User pace nielsen - MathOverflow

Noncommutative computational package

2012-09-25T20:23:58Z

I am wondering if there is a program which can do simple operations over noncommutative rings, like expand products and substitute one expression for another.

To clarify, consider the following situation. I have two reductions $ab\mapsto 1$ and $ca\mapsto c-1$. If I consider the monomial $cab$ I can reduce it in two ways: $cab=c(ab)=c$ or $cab=(ca)b=(c-1)b=cb-b$. I can combine these computations to arrive at a third reduction $cb\mapsto b+c$.

I'm in a situation where I have upwards of twelve reduction rules, and it gets very complicated doing the reductions. I find myself making small errors. Thus, the need for a machine to do these computations for me.

To make this more precise, is there a program where I can first input a number of reductions, and then second have it work on a monomial and spit out a reduced form?

Answer by Pace Nielsen for How do we recognize an integer inside the rationals?

2010-11-23T16:50:12Z

I just ran across the following paper: Defining Z in Q

It uses another characterization of the integers inside the rationals that none of us listed, perhaps because it is so trivial. Namely, the integers are the complement of the set $\mathbb{Q}\setminus\mathbb{Z}$! Apparently this basic fact is put to good use in the paper.

How do we recognize an integer inside the rationals?

2010-03-30T15:24:43Z

My question is fairly simple, and may at first glance seem a bit silly, but stick with me. If we are given the rationals, and we pick an element, how do we recognize whether or not what we picked is an integer?

Some obvious answers that we might think of are:

A. Write it in lowest terms, and check the denominator is 1.

B. Check that the p-adic valuation is non-negative, for all p.

C. Decide whether the number is positive (or negative) and add 1 to itself (or -1 to itself) until it is bigger than the rational you picked. (If these multiples ever equaled your rational, then you picked an integers.)

Each of these methods has pluses and minuses. For example, in option A we presuppose we know how to write an arbitrary rational number q as a quotient of integers and reduce. In C, we have issues with stopping times. etc..

To provide some context for my question: We know, due to the work of Davis, Putnam, Robinson, and Matijasevic, that the positive existential theory of $\mathbb{Z}$ is undecidable. The same question for $\mathbb{Q}$ is not entirely answered. One approach to this new question is to show that that, using very few quantifiers, one can describe the set of integers inside the rationals; and then reduce to the integer case. For example, see Bjorn Poonen's paper "Characterizing integers among rational numbers with a universal-existential formula." There, he finds a way to describe the p-valuation of a rational number (i.e. he finds a way to encode option B in the language of quantifiers and polynomials on the rationals). I'm wondering if there are other characterizations of the integers which would follow suit.

Official names for specific compound sentences

2012-05-17T16:55:02Z

This question is, admittedly, a little less mathematical than what I normal ask. I seemed to remember that the compound sentence $A\wedge \neg A$ has an official name (maybe even "contradiction" but I seem to remember a different name) just like $A\wedge B$ is "conjunction" and $\neg A$ is "negation".

So, first, what is the classical name of $A\wedge \neg A$?

Second, is there a nice resource to find a list of all classical names for compound sentences?

Answer by Pace Nielsen for Please help me out

2012-04-13T21:09:47Z

That sounds like a personal decision you will have to answer on your own, but here are just a few things to consider as you make your decision.

Have you already obtained a Master's degree? If not, you could finish that and leave for the PhD program elsewhere with fewer problems.
It is okay to change advisers if one is not working out. You don't necessarily need to leave the school you are at to do that.
If you are comfortable with it, talk with your department chair. Explain what is happening, and get his/her advice. Another alternative would be the graduate committee chair. They want you to succeed, and can offer advice on ways to change advisers.
It is a normal part of research to find out others have done the work before you. Some options include: generalizing what was done, researching similar but different questions, communicating with the authors of those other papers about other research avenues, etc...

Whether or not your adviser will be angry with you depends a lot on how you deal with this situation, and what your adviser is like. My advice: be honest and talk with your adviser.

Density in van der Waerden's theorem

2012-01-13T18:56:54Z

Color the positive integers using just two colors. By van der Waerden's theorem, we can find a $k$-term arithmetic progression as long as we consider a long interval.

