mm-2469 === Subject: Re: Divine Proportions: Rational Trigonometry to Universal Geometry? not bad coinage. I thought of avoiding roots like that, some time ago (on another list, along with a couple of other related math areas), but did not try to develop it for trigonometry. probably very worthwhile, in the later chapters. > I wonder how most secondary students would fare if that approach were > their first contact with trigonometry. --Hemp for Haemarrhoids (Bogart that Poultice, Friend) !! http://members.tripod.com/~american_almanac === Subject: Re: Mathematical ASL? Well, as I say, I attended and took notes in many classes over the years. I would write down everything on the chalkboard and write down any oral comments that seemed worth writing down. Looking back at the notes after reading the book, it is apparent how much information is missing, or is simply a 2nd rate rehashing of the book without complete sentences, references to previous material, or an index! Very rarely there would be some insightful comment, but that was very rare. So if the information isn't there, it doesn't matter what the learning style of the student is. I wouldn't want to encourage anyone who is lecturing to give up and not prepare the best lectures they can, but rather to realize that the lectures are not the be-all, end-all of learning math. In my opinion, they are frequently a hinderance, in fact. The worst offender of all is the graduate seminar. To this day I still think that there has never been a seminar that was actually worth attending, but that is another topic... --Jeff === Subject: Re: Sphere packing - Why so difficult? <430254ee$0$566$b45e6eb0@senator-bedfellow.mit.edu> so, the above two explanations refer to the same, goofy packing?... I think I have the idea, now. --Hemp for Haemarrhoids (Bogart that Poultice, Friend) !! http://members.tripod.com/~american_almanac === Subject: Re: Sphere packing - Why so difficult? <11949713.1124980929071.JavaMail.jakarta@nitrogen.mathforum.org> well, the usual phrase is not even wrong, since you didn't bother to give any *reason* for what ever you're trying to do. thus quoth: There actually is 32 half's with 28 half's assembled and 4 half's unassembled to give the tightest fit possible for a rectangular box 6*21*24 holding 16 spheres. These are soled half spheres that when glued together by glue or magnetically attracted for a perfect sphere 6 in diameter. I could be wrong here! --Hemp for Haemarrhoids (Bogart that Poultice, Friend) !! http://members.tripod.com/~american_almanac === Subject: Re: density of a sphere - thank you === Subject: How to integrate this by pen and pencil? Hi all, how do you integrate this indefinite integral? Integrate(exp(-2x)*(sec(x))^2) === Subject: Re: How to integrate this by pen and pencil? > Hi all, > how do you integrate this indefinite integral? > Integrate(exp(-2x)*(sec(x))^2) It is not expressible in terms of elementary functions. If you integrate by parts, the non-trivial parti becomes int(exp(-2x)*tan(x) dx). The two transcendental functions here are e(x) = exp(-2x) and tan(x). Liouville's principle states that if an elementary antiderivative exists, then it's going to be of the form R(x) + ln Q(x), where R and Q are rational functions in e(x) and tan(x). If you pick one of the transcendentals, say e(x), you can express both R and Q in canonical form with respect to it. The canonical form is obtained by a partial fraction decomposition of R and a factorization into irreducibles of Q. R = P + F, and Q = G1*G2*G3..., where P is a polynomial in e(x), F is a sum of irreducible fractions, and G1, G2, ... are irreducible polynomials in e(x), with the coefficients being rational functions of tan(x). This gives a complete parametrization of possible antiderivatives. Now you just take a derivative and try to match up coeffients with the integrand in the same canonical form. Basically, since the integrand is a polynomial in e(x), the antiderivative A has to be a polynomial in e(x) as well. You can further narrow it down to being a first degree polynomial in e(x), A = f(x)*e(x) + g(x). But then, since A' = tan(x)*e(x), g(x) has to be a constant and f(x) has to satisfy tan(x) = -2f(x) + f'(x). Now, you repeat the above procedure replacing e(x) by tan(x) and allow the coefficients to be only real numbers. Since tan'(x) = 1 + tan(x)^2, the only way to get something of degree 1 in the expression -2f(x) + f'(x) is to have term of the form ln(1+tan^2(x)) in f(x). But then f'(x) expression -2f(x) + f'(x), which contradicts its equality to tan(x). This shows that f(x) cannot be a rational function of tan(x). Which implies that the antiderivative of exp(-2x)*tan(x) cannot be an elementary function. This kind of proof is an application of the transcendental version of the Risch algorithm for integration of elementary functions. Luckily, it is programmed into most computer algebra systems, so they can do the work for you if you need to prove that some function has no elementary antiderivative. Igor === Subject: Re: How to integrate this by pen and pencil? > Hi all, > how do you integrate this indefinite integral? > Integrate(exp(-2x)*(sec(x))^2) You don't. You can't express this in a closed form without using hypergeometric functions (http://en.wikipedia.org/wiki/Hypergeometric_function if interested) or something similar. === Subject: Re: How to integrate this by pen and pencil? Hi Daniel, I read the webpage you've provided... but I still don't understand your points... How to express the result of the integration? === Subject: Re: How to integrate this by pen and pencil? >Hi Daniel, >I read the webpage you've provided... but I still don't understand your >points... >How to express the result of the integration? A function is called an elementary function if it can be obtained from the basic elementary functions by any combination of arithmetic operations (+, -, *, /, ^) and composition. As far as I remember, the basic elementary functions are constant functions, x, the standard trig and inverse trig functions, e^x, ln(x). Using these you can get polynomials, rational functions, powers and roots, and these can in turn be reapplied to get more elementary functions. The derivative of an elementary function is always an elementary function, however for integration this is not the case. It can be proved that for many (in fact most) elementary functions, the antiderivative is not expressible as an elementary function. A few simple examples of this phenomenon: int(x^x) int(sqrt(sin(x))) int(1/(x^3+1)) For the above examples, an antiderivative exists -- it's just not expressible in elementary form. The values of such an antiderivative can always be obtained by numerical integration, If an antiderivative is not elementary but is used a lot, then in some cases, a new function name is invented to represent the antiderivative. A power series representations is another alternative. Determining which elementary functions have elementary antiderivatives requires a specialized theory. The theory uses methods of modern algebra related to Galois theory to algorithmically characterize the class of elementary functions. In practice, no one really uses this theory when actually given an indefinite integral to work out. Instead, the usual methods of calculus are applied if the integrand appears to be of the form where one of the methods will work. If all attempts fail, or if the integrand matches the form of a function which is known to have no elementary