common convention) or as symbols denoting abstract objects. If you apply either convention consistently you'll see that there are no issues. Morana 08:04, 13 October 2007 (UTC)

Indeed. For myself, I would like to stick to the convention that x and y are variables (symbols, if you like) and that $\{x,y\}$ is the set containing the two variables x and y and nothing more. With that convention there is no confusion. It is Meni who is insisting that $\{x,y\}$ contains the values of the undefined variables x and y - although he cannot say what those values are, or even whether $\{x,y\}$ contains one member or two. My contention is that Meni's $\{x,y\}$ is not a set, and my questions were aimed at highlighting the confusion inherent in his position when he tries to treat his $\{x,y\}$ as if it were a set. For my part I am not at all confused, but thank you for your concern. Gandalf61 08:25, 13 October 2007 (UTC)

And "I" (which here means the mathematical community as I know it) would like to stick to the convention that x and y are sets, and $\{x,y\}$ is the set containing x, containing y and nothing else. We do need to know more about x and y if we wish to know more about $\{x,y\}$, but that is obvious - the same way we need to know about some number x if we wish to know about $x+1$, and that we also need to know about the number x if we wish to know whether $x>0$, and we need to know the location of Spain if we wish to get there, and we need to know the individual digits of the decimal expansion of the smallest prime factor of RSA-2048 if we wish to get 200000$.

Your highly unusual interpretation might be consistent (I still have no idea how one can do any mathematics this way, though you are a living example that it is possible), but I can't accept you saying that "my" interpretation is inconsistent. -- Meni Rosenfeld (talk) 11:36, 13 October 2007 (UTC)

To back up my claim that your interpretation is unusual, I can offer a little game - I will find literary references reading "let x be a real number", and you will find "let x be a variable representing a real number". -- Meni Rosenfeld (talk) 11:41, 13 October 2007 (UTC)

<rmv indent>Oh, I have no argument with usage. "Let x be an integer" is obviously a pragmatic and standard shorthand for "let x be a variable representing an integer". The variable x is not an integer, but this is usually obvious from context. The student in their first algebra class who responds with "So is x even or odd ?" is actually asking a perceptive question, because they are examining the consequences of treating the shorthand literally - altough they are likely to be rewarded with the non-answer "I don't know. It could be either. It doesn't matter". Over time we become accustomed to the informal usage and cease to notice it. Note, however, that this is a different usage from "let x be 7" or "let x be the smallest prime factor of RS-2048" - in these cases x certainly is an integer. Even though in the second example we don't know the exact value of x, we are confident that it is sitting up there somewhere in Platonic heaven. The distinction between a variable and its value is not often significant in mathematics. However, it is an important distinction for computer programmers - "let x be an integer" is the equivalent of assigning a data type to x, but x has no value at this point, and the programmer will get an undefined variable error if they forget this. And the distinction has a long and venerable history in philosophy - see our article on sense and reference.

I have another question for you. You now say that x and y in your $\{x,y\}$ are sets. Do you mean they are literally sets, or do you mean they are variables representing sets ? If x is an undefined variable representing an integer (or, in shorthand, x is an integer but we haven't decided which integer it is) is x literally a set, and if so, then what set ? Gandalf61 14:29, 13 October 2007 (UTC)

I've said this before. The English letter "x" plays here the role of a variable representing a set. As such, when I write x, I mean the set represented by the variable denoted with the English letter "x". So x is the set represented by "x", and the set represented by "x" is a set, so x is a set.

Now that you have mentioned computer programming, we can talk about pointers. This depends on the specific language, but I will try to describe the general ideas. If I define the variable "x" to be of the data type (4-byte) Integer, then there is effectively an arrow from the letter "x" to a box in memory of size 4 bytes, and that box can contain integer values. By specifying the data type I effectively choose the size of the box, but I have not yet specified its content.

