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June 15

Middle factor not a zero divisor

Is there a (noncommutative) ring with an element b that is neither a left nor right zero divisor but such that there exist elements a and c with ac ≠ 0 such that abc = 0? GeoffreyT2000 (talk) 01:05, 15 June 2016 (UTC)[reply]

How about making a monoid

\langle x,y,z,0|xyz=0,x0=0x=y0=0y=z0=0z=0\rangle

and then taking formal sums, but identifying 0 of the monoid with 0 of the sums? Then y is such an element.--130.195.253.10 (talk) 06:27, 15 June 2016 (UTC)[reply]

Put another way, take the formal sums of strings in x, y and z (the monoid ring constructed from the free monoid on 3 generators), and mod out by the ideal

(xyz)

.--130.195.253.10 (talk) 06:30, 15 June 2016 (UTC)[reply]

One example is matrices. E.g.

A={\begin{pmatrix}1&0\\0&0\end{pmatrix}},B={\begin{pmatrix}0&1\\-1&0\end{pmatrix}},C={\begin{pmatrix}1&0\\0&0\end{pmatrix}}

Basically A as a transformation projects onto the x-axis by zeroing y values. Done twice (AC = AA = A) it produces the same result. But introducing a 90° rotation between the projections causes it to zero both x any y so ABC = 0.--JohnBlackburne^words_deeds 08:54, 15 June 2016 (UTC)[reply]

Cute!--JBL (talk) 14:43, 15 June 2016 (UTC)[reply]

Expressibility of Peano arithmetic and the Arithmetical Hierarchy

First-order Peano arithmetic has no non-logical symbols other than S, +, *, < and variables. One allows finite quantification over predicates such as: $\forall k<n:\phi (k)$ where $\phi (k)$ is a logical formula contatining some non-logical symbol such as A_k. This is because it is understood as a shorthand for $\phi (1)\land \phi (2)...\land \phi (n)$ . That way the A_k-s are nothing but n different variables. For example:

(A_{0}=1)\land \forall k<3:(A_{k+1}=A_{k}*(k+1))

is actually:

(A_{0}=1)\land (A_{1}=A_{0}*1)\land (A_{2}=A_{1}*2)

.

This way we can deal with primitive recursive functions.

However, situation is different when moving up in the arithmetical hierarchy. For example, in $\Sigma _{1}^{0}$ it is no longer possible to make the above substitution, because the number of variables will now depend on a non-fixed variable on which we quantify in a non-bounded way. Taking the above example, we now have:

\exists n:(A_{0}=1)\land \forall k<n:(A_{k+1}=A_{k}*(k+1))

which we obviously cannot turn into a formula with no new non-logical symbols.

So my question is: do we actually allow other non-logical symbols in first-order Peano arithmetic in certain conditions? or do the arithmetical hierarchy formulas actually not belong to first-order Peano arithmetic? or am I missing something?

Thanks! Dan Gluck (talk) 08:37, 15 June 2016 (UTC)[reply]

Nit: Where you wrote Σ¹₀, you have the superscript and subscript reversed. A superscript 1 means you're quantifying over sets of naturals (or "equivalently", at least in a lot of contexts, over reals).

As for the substantive question, this is one of those coding details that most people struggle through once (if that) and then promptly forget. See course-of-values recursion. You can code up the sequence of A's into a single natural number, and write a formula that effectively decodes it and recovers your meaning. I don't remember the details of how to do it, assuming I ever knew them. I have a fuzzy memory that the Chinese remainder theorem is somehow relevant. --Trovatore (talk) 19:20, 15 June 2016 (UTC)[reply]

Thanks, I corrected the superscript/subscript error. What still puzzles me, is that the esatblishment of Goedel's numbering seem to require the use of primitive recursive functions, which again require the addition of non-logical predicates... I guess it can be done somehow without that, but I still don't understand how.Dan Gluck (talk) 05:34, 16 June 2016 (UTC)[reply]

Let me rephrase my question. Suppose that we want to construct a_n = 2ⁿ.