antiderivative, then numerical integration is the usual fall back. quasi === Subject: Re: How to integrate this by pen and pencil? >Determining which elementary functions have elementary antiderivatives >requires a specialized theory. The theory uses methods of modern >algebra related to Galois theory to algorithmically characterize the >class of elementary functions. I meant to say: Determining which elementary functions have elementary antiderivatives requires a specialized theory. The theory uses methods of modern algebra related to Galois theory to algorithmically characterize the class of elementary functions which have elementary antiderivatives. quasi === Subject: Re: How to integrate this by pen and pencil? >Hi all, >how do you integrate this indefinite integral? >Integrate(exp(-2x)*(sec(x))^2) If integration by parts doesn't work (and it doesn't look like it does), I would give up and integrate numerically. quasi === Subject: Re: Tools and software to help new math student? >Can you point me to some god emulators for HP and the >TI line of graphing calcs then? GOD EMULATOR version 2.0 (c) Amalgamated Deities Inc. Enter your prayers: > SMITE CRANKS The cranks evaporate in a greasy cloud of grey smoke. Enter your prayers: > SOLVE RIEMANN HYPOTHESIS All the zeros of zeta off the critical line and the real axis suddently vanish. You have been awarded a Fields Medal. Current score: 10 points. Enter your prayers: > MAKE MATH POPULAR The throngs of starlets throwing themselves at your feet cause you to stumble. You have fallen into a deep pit. Game over. You scored a miserable 10 points. Would you like to play again? === Subject: Re: Exhaustive list of component sets? Try searching on set partitioning and you should pick up various studies of problems like this. Most commonly the the goal is to find a cheapest partitioning, though; you'll have to look harder to find work on the problem of enumerating all possible partitions. Bob Fourer 4er@iems.northwestern.edu > Hello all, > I am struggling with a problem that may be familiar to some readers of > this group. > I have a collection of sets (aka linear lists). Members in each set > are unique. There are no empty sets. I have established a procedure > than ensures that there are no duplicate sets in the collection. > For a given set in the collection, I'm trying to determine which > combinations of remaining sets contain (exactly, with no duplications > and no leftovers) the same members. I need an exhaustive list of > solutions, rather than an optimal (or first) solution. > Is there a recognized name for this problem? Is there a solution other > than by brute force? Are there web or literature resources available? > Tom === Subject: Re: Exhaustive list of component sets? Actually, this problem is something like the inverse of partitioning. A partition divides a set into subsets. This is a search among existing subsets for a cover (in fact, all covers). === Subject: Re: Exhaustive list of component sets? > Actually, this problem is something like the inverse of partitioning. > A partition divides a set into subsets. This is a search among > existing subsets for a cover (in fact, all covers). The set partitioning problem is a combinatorial problem (usually modeled as an IP) in which a collection of sets is given and the mission is to find among them an exact cover for the universal set that optimizes some weight function. A solution to a set partitioning problem picks up every item exactly once. The problem here is to pick up every item in the target set exactly once (and no item not in the target set), so the first step would have to be ruling out any sets that are not subsets of the target. After that, using the reduced population of sets, a set partitioning problem (with arbitrary objective function) would find one of the desired solutions. One could find other solutions by tinkering with the objective function. For instance (assuming a minimization formulation), solve the SPP with all objective coefficients zero to get a first partition. Then change the objective weight of every set used to 1 and solve again, hopefully getting a new partition. Unfortunately, I don't hold out much hope that systematic tinkering with the objective would be guaranteed to produce all possible partitions. I suspect that, given the need to find all partitions, a brute-force approach may be best. Paul === Subject: Re: Symbolic complex matrix inverse > Sure. Invertible matrices over a PID form a PID, so inversion is > always unique. Stupid me mixing up the English terms again: Replace PID by domain in the above, please. Christopher Creutzig === Subject: Re: Open Ball vrs Neighborhood > In analysis, I'm confused about the distinction between an open ball > and a neighborhood. Could someone please point out what makes one, > rather than the other? A neighbourhood of a point, x, can be defined as a set containing x, and also containing some open ball centered on x. Sometimes a neighbourhood of x is by definition an open set containing x; in a metric space, such a neighbourhood is always a neigbourhood in the previous sense, but the converse needn't hold. In each case, a neigbbourhood is a generalisation of the open ball round a point. Thinking in terms of neighbourhoods is what you have to do when you are no longer in a metric space, just a topological space. === Subject: Re: Open Ball vrs Neighborhood >set containing x, and also containing some open ball >centered on x. But I am still left with questions about open balls. Your defn seems to imply that maybe an open ball centered on x need not contain x. I'm guessing that this is wrong, and you simply overdid the defn by including that middle phrase. Maybe the problem I have is that you are using the term 'point'. This is why I have stopped consulting books an analysis to understand this stuff. They assume the set X corresponds to the real line or plane or R**n). So they use the term 'point'. But then I have a hell of a time trying to figure out how to interpret the defns and theorems in terms of finite sets, like X = {1,2,3}. Is it true that an open ball of radius r around point 'a' MUST include EVERY element of X for which d(x, a) < r ? If so, how can you have an open ball that has no elements? My text has a theorem that A subset of a metric space is an open set if and only if it is a union of open balls. But the null set is an open set. How can this be, since it has no elements? If it is true that an open ball of radius r around point 'a' MUST include EVERY element of X for which d(x, a) < r , then, given X, a, d(x,a) and r, is it true that there only one unique open ball? It X={1,2,3,4,5} , a = 1, r=0.5 then the open ball around 'a' is {a}. Is this correct? Interestingly, the text defines an open set this way. A subset of a metric space is said to be open if it is a neighborhood of each of its points. [ Here we go with 'point' again! ] If my thinking above is correct, and my text is correct, then I'm confused about this: How can a subset of a metric space not be open? It seems that you can always find an open ball around each element, and that can be the neighborhood for that element. Alan p.s. the text I am getting most of this from is Mendelson Introduction to Topology. === Subject: Re: Open Ball vrs Neighborhood >>set containing x, and also containing some open ball >>centered on x. >But I am still left with questions about open balls. >Your defn seems to imply that maybe an open ball