I imagine it is possible to find a $k$-term arithmetic progression so that the terms in the progression have minimal gaps by possibly taking an even longer interval (of some fixed size depending only on $k$). If so, how do these minimal gaps behave as a function of $k$?

Sequences without long arithmetic progressions

2012-02-08T19:08:06Z

First, a bit of notation. If we have an arithmetic progression $a, a+k, a+2k, \ldots, a+(n-1)k$ we will call $k$ the distance, and $n$ the length.

While trying to find an example for a paper I'm writing in ring theory, I was led to ask the question: Is there a sequence of 0's and 1's for which if there is an arithmetic progression in the sequence which is constantly 0 (or 1), is there a bound for the length in terms of the distance?

I found that the answer is yes, and the Thue-Morse sequence works. Modifying the ideas of Corollary 2 in "Thue-Morse at multiples of an integer" (available here), we see that the length of a constant arithmetic progression on the Thue-Morse sequence of distance $k$ is bounded by $32k^3$.

So, here are my questions for you experts.

(1) Is there an easier sequence where one can prove this is true (possibly with citation in the literature)?

(2) If not, is there a straightforward citation for this fact for the Thue-Morse sequence? (The reference I gave above works for arithmetic progressions which start near the front of the sequence. But to get an arbitrary arithmetic progression, you need to increase the bound given in the paper a little, and also give a supplementary argument.)

Does a "composite field" always exist?

2012-02-06T19:54:53Z

Suppose $F$ is a field, and $F_1, F_2$ are two extension fields of $F$. Is it always the case that there is a field $L$, containing three subfields $F, K_1, K_2$ and two ring isomorphisms $\varphi_{i}:F_i\rightarrow K_1$ fixing $F$?

Note 1: We lose no generality assuming $F$, rather than an isomorphic copy of $F$, is a subfield of $L$.

I ask this because I was wondering if there is a way to combine the reals and the $p$-adic numbers into a single extension of $\mathbb{Q}$.

Note 2: I seem to recall someone telling me this couldn't be done (perhaps with additional topological data preserved). But I cannot seem to remember the reason why. In any case, I want to know if there is something other than topology which prevents it.

Answer by Pace Nielsen for On the prime $k$-tuple problem

2012-02-03T19:53:43Z

It is currently believed that the second conjecture is likely false, but it hasn't been proven quite yet. There is an interval of size 3159 which is not prevented from having more primes than the initial segment of 3159 integers, which is how the first Hardy-Littlewood conjecture would refute the second one. See this wiki article for more information.

Answer by Pace Nielsen for Omniscient bots gathering on $\mathbb{Z}^2$

2012-01-20T20:04:12Z

This should formalize Koen S's answer: which fully answers Q2.

Stage 1: Every bot scans to see if they are the "lowerleftmost" bot; the bot which satisfies that condition calls itself Alice. Of the remaining bots, each bot on the left end of its row calls itself Bill, and if a bot is not the left-most one on its row it calls itself Carl.

Stage 2: Alice moves down once, and then left until there are no bots to her left, nor directly above. Alice moves left once more for good measure. (Note: If there was another bot on the same initial row as Alice, the left-most one changes its name from Carl to Bill.)

Stage 3: Whenever Alice is on an empty row, she moves up one row (unless there are no bots on rows above her). If Bill sees Alice move onto his row, he moves down once on the next turn. If Carl sees the bot directly to his left (on the same row) move down, he moves down on the next turn.

This allows Alice to count how many bots there are, on each row, and in which positions. Each bot, one by one, simply moves down one square.

Stage 4: If Alice is on a row with no other bots (but at least one bot on a row below her) she moves down. If she is on a row with other bots, she communicates with the left-most bot. She does this by left-right movements (which allow her to encode the information in binary). She communicates where that bot should move into a square (far above the original configuration). After she tells the bot where to go, she waits until that bot will have gotten into position, and then communicates with the next bot on the row, moves down, or (at the very end) takes her own position in the final square.

Answer by Pace Nielsen for Should I write to the referee?

2012-01-19T19:05:01Z

If the correction will help the referee understand your paper better, you should definitely write it up and send it to the editors, who will pass it on to the referee. Having served as a referee I can honestly say that I would always appreciate receiving clarifications, and would not hold that against anyone. You might also send a very short note, explaining the changes.

Answer by Pace Nielsen for Is this Ramsey-type problem an open problem?

