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June 17

Map pinpont problem

Say a friend chooses a secret point on a map, and gives you a random starting point. When you choose successive points, he gives you the hints "warmer," "colder," and "neither" to indicate if you've moved radially closer to the secret point or not. Given that you want to find the secret point as quickly as possible, what would be your optimal scheme? One possibility would be to choose points until you are neither "warmer" nor "colder" with respect to your random starting point, and thus form a curve of constant radius about the secret point; you could then extrapolate the center of the circle this begins to form. Is there a more efficient way? —Preceding unsigned comment added by 24.117.105.163 (talk) 05:18, 17 June 2010 (UTC)[reply]

Let point B be warmer/colder/neither than point A. The segment bisector between A and B divides the map into a warm region and a cold region, the line itself being neither. Once the warm region is the inside of a polygon, you should choose your next point such that the polygon is divided into (almost) equal areas, finding X by the bisection method. Bo Jacoby (talk) 06:21, 17 June 2010 (UTC).[reply]

The problem only talks in terms of 'warmer' and 'colder' not 'warm' or 'cold', one can only directly compare with the last point. I'm wondering what 'neither' means and I don't think it is necessary and it may cause problems if it is a range, also there must be an area which is counted as the target rather than it being a point. Anyway it sounds an interesting problem, somebody else very possibly has investigated it before but I'll not look at google yet! ;-) Dmcq (talk) 13:39, 17 June 2010 (UTC)[reply]

You can always (except maybe pathological cases) find a new point so that its segment bisector with the last point will halve the area. "Neither" probably means just that, that the new and last points are equally distant from the target. This won't happen very often, but if it does you're very lucky because the search will then be restricted to a single line. -- Meni Rosenfeld (talk) 13:49, 17 June 2010 (UTC)[reply]

Anyway my first solution is to use grid steps one way until I get colder, then do this at right angles, then choose a smaller grid and do it from the new point I'm at till I'm sufficiently close. Dmcq (talk) 13:43, 17 June 2010 (UTC)[reply]

I must obviously explain more carefully to Dmcq. Let the given point be A=(0,0). Choose the point B=(1,0). If B is warmer than A then the unknown point X=(x,y) satisfies x>1/2, and if B is colder than A then x<1/2, and if B is neither warmer nor colder than A then x=1/2. Next choose the point C=(1,1). Now you also know if y>1/2 or y<1/2 or y=1/2. The 'warm' region is the convex area in which the unknown point is known to be located. Now choose a point D far away in the warm region, hoping that it is colder than C. Then the warm region is a finite triangle, otherwise it is infinite. Bo Jacoby (talk) 15:57, 17 June 2010 (UTC).[reply]

Sorry Bo Jacoby, silly me, you're quite right. I really should stop myself when I have the urge to rush something off before I have to go somewhere. It just means I keep thinking how silly I am till I get back :) Dmcq (talk) 16:57, 17 June 2010 (UTC)[reply]

Don't torment yourself. Communication is a joint effort. Bo Jacoby (talk) 11:07, 18 June 2010 (UTC).[reply]

Thanks for the responses everyone! The bisection method is definitely a good one. Is it also the most efficient? It seems that perhaps some clues may be gleaned from previous hints that allows you to cut out more than half of the remaining area. 24.117.137.23 (talk) 22:59, 18 June 2010 (UTC)[reply]

No, all information is coded in the 'warm' polygon.Bo Jacoby (talk) 20:02, 23 June 2010 (UTC).[reply]

Proportional_navigation may be helpful. Zoonoses (talk) 14:45, 18 June 2010 (UTC)[reply]

It seems to me that if we are really talking about mathematical points, then the expected number of steps will always be infinite no matter what the algorithm. The best you can do is bisect the remaining region, but there are as many points in half of it as there were originally (i.e. an infinite number). Therefore there isn't really a "best" algorithm, or to put it another way, "The only way to win is not to play". --Anonymous, 19:33 UTC, June 18, 2010.