So we find the Goedel number G of the formula:

\forall n:a_{n+1}=2*a_{n}

.

Now, we just have to show there is some formula

\phi (G,n,m)

(informally meaning m = sⁿ such that:

\forall n:\phi (G,n,m)\rightarrow \phi (G,n+1,2*m)

and:

\phi (G,0,1)

.

While it is clear to me such a formula can be constructed by using new non-logical symbols, it is unclear how to do that without such symbols.Dan Gluck (talk) 09:32, 16 June 2016 (UTC)[reply]

I kind of convinced myself that all the steps I was thinkg about don't really necessitate the use of new non-logical symbols, so I guess that's it. Thanks again.Dan Gluck (talk) 11:00, 16 June 2016 (UTC)[reply]

Number of De Bruijn sub-sequences of a given length

I'm trying to a write a program that generates arbitrary subsequences of De Bruijn sequences. The first step is to figure out exactly how many of such subsequences are there.

From De Bruijn sequence:

...a De Bruijn sequence is a cyclic sequence of a given alphabet A of size k that contains every possible subsequence of length n in A exactly once. Such a sequence is written B(k, n) and has length kⁿ, which is also the number of distinct subsequences of length n in A; de Bruijn sequences are therefore optimally short. There are ${\dfrac {\left(k!\right)^{k^{n-1}}}{k^{n}}}$ distinct De Bruijn sequences B(k, n).

Given j, where kⁿ - n < j <= kⁿ, each B(k, n) has ${k^{n} \choose j}$ subsequences of length j. In each such subsequence the useful property of "every possible subsequence of length n occur only once" still remains true. We denote these subsequences by B_j(k, n, j).

The Problem

There are kⁿ sequences of length n (given alphabet size k), of which ${\dfrac {\left(k!\right)^{k^{n-1}}}{k^{n}}}$ are de Bruijn sequences.

There are k^j sequences of length j, how many of these are B_j(k, n, j)?

My solution so far

Each B(k, n) has ${k^{n} \choose j}$ subsequences of length j. Since j > n, each B_j(k, n, j) of the same de Bruijn sequence are necessarily unique.

If we suppose that each B_j(k, n, j) of different de Bruijn sequences are also unique, then the answer to the above question is ${\dfrac {\left(k!\right)^{k^{n-1}}}{k^{n}}}{k^{n} \choose j}$ .

Is this correct? Johnson&Johnson&Son (talk) 09:18, 15 June 2016 (UTC)[reply]

Three comments: (1) Do you really mean "kⁿ − n < j ≤ kⁿ" and not "n < j ≤ kⁿ"? (2) When you introduce the binomial coefficient

{\binom {k^{n}}{j}}

, you are using a different notion of "subsequence" than is usual in this context: De Bruijn sequences are about consecutive subsequences. (3) Re: your final "if we suppose" -- there is no chance that this supposition is typically valid for arbitrary j > n, the numerics will often be terribly wrong (i.e., the number you've written down will be vastly larger than k^j). --JBL (talk) 14:50, 15 June 2016 (UTC)[reply]

Thanks! I've made a number of huge mistakes. It's better if I just re-write the question and submit later rather than correcting the current one. Johnson&Johnson&Son (talk) 03:44, 16 June 2016 (UTC)[reply]

Intuitionistic proofs about strict orders.