2011-10-18T15:51:56Z

The question you pose is (currently) still open, but there was an interesting result by my friend Peter Blanchard which proves a "divisible" version of the problem. Namely, given a finite coloring, there are $x$, $y$ and $x+y$ all with the same color, and $x|y$.

Pseudo-arithmetic sets and Ramsey theory

Answer by Pace Nielsen for Are quivers useful outside of Representation Theory?

2011-10-12T16:45:45Z

The formation of quiver algebras is useful in ring theory in attempts to construct examples/counter-examples. One can often classify when a quiver algebra has some property in terms of some intrinsic property of the quiver itself, and then finding counter-examples boils down to forming a quiver with a certain easy-to-see property.

Answer by Pace Nielsen for Geometric interpretation of matrix minors

2011-09-12T18:53:24Z

One graph-theoretic interpretation of minors is given by Linstrom's lemma. http://geomblog.blogspot.com/2004/06/lindstroms-lemma.html

Answer by Pace Nielsen for Other Ring Structures on $\mathbb{Q}$

2011-08-31T18:38:44Z

The answer is yes (unless I made a mistake somewhere).

For example, you can replace addition with the operation $a\oplus b=a+b-1$. This is a commutative, associative binary operation with identity $1$ and the inverse of $a$ is given by $2-a$.

You replace multiplication by $a\odot b= a+b-ab$. This is a commutative, associative binary operation with identity $0$.

All that remains is to show that the distributive laws hold.

Answer by Pace Nielsen for Proofs that require fundamentally new ways of thinking

2011-08-29T14:45:27Z

The use of ideals in rings, rather than elements (in terms of factorization, etc...).

This was followed by another revolutionary idea: using radical (Jacobson radical, etc...) instead of simple properties on elements.

Is there an elementary way to find the integer solutions to $x^2-y^3=1$?

2010-09-22T00:04:26Z

I gave this problem to my undergraduate assistant, as I saw that Euler had originally solved it (although I am having trouble finding his proof). After working on it for two weeks, we boiled the hard cases down to showing that (1) in $\mathbb{Z}[w]$ the fundamental unit is $1+w+w^2$ (where $w$ is the real cube-root of 2) [which I'm sorry to say, I'm not certain I know off the top of my head how to prove], and (2) using that fundamental unit, I found a crazy ad hoc computation to show that there are only the obvious solutions.

So I'm wondering if someone else out there is more clever, knows where I can find Euler's proof, or if there is another nice elementary proof in the literature.

Answer by Pace Nielsen for What is a "best" transcendence basis for R/Q ?

2011-06-13T21:26:14Z

No, you cannot choose $T$ so nicely. For cardinality reasons, we can choose some $s\in S$ with $s\notin \overline{\mathbb{Q}}$. Now, consider the element $\sqrt{s}$. I'll leave it to you to consider how large this makes $T$.

Does the following Diophantine equation have nontrivial rational solutions?

2011-05-16T22:40:53Z

Are there any solutions to the equation $s^{2}(1+t^{2})^{2}+t^{2}(1+s^{2})^{2}=u^2$ where $s,t,u\in \mathbb{Q}$ and $0 < s,t<1$? If so, is there a simple way to parametrize them all?

If I am understanding the geometry behind this problem, even if we pick a specific value for $t$ we are left with an elliptic curve, and it is possible there could be infinitely many solutions.

This question arises from some related (and somewhat esoteric) questions I have about rational points on unit circles. But this question looked "pretty" enough, I thought I'd ask the experts here.

Answer by Pace Nielsen for Philosophical Question related to Largest Known Primes

2011-04-25T20:20:46Z

I've always understood "the largest known primes" to colloquially refer to primes which (i) have decimal expansions which can be written down in a short amount of time and (ii) have primality proofs which have been double/triple checked.

I imagine that if the sequence you suggest were shown to always be prime, people would stop talking about the largest known primes. They might continue to search for ways to find lots of primes of the same large size, and improve the bound on that size.

Answer by Pace Nielsen for Factorizing polynomials in $\mathbf{Z}[[x]]$

2011-02-22T19:43:51Z

Edited due to mistakes pointed out in the comments:

I think the answer to the problem might be yes. Here are some preliminary thoughts.