True, but surely there are more efficient ways to get "close enough." 24.117.137.23 (talk) 22:59, 18 June 2010 (UTC)[reply]

Arithmetic sum question

The question says: $S\sim Geo\left({\frac {1}{12}}\right)$ so $P\left(S=s\right)={\frac {1}{12}}{\left({\frac {11}{12}}\right)}^{s-1}$ where $1\leq s<\infty$ and so $P\left(S\leq s\right)=\sum _{i=1}^{s}{\frac {1}{12}}{\left({\frac {11}{12}}\right)}^{i-1}=1-{\left({\frac {11}{12}}\right)}^{s}$ , but I'm having trouble following the logic of that last summation. So far I've got: $\sum _{i=1}^{s}{\frac {1}{12}}{\left({\frac {11}{12}}\right)}^{i-1}={\frac {1}{12}}\left[{\left({\frac {11}{12}}\right)}^{0}+{\left({\frac {11}{12}}\right)}^{1}+\dots \right]={\frac {1}{12}}\left[{\left(1-{\frac {11}{12}}\right)}^{-1}\right]=1$ because ${\left(1-x\right)}^{-1}=x+x^{2}+\dots$ ; where am I going wrong in this? Thanks in advance. It Is Me Here ^{t / c} 11:11, 17 June 2010 (UTC)[reply]

You have confused

\sum _{i=1}^{s}{\frac {1}{12}}{\left({\frac {11}{12}}\right)}^{i-1}

with

\sum _{i=1}^{\infty }{\frac {1}{12}}{\left({\frac {11}{12}}\right)}^{i-1}

. For the summation formula of a finite geometric series, see Geometric series. -- Meni Rosenfeld (talk) 11:26, 17 June 2010 (UTC)[reply]

Sorted it now, cheers. It Is Me Here ^{t / c} 12:00, 17 June 2010 (UTC)[reply]

Given a family of sets, I can generate a partition of their union set, by looking at their overlapping...

... as an example, let us consider 3 sets A, B and C; this sets generate 7 disjoint nonempty subsets, that is: A.B.C (elements belonging to the intersection of the 3 sets) A.B.notC (elements belonging to A and B, but not to C) A.notB.C (... and so on) A.notB.notC notA.B.C notA.B.notC notA.notB.C Now, given a family of sets, I can always build such a partition. Which is the name of one of such part? In a partition, one of these disjoint subset is called "part" or "block" or "cell"; but it is a more general concept, it does not imply that it comes from a family of sets. Does a name already exist? —Preceding unsigned comment added by 130.88.195.163 (talk) 18:08, 17 June 2010 (UTC)[reply]

Eight - you missed out the set excluding all 3. Dmcq (talk) 18:20, 17 June 2010 (UTC)[reply]

And you missed "partition of their union set". Algebraist 18:23, 17 June 2010 (UTC)[reply]

True, seven then. I think I call them clause (logic), except they don't have to contain every literal so A on its own would be a clause. I havve comae across a term from programmable logic devices but I couldn't find it now. Canonical form (Boolean algebra) uses 'minterm' to refer to them but that's not what I was trying to think of. Dmcq (talk) 18:38, 17 June 2010 (UTC)[reply]

Well, although I need a definition like this for decomposing a set of logical formulae, I would like to have a name in set theory; something like "the partition of the union set given by the family". —Preceding unsigned comment added by 130.88.195.163 (talk) 18:45, 17 June 2010 (UTC)[reply]

I think full disjunctive normal form in Disjunctive normal form where are the clauses contain every literal is the closest I'm going to get but that refers to the full logical formula. Dmcq (talk) 18:50, 17 June 2010 (UTC)[reply]

The following notation is not described in wikipedia because it is original research on my part, but I guess I can confidently inform you. The numbers are called ordinal fractions. The first set A=100, and notA=200. B=010, and notB=020. C=001, and notC=002. This defines the 27 sets 000 001 002 010 011 012 020 021 022 100 101 102 110 111 112 120 121 122 200 201 202 210 211 211 220 221 222. The 8 sets 111 112 121 122 211 212 221 222 are mutually disjoint. (For example 121=A.notB.C). Bo Jacoby (talk) 23:56, 17 June 2010 (UTC).[reply]