Let S be a set of natural numbers, such that the strict order "<" [ i.e the standard strict order ] has in S a number m - which has in S no natural number n satisfying: m<n. Is it intuitionistically provable that, not only [ the standard ] "<" - but also every other strict order R, has in S a number m - which has in S no natural number n satisfying: mRn ? HOTmag (talk) 16:05, 15 June 2016 (UTC)[reply]

I'd say no: it is easy to define a strict order on the natural numbers (or any infinite subset) in which there is no lower bound. Just put R as >. Since the statement is false, we would not be able to prove it in any system of logic, including intuitionistic logic. —Quondum 18:19, 15 June 2016 (UTC)[reply]

~~Oops, sorry, I've just fixed up my mistake, please see again my updated question, above, which now refers to both the lower bound and the upper bound.~~ ">" is not a counterexample, because it has an upper bound (being a lower bound if we replace back - the ">" - by the original "<"), because S is a set of natural numbers. HOTmag (talk) 19:07, 15 June 2016 (UTC)[reply]

The axioms on a strict order in Partially ordered set § Strict and non-strict partial orders do not appear to exclude the strict order "<" (not our classical "<") for which there exist elements m and n such that ¬(m=n) ∧ ¬(n<m) ∧ ¬(m<n). Thus, we could have a strict order "<" with both lowest and highest elements (bounds) on the set of natural numbers, despite the set being infinite, and another strict order R without such elements. —Quondum 21:19, 15 June 2016 (UTC)[reply]

So you claim there isn't such an Intuitionistic proof I'm looking for, don't you? HOTmag (talk) 21:44, 15 June 2016 (UTC)[reply]

The statement you seek to prove seems to me to be false mathematically (a counterexample seems easy to construct) – even if you restate your question replacing "strict order" with "strict total order". Only once you have a mathematically true statement does it make sense to ask whether a proof exists in any specific system of logic. Perhaps you need to review how you've stated the question, even ignoring the intuitionist aspect? —Quondum 03:34, 16 June 2016 (UTC)[reply]

"a counterexample seems easy to construct ". Like what? (assuming we're talking about strict total orders). HOTmag (talk) 08:27, 16 June 2016 (UTC)[reply]

Just to check my interpretation, your question asks: given a strict order on a subset of the naturals with both an upper and a lower bound, can we prove (intuitionistically) that every strict order on this subset necessarily also has two such bounds? I am subject to confusing simple things at times.

Construct a relation ◅ ("<" in your question) on N that is the standard < except for with the element 0, and we define n◅0 whenever n≠0. In your presentation, the bounds are m=1 (lower) and n=0 (upper). This constructed relation is a strict total order. The standard < on N can be taken as a strict total order R with only one bound, and we have our counterexample. An example with no bounds can be constructed through a 1-to-1 map N→Z. —Quondum 15:15, 16 June 2016 (UTC)[reply]

When I wrote "<", I meant the standard "<", so your example does not disprove my original statement about the strict orders. Further, now I see that the original version of my question shouldn't have been fixed up: I didn't have to mention a lower bound nor a total order (See below why). So I'm reverting again to the original version of my question, which has been about whether there exists an intuitionistic proof - for a true statement mentioning the upper bound only.

As for the strict order ">" mentioned in your first response: It's not a counterexample, because it has an upper bound (being a lower bound if we replace back - the ">" - by the original "<"), because S is a set of natural numbers. HOTmag (talk) 17:48, 16 June 2016 (UTC)[reply]

With the clarification on the order "<", this makes sense. By implication, the set S is finite and nonempty. —Quondum 06:11, 17 June 2016 (UTC)[reply]

Trivial.

But I'm still asking, whether there is an intuitionistic proof for my original statement. HOTmag (talk) 07:32, 17 June 2016 (UTC)[reply]

Understood, but I can probably not help you. I can see what seems to me to be a straightforward (apparently constructive) approach, but it does not use symbolic logic and you'll have to decide whether it would be even a hint at a direction. A strict order appears to partition a set into a several disjoint sets (which we can consider to be equivalence classes), and if we select and arbitrary element from each, we have a total order over the resultant subset. Since the original set is finite, the largest element m (under this order) exists, and proves the statement. But I must make the disclaimer that I'm completely out of my depth here. —Quondum 14:44, 17 June 2016 (UTC)[reply]

"Since the original set is finite, the largest element m (under this order) exists ". How do you base intuitionistically, this "since "? I agree with you that the set is finite - under the [standard old strict] order, but I can't see the intuitionistic connection between this fact and the existence of "the largest element m (under this [new strict] order)". HOTmag (talk) 18:38, 18 June 2016 (UTC)[reply]

Are the numbers in mathematical operations real or unreal??