First, you know $f_{\nu}(x)$ divides $f(x)$, since $f(x)$ is always in the kernel, the kernel is principal, and $\mathbb{Z}[[x]]$ is a UFD.

Second, you know $f_{\nu}(x)$ is irreducible, else you don't map into a domain.

Third, I think you can probably prove that these polynomials (the $f_{\nu}(x)$) are distinct. If so continue:

Fourth, this all says that $\prod_{\nu}f_{\nu}(x)$ divides $f(x)$ in $\mathbb{Z}[[x]]$. So, to get the result you just have to prove that the constant coefficients agree up to a unit (i.e. up to $\pm 1$). This is equivalent to showing that the constant coefficient of $f_{\nu}(x)$ is $p^{f_{P}\nu_{P}(\alpha)}$ where $P$ is the prime (above $p$) associated to $\nu$ and $f_{P}$ is the inertial degree of $P$ in the ring of integers over $K$. (Sorry for the double use of $f$--for the polynomial and for the inertial degree.)

Answer by Pace Nielsen for Rig of fractions, including zero denominators

2011-01-13T22:16:50Z

Although this was said in previous answers and comments, I like to think of it this way.

I imagine you want the element $1/0$ to be the inverse of the element $0/1$. But $0/1=0$ is the additive identity, and multiplying by the additive identity always gives $0$.

So you have something like $1=1/0 * 0/1 = 0$ (depending on which way you think about the multiplication). Note: This does give you a ring, it is called the zero ring.

Answer by Pace Nielsen for An equation in the free associative ring

2011-01-05T17:52:50Z

If I understand your question, the answer to your easier problem is yes. (I'm assuming that by the free associative ring over $X$ you mean to take coefficients of words over the integers.)

First, you can order words in the alphabet, initially by degree, and then by ordering elements in $X$ and using a lexicographical ordering on words of equal degree.

Case 1: $u=v$. Then if $p=q$ you can let $x,y$ be arbitrary. If $p\neq q$ then either $x=0$ or $y=0$.

Case 2: $u\neq v$. Clearly $p\neq q$. Without loss of generality, we may assume that $u$ is of larger order than $v$, and also that $p$ is of larger order than $q$ (replacing $x$ by $-x$ if necessary).

Let $x'$ be the term from $x$ with largest order, and let $x''$ be the term (with non-zero support) with smallest order (these might agree), and similarly define $y',y''$. We must then have $x'py'=u$ and $x''qy''=v$. There are thus only finitely many choices for $x',x'',y',y''$.

But then there are only finitely many choices for terms between $x'$ and $x''$ if we limit ourselves to words in elements of $X$ appearing in $u,v,p,q$. If $x$ involved a term with an element of the alphabet $X$ not appearing in those four words, a simple argument tells us that $x(p-q)y$ would have a term that cannot cancel involving that variable, hence could not equal $u-v$. Thus, there are only finitely terms to try (and letting the coefficients be arbitrary constants, you get a system of linear equations).

Answer by Pace Nielsen for On Sorli's Conjecture Re: OPNs (Circa 2003)

2010-12-04T00:12:46Z

As far as I know, there are no such effective bounds. In fact, even if $p=5$ and $k=1$, there are no known effective bounds on $N$. (There are bounds on $N$ in terms of the number of distinct factors.)

Answer by Pace Nielsen for Does $S_4$ inject into $SL(2,R)$ for some commutative ring $R$?

2010-11-29T23:04:20Z

Over a field, the answer to the first question is no. An element of $A$ of order 4 in SL(2,R) must satisfy the equation $x^{4}-1$. But it also satisfies the characteristic polynomial $x^{2}-{\rm Tr}(A) x+1$. Taking the gcd of these two polynomials, we quickly see that $A$ also satisfies $({\rm Tr}(A)^{3}-2{\rm Tr}(A)x+{\rm Tr}(A)^{2}$. If the coefficient of $x$ is nonzero then $A$ satisfies a linear polynomial, so is diagonal, thus has 4th roots of unity along the diagonal. There are only finitely many such matrices. If the coefficient of that linear polynomial is zero then ${\rm Tr}(A)=0$ and so $A$ is a root of $x^{2}+1$, which means any two elements of order 4 have equal squares (except for the finitely many possibilities expressed in the previous case).