Moment-generating function of the Binomial distribution

You are told that $X\sim B\left(n,p\right)$ and so $M_{X}\left(\theta \right)={\left(q+p{e}^{\theta }\right)}^{n}$ where $q=1-p$ ; then you are told that $Z={\frac {X-\mu }{\sigma }}$ and are asked to use the linear transformation result to show that $M_{Z}\left(\theta \right)={\left(qe^{-{\frac {p\theta }{\sqrt {npq}}}}+pe^{\frac {q\theta }{\sqrt {npq}}}\right)}^{n}$ . Now, I have managed to get to $e^{-\theta {\sqrt {\frac {np}{q}}}}{\left(q+pe^{\frac {\theta }{\sqrt {npq}}}\right)}^{n}$ , but don't know how to turn that into the final answer (or 'break open' that final ${\left(a+b\right)}^{n}$ bracket). Any help would be appreciated. It Is Me Here ^{t / c} 19:48, 17 June 2010 (UTC)[reply]

-\theta {\sqrt {\frac {np}{q}}}=n\left(-{\frac {p\theta }{\sqrt {npq}}}\right)

Dmcq (talk) 20:10, 17 June 2010 (UTC)[reply]

Ah, OK, thanks. It Is Me Here ^{t / c} 20:19, 17 June 2010 (UTC)[reply]

some controversial questions

We have some controversial questions on Talk:Exponentiation#0.E2.81.B0 .E2.80.94 limits and continuity.

Is the set of integers a subset of the set of real numbers?
Is 5+0i>3+0i ?
Is 0=0.0 ?

You comments are appreciated. Bo Jacoby (talk) 23:37, 17 June 2010 (UTC).[reply]

To which my answer in all three cases is no. Dmcq (talk) 23:43, 17 June 2010 (UTC)[reply]

The answer to the first question is no, strictly speaking, although the point is pedantic. The second is really a matter of notation and convention. I would be interested in how you justify a negative answer to the last question. "0.0" is just notation. For all you know, it could be a placeholder I use for the "integer" 0. Phils 23:49, 17 June 2010 (UTC)[reply]

Not that I agree with it, but I guess the justification is that 0 is an integer and 0.0 is a real number. I think that's hypercorrect, personally... --Tango (talk) 23:55, 17 June 2010 (UTC)[reply]

Well I'd only distinguish between them when I though it was important to distinguish between integers and real numbers (or in fact any other thing like a complex number or zero matrix). Dmcq (talk) 23:59, 17 June 2010 (UTC)[reply]

I would distinguish them in a clearer way, though.

0_{\mathbb {Z} }

and

0_{\mathbb {R} }

, perhaps. --Tango (talk) 00:02, 18 June 2010 (UTC)[reply]

Mathematics is not programming or engineering (and besides, many high level programming languages would interpret the number as a real or as an integer correctly, depending on context). Unless the authors specified it explicitly, nobody reading "0.0" in a math journal would assume that it means the zero element of the reals as opposed to an integer. Phils 00:04, 18 June 2010 (UTC)[reply]

Could someone expand on the answers for the curious few in the peanut gallery? Why the distinction between "real zero" and "integer zero"? Does this imply that the combined length of three identical line segments is technically not equal to the length of the original line segment after being scaled by a factor of three (e.g. 3 × 2.6 ≠ 3.0 × 2.6) ? -- 140.142.20.229 (talk) 01:01, 18 June 2010 (UTC)[reply]

Well, technically you can't multiply elements in different rings/groups, so you couldn't multiply the integer 3 by the real number pi, but you could multiply the real number 3 by the real number pi, and you can set up an injection from Z to R by f(x) = x. In practice there's no difference, because by convention we assume that such an injection is made, so in 3*pi we're multiplying f(3) by pi. -mattbuck (Talk) 01:27, 18 June 2010 (UTC)[reply]

For question 2, it's again a technicality - 5+0i = 5, and 3+0i = 3, but when working in the complex plane (which is implied by the i) you have the issue that C is not a well-ordered set - you have numbers a,b where a !< b and b !< a and further a != b.

For question 3, 0 multiplied by itself is technically undefined, but by convention we say that anything multiplied by 0 is 0 (or at least for finite values) and so we say that 0.0 is 0. -mattbuck (Talk) 01:32, 18 June 2010 (UTC)[reply]

Well, R isn't a well-ordered set either... but presumably you meant a total order.