For clarification, "real" does not have the sense used in the real number article. It's being used in a way analogous to the following URL:

http://www.uhuh.com/unreal/moncur.htm

In addition, both a and b (in a+b=c) are real. There's no doubt on this.

In multiplication, a*b is a short way of saying a + a + a + a... = c. Thus, one point of view suggests that a is real and b is not, but because of multiplication's commutative property, both a and b are still real. This property can be demonstrated by rotating an a by b rectangle 90 degrees and noting that its size doesn't change.

But how about in exponentiation?? This is when a is real and b is not. For example, in 2^3 is 8, the 2 is real but the 3 is not. Any disagreement?? Georgia guy (talk) 20:00, 15 June 2016 (UTC)[reply]

I think it's a narrower question to ask whether there's any agreement.

This is a novel ontological view that I have never heard of, and do not really understand. I cannot think of any philosophical perspective from which your claims would make sense. Both the 2 and the 3 in 2³ are abstract objects, and some people consider abstract objects to have real existence, some consider that they do not, and some consider that it depends. But I can't think of any reason it should depend on whether the number is a base or an exponent. --Trovatore (talk) 20:12, 15 June 2016 (UTC)[reply]

By "unreal" number you mean a number which can be replaced by "pure numbers" only (probably this is the article you're looking for), whereas by "real" number you mean a number that can be replaced by an apple or by a meter and likewise. So yes, the 2 in the term 2^3 is really real - because you can replace it by a meter (thus getting the volume of a cube the length of whose sides is a meter), but the 3 is not real: You cannot replace it by a meter, nor by an apple, nor by any real object. So 3 must really be a pure number. HOTmag (talk) 20:17, 15 June 2016 (UTC)[reply]

Well deciphered! I would not have figured out that that was what GG was talking about, but it seems likely that that's it. I don't really see the connection to the "money" webpage, though. --Trovatore (talk) 22:40, 15 June 2016 (UTC) [reply]

Why can't I replace 3 by an orthonormal basis, thus getting the volume of a box with side-length 2 in every basis direction? Or is a basis somehow not a sufficiently concrete thing, while a meter is? I'm skeptical of any such classification that separates the two.130.195.253.9 (talk) 23:43, 15 June 2016 (UTC)[reply]

You could have mentioned simpler examples, like imaginary numbers. However, please note that a number is "real" (in the sense meant by the OP), if and only if it can be replaced by a non-constructible object. Meters and apples are not constructible, while orthonormal bases - imaginary numbers - and natural numbers, are. Anyways, physics is full of non-constructible objects (like physical lengths, segments of time, electric charges and likewise) that can be multiplied by themselves, but none of them can be raised to the power of themselves. That's why the exponential is always considered to be a pure number (whether it is a natural number, or an irrational number, or an imaginary number, or an orthonormal basis). HOTmag (talk) 00:04, 16 June 2016 (UTC)[reply]

You want to make this distinction based on whether something is in L (or "can be replaced by" something in L)? That doesn't sound right to me. That means you're putting 0# in the same category as meters? --Trovatore (talk) 06:42, 16 June 2016 (UTC)[reply]

Instead of mentioning non-constructible objects, I'd better mentioned the indiscernibles, and...yes: I'm putting them in the same category as meters. HOTmag (talk) 08:10, 16 June 2016 (UTC)[reply]

That's a viewpoint I've never heard of before. Intuitively it strikes me as very odd indeed. Do you have more to say about it, or can you point us to anyone who categorizes things this way?

By the way, 0# is (or, if you prefer, can be coded by) a real number (in the mathematical sense of "real number"), so you seem to be saying some real numbers are like meters and some are not. --Trovatore (talk) 14:52, 16 June 2016 (UTC)[reply]

Do you consider 0# to be an indiscernible? I do, and I make a clear distinction between 0# itself - being an indiscernible (like a meter), and its coding - being a natural number (rather than an indiscernible).