centered on x need >not contain x. I'm guessing that this is wrong, and you simply >overdid the defn by including that middle phrase. Yes, it's redundant. For a given point x (element of the space) and a given real r, r>0, the open ball B(x,r) centered at x with radius r is defined by: B(x,r)={y| d(x,y)0 and d(x,x)=0. In other words, an open ball always contains it's center, so in particular, an open ball is always nonempty. An open set A is a set such that, for each element x of A, there is an open ball containing x and contained in A (that is, the open ball is a subset of A). As was pointed out by in a previous reply, in the space R^2, the interior of a square is an open set but not an open ball since an open ball must be precisely the interior of a circle centered at some x, with some radius r, r>0. Also, as was pointed out in another reply, the empty set is an open set since it contains an open ball around each of its points. The truth of that is just pure logic. For the empty set not to be open, there would have to be some point in it for which there was no open ball contained as a subset, but since it has no points, you can't find such a point, so the empty set (vacuously) satisfies the definition of open set. So the empty set is an open set, although it is not an open ball. The definition of neighborhood is not completely standard so you need to be careful here. There are 2 definitions that are common, depending on which textbook you are using. Definition #1: An neighborhood of a point x is any set which contains an open set containing x. Note that in this definition, a neighborhood of x doesn't have to be open, as long as it contains some open set which contains x. An open neighborhood of x is a neighborhood of x which is also an open set. Definition #2 A neighborhood of x is an open set containing x. So if your text is using definition #2, then a neighborhood of x is the same as what is called an open neighborhood in definition # 1. But note that with either definition, a neighborhood of x contains an open ball which contains x. The idea is that a neighborhood contains all points which are sufficiently close to x, but it need not be in the shape of an open ball (with either definition) and for definition # 2, the neighborhood itself need not even be open. quasi === Subject: Re: Open Ball vrs Neighborhood quasi >Would you mind answering the following questions, just so I have >things clearer in my mind? >In the following questions, I am assuming a common-sense, non-aberrant >metric, such as d(x, y) = |x-y|. >I understand that given X = the reals, that G = [0, 1] is not an open >set, since every open ball around 1 contains elements of X which are >not elements of G. Right. >But if I define X=G=[0, 1], then X itself is an open set, as I am >excluding numbers greater than 1 from my consideration. So no open >ball around 1 contains elements of X which are not elements of G. >Therefore G = X is an open set. >Right or wrong? Right again. >Every set of one element is an open set. >Right or wrong? Wrong. The radius r for an open ball is required to be positive, so an open ball will typically be an infinite set. However, it all depends on the space and the metric. For example, take any finite subset of a metric space and view it as a space in its own right keeping the same distance function. Then sufficiently small open balls are in fact singleton sets consisting of just the center. >Every finite set of elements {x_1, x_2, . . . x_n} is an open set. >Right or Wrong? Wrong, same explanation as above. For example, over the reals, any open ball is actually a bounded open interval (x-r,x+r) for some center x and some radius r>0, so open balls are clearly infinite, in fact, uncountable. quasi === Subject: Re: Open Ball vrs Neighborhood > In the following questions, I am assuming a common-sense, non-aberrant > metric, such as d(x, y) = |x-y|. > I understand that given X = the reals, that G = [0, 1] is not an open > set, since every open ball around 1 contains elements of X which are > not elements of G. Correct. > But if I define X=G=[0, 1], then X itself is an open set, as I am > excluding numbers greater than 1 from my consideration. So no open > ball around 1 contains elements of X which are not elements of G. > Therefore G = X is an open set. > Right or wrong? Right. > Every set of one element is an open set. > Right or wrong? Do you mean |X| = 1? If so, then right. Do you mean |G| = 1 with X = R or X = [0,1]? If so, then wrong. There isn't really such a thing as an unqualified open set; it's always an open set within some particular space. Often the space is obvious from context and the qualification need not be explicitly mentioned. {1} is not open in the reals with the usual metric, but is open in {1}. - Tim === Subject: Re: Open Ball vrs Neighborhood >set containing x, and also containing some open ball >>centered on x. > Your defn seems to imply that maybe an open ball centered on x need > not contain x. It doesn't imply that. > So they use the term 'point'. But then I have a hell > of a time trying to figure out how to interpret the defns and theorems > in terms of finite sets, like X = {1,2,3}. For 'point' think 'element of the set X'. > Is it true that an open ball of radius r around point 'a' MUST include > EVERY element of X for which d(x, a) < r ? Yes: the definition of the open ball of radius r centred on is that it is the set of all points of distance less than r from a. > If so, how can you have an open ball that has no elements? You can't: unless you think of radius 0, because there are no points of distance less than 0 from a. > My text has a theorem that A subset of a metric space is an open set > if and only if it is a union of open balls. > But the null set is an open set. How can this be, since it has no > elements? You can get this by taking the union of no open balls. (Just as the sum of no numbers is 0, and the product of no numbers is 1, the union of no sets is the empty set.) > If it is true that an open ball of radius r around point 'a' MUST > include EVERY element of X for which d(x, a) < r , then, given X, a, > d(x,a) and r, is it true that there only one unique open ball? Yes. > It X={1,2,3,4,5} , a = 1, r=0.5 > then the open ball around 'a' is {a}. Is this correct? You haven't said what the metric on X is. If it's the metric induced from R, so d(a,b) = |a-b|, yes that is correct. > How can a subset of a metric space not be open? It seems that you can > always find an open ball around each element, and that can be the > neighborhood for that element. Consider the set {0} in R. Any open set containing 0 contains some other points too, so {0} is not open. In general, a subset of R is open precisely when it is the union of sets of the form (a,b) (meaning {x in R : a >>set containing x, and also containing some open ball > >>centered on x. > > Your defn seems to imply that maybe an open ball centered on x need > > not contain x. > It doesn't imply that. Technically, no. But it's not surprising the poster wonders about it, because it's a superfluous thing to write. This isn't a criticism. The standard for interpreting Usenet posts should be, they're equivalent to oral conversations: and we often add superfluities in conversation, to help guarantee understanding. --Ron Bruck --Ron Bruck === Subject: Re: Open Ball vrs Neighborhood >set containing x, and also containing some open ball >centered on x. Your defn seems to imply that maybe an open ball centered on x