For question 3, I think the OP was referring to the real number 0.0, where the period is interpreted a decimal point. So then the question is just whether we equate the integer 0 with the real number 0.0, to which we would usually answer yes, unless we have a good reason to be pedantic about the distinction. --COVIZAPIBETEFOKY (talk) 01:42, 18 June 2010 (UTC)[reply]

What is the distinction to be pedantic about? Bo Jacoby (talk) 07:59, 18 June 2010 (UTC).[reply]

As others have said, it depends on our specific constructions of the integers and real numbers, but usually, the real numbers that correspond to integers are distinct objects from the integers. So the distinction is between the set of integers, which are often constructed, eg, as an equivalence class on ordered pairs of finite ordinal numbers, or the set of real numbers that correspond to some integer. --COVIZAPIBETEFOKY (talk) 11:24, 18 June 2010 (UTC)[reply]

You can multiply an integer, specifically, by any member of an additive group. It is defined as repeated addition in that group.

If you consider Z not to be a subset of R, then you can't have

f(x)=x

. Writing this just muddies the issue. What you can have is "

f(x)

is the real number corresponding to x".

Of course 0 multiplied by itself is defined. That is, in any structure where there is a member called "0" and an operation called "multiplication", 0*0 is defined (and in all such structures I can think of, it is equal to 0). -- Meni Rosenfeld (talk) 08:46, 18 June 2010 (UTC)[reply]

Claiming that the sentence the integers are a subset of the real numbers is false with no definition of integer and real numbers around makes little sense. If you look at the "construction" sections of Integer and Real number, you will notice that they are formally disjoint sets. They consist of different objects, whatever approach you take to construct the real numbers. Of course, in almost every context, the integers are identified with their image inside the reals. Regarding the three segments, the only physically reasonable answer is that the two lengths are the same. Formalism and abstraction exist to make complicated things intelligible, not obfuscate the obvious. Phils 01:36, 18 June 2010 (UTC)[reply]

I have a better question.

Is 1+1=0?

The answer is "it depends on which 1, 0 and + we mean". If we mean the integers 1, 0 and integer addition, then this is false. If we mean the

\mathbb {Z} _{2}

elements and

\mathbb {Z} _{2}

addition, this is true. We use integers more often than

\mathbb {Z} _{2}

, so by convention, without a context explicitly specified, the notation refers to integers and we say the statement is false. But this does not mean that "1+1=0" is really, truly false, only that conventionally we use it to mean something that is false.

People speak of "the set of integers", but there are actually many sets which can be referred to as "the set of integers". There's a set of equivalence classes of ordered pairs of finite ordinals with the obvious relation; to this set I shall refer as "Integers1". Then we go on to construct the set of real numbers, and in it we identify elements which correspond to the elements of Integers1. We call the set of those elements Integers2. Integers1 and Integers2 are of course disjoint. The question "Is the set of integers a subset of the set of real numbers?" reduces to the question "which set of integers do we mean?". If we mean Integers1, no. If we mean Integers2, yes.

In most cases it doesn't matter which set we mean when we say "integers". I suspect that usually people actually think about Integers2, but maybe that's just me. When it comes to questions where it does matter, since apparently we don't have a clear convention on what we mean by default, we just need to specify what we mean. For 0^0, if the exponent is considered an element of Integers1, it follows from the natural definition of integer exponents that the answer is 1. If the exponent is considered an element of Integers2 - a real number which happens to be an integer2 - it follows from the natural definition of real exponents that this is indeterminate.

For question 2 - it's again a question of Reals1 vs. Reals2 (a subset of Complexes). But even if we do mean Reals2, we can and should define a relation > on the complexes which is not total, and holds only for unequal real2 numbers.

Question 3 - Like I said, I think 0 and 0.0 both refer to the Integers2-0, so they're equal. If it turns out that the convention is that 0 refers to Integers1 and 0.0 to Integers2, then no. -- Meni Rosenfeld (talk) 08:36, 18 June 2010 (UTC)[reply]

Of course, 0.0 could represent the zero element in the field of rational numbers, and "set of integers" could mean algebraic integers (in algebraic number theory texts it is common to use "integer" to mean any algebraic integer, and to distinguish the integers in

\mathbb {Z}

or

\mathbb {Q}

by calling them "rational integers"). As several editors have already said, the answers to all three questions depend on context and interpretation. Gandalf61 (talk) 09:40, 18 June 2010 (UTC)[reply]

Apart from $f(x)=0^{x}$ , are there any examples in mathematics where $\scriptstyle f(0)\neq f(0.0)$ ? Bo Jacoby (talk) 11:18, 19 June 2010 (UTC).[reply]