As for your first question: I'm talking from the OP's point of view: I think they put the indiscernibles in the same category as meters, as far as the "real" numbers is concerned ("real" in the sense meant by the OP). HOTmag (talk) 16:30, 16 June 2016 (UTC)[reply]

0# is not an indiscernible. Exactly what 0# is depends on the viewpoint you take, but the usual definition is that it's the theory of L together with parameters that form an uncountable set of indiscernible ordinals. This theory can be coded as a set of natural numbers, or as a single real number. Alternative viewpoints would consider 0# to be a mouse, or perhaps an elementary embedding from L to L. I can't make sense of the claim that 0# is an indiscernible in and of itself.

Anyway, this whole approach seems to be singling L out in an unmotivated way. L is a very special model of set theory with a lot of interesting properties, but you seem to be using it to make ontological distinctions for which I can't figure out the basis. --Trovatore (talk) 17:43, 16 June 2016 (UTC)[reply]

Our article indiscernibles (roughly) defines: "indiscernibles are objects which cannot be distinguished by any property or relation defined by a formula ". How can you define/distinguish 0# by a formula? Please note that such a formula must contain the very concept of indiscernibles, and I can't see how this concept can appear in any formula.

Anyways, let's assume you're right, so that 0# is really not an indiscernible. All right. So what did you mean by your saying: "you seem to be saying some real numbers are like meters and some are not "? How did you conclude that about me? Had I ever said anything about 0#, before you concluded that about me?

As for L: I can't figure out why you keep mentioning it, after I'd explicitly written: "Instead of mentioning non-constructible objects, I'd better mentioned the indiscernibles, and...yes: I'm putting them in the same category as meters ". HOTmag (talk) 18:21, 16 June 2016 (UTC)[reply]

(ec) Answering the second paragraph first, I took your statement that ends in yes: I'm putting them in the same category as meters to be talking about 0#. Perhaps I misinterpreted here.

The first paragraph is a bit more involved. I'm not sure you've quite grasped the concept of indiscernibles in the model-theory sense; it might make more sense if you went and spent a little time on that. It doesn't make (much) sense to say that a single object is "indiscernible". The first place the concept becomes nontrivial is with two objects, let's say a and b, which are indiscernible (over some given model M) if M satisfies exactly the same formulas about a that it does about b.

In the 0# case, we're really talking about order-indiscernibles, not full indiscernibles. Here a, b, c are order-indiscernible over M if M always has the same truth value for φ(a,b) that it has for φ(a,c) and φ(b,c), but the truth value of φ(b,a) might be different.

Indiscernibility over M is certainly definable by a formula; just not, usually, by a formula interpreted in M.

In any case, you don't necessarily have to define the indiscernibles to define 0#; as it happens, 0# can be characterized in a large number of equivalent ways, only some of which mention indiscernibles. (I think I never really did finish that homework question; there were just too many of them.) --Trovatore (talk) 18:49, 16 June 2016 (UTC)[reply]

Post-edit-conflict: OK, I have a slightly better idea what you're talking about now. This is not really what we mean by "indiscernible" in the model-theory sense. See remarks above; they may be helpful. --Trovatore (talk) 18:50, 16 June 2016 (UTC)[reply]

Thank you for your clarifications. All right, so 0# is not an indiscernible. Anyways, instead of non-constructible objects or indiscernibles, I'd better talked about objects not definable by any well-formed formula (of Set theory), and yes: I'm putting such objects in the same category as meters (from the OP's viewpoint). HOTmag (talk) 19:52, 16 June 2016 (UTC)[reply]

I'm not seeing how this is a response to my criticism. The set {north, west} is a nonconstructible orthonormal basis.--2406:E006:7C7:1:35F4:BD7F:B65E:4D48 (talk) 10:23, 16 June 2016 (UTC)[reply]