As for the original question: $S_{4}$ has a presentation $\langle s,t| s^{2}=t^{3}=(st)^{4}=1\rangle$. I would suggest looking at the ring $R=\mathbb{Z}[a,b,c,d,e,f,g,h]/I$ where $I$ is the set of relations forcing the 2x2 matrices give by s=((a,b),(c,d)) and t=((e,f),(g,h)) to satisfy the relations for $S_4$ and have determinant 1. Finding a Grobner-type basis for $I$ should demonstrate that the resulting structure has at least 24 distinct matrices in the group generated by s and t (or it doesn't and you cannot embed $S_{4}$ in such a ring).

Answer by Pace Nielsen for Maximal ideal that annihilates entire ring

2010-11-04T23:52:43Z

Ok, I think I worked out the bugs in my previous answer.

Let $F$ be the field with two elements (for simplicity). Let $T=F< a_{i}, x_{k,i},y_{k,i}:i\in \mathbb{N}, k\in K>$ be the free algebra over $F$, generated by the non-commuting variables $a_{i}$, $x_{k,i}$, and $y_{k,i}$ (where $k$ runs over an indexing set $K$ we will describe later, and $i$ runs over the non-negative integers), with no constant term. Let $J$ be the ideal generated by the relations $a_{i}=a_{2i+1}a_{0}a_{2i+2}$, $x_{k,i}=x_{k,2i+1}a_{0}x_{2i+2}$, $y_{k,i}=y_{k,2i+1}a_{0}y_{k,2i+2}$ and $a_{0}^{2}a_{i}=a_{0}^{2}x_{k,i}=a_{0}^{2}y_{k,i}=a_{i}a_{0}^{2}=x_{k,i}a_{0}^{2}=y_{k,i}a_{0}^{2}=0$ for each $i$. Let $S=T/J$ and identify each variable with its image in the quotient (and continue this practice below). Notice that $S$ is generated by $a_{0}$ and $S^{2}=S$. Also note that $a_{0}^{2}$ is a universal zero-divisor. The ring $R$ we want will be a quotient of $S$, so will still be generated by $a_{0}$, we will still have $R^{2}=R$, and $a_{0}^{2}$ will still be a universal zero-divisor. We will construct $R$ so that $a_{0}^{2}$ remains nonzero and generates a maximal ideal.

Let $I'$ be the set of words $w\in T$, $w\notin a_{0}^{2}+J$, such that there is a word $w_{1}\in T$ of length 1 (i.e. a variable) with $w_{1}w-a_{0}^{2}\in J$ or $ww_{1}-a_{0}^{2}\in J$ but no words $w_{2},w_{3}\in T$ with $w_{2}w-a_{0}\in J$ or $ww_{3}-a_{0}\in J$ or $w_{2}ww_{3}-a_{0}\in J$. [For example, $a_{0}a_{1}a_{0}a_{5}a_{0}$ is such a word. Multiplying on the right by $a_{6}$ it reduces to $a_{0}^{2}$, but it does not equal $a_{0}^{2}$ modulo $J$, and can never be multiplied to $a_{0}$.] If we make words in $I'$ zero (or even zero divisors) that will possibly make $a_{0}^{2}$ zero. So, let $I$ be the ideal generated by the following relations: $x_{w,0}wy_{w,0}=a_{0}$ if $w\in I''$, and take $S_{1}=S/I$.

At this point, we repeat the argument in the previous paragraph on $S_{1}$ (our new set $I_{1}'$ will have new words in it). We do this countably many times, the resulting ring is $S_{\infty}$.

Next, mod out by the ideal generated by words $w\neq a_{0}^{2}$ such that there exist no words $w_{1}, w_{2}$ so that $w_{1}w=a_{0}^{2}$ or $ww_{2}=a_{0}^{2}$ or $w_{1}ww_{2}=a_{0}^{2}$ (in $S_{\infty}$).

The resulting quotient ring should be the structure you are looking for. (The index set $K$ could initially be all words in $T$, then cut down to words that appeared in any of the $I'$'s above.)

This is complicated enough that I might have made a mistake somewhere--this is just a sketch of my ideals. One of the main points that should be noted is that (besides $a_{0}$) each variable appears in exactly two relations (except those when a word is made equal to 0): one where it is used to reduce the size of a word, and one where the size expands.