There is a similar issue when you move from the real numbers to the complex numbers; ln(2.0) is a single real number, but you have to choose a branch cut to define ln(2.0 + 0i). This is because "ln" has different meanings in the two contexts. — Carl (CBM · talk) 11:36, 19 June 2010 (UTC)[reply]

It sounds unlikely that there's any other straightforward thing like that, but here's a similar conundrum - what's the difference between zero and nothing? In fact in some definitions of 0 there's no difference but it's a good one to ponder. Dmcq (talk) 11:48, 19 June 2010 (UTC)[reply]

Can I be an idiot and ask why the set of integers is not a subset of the set of real numbers? I would have no problem writing ℤ ⊂ ℝ. After all, n ∈ ℤ ⇒ n ∈ ℝ and so ℤ is a subset of ℝ. In fact, don't we have a sequence of nested proper subsets given by ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ ⊂ ℂ ⊂ ℍ? •• Fly by Night (talk) 12:25, 19 June 2010 (UTC) [reply]

With the usual definitions of those sets, none of your proposed chain of subsethoods holds. Each set does contain a canonical isomorphic copy of the previous one, though, so you can if you wish redefine ℕ to be its canonical copy in ℤ (and so on for the rest), and then ℕ will indeed be a subset of ℤ. Meni has already explained this above. Algebraist 12:31, 19 June 2010 (UTC)[reply]

Are there industry standard definitions for the integers, real numbers, etc? The definitions of ℂ and ℍ are built on ℝ. (I didn't see Meni's post; there was far too much thread to read. Sorry.) •• Fly by Night (talk) 13:04, 19 June 2010 (UTC)[reply]

P.S. The more I think about it, the less it makes sense. If the integers are not, by defintion, a subset of the real numbers then it must mean that there is a definition of the real numbers for which not all integers are considered as real numbers. What possible definition of the real numbers could there be for which not every integer is considered to be a real number? Let's say there exists an integer, n, which is not a real number; then what is n? •• Fly by Night (talk) 14:11, 19 June 2010 (UTC)[reply]

Let me give you some examples. The natural number 2_N is defined as the ordinal

\{\varnothing ,\{\varnothing \}\}

. Depending on the construction, the integer 2_Z may be the ordered pair

\langle \varnothing ,2_{\mathbb {N} }\rangle

(if I choose to represent the + sign by the empty set), or the set

\{\langle a,b\rangle \mid a,b\in \mathbb {N} ,a=b+2_{\mathbb {N} }\}

, or something else. The rational number 2_Q may be the ordered pair

\langle 2_{\mathbb {Z} },1_{\mathbb {N} }\rangle

or the set

\{\langle a,b\rangle \mid a,b\in \mathbb {Z} ,a=2_{\mathbb {Z} }b\}

. The real number 2_R may be the set

\{a\in \mathbb {Q} \mid a<2_{\mathbb {Q} }\}

or the set of all rational sequences which converge to 2 etc. The complex number 2_C can be the pair

\langle 2_{\mathbb {R} },0_{\mathbb {R} }\rangle

. All of these 2's are different objects, even though 2_Z plays in Z the same role as 2_R plays in R; there is a canonical embedding of Z into R which maps 2_Z to 2_R. There are no industry standard definitions of these mathematical structures, or rather, there are too many of them, but essentially none of the usual constructions make, say, the integers a subset of the reals. Pretty much the only reasonable way how to achieve that would be to make explicit exceptions in the construction, such as: a real number is either a Dedekind cut on the rationals with no rational limit, or a rational number itself. This is awkward to work with, and it's much easier to go with the usual strategy where the canonical embeddings are not literal inclusions.—Emil J. 15:58, 19 June 2010 (UTC)[reply]

I'm sorry but, to me, this seems to be totally over the top. You have the real numbers, then most people would, in your notation, make the following definitions: ℕ = {0_R, 1_R, 2_R,…}, ℤ = {…, –2_R, –1_R, 0_R, 1_R, 2_R,…}, ℚ = {x/y : x, y ∈ ℤ ∧ y ≠ 0_R}, and ℂ = {x + iy : x, y ∈ ℝ}. Thus ℕ ⊂ ℤ ⊂ ℝ ⊂ ℂ. I guess the problems come from the fact that when you try to construct the integers or the reals from nothing then you end up having to use different techniques because of the different nature of the integers and the reals. The results are then incompatible. So as sets ℕ ⊂ ℤ ⊂ ℝ ⊂ ℂ, but the construction of one doesn't fit with the construction of the next. •• Fly by Night (talk) 18:37, 19 June 2010 (UTC)[reply]

The problem is that there's no good way to construct real numbers without using rational numbers. And there's no good way to construct the rational numbers without integers, and so on. So you have to start with an elementary definition of natural numbers and move up from there.