Now, let us assume at first that our compiler uses the symbol + to denote following an arrow (this is mostly hypothetical, I have not seen just such a compiler). I can then give the command 'Print x' and the literal letter "x" (without the quotation marks) will be printed. I can also give the command 'Print x+' and the contents of the box to which "x" points will be printed.

Now, assume that "x" is a variable of type Pointer. Then the box to which "x" points contains a memory address, an arrow to another box. I can then type 'Print x' and the letter "x" will be displayed; I can type 'Print x+' and the address in the box pointed to by "x" will be displayed; And I can type 'Print x++', and the contents of the box in the address appearing in the box to which "x" points will be displayed. This can be generalized to higher depths of pointers.

But it is plausible to assume that if we use pointers a lot, it will be uncomfortable to use all those +'s. It is sensible to give the compiler a different directive; Whenever it encounters an arrow, it automatically follows it, and continues following arrows until it finds something that is not an arrow. If we do want to work with the addresses themselves, we can use the minus symbol to tell the compiler to hold back one level of following arrows. In this setting, 'Print x' will follow arrows all the way until the contents of the box in the address appearing in the box to which "x" points; 'Print x-' will give the address in the box to which "x" points, and 'Print x--' will print the literal letter "x".

I could then describe what happens when you perform assignments, but let us return to mathematics, where:

• Multiple levels of pointers are not generally used;
• The second approach is used.

So, when I speak about x, I am speaking about what would be called x+ in the first convention, or x in the second convention. When I say $x=3$, I don't mean that the English letter "x" will henceforth be the number 3, but rather that x, the real number represented by "x", is 3. If I want to discuss the letter "x" or the variable "x" themselves, I use the quotation marks, which have the same effect as the minus symbol above.

And no, this is not a notational shortcut. This is what those things are really supposed to mean. -- Meni Rosenfeld (talk) 15:05, 13 October 2007 (UTC)

Oh, I have just reread your comment 'one man's precision is the next man's "silly quibble"'. There is nothing imprecise about the common perspective, and I am not sure there is anything precise about your own perspective. Perhaps your personal perspective is just the first approach I have described to pointers in disguise, but I am not sure I have seen you use it consistently, and I have not seen you provide any notation (which is absolutely mandatory) for "following arrows". Without such a notation, you can't even discuss the values which variables are capable of representing. -- Meni Rosenfeld (talk) 15:32, 13 October 2007 (UTC)

Meni, I am glad that you understand programming terminology, because the following analogy should make sense to you. A variable is like a pointer. An undefined variable is like a null pointer. Making a statement about the value of an undefined variable is like trying to dereference a null pointer, which generates a run time error. The set $\{x,y\}$ where x and y are undefined variables is like a set of two null pointers. It cannot contain anything referenced by these pointers because they are null pointers, so they cannot be dereferenced. The set $\{x,y\}$ always contains exactly two members, x and y - I cannot say "but it only contains one member if x=y" because that test involves dereferencing the null pointers, which I cannot do.

This might also clear up the issue about the difference between equality and identity that we encountered earlier in this discussion. If I assign values to x and y then I can now test whether x=y. If x and y dereference to the same value then x=y is true - x and y are now equal. However, they are not identical - they are still two separate pointers, and I can assign a new value to x without affecting y. Some programming languages support a test specifically for identity as distinct from mere equality. Does all that make sense ? Gandalf61 17:48, 13 October 2007 (UTC)

Yes and no. Mostly no, because I disagree on at least two points:

• If I state that x is a real number but have not stated any specific value for it, it is not like a null pointer - it is a pointer that points to a specific address in memory, and in that address appears a real number - I just don't know what that real number is.
• It looks like you are saying, 'x and y are equal if they point to the same value, and identical if they are the same pointer'. I find this to be unnecessarily confusing terminology, since in both cases you write x and y the same way, and yet one time you speak about the value to which they point, and one time you speak about the pointer itself. It makes much more sense to consistently mean one or the other, and use some specialized notation to refer to the other as necessary. For example, you can, as is usual in mathematics, write x when you mean the referenced value, and should the need arise (but the need never arises), use x- or "x" to refer to the variable, letter, pointer or whatever. Another example is to use x to mean the variable and x+ to mean the value. But, since mathematical statements always speak about the values (mathematics is about abstract objects, not about the letters used to represent them), the result is simply adding + to every appearance of a letter in the mathematical literature. (For example, if x and y are just pointers to the real numbers say 2 and 3, then what is xy? How can you "multiply" pointers? You can only multiply the real numbers referenced by those pointers, which should, in your system, be denoted x+y+ or whatever).

So what you call 'x and y are equal', I would call 'x and y are equal\identical' (x is the value referenced by x-, y is the value referenced by y-, and in this case they are the same value). As for your typographical observation, 'x and y are not identical' (as in, not the same variable, letter, pointer, symbol), I would have said 'x- and y- are nonequal\non-identical'. But there is never a need to make such a trivial, non-mathematical comment. -- Meni Rosenfeld (talk) 18:28, 13 October 2007 (UTC)

In fact, we can say more. I am assuming we are working in ZFC. Even without saying anything, we know that "x" is a pointer referencing a specific box (since we know that English letters are used to denote variables), and that in this box is a set. We also know that "y" is another pointer, referencing another box, within which is also a set (since we know that "x" and "y" are different English letters). But we currently know nothing about the set in the box pointed to by "x". But since mathematical theorems start with assumptions and end with conclusions, we can learn more about what, in our current context, is the set in that box. For example, if we say "let x be a real number", we know that the set in this box is an equivalence class of Cauchy sequence of rational numbers. If we say "let x be a positive real number", we know that this number is also positive. If at some point we say "from our previous calculations we can conclude that $x<2$", we also know that it is less than 2. We still don't know exactly what that set is, and we still can't answer questions like "is this set a real number greater than 1", but we know a lot more than when we started, and we can answer questions like "is this set a real number less than 3". -- Meni Rosenfeld (talk) 18:58, 13 October 2007 (UTC)

Okay, I think I understand where you are coming from. For you x always has a value even if we don't know what it is, and the question "is x positive" has an answer in Platonic heaven, even if we don't know what the answer is. For me an undefined x has no value, and the question "is x positive" is meaningless. For me, an undefined variable is a null pointer; an uninstantiated class; an abstraction of a number, but definitely not a number. It's like the photon in the double-slit experiment - it's not just that we don't or can't know which slit it went through, we can't even meaningfully ask which slit it went through. It Well, this thread has become over long and its probably time to close the discussion, but it's been interesting. Gandalf61 21:37, 13 October 2007 (UTC)

Apart from the answer given at the beginning - being the obvious one - doesn't it help to define what exactly the sets are containing eg numbers or ratios of n/m, .. and in the second case defining explicitly whether identical ratios eg 2/3 4/6 are counted as equivalent.. otherwise as many have said the answer is seems trivially simple.87.102.87.36 13:28, 12 October 2007 (UTC)

Pity the poor high school students! Their teachers teach them

• Natural Numbers
• Integers
• Rational Numbers
• Real Numbers

But their teachers did not teach them about the difference between the representation and the value. The map is not the territory, see Map-territory_relation

• ${\frac {0}{1}}$ is a representation, it has a value of $R_{0}$ which is a rational number.
• ${\frac {0}{2}}$ is a representation, it has a value of $R_{0}$ which is a rational number.
• ${\frac {0}{3}}$ is a representation, it has a value of $R_{0}$ which is a rational number.

So we have three different representations but they all have the same value $R_{0}$.

So there are two possible questions

• Do we have a set of 3 different representations (of a rational number) or a set of 1 representation (of a rational number)?
• Do we have a set of 3 different values (of rational numbers) or a set of 1 value (of rational number)?