Answer by Pace Nielsen for Algebraic Attacks on the Odd Perfect Number Problem

2010-09-20T21:46:27Z

This is a problem I have thought alot about. I have not seen any of the modern techniques in your list applied to the problem. Part of the issue is that if you represent $\sigma(n)=2n$ as a Diophantine equations in $k$ variables (corresponding to the prime factors--but allowing the powers to vary) then there are lots of solutions (just not where all the variables are simultaneously prime). So the usual methods of trying to show non-existence of solutions just don't cut it. Historically, this multiplicative approach is the one many people have taken, because at least some progress can be made on the problem. My personal feeling is that maybe someday these bounding computations will be tweaked to the point that they lead to the discovery of some principle that will solve the problem. For example, in one of my recent papers, I was led to consider the gcd of $a^m-1$ and $b^n-1$ (where $a$ and $b$ are distinct primes). I would conjecture that this gcd has small prime factors unless $m$ or $n$ is huge. If that happens, many of the computations related to bounding OPNs become much easier.

I have occasionally thought about whether modular forms might say something about this topic (which is why I'm currently sitting in on my colleague's course). Instead of $\sigma(n)$, the `right' function to consider is $\sigma_{-1}(n)=\sigma(n)/n$ and I don't know off the top of my head if it appears in connection with (weakly holomorphic) modular forms. But I know there are some nice techniques about multiplicative functions that decrease over the primes, etc...

Answer by Pace Nielsen for "Oldest" bug in computer algebra system?

2010-09-15T16:03:33Z

In mathematica, if you look at the dirichlet characters modulo 4, you don't actually get the characters.

Character sums over prime arguments

2010-06-09T21:19:48Z

Let $f$ be a monotone decreasing, continuously differentiable function with $\lim_{x\rightarrow \infty}f(x)=0$. Let $\chi$ be a non-principal Dirichlet character. It is standard to show that $\sum_{n\geq x}\chi(n)f(n)=O(f(x))$, where the big-O constant is easily computable and depends only on $\chi$. In particular, we have $\sum_{n\leq x}\chi(n)f(n)=A+O(f(x))$ where $A=\sum_{n\in \mathbb{N}}\chi(n)f(n)$ is a constant.

When $f(x)=\log(x)/x$, Mertens used the fact that $\log(ab)=\log(a)+\log(b)$ to re-express the sum in terms of a sum over primes. He showed that $\sum_{p\leq x}\chi(p)\log(p)/p$ is bounded, in absolute value, by a computable constant. Then, by partial summation techniques, he removed the $\log(p)$ from the numerator and obtained bounds of the form $$\left|\sum_{p\leq x}\chi(p)/p- C \right| < D/\log(x)$$ where $C$ and $D$ are easily computable constants (possibly depending on $\chi$).

My question is two-fold. First, what conditions on a function $f$ (satisfying any of the nice properties above, or more) guarantees that $\sum_{p\in \mathbb{N}}\chi(p)f(p)$ exists?

Second, since $-L'(s;\chi)/L(s;\chi)$ is analytic in a neighborhood of $s=1$, we know that $\sum_{p\in\mathbb{N}}\chi(p)\log(p)/p$ converges, say to a constant E. Is there an easy way to obtain explicit bounds of the form $$\left|\sum_{p\leq x}\chi(p)\log(p)/p - E \right| < o(1)$$ where the $o$-function is fairly simple, etc...?

The reason I ask is that I want to find an effective formula for $\sum_{p\equiv a\pmod{k},\, p\leq x}\log(p)/p$, where the error term is small. If anyone has such a reference that would also be appreciated.

Answer by Pace Nielsen for Subtle counterexample to $m\neq n$ but $R^m=R^n$ for some ring $R$ ?

2010-08-24T17:53:10Z

There are examples of exotic behavior like that which you propose. The specific objects which you should look for are the Leavitt algebras of type (2,2). A very good source on how to create many such examples is George Bergman's paper "Coproducts and some universal ring constructions" although there are easier methods for the specific example you seek.

Comment by Pace Nielsen

2012-09-05T15:28:09Z

Xander, your proof of the exercise is absolutely correct, and Qiaochu's comment explains why the map $\varphi$ is not an $A$-module isomorphism. However, I wanted to point out that for non-commutative rings this result fails miserably. For example, if $A$ is the endomorphism ring of an infinite dimensional vector space then $A^2\cong A$. The idea in proving this isomorphism is very similar to the map you took above!