Two comments:

The set {north, west} is really "real" (as a meter is), in the sense meant by the OP.
The 3 in the term 2^3 (i.e. in the operation of multiplying two by itself three times) cannot be replaced by the set {north, west} without replacing the operation itself as well: Really, the name of the operation is the same, but the definition is different. Please note that the classical definition of the operation which raises a to the b-th power - is the operation of multiplying a by itself b times, or (for the general case when b is unnecessarily a natural number): the entire function satisfying for every a,b,c (in the domain) both a^1=a and a^(b+c)=(a^b)(a^c). This cannot be applied to pseudo-exponentials like the set {north, west}, so when we are talking about the operation of raising a given number to the power of {north, west}, we are talking about another operation - despite the common name. HOTmag (talk) 11:07, 16 June 2016 (UTC)[reply]

I am again skeptical of the alluded to classification. In what sense is 3^2 a different operation from 3^{north, west}, yet 3^2 and (meter)^2 are the same operation?

Unless the intention is that an argument to an operation is real if the operation was developed to generalize a real world phenomenon in which said argument was replaced by a real (in the sense of OP) object (e.g. integer addition was developed to generalize combining piles of pebbles). In that case, I think it's placing too much weight on historical accident.--2406:E006:7C7:1:35F4:BD7F:B65E:4D48 (talk) 11:40, 16 June 2016 (UTC)[reply]

If (as you suggest) the concept of "a meter multiplied by a meter" is not regarded as a legitimate classical multiplication (that expresses a two-dimensional area), then not only the 2 in the operation 3^2 but also the 3 in that operation - should be regarded as "unreal" (in the sense meant by the OP), because not only the 2 but also the 3 - cannot be replaced by a non-constructible object - without replacing the very (classical) operation as well, so I wonder why the OP claimed the opposite (regarding the term 2^3). HOTmag (talk) 16:21, 16 June 2016 (UTC)[reply]

I had a look at that link. It seems the author is using the name “unreal number“ to characterise money, as a preparation for making a political point. It does not have a mathematical meaning. Mathematically, or logically, it’s more normal to talk about tangible things (apples, people, cars) and intangibles such as money. But the same sorts of numbers apply to both. In fact that is the power of mathematics and numbers, it can be used with tangible things, with intangibles and imaginary things, and with entirely mathematical constructs. But the rules of mathematics don't change, and in particular numbers work the same way in all cases.--JohnBlackburne^words_deeds 20:20, 15 June 2016 (UTC)[reply]

It's worth noting that the link presented by the OP does not in fact consider ANY numbers to be "real". To quote: "Plain, old numbers are unreal. When attached to something real, they take on more meaning, but the numbers are still unreal. When attached, numbers represent a certain amount of something real or unreal. [...] Let’s consider 3 apples. The apples are real, but the 3 is unreal. Instead of saying apple, apple, apple, we can say 3 apples. That is convenient. We are using an unreal idea to make communication easier." So all numbers are unreal), but some numbers can be applied to real things (there's a later sentence which questions whether real numbers (mathematical term) are real (exist), but it's left open, and the author's opinion seems pretty clear from the above quote).

There's a later sentence which says "We understand how to use unreal numbers to find real (?) numbers.", but in this case "real numbers" is either referring to the normal mathematical definition (i.e. we can use imaginary numbers to find real numbers), or possibly as a shorthand to "a number of real things" (i.e. we can use abstract numbers to work out how many apples there are). The (?) seems to indicate that even the author does not agree with the use of the term here.

It's also worth noting that even when the author uses the actual mathematical term "real number", they get it wrong, restricting to only positive numbers.MChesterMC (talk) 08:41, 16 June 2016 (UTC)[reply]

Topology question

Assume an $\mathbb {R} ^{2}$ universe with a metric topology. Assume $A$ is a closed set in the universe. I'm fairly sure that $A$ is connected if it's not possible to separate it into two closed subsets $A_{1}$ and $A_{2}$ so that $A_{1}\cup A_{2}=A$ , $A_{1}\cap A_{2}=\emptyset$ , and both subsets are closed. Is this right?