These are two different questions.

As a further exercise, consider this:

$7=7.0=7.00000\,$

Now if we have a set of

$\left\{7,7.0,7.00000\right\rbrace$.

Do we have a set of 3 real numbers or a set of 1 real number?

211.28.129.8 14:45, 12 October 2007 (UTC)

Ultimately the question can be ambiguous, but using a little bit of common sense and judgement will usually save the day. If you are in high school or earlier, or any other math class, that hasn't bothered with the appropriate formal distinctions then you can safely assume the question is asking about values, not representations. If you are in the sort of math class that takes care over such distinctions... the instructor will be careful enough not to leave such ambiguity in the question. As for your final question: take note what you are asking: is a set of 3 real numbers or a set of 1 real number; you clearly refer to the elements as real numbers, not distinct representations. As all the given presentations are all in the same equivalence class of real numbers, the answer is that it is a set of just 1 real number. -- Leland McInnes 15:02, 12 October 2007 (UTC)

This question seems to have instigated a fairly long and involved debate, which is ok since we are trying to determine the best answer. From my thinking and having read somemore above, I think in modern ZFC set theory (or axiomatic set theory) it becomes {0}, but in the original naive set theory it is still ${\big \{}{\frac {0}{1}},{\frac {0}{2}},{\frac {0}{3}}{\big \}}$. A math-wiki 00:45, 13 October 2007 (UTC)

$\left\{{\frac {0}{1}},{\frac {0}{2}},{\frac {0}{3}}\right\rbrace$ is a set of three representations of a number, or a set of one number (in this case zero). Without being given a proper superset (the set of representations of numbers, the set of numbers), this "set" is not well-defined. Consider the "set" {"three", "tres", "drei", "3"}. This "set" contains either four representations of numbers, or just one number. Context is needed to supply an answer. The problem here is deciding whether the symbols used to designate numbers should be considered as well. Suppose the "set" contained "III" (Roman numeral three).--Mostargue 14:07, 13 October 2007 (UTC)

If you believe that the set {0/1, 0/2, 0/3} contains three elements, then you must also believe that 0/1 ≠ 0/2, 0/1 ≠ 0/3, and 0/2 ≠ 0/3. Otherwise, there is no bijection with the set {1, 2, 3}, and then the two sets do not have the same size. Whether 0/1 = 0/2 or not may depend on your definition of the operation denoted by the slash, but under the usual definition of the operation in the field of the rational numbers, 0/1 = 0/2.  --Lambiam 11:51, 14 October 2007 (UTC)

dot and cross product

(redirected)

why we use cos in dot product and sin in cross product? —Preceding unsigned comment added by 202.69.33.15 (talk) 06:43, 12 October 2007 (UTC)

Dot_product#Proof_of_the_geometric_interpretation answer to first part given.87.102.87.36 13:33, 12 October 2007 (UTC)

It is probably easiest to see in terms of geometric interpretations of dot and cross products, so perhaps compare Dot_product#Geometric_interpretation and Cross_product#Geometric_meaning to get some idea. If you wish to think in terms of right triangles you can think of the cross product projecting down, thus we are considering the side adjacent to the angle formed by the vectors, while the cross product (in calculating the area of the spanned parallelogram) considers the height of the vertical, thus we are considering the side opposite to the angle formed by the vectors. -- Leland McInnes 14:55, 12 October 2007 (UTC)

64,000,000 sided dice

A friend of mine was talking about rolling a dice to win the lottery. This lead to a discussion about a 64,000,000 sided dice. Our question was how many sides would each face have if each face had the same number of side and if this was even feasible. —Preceding unsigned comment added by 66.168.122.79 (talk) 19:59, 12 October 2007 (UTC)