need not contain x. >>It doesn't imply that. > Technically, no. But it's not surprising the poster wonders about it, > because it's a superfluous thing to write. I don't see what's superfluous about it. [0,1] isn't an open subset of R, and although [1/2,1] is a set containing 1 and containing open balls, it doesn't contain an open ball centred on 1: that's why it isn't a neighbourhood of 1. Or maybe I'm misunderstanding which bit is meant to be superfluous. If you can give a more streamlined definition, I'll be happy to see it. === Subject: Re: Open Ball vrs Neighborhood >> set containing x, and also containing some open ball >> centered on x. > Your defn seems to imply that maybe an open ball centered on x need > not contain x. It doesn't imply that >> Technically, no. But it's not surprising the poster wonders about it, >> because it's a superfluous thing to write. > Or maybe I'm misunderstanding which bit is meant to be > superfluous. Ah, the penny just dropped. It's saying 'set containing x, and also' that's superfluous. I thought it was the open ball having to contain x that was being questioned. Yes, it would have been neater to say 'a neighbourhood of x is a set containing an open ball centred on x'. === Subject: Re: Open Ball vrs Neighborhood >In analysis, I'm confused about the distinction between an open ball >and a neighborhood. Could someone please point out what makes one, >rather than the other? In a metric space (M,d), an open ball is explicitly a set of the form {y in M: d(x,y) < r} where x is some fixed point in M and r is some fixed positive number. A neighborhood is any open set. Thus, any open ball is a neighborhood, but not conversely. For example, let M = R^2 and d(x,y) = sqrt((x1-y1)^2 + (x2-y2)^2). Let U = {(x,y): x^2 + y^2 < 1} and V = {(x,y): maj(|x|, |y|) < 1}. [maj denotes larger.] Both U and V are neighborhoods of 0. U is the open ball B(0,1), whereas V is not an open ball in (M,d). However, e(x,y) = maj(|x1-y1|, |x2-y2|) is also a metric on M. (In fact, e induces the same topology as d). U is not at open ball in (M,e); V is the open ball B(0,1) in (M,e). Sundry facts: - Any neighborhood is equal to the union of open balls. - Neighorhoods and open sets generalize to general topological spaces; open balls do not. - One usually refers to the neighborhood *of a point*. When no point is specified, one usually refers to open sets. - Some authors (notably, Kelley) define a neighborhood of a point to be any set whose interior contains the point. E.g., for Kelley, the closed ball {y: d(x,y) <= 1} is a neighborhood of x. I think this usage is less common nowadays (though I kind of like it). -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Open Ball vrs Neighborhood following is wrong: A neighborhood can be seen as an open ball with possibly some of the elements of the set missing. Ex: X = {1,2,3,4,5,6} THE (unique) open ball around x=1, with r=4.5 is {1,2,3,4} A neighborhood around x=1, with r=4.5 is {1,3}, Another neighborhood is {1,2}, Yet another is {1,2,3,4} = the open ball. Alan === Subject: Re: Open Ball vrs Neighborhood >A neighborhood can be seen as an open ball with possibly some of the >elements of the set missing. I would not say that at all. The interval (-1,1) is the open ball B(0,1) on the real line. The set {-1/2, 0, 1/2} is this open ball with ... some of the elements of the set missing. Yet the latter set is not a neighborhood of 0. Also, the set {(x,y) in R^2: |x| < 1} is a neighborhood of the origin. However, it is contained in no open ball under the Euclidean distance metric. Why did the comparison of the open disc and the open square not clear things up for you? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Open Ball vrs Neighborhood >Why did the comparison of the open disc and the open square not clear >things up for you? Because, for me, math things take a long time to sink in. Happily, once I have learned them, they usually appear obvious. I guess that the problem is that math authors and teachers (you, for example) don't know in which ways I am confused about a certain point, and so they don't know what points to concentrate on. rereading of the texts I have. Let me try again to put it into my words. Given a set X of elements x and selected element 'a' and metric d and distance r: The unique open ball around 'a' consists of every element of X for which d(a, x) < r. This obviously includes the element 'a'. [Wait a minute. This can't be right, since the null set is an open set (so, by defn, it must be a union of open balls). So it would need to include an element 'a'.] A neighborhood of 'a' consists af any subset of X which contains an open ball of 'a'. [But doesn't this mean that some neighborhoods are not open?] Question: Is the set {a} an open ball around 'a'? A neighborhood of 'a'? Alan === Subject: Re: Open Ball vrs Neighborhood >>Why did the comparison of the open disc and the open square not clear >>things up for you? > Because, for me, math things take a long time to sink in. Happily, > once I have learned them, they usually appear obvious. I guess that > the problem is that math authors and teachers (you, for example) don't > know in which ways I am confused about a certain point, and so they > don't know what points to concentrate on. > rereading of the texts I have. Let me try again to put it into my > words. > Given a set X of elements x and selected element 'a' and metric d and > distance r: > The unique open ball around 'a' consists of every element of X for > which d(a, x) < r. This obviously includes the element 'a'. > [Wait a minute. This can't be right, since the null set is an open > set (so, by defn, it must be a union of open balls). So it would need > to include an element 'a'.] The null set is an empty union of open balls. Another way to look at this is to say a subset U (of a metric space X) is open if and only if for all x in U there is an r > 0 s.t. the open ball B(x,r) = {y in X : d(y,x) < r}is a subset of U. if you look at this definition closely, you will see that the empty set satisfies the condition vacuously, since the statement x in empty set is always false. > A neighborhood of 'a' consists af any subset of X which contains an > open ball of 'a'. not quite. A neighborhood (the usual definition at least, someone pointed out that in Kelley's Topology he varies from this a bit) of a point x is simply an open set that contains x. You have seen that your condition is not the same as this condition. > [But doesn't this mean that some neighborhoods are not open?] > Question: > Is the set {a} an open ball around 'a'? A neighborhood of 'a'? Depends on the metric. If the discrete metric is used, yes. If the usual metric on R is used, no. > Alan === Subject: Re: Open Ball vrs Neighborhood >>Why did the comparison of the open disc and the open square not clear >>things up for you? >> >Because, for me, math things take a long time to sink in. Happily, >once I have learned them, they usually appear obvious. I guess that >the problem is that math authors and teachers (you, for example) don't >know in which ways I am confused about a certain point, and so they >don't know what points to concentrate on. You need to help us figure out what that certain point is. Thus, my question. After all, we are not mind readers. (By the way, your parenthetical remark reads to me like a dig - am I being too sensitive? My question was sincere and not meant to chastise.) >rereading of the texts I have. Let me try again to