But what I'm actually wondering about, is it possible to define the concepts of a similar closed set "having a hole in it" or "not having a hole in it" using only the notions of openness, closedness, exterior, interior, boundary, union, intersection and complement? Or do we have to bring metrics into it, or even have even more elaborate constructs? Would the notion of ${\bar {A}}$ being disconnected work? JIP | Talk 21:18, 15 June 2016 (UTC)[reply]

"Not having a hole" probably means simply connected. You don't need a metric, but you do need the notion of homotopy.

There's also a chance that what you're thinking of when you say "not having a hole" is that it's a regular open set, but that seems less likely. --Trovatore (talk) 22:01, 15 June 2016 (UTC)[reply]

I think you have it, with the exception of maybe "connected space". I doubt OP was picturing those funny things that are connected but not simply connected, but they are still neat to be aware of. Punctured disc might be good to check out too. SemanticMantis (talk) 14:06, 16 June 2016 (UTC)[reply]

It doesn't have to be all that unusual to be connected but not simply connected. An annulus or a torus does the job, or even a circle (with the understanding that a circle is different from a circular disk). Maybe you're thinking of path-connected? It does take a little massaging of the intuition to grasp something that's connected but not path-connected. --Trovatore (talk) 15:50, 16 June 2016 (UTC)[reply]

Of course, thanks, it's been a while. Yes I meant connected but not path connected, like the graph of y=sin(1/x) unioned with the [-1,1] bit of the y axis. I thought User:JIP might be interested in those, here's a nice brief write up [1] SemanticMantis (talk) 15:44, 17 June 2016 (UTC)[reply]

The 2D image with multiple holes in simply connected is what I had in mind, but the article says a hollow ball in

\mathbb {R} ^{3}

is simply connected, but I consider it as having a hole in it. I consider "having a hole", informally, as "there is a point in the set's complement that is fully enclosed in the set, in other words, it's not possible to reach the rest of the universe via a path without crossing the set". Would "the set's boundary is disconnected" count as a definition of "having a hole", providing the set has a non-zero interior and thus is itself not just a boundary? JIP | Talk 18:48, 16 June 2016 (UTC)[reply]

Right, so a 2d sphere in R^3 has no "hole" in the topological sense, since any loop on the ball can be shrunk to a point. In an annulus or punctured disc, that is not true. What you're talking about with the sphere is related to Contractible_spaces, which are informally spaces that can be smoothly shrunk down to a point in that space. A sphere is connected and path connected, but its complement is disconnected. It is not contractible because of the "hole" that you mention. A solid disc can be sort of smoothly deformed to a point, but a sphere cannot. Weirdly, a finite dimensional sphere is not contractible, but the unit sphere in Hilbert space is! So an infinite dimensional sphere does not have the same kind of "hole" that a 2d sphere does... Jordan_curve_theorem is also somewhat relevant. SemanticMantis (talk) 15:47, 17 June 2016 (UTC)[reply]

Number of ratios/quotients

How many ratios or quotients can be formed from n numbers b1, b2, ..., bn?--82.137.15.5 (talk) 22:38, 15 June 2016 (UTC)[reply]

If you're referring to irreducible fractions only, then it's impossible to answer (depends what numbers we're talking about). But if you don't exclude any kind of fraction, then the answer is quite easy:

n*n

. HOTmag (talk) 22:55, 15 June 2016 (UTC)[reply]

{\frac {b_{i}}{b_{i}}}=1

for all i (assuming non-zero numbers), so the largest possible number of distinct values is n × n − n + 1. I guess b_i = 2ⁱ gives the smallest possible number for distinct non-zero numbers. PrimeHunter (talk) 12:13, 16 June 2016 (UTC)[reply]