Comment by Pace Nielsen

2012-07-25T19:44:40Z

With the most recent edit, doesn't a cyclic group of prime order work? (Or, more generally, any F_p vector space, thought of as an additive group?)

Comment by Pace Nielsen

2012-05-30T19:18:05Z

You question seems to have many easy intuitive answers. For example, fix the point $x$ you want to decide is either inside or outside. Bound the given diagram by a large enough circle, and fix a point $y$ outside the circle. Connect $x$ and $y$ by a straight line. Now just count the number of times this line crosses the boundary of the figure. If it is odd, the point x lies inside. If it is even, the opposite. (Of course, this only works because your curve is piece-wise linear.)

Comment by Pace Nielsen

2012-05-17T22:49:02Z

I wouldn't be opposed to Latin terms, but I wouldn't require it either. (By the way, the closest thing I could find so far is "antinomy".)

Comment by Pace Nielsen

2012-04-13T22:13:54Z

@Daniel: Fixed I hope. @Fernando Muro: Good points.

Comment by Pace Nielsen

2012-02-14T22:03:58Z

You could make "most" less fuzzy by asking: Does the ring of integers satisfy the property that if there is a subring $S$ of $L$ in which $x\in S$ is irreducible, then $x\in R$ and is irreducible or a product of irreducibles. One problem is that, unless the class number is 1, the ring of integers is not a UFD. So irreducibles behave badly. This is why, historically, one uses prime ideals rather than irreducibles. We have unique factorization of prime ideals.

Comment by Pace Nielsen

2012-02-09T16:32:46Z

If I read Andreas correctly, I think he was saying that in his philosophically motivated "model" of mathematics, there are no proper classes. That is not to say there are not other ways of thinking about mathematics, which are consistent, and make use of proper classes.

Comment by Pace Nielsen

2012-02-08T19:56:20Z

If the margin is too small, how about giving me a reference? :-)

Comment by Pace Nielsen

2012-02-06T16:58:18Z

Brevity. After your comment I did try to expand my answer a little. Sorry I was so brief. I perhaps should have mentioned all of the results which tend to validate the first Hardy-Littlewood conjecture (such as the existence of infinitely many almost prime pairs).

Comment by Pace Nielsen

2012-02-03T20:09:40Z

There is current numerical work trying to show that 3159 is the smallest size where this happens: opertech.com/primes/k-tuples.html

Comment by Pace Nielsen

2012-02-03T20:07:31Z

Yes. And people currently strongly believe the first conjecture. There is all sorts of evidence towards it, but there doesn't seem to be any reason to believe the second conjecture.

Comment by Pace Nielsen

2012-02-02T16:51:31Z

Gerry, I interpreted his equation $x\neq 0$, to mean that $x$ is nonzero in some context (such as, $5$ is a nonzero integer). In other words, I interpreted his question as using the symbols $=$ and $\neq$ in different contexts. (From how the question was asked, I truly think that he needed to hear about modular arithmetic before we sent him looking at semigroups or ordinal arithmetic.)

Comment by Pace Nielsen

2012-01-25T19:03:49Z

@Barry: Thanks! (Just for that I gave your answer an upvote. Cheers.) @Joseph: Barry got it. After that first step bot $a$ is then the lowerleftmost, but only after that first step and he realizes this fact, so resumes his place and calls himself Bill. Bot $c$ always knew she was Alice, and bot $b$ always knew he was Carl.

Comment by Pace Nielsen

2012-01-24T17:10:02Z

A couple clarifications: In stage 1, there might be multiple bots which scan and think they might be the "lowerleftmost" bot, but their view is obstructed. So, they each can take a step down one spot, and then rescan. If they were not the lowerleftmost, they still are not, and they move directly back to where they were. Stage 4 starts when Stage 3 finishes, in which case Alice is alone on a row above all the other bots, and starts working her way back down through the ranks.

Comment by Pace Nielsen

2012-01-24T17:01:58Z

@Barry: In stage 1 how do the bots know when Stage 1 is done? For example, suppose that they are all on the same row, all spaced at least N columns away from each other? Then each bot can only see the bot to his immediate left or right, and thus doesn't know what is going on, and if stage 1 has completed. @Joseph: What did you find wrong with my solution? I thought it was fairly detailed.