put it into my >words. >Given a set X of elements x and selected element 'a' and metric d and >distance r: >The unique open ball around 'a' consists of every element of X for >which d(a, x) < r. This obviously includes the element 'a'. >[Wait a minute. This can't be right, since the null set is an open >set (so, by defn, it must be a union of open balls). So it would need >to include an element 'a'.] No. The empty set is the union of of the empty set. To make that clearer: x is in the union of a set A iff there exists Y in A such that x in Y. (You are probably more used to the notation U_(X in A) X.) Since there exists no Y in {}, there exists no element in the union of {}. >A neighborhood of 'a' consists af any subset of X which contains an >open ball of 'a'. >[But doesn't this mean that some neighborhoods are not open?] No. A neighborhood of a is any *open* set which contains a (for most authors). >Question: >Is the set {a} an open ball around 'a'? A neighborhood of 'a'? That depends on the topology / metric. For example, consider a set M; define the metric d(x,x) = 1, d(x,y) = 0 if x differs from y. Then for each x in M, {x} = B(x, 1/2) is a neighborhood of x. In fact, in any metric space, a singleton (i.e., a set with exactly one element) is an open ball if and only if it s a neighborhood of its sole element. Such points are called isolated points of the space. In most spaces you will consider (e.g., R^n), singletons are neither open balls nor neighborhoods of the element. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Open Ball vrs Neighborhood >You need to help us figure out what that certain point is. Thus, my >question. After all, we are not mind readers. (By the way, your >parenthetical remark reads to me like a dig - am I being too sensitive? >My question was sincere and not meant to chastise.) Of course, no dig intended!! I appreciate that you are willing to try to help me and others out. I was trying to point out that the way I think and the way mathematicians think seems so different. [student to bartender: My math teacher doesn't understand me!] [ I am reminded of a Feynman story. showing such-and-such, and just couldn't get it. Finally, in frustration, he just went and figured it out himself. And he thought. This isn't difficult at all! If only the author had written the way I think! ] In fact, Stephen, since I have this trouble studying math books, it is people like you who actually solve this problem by answering my questions. Bravo! Alan === Subject: Re: Points and numbers. <4310926D.4010707@netscape.net >I have read several times about the difficulty of ordering the real numbers (though I don't really understand why it is difficult... perhaps someone could explain in layman's terms). But, anyway, doesn't mapping the real numbers to a line necessarily impose an order and solve the problem? > I suspect you are confusing the usual order of the reals with a > well-ordering of the reals. That could well be the case. > The former is well-defined and just what > you know it to be. > An ordered set is by definition well-ordered iff each of its nonempty > subsets has a least element. Perhaps I would understand this better if I understood the choice of the term well-ordered. Though I kind of understand the definition, I don't really see why this property makes a set any better ordered than one without it. > Thus, the rationals are not well-ordered > under the usual ordering. Could you give an example of a set of rational numbers (I assume it has to be an infinite set) which does not have a least element under the natural ordering? Would it be as simple as something like all the rationals greater than zero, or am I missing the point? (No pun intended!) > However, let f be be a bijection from the > rationals to the natural numbers. Define a relation R on the > rationals by saying q R r iff f(q) < f(r). Then the rationals are > well-ordered by the relation R. OK. This is the same bijection as would show that the rationals are countable, right? > The reals are not well-ordered by the usual order. The Axiom of Choice > implies that there exists a well-ordering of the reals. However, it is > impossible to expliclty present a well-ordering as I did with the > rationals above. This bothers some people's intuition and leads them to > be suspect of the Axiom of Choice. Let's say I manage to map the real numbers to points on the number line. The points on the line are ordered according to their natural order, yes? Any subset of the reals is a subset of the points on the line, so you are essentially saying that some sets have no leftmost point, yes? Does this mean that continuous line segments exist that have no leftmost point, or does it only apply to pathological specially-constructed sets of points? I suppose all points on the line greater than zero and less than one might be such a continuous set. This line segment has no leftmost point? === Subject: Re: Points and numbers. > > >I have read several times about the difficulty of ordering the real > > >numbers (though I don't really understand why it is difficult... perhaps > > >someone could explain in layman's terms). But, anyway, doesn't mapping the > > >real numbers to a line necessarily impose an order and solve the > > >problem? > > > I suspect you are confusing the usual order of the reals with a > > well-ordering of the reals. > That could well be the case. > > The former is well-defined and just what > > you know it to be. > > An ordered set is by definition well-ordered iff each of its nonempty > > subsets has a least element. > Perhaps I would understand this better if I understood the choice of > the term well-ordered. Though I kind of understand the definition, I > don't really see why this property makes a set any better ordered > than one without it. When one has a well-ordered set then it is order isomorphic to some ordinal number, and is said to have that ordinal number. The finite ordinals are first,second, third, etc.,in contrast to the cardinal numbers one Two, three etc. For finite numbers there is little significant difference, but with infinite numbers there is considerable difference, since there are already infinitely many different infinite ordinals that all have the same cardinality as the set of natural numbers. But that is may be going beyond where you are immediately interested in going. Transfinite ordinals and cardinals require some getting used to. > > Thus, the rationals are not well-ordered > > under the usual ordering. > Could you give an example of a set of rational numbers (I assume it has > to be an infinite set) which does not have a least element under the > natural ordering? Would it be as simple as something like all the > rationals greater than zero, or am I missing the point? (No pun > intended!) Your example is quite correct. The set of rationals strictly greater than any particular real number will have no smallest member. > > However, let f be be a bijection from the > > rationals to the natural numbers. Define a relation R on the > > rationals by saying q R r iff f(q) < f(r). Then the rationals are > > well-ordered by the relation R. > OK. This is the same bijection as would show that the rationals are > countable, right? Yes! > > The reals are not well-ordered by the usual order. The Axiom of Choice > > implies that there exists a well-ordering of the reals. However, it is > > impossible to expliclty present a well-ordering as I did with the > > rationals above. This bothers