Of course, you can also leave the real numbers undefined, and add some axioms to give them the properties you want. Then you can define integers as specific real numbers. But most people prefer to have the amount of undefined concepts and axioms at a minimum. Since you can do all standard mathematics with just sets, and you can't do much without sets, the standard approach is to axiomatize set theory and define everything else in terms of sets.

Until now I thought it was also standard to discard the initial definitions of integers etc. once they're no longer needed, and redefine everything as subsets of complex numbers. -- Meni Rosenfeld (talk) 18:55, 19 June 2010 (UTC)[reply]

Yeah, that's not common in set theory at least. The number 2 will almost always mean {0,1} to a set theorist, where 0 = {} and 1 = {0}. — Carl (CBM · talk) 19:00, 19 June 2010 (UTC)[reply]

But isn't this all a problem with the construction and formalisation of the different number systems? If you treat the real numbers intuitionally and geometrically (see the the number line picture above) then there really is not problem at all. Aren't all of these problems coming from our own inability to formally construct and define the different number systems in a cohesive way? It all feels like it's been cobbled together like a patchwork quilt. If there are so many technical hurdles to overcome when formalising the real numbers then how on earth can we make sense of Riemannian manifolds? (Never mind more exotic creatures like Kähler manifolds.) •• Fly by Night (talk) 19:10, 19 June 2010 (UTC)[reply]

This is getting like the argument amongst educationalists over whether to introduce multiplication as repeated addition and what to do when measurements are multiplied together when repeated addition doesn't work. The reals just are different from the integers. Dmcq (talk) 19:55, 19 June 2010 (UTC)[reply]

Right. To FlyByNight: the reals are conceptually and qualitatively different from the natural numbers. The natural numbers are about counting things. You can't count things with the reals, not even the ones that correspond naturally to the naturals.

Remember the scene in Terminator 2 where Arnold has been ordered not to kill anyone, and after an intervention of surpassing violence tallies up the deaths, and they come to 0.0? It's funny, not just because of unnecessary precision, but because it's a category error. There aren't any .5 people, and there aren't any .0 people either. --Trovatore (talk) 23:22, 19 June 2010 (UTC)[reply]

The reals are indeed conceptually and qualitatively different to the integers; but since when does a subset have to be conceptually and qualitatively the same as one of its supersets? For example, the empty set ∅ is a subset of any given set; but not every set is conceptually and qualitatively the same as the empty set. •• Fly by Night (talk) 22:22, 20 June 2010 (UTC)[reply]

There is no bar to including objects of disparate type in the same set. For example, even if you don't consider the natural numbers to be real numbers, you could always take the union of the naturals and the reals, and the naturals would then be a subset of that union.

So it's not a question about the subset relation in general. It's a question of whether the natural numbers are naturally considered a subset of the reals. Are they really the same sort of thing? I think there is a reasonably clear distinction even at an intuitive level. The real number 1.0, for example, is a position along a continuous line; the natural number 1 is more like "what all unique objects have in common". These are different things.

So when it turns out that the most convenient codings turn out quite different (1 being coded as the singleton of the empty set, 1.0 as a collection of sequences of rationals), rather than see that as a defect of the coding, we can embrace that as a clear reflection of the underlying typedness of the objects.

Of course, there are plenty of contexts where the distinction doesn't matter, or even where it's actively useful to ignore it, and in those contexts we happily ignore it without comment, counting on the reader to make the necessary allowances. --Trovatore (talk) 22:50, 20 June 2010 (UTC)[reply]