{\frac {b_{i}}{b_{i}}}

is not an irreducible fraction (unless

b_{i}=1

), and I have already indicated that: "If you're referring to irreducible fractions only, then it's impossible to answer (depends what numbers we're talking about) ". Anyways: when I wrote "if you don't exclude any kind of fraction - then the answer is quite easy: n * n ", I meant that 1/2 and 2/4 should be counted twice, and I don't see any difference between - the case of the ordered pair (1/2, 2/4) - and the case of the ordered pair (1/1, 2/2): In both cases the ordered pair has two elements. HOTmag (talk) 15:55, 16 June 2016 (UTC)[reply]

I'm not arguing with your

n*n

for that interpretation. I was just discussing upper and lower limits for the other interpretation where the fractions must have distinct rational values. It's impossible to give a precise answer but it's still possible to discuss aspects of the question. PrimeHunter (talk) 16:15, 16 June 2016 (UTC)[reply]

Of course distinct non-zero values are of interest. I thought that the number of distinct ratio is given by 2(k)-permutations of n (n numbers) or n(n-1)(n-2)...(n-3) ratios, because the ratios are formed by taken ordered subsets of 2 elements from n numbers. For instance, for 3 numbers 6 ratios are possible:

{\frac {b_{1}}{b_{2}}},{\frac {b_{2}}{b_{3}}},{\frac {b_{1}}{b_{3}}}

and their inverses are possible.--82.137.9.188 (talk) 15:43, 16 June 2016 (UTC)[reply]

(edit conflict) I think any set of n distinct primes gives the maximum number of distinct ratios n²–n+1. Loraof (talk) 15:50, 16 June 2016 (UTC)[reply]

I don't quite understand from where +1 in n²–n+1 comes from. I see that n² is the cardinality of the cartesian product of set of b_i with itself and n the number of ratios of value 1.--82.137.12.243 (talk) 00:49, 17 June 2016 (UTC)[reply]

In what sense is the number of irreducible ratios impossible to be specified? Due to non-specification of concrete numbers? Perhaps in absence of concrete numbers, a probabilistic analysis of the problem is welcome, possibly using techniques from probabilistic number theory. Thoughts?--82.137.8.81 (talk) 16:06, 16 June 2016 (UTC)[reply]

The whole issue has something to do, with the absence of a well formed formula - counting the exact number of primes up to a given prime. HOTmag (talk) 17:03, 16 June 2016 (UTC)[reply]

It seems interesting the absence of a well formed formula for prime-counting function and prime gap, altough prime number theorem tries to establish some bounds.--82.137.12.243 (talk) 00:25, 17 June 2016 (UTC)[reply]

Along with the absence of a well-formed function (for every natural n), giving the n-th prime. HOTmag (talk) 09:29, 17 June 2016 (UTC)[reply]

What is the form of ratios included in the difference between n²–n+1 and 2-permutations of b_n?--82.137.8.254 (talk) 01:03, 17 June 2016 (UTC)[reply]

There are n²–n 2-permutations of b_n. The difference is ratios of form

{\frac {b_{i}}{b_{i}}}

. If we assume such fractions with the same numerator and denominator are allowed then they contribute 1 new ratio in total since they give the same ratio 1 for any i. PrimeHunter (talk) 01:40, 17 June 2016 (UTC)[reply]

One can notice the equivalence between the cartesian product approach and the 2-permutations approach to the number of distinct ratios in a division table. The distinct ratios are those outside main diagonal of an operation table. The value 1 of the multiple non-distinct ratios may be uninteresting (and hence excluded) to further use of these distinct ratios of bn numbers or physical quantities.--82.137.9.22 (talk) 12:09, 17 June 2016 (UTC)[reply]

System of linear equations with a sum and n-1 ratios

How can these distinct ratios can used in a system of n linear equations with the sum of bn numbers and n-1 distinct ratios chosen from n²–n possible distinct ratios? (See the thread below)--82.137.10.68 (talk) 17:34, 17 June 2016 (UTC)[reply]