some people's intuition and leads them to > > be suspect of the Axiom of Choice. > Let's say I manage to map the real numbers to points on the number > line. The points on the line are ordered according to their natural > order, yes? Any subset of the reals is a subset of the points on the > line, so you are essentially saying that some sets have no leftmost > point, yes? Does this mean that continuous line segments exist that > have no leftmost point, or does it only apply to pathological > specially-constructed sets of points? If A and B are distinct points on a line running from leaf to right then the set of points strictly between A and B will have neither a leftmost nor a rightmost point. > I suppose all points on the line > greater than zero and less than one might be such a continuous set. > This line segment has no leftmost point? Correct! === Subject: Well-ordering; was: Re: Points and numbers. >> I have read several times about the difficulty of ordering the real numbers (though I don't really understand why it is difficult... perhaps someone could explain in layman's terms). But, anyway, doesn't mapping the real numbers to a line necessarily impose an order and solve the problem? >>I suspect you are confusing the usual order of the reals with a >>well-ordering of the reals. >> >That could well be the case. >>The former is well-defined and just what >>you know it to be. >>An ordered set is by definition well-ordered iff each of its nonempty >>subsets has a least element. >> >Perhaps I would understand this better if I understood the choice of >the term well-ordered. Though I kind of understand the definition, I >don't really see why this property makes a set any better ordered >than one without it. You can do certain things, e.g., proofs and definitions by induction, with well-ordered sets. But it's just the choice of words; don't read to much into it. What gets me is the use of the term well order as a noun. Shouldn't that be good order (which one never sees)? Fortunately, one sees well-ordering much often than than well order >>Thus, the rationals are not well-ordered >>under the usual ordering. >> >Could you give an example of a set of rational numbers (I assume it has >to be an infinite set) which does not have a least element under the >natural ordering? Would it be as simple as something like all the >rationals greater than zero, or am I missing the point? (No pun >intended!) Yes, your example works: The set of positive rationals has no least element. And, yes, any totally ordered finite set is well-ordered. >>However, let f be be a bijection from the >>rationals to the natural numbers. Define a relation R on the >>rationals by saying q R r iff f(q) < f(r). Then the rationals are >>well-ordered by the relation R. >> >OK. This is the same bijection as would show that the rationals are >countable, right? Yes. I prefer Virgil's improvement of my example. >>The reals are not well-ordered by the usual order. The Axiom of Choice >>implies that there exists a well-ordering of the reals. However, it is >>impossible to expliclty present a well-ordering as I did with the >>rationals above. This bothers some people's intuition and leads them to >>be suspect of the Axiom of Choice. >> >Let's say I manage to map the real numbers to points on the number >line. The points on the line are ordered according to their natural >order, yes? Any subset of the reals is a subset of the points on the >line, so you are essentially saying that some sets have no leftmost >point, yes? Does this mean that continuous line segments exist that >have no leftmost point, or does it only apply to pathological >specially-constructed sets of points? I suppose all points on the line >greater than zero and less than one might be such a continuous set. >This line segment has no leftmost point? Yes, the interval (0,1) has no least element. Neither does Q(0,1), the set of rationals strictly between 0 and 1. It is not hard to come up with subsets of Q or R which have no least element. The entire sets themselves have have no least element. The set {1/n: n a positive integer} has no least element. On the other hand, the set N of natural numbers (i.e., nonnegative integers) under the usual ordering is an example of a well-ordered set. What some people find hard to accept is that the uncountable set R can be well-ordered. However, one does not need the axiom of choice to establish the existence of a well-ordered uncountable set. It's just one cannot well-order as set with the same cardinality as R without some version of choice (I think). -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Well-ordering; was: Re: Points and numbers. > > > > >> > > I have read several times about the difficulty of ordering the real > numbers (though I don't really understand why it is difficult... perhaps > someone could explain in layman's terms). But, anyway, doesn't mapping the > real numbers to a line necessarily impose an order and solve the > problem? > >> > >> >>I suspect you are confusing the usual order of the reals with a > >>well-ordering of the reals. > >> > > >That could well be the case. > > > >>The former is well-defined and just what > >>you know it to be. > > >>An ordered set is by definition well-ordered iff each of its nonempty > >>subsets has a least element. > >> > > >Perhaps I would understand this better if I understood the choice of > >the term well-ordered. Though I kind of understand the definition, I > >don't really see why this property makes a set any better ordered > >than one without it. > > > You can do certain things, e.g., proofs and definitions by induction, > with well-ordered sets. But it's just the choice of words; don't read > to much into it. > What gets me is the use of the term well order as a noun. Shouldn't > that be good order (which one never sees)? Fortunately, one sees > well-ordering much often than than well order > > > >>Thus, the rationals are not well-ordered > >>under the usual ordering. > >> > > >Could you give an example of a set of rational numbers (I assume it has > >to be an infinite set) which does not have a least element under the > >natural ordering? Would it be as simple as something like all the > >rationals greater than zero, or am I missing the point? (No pun > >intended!) > > > Yes, your example works: The set of positive rationals has no least > element. And, yes, any totally ordered finite set is well-ordered. > > > >>However, let f be be a bijection from the > >>rationals to the natural numbers. Define a relation R on the > >>rationals by saying q R r iff f(q) < f(r). Then the rationals are > >>well-ordered by the relation R. > >> > > >OK. This is the same bijection as would show that the rationals are > >countable, right? > > > Yes. I prefer Virgil's improvement of my example. Actually, I cribbed it from someone else, whose Identity I have lost track of. But I liked it so much, I had to pass it on. > > > >>The reals are not well-ordered by the usual order. The Axiom of Choice > >>implies that there exists a well-ordering of the reals. However, it is > >>impossible to expliclty present a well-ordering as I did with the > >>rationals above. This bothers some people's intuition and leads them to > >>be suspect of the Axiom of Choice. > >> > > >Let's say I manage to map the real numbers to points on the number > >line. The points on the line are ordered according to their natural > >order, yes? Any subset of the reals is a subset of the points on the > >line, so you are essentially saying that some sets have no leftmost > >point, yes? Does this mean that continuous line segments exist that > >have no leftmost point, or does it only apply to pathological > >specially-constructed sets of points? I suppose all points on the line > >greater than zero and less than one might be such a continuous set. > >This line segment has no leftmost point? > Yes, the interval (0,1) has no least element. Neither does Q(0,1), > the set of rationals strictly between 0 and 1. It is not hard to come > up with subsets of Q or R which have no least element. The entire > sets themselves have have no least element. The set {1/n: n a > positive integer} has no least element. > On the other hand, the set N of natural numbers (i.e., nonnegative > integers) under the usual ordering is an example of a well-ordered set. > What some people find hard to accept is that the uncountable set R can > be well-ordered. > However, one does not need the axiom of choice to establish the > existence of a well-ordered uncountable set. It's just one cannot > well-order as set with the same cardinality as R without some version > of choice (I think). === Subject: Re: Well-ordering; was: Re: Points and numbers. <4310926D.4010707@netscape.net> <4310D38C.4070402@netscape.net> In the Finlayson reals, R-bar-umlaut, the natural ordering is the well-ordering, because the reals are at once complete ordered field and partially ordered ring. Then again I call those the real numbers. infinite sets are equivalent and Cantor's first, it is as I say. So, are the reals a set in ZFC? Some people accept the well-ordering principle as self-evident. Biject any set to an ordinal, it's well-ordered. N/U EF: a function with utility. I've written a variety of things about points, lines, and numbers. I'm serious, and honest, and extend Cantorian and Goedelian results towards deep foundations and revival of the Hilbert program. Ross -- Ross A. Finlayson === Subject: Re: Points and numbers. <4310926D.4010707@netscape.net > >I have read several times about the difficulty of ordering the real > > >numbers (though I don't really understand why it is difficult... perhaps > > > someone could explain in layman's terms). But, anyway, doesn't mapping > > >the real numbers to a line necessarily impose an order and solve the > > > problem? > > > I suspect you are confusing the usual order of the reals with a > > well-ordering of the reals. > That could well be the case. > > The former is well-defined and just what > > you know it to be. > > An ordered set is by definition well-ordered iff each of its nonempty > > subsets has a least element. > Perhaps I would understand this better if I understood the choice of > the term well-ordered. Though I kind of understand the definition, I > don't really see why this property makes a set any better ordered > than one without it. A set is totally ordered, which is usually just called ordered, if every two members can be compared; i.e., given x and y, exactly one of xy is true. We could then say that every set of two distinct elements {x,y} has a least element; and I think you can see that every finite set of elements then has a least element. A well-order does even better than that;it is a total order, such that /every/ set of elements, finite or not, has a least element. > > Thus, the rationals are not well-ordered > > under the usual ordering. > Could you give an example of a set of rational numbers (I assume it has > to be an infinite set) which does not have a least element under the > natural ordering? Would it be as simple as something like all the > rationals greater than zero, or am I missing the point? (No pun > intended!) The usual order on the rationals is a total order - every finite subset of Q has a least element. But it is not a well-order - as you note the setof all strictly positive rationals doesnt have a least element. > > However, let f be be a bijection from the > > rationals to the natural numbers. Define a relation R on the > > rationals by saying q R r iff f(q) < f(r). Then the rationals are > > well-ordered by the relation R. > OK. This is the same bijection as would show that the rationals are > countable, right? Yep. Since the naturals are well-ordered (every set has a smallest natural), then we just map that ordering onto the rationals, and voila! > > The reals are not well-ordered by the usual order. The Axiom of Choice > > implies that there exists a well-ordering of the reals. However, it is > > impossible to expliclty present a well-ordering as I did with the > > rationals above. This bothers some people's intuition and leads them to > > be suspect of the Axiom of Choice. > Let's say I manage to map the real numbers to points on the number > line. The points on the line are ordered according to their natural > order, yes? Any subset of the reals is a subset of the points on the > line, so you are essentially saying that some sets have no leftmost > point, yes? Exactly. For example consider the set of points in the interval [0,1]. Remove 0 from this set; the resulting set has no leftmost point (just as with the rationals). > Does this mean that continuous line segments exist that > have no leftmost point, or does it only apply to pathological > specially-constructed sets of points? I suppose all points on the line > greater than zero and less than one might be such a continuous set. > This line segment has no leftmost point? Suppose such a point p existed. It would correspond to a real number r 0 < r < 1. Where would you expect to find the point which corresponds to r/2? To the left or the right of p? Not at all; you are correctly answering your own questions! === Subject: Re: Points and numbers. <4310926D.4010707@netscape.net > > > >I have read several times about the difficulty of ordering the real > > > >numbers (though I don't really understand why it is difficult... perhaps > > > > someone could explain in layman's terms). But, anyway, doesn't mapping > > > >the real numbers to a line necessarily impose an order and solve the > > > > problem? > > > > > > I suspect you are confusing the usual order of the reals with a > > > well-ordering of the reals. > > That could well be the case. > > > The former is well-defined and just what > > > you know it to be. > > > > An ordered set is by definition well-ordered iff each of its nonempty > > > subsets has a least element. > > Perhaps I would understand this better if I understood the choice of > > the term well-ordered. Though I kind of understand the definition, I > > don't really see why this property makes a set any better ordered > > than one without it. > A set is totally ordered, which is usually just called ordered, if > every two members can be compared; i.e., given x and y, exactly one of > xy is true. > We could then say that every set of two distinct elements {x,y} has a > least element; and I think you can see that every finite set of > elements then has a least element. > A well-order does even better than that;it is a total order, such > that /every/ set of elements, finite or not, has a least element. > > > Thus, the rationals are not well-ordered > > > under the usual ordering. > > Could you give an example of a set of rational numbers (I assume it has > > to be an infinite set) which does not have a least element under the > > natural ordering? Would it be as simple as something like all the > > rationals greater than zero, or am I missing the point? (No pun > > intended!) > The