It is theoretically possible to construct such a die (singular of dice), probably even feasible given the right budget (tossing it is a different matter...). However, the faces will not be regular polygons, and there are multiple ways to construct it - I suspect that pretty much any (small enough) side number is possible. One construction which is certainly possible is as a bipyramid, composed of triangles (the AD&D d10 is a (distorted) pentagonal bipyramid). -- Meni Rosenfeld (talk) 20:12, 12 October 2007 (UTC)

For most purposes it would be indistinguishable from a spherical ball. Just buy a marble, and say it's a 64000000-sided die. --Trovatore 20:15, 12 October 2007 (UTC)

You might also find the article on platonic solids interesting. And as an aside, "dice" is plural. The singular is "die." Donald Hosek 20:17, 12 October 2007 (UTC)

Trovatore! Just the man I was looking for. Perhaps you can add anything to the discussion two threads above (I suspect that in fact, you were attracted here by it)?

Donald, I think I have already commented on the singular\plural issue... -- Meni Rosenfeld (talk) 20:28, 12 October 2007 (UTC)

Oops, that short posting kept getting hit by edit conflicts and I missed your note when I did a quick scan to see if you'd covered the platonic solids... Curse you, chaos butterfly! Donald Hosek 00:27, 13 October 2007 (UTC)

I like simply adding [edit conflict], this way, no one will hold it against me if I repeat something :). The assumption is that if someone has spent time working on a reply, he should be able to post it even if ec's made it irrelevant. -- Meni Rosenfeld (talk) 11:44, 13 October 2007 (UTC)

Indeed a bipyramid – or a kite-faced figure like the common D10, or something in between – is, for most N, the only way to make a die whose N faces are all alike. See http://www.mathpuzzle.com/Fairdice.htm —Tamfang 01:46, 13 October 2007 (UTC)

Since flaws in the making of such a die would likely be have more effect than its edges, you'd be better off using six icosahedral dice to get a number in base 20. —Tamfang 02:24, 13 October 2007 (UTC)

Holy shit, Trovatore says " I hold a PhD in mathematics (set theory) from UCLA. " on his talk page!!! Please please please straighten up the sets discussion two above this one!!!! —Preceding unsigned comment added by 84.0.126.201 (talk) 21:27, 12 October 2007 (UTC)

I believe that many of the most frequent posters here (including Meni) are practicing mathematicians, not just Trovatore. Tesseran 22:19, 13 October 2007 (UTC)

Possibly, but Trovatore's specialization in set theory, interest in foundations and superior intellect are especially relevant to said discussion. Though the discussion has diverged from sets per se, it started with them, so anon only needed to emphasize "set theory" to justify his enthusiasm. -- Meni Rosenfeld (talk) 12:43, 14 October 2007 (UTC)

See also: Barrel die, Teetotum. —Ilmari Karonen (talk) 10:56, 16 October 2007 (UTC)

what do mathematicians do?

Say I disagree with Mathematics (I think it's a 'lie') and consider teaching mathematics as not actually doing anything but spreading the lie (from your perspective, maybe you could think of divinity school). So, what else does a mathematician do besides propagate the lies?

I can think of: - Actuary (statistics) - Computer science proofs of code correctness (discrete mathematics)

[Do I have the terminology right?]

Anyway, is that all, or is there something else mathematicians do besides propagate mathematics? (For the divinity analogy, the other thing trained priests do is run a parish, community stuff, counseling, weddings by the power vested in them by the state, etc).

Thank you! —Preceding unsigned comment added by 84.0.126.201 (talk) 21:44, 12 October 2007 (UTC)

I forgot to add other than publishing in journals run by other mathematicians. —Preceding unsigned comment added by 84.0.126.201 (talk) 21:45, 12 October 2007 (UTC)

That depends on what you mean by "mathematicians". I think a fairly common definition is people who conduct mathematical research. Of course, the research results are published in journals, making this (as opposed to teaching) the primary occupation of mathematicians.