Thank you, everyone, for comments and answers. The article on isomorphism has reference to a relevant article: Mazur, Barry (12 June 2007), When is one thing equal to some other thing? (PDF). The ring of integers is defined abstractly as an axiomatic system. The axioms are sufficient for calculations, but two questions remain unanswered: Are the axioms consistent, such that at least one realization exists? and: are the axioms sufficient, such that two realizations are necessarily isomorphic? In order to settle the first question a set theoretical construction of integers is made, such as described above. When such an example satisfy the axioms, then the axioms cannot be contradictory. To me this is the sole purpose of the construction. The ring of integers is defined abstractly modulo isomorphism by the axioms, and a concrete realization is defined by the set theoretical construction. In the abstract sense, the ring of reals contains a unique subring of integers, and we write ℤ ⊂ ℝ and 0=0.0 . In the concrete sense, different, (albeit isomorphic), constructions can be made, and the zero of one construction is not the same mathematical object as the zero of some other construction and then 0≠0.0 . So it seems to me that the controversy is about differentiating between the abstract and some concrete interpretation of the words 'integer' and 'real'. In the abstract sense the three original questions are answered affirmatively, as Fly by Night and I myself would do. Bo Jacoby (talk) 05:55, 21 June 2010 (UTC).[reply]

No one doubts that the integers and, for lack of a better term, "integer reals", are isomorphic as rings. So for the purposes of ring theory, you can consider them the same, at least when not considering them as embedded in some larger ring.

But that's only for the purposes of things that can be expressed in the language of rings. Your original question was motivated by exponentiation, which is not a ring operation, and is not definable in the language of rings. There is no notion of raising an element of a ring to the power of another element of the ring. --Trovatore (talk) 06:29, 21 June 2010 (UTC)[reply]

Rings don't even have greater than and less than so that lets out question 2. Dmcq (talk) 14:16, 21 June 2010 (UTC)[reply]

Abstract ring elements do have well defined powers to nonnegative integer exponent. So if x is a real number and n is a nonnegative integer, then xⁿ (including 0⁰ = 1) is defined by ring operations. Abstract integers also have a well defined total order, which is the restriction of the order of the reals. Bo Jacoby (talk) 15:46, 21 June 2010 (UTC).[reply]

Abstract ring elements do indeed have well-defined powers to natural-number exponents. But that just makes my point for me! When you're defining ring-element-to-natural-number, the ring can be anything you want, but the natural numbers are always the same. The natural numbers are therefore something external to the ring. --Trovatore (talk) 18:27, 21 June 2010 (UTC)[reply]

Is the set of integers a subset of the set of real numbers? It depends on exactly how real numbers are defined, and so can only be answered in the context of a particular construction of the real numbers. What is true is that there is a subset of the real numbers Z' = {..., -3.0, -2.0, -1.0, 0.0, 1.0, 2.0, 3.0, ...} which is isomorphic to the integers Z = {..., -3, -2, -1, 0, 1, 2, 3, ...}. Because Z' behaves exactly like Z does, and because it satisfies the usual axioms, most mathematicians will say "the integers are a subset of the reals", although in certain technical contexts the difference between Z and Z' may become important. -- Radagast 3 (talk) 10:13, 22 June 2010 (UTC)[reply]

Why say that 1.0 is different to 1? I was always under the impression that 1 = 1.0 = 1.00 = 1.000 = 1.0000 = …; after all, for any integer n

1.0\dots 0=1+\left(\sum _{k=1}^{n}{\frac {0}{10^{k}}}\right)=1+\left(\sum _{k=0}^{n}0\right)=1+0=1.

If 1 and 1.0 are different then we're in real trouble. Why is there a focus on using decimal notation anyway? I don't think we can write every real number as a decimal, can we? Only the rationals can be written using decimals and bars (finite or repeating decimal expansions). I mean, how do we write √2 in decimal notation (base 10)? (But maybe that's a different question for a new thread) •• Fly by Night (talk) 19:03, 22 June 2010 (UTC)[reply]

Of course the real number 1.0 is the same as the real number 1. Apparently people use "1" to mean the integer 1, and "1.0" to mean the real 1. This is because decimal expansions are suitable for real numbers, but not at all for integers. Any real number has a (possibly infinite) decimal expansion. But the focus is not on the decimal expansions, here they're just used as a notational shortcut to distinguish integers from reals. -- Meni Rosenfeld (talk) 07:39, 23 June 2010 (UTC)[reply]

Is 5+0i>3+0i? It depends on what comparison operator is intended. The notation indicates that two complex numbers (3 and 5) are being compared, so it's not clear what the term "5+0i>3+0i" means. -- Radagast 3 (talk) 10:18, 22 June 2010 (UTC)[reply]