usual order on the rationals is a total order - every finite subset > of Q has a least element. But it is not a well-order - as you note the > setof all strictly positive rationals doesnt have a least element. > > > However, let f be be a bijection from the > > > rationals to the natural numbers. Define a relation R on the > > > rationals by saying q R r iff f(q) < f(r). Then the rationals are > > > well-ordered by the relation R. > > OK. This is the same bijection as would show that the rationals are > > countable, right? > Yep. Since the naturals are well-ordered (every set has a smallest > natural), then we just map that ordering onto the rationals, and voila! that is bothering me a little. Take the negative integers. The (infinite) set of ALL negative integers doesn't have a least member under natural ordering? Does this mean that the negative integers are not well-ordered under their natural ordering? Seems wrong. === Subject: Well-ordering; was: Re: Points and numbers. >There is something else that I just thought of >that is bothering me a little. Take the negative integers. The >(infinite) set of ALL negative integers doesn't have a least member >under natural ordering? Does this mean that the negative integers are >not well-ordered under their natural ordering? Seems wrong. You are correct; the set of negative integers is not well-ordered under the usual ordering. Why does that seem wrong to you? The set of all integers is not well-ordered, either. However, let X be the set of negative integers, and define the relation R on X by x R y iff x > y. The R well-orders X, for every subset of X has an R-least element. (Think about it.) We also say that (X, >) is a well-ordered set, or that > well-orders X: Every subset of X has a >-least element! I understand that the last statement takes some mind-wrapping, and I doubt whether one would use it much. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Well-ordering; was: Re: Points and numbers. <4310926D.4010707@netscape.net> <4310D5A6.60206@netscape.net >There is something else that I just thought of > >that is bothering me a little. Take the negative integers. The > >(infinite) set of ALL negative integers doesn't have a least member > >under natural ordering? Does this mean that the negative integers are > >not well-ordered under their natural ordering? Seems wrong. > You are correct; the set of negative integers is not well-ordered under > the usual ordering. Why does that seem wrong to you? The set of all > integers is not well-ordered, either. > However, let X be the set of negative integers, and define the > relation R on X by x R y iff x > y. The R well-orders X, for > every subset of X has an R-least element. (Think about it.) We also > say that (X, >) is a well-ordered set, or that > well-orders X: > Every subset of X has a >-least element! I understand that the last > statement takes some mind-wrapping, and I doubt whether one would use it > much. something, which at my age is an achievement! === Subject: Re: Points and numbers. >I just forgot to mention field structure but it would be very nice if >one could prove that there was just one field with continuous operations >on the order structure. Do you think it can be done? I was thinking of something like Hilbert's 5th problem. Gleason (1952) and Montgomery-Zippen (1955) proved that locally Euclidean groups are all Lie groups, that is, all manifolds with continuous group operations on them have a smooth structure, one in which the group operations are also smooth. Whether this result is relevant or not, I will have to think about, now that you've asked. dave === Subject: Re: Points and numbers. > I was thinking of something like Hilbert's 5th problem. > Gleason (1952) and Montgomery-Zippen (1955) proved that > locally Euclidean groups are all Lie groups, that is, > all manifolds with continuous group operations on them > have a smooth structure, one in which the group operations > are also smooth. Doesn't that actually go through to analytic? (Or am I remembering something different...) === Subject: Re: Points and numbers. >> Gleason (1952) and Montgomery-Zippen (1955) proved that >> locally Euclidean groups are all Lie groups, that is, >> all manifolds with continuous group operations on them >> have a smooth structure, one in which the group operations >> are also smooth. >Doesn't that actually go through to analytic? Yes. The manifold is analytic (i.e. the coordinate transformations on the intersections of the coordinate patches may be chosen to be analytic) and the group operations inv : G --> G and mult : G x G --> G are analytic (i.e. locally expressible in power series.) dave === Subject: Re: Points and numbers. > An ordered set is by definition well-ordered iff each of its nonempty > subsets has a least element. Thus, the rationals are not well-ordered > under the usual ordering. However, let f be be a bijection from the > rationals to the natural numbers. Define a relation R on the > rationals by saying q R r iff f(q) < f(r). Then the rationals are > well-ordered by the relation R. > The reals are not well-ordered by the usual order. The Axiom of Choice > implies that there exists a well-ordering of the reals. However, it is > impossible to expliclty present a well-ordering as I did with the > rationals above. This bothers some people's intuition and leads them to > be suspect of the Axiom of Choice. Is it known to be impossible to give an explicit well-ordering of the reals or is it just something that no one has been able to do as yet? === Subject: Re: Points and numbers. > Is it known to be impossible to give an explicit well-ordering of the > reals or is it just something that no one has been able to do as yet? I'm pretty sure (i.e. people I generally trust told me) that you need the axiom of choice to get a well-ordering on the reals. It is therefore not possible to give an explicit description of such a well-ordering. === Subject: Well-ordering the reals; Was: Re: Points and numbers. >>The reals are not well-ordered by the usual order. The Axiom of Choice >>implies that there exists a well-ordering of the reals. However, it is >>impossible to expliclty present a well-ordering as I did with the >>rationals above. This bothers some people's intuition and leads them to be suspect of the Axiom of Choice. >> >Is it known to be impossible to give an explicit well-ordering of the >reals or is it just something that no one has been able to do as yet? Good question. Here are two others which might answer it. 1) Is it consistent with the ZF than there exists no well-ordering of the reals? (My guess is that the answer is known to be yes.) 2) Is there some meta-theorem along the lines of the following? Let p be some statement that is known to be consistent with ZF, and suppose that not AC is a theorem of ZF + p. Then it is impossible to explictly construct a set demonstrating property p. I know, this question probbably requires a more precise statement, and, as stated, the hypothesis may be too general. But I hope the reader can infer what I am trying to get at. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Points and numbers. > >I have read several times about the difficulty of ordering the real numbers > >(though I don't really understand why it is difficult... perhaps someone > >could explain in layman's terms). But, anyway, doesn't mapping the real > >numbers to a line necessarily impose an order and solve the problem? > I suspect you are confusing