You can generalize the definition to include anyone whose education is primarily in the subject of mathematics and that uses this knowledge in his work. Even with such a definition, you may not find a lot of mathematicians doing practical work, as this usually requires not only mathematics, but rather mathematics combined with a specific field (such as engineering, economics and so on), and people trained in this field might not satisfy this definition.

However, I think the better question is "where is mathematics used", and the answer is "everywhere". Any sort of occupation requires using varying degrees of mathematics (sometimes at a very high level), using the knowledge gained from mathematics teachers and, at the advanced level, the results in published articles. The job of mathematicians is to make sure that everyone has the knowledge needed to do whatever they do.

The fact that mathematics is successfully applied everywhere seems to contradict the hypothesis that "mathematics is a lie". -- Meni Rosenfeld (talk) 21:56, 12 October 2007 (UTC)

Another point: The journals may be run by mathematicians, but are read by others as well, improving the quality of their work. So no matter how you look at it, mathematics is not a closed system. -- Meni Rosenfeld (talk) 21:59, 12 October 2007 (UTC)

• Cryptography, Accounting, Mathematical finance, Actuarial science, Computer science, Game theory, Logic, and Computational engineering are all applied mathematics. People with careers in those fields generally may consider themselves to be mathematicians. --M@rēino 21:58, 12 October 2007 (UTC)

This is tricky territory. The examples you mention are disciplines that map mathematics onto the real world. Statistics does this for processes where probability is involved, and CS does it for for computers (obviously). The main point is that in these fields, the results of mathematics become observable. Sidestepping the question of whether math is inherent in the universe, pure mathematics deals with non-observable phenomena. Once you start looking at processes that can be observed, as well as described mathematically, you're leaving the field of pure mathematics and slowly entering the world of science. If you look at the problem this way, all science (specifically fields like physics and chemistry) can be seen as something that mathematicians 'do'. No mathematician will actually be involved with physics (or he would be a physicist), but if you call a statistician a mathematician, then, calling a physicist one too is just the next step. And even pure mathematicians are often the ones that develop the models required for physics, before the physicists know they need them.

There are other outflows of mathematics. Consider the field of Artificial Intelligence. This field is not strictly involved in finding truth in the scientific sense. Rather, it attempts to create something, a model that doesn't predict the world, it just has to behave a certain ill-defined way. AI has a strong mathematical side to it, in attempts to describe the 'rules' of language formally, finding formal descriptions for reasoning (ie. logic), stuff like that. It's not strictly science based on observation, and it's not strictly the propagations of mathematics. In fact, all engineering can be considered an answer to your question, in this light. It makes strong use of mathematics, to create rather than explain, the development of these models has a directly applicable use and the people developing these model do qualify for the title of mathematician.

Of course, there is a problem if you don't accept these answers. If you restrict the definition of mathematician to 'someone who works on pure mathematics' without considering the relation to the observable world', the reasoning becomes circular, as you're defining mathematicians as people who do nothing but propagate mathematics, and the ask if they do anything else.

Finally, they may also make pretty pictures (or is that a side effect?)

risk 22:10, 12 October 2007 (UTC)

I take it you're a fictionalist. Personally, I'm an embodied mind theorist. — Daniel 23:21, 12 October 2007 (UTC)

If the above description by Daniel, isn't making much sense, think about mathematics as it's own logical framework. Mathematics was started when (presumably) ancient man started to name the whole numbers to represent the number of some object (animate or not) in it's environment. At this point the objects are no longer really relevant, the logical idea is independent of the real world at that point. Mathematics simply isn't based on 'the real world' rather it's own world, which is at the most fundamental level the concept of quantity itself. From that point mathematics development took off, and man discovered that the results of pure mathematical thinking seem to have very stong and very precise relationships to things in the 'real world.' That use of mathematics to solve real world problems is application (applied mathematics), pure, theoretical mathematics is unconcerned with those relationships. A math-wiki 00:52, 13 October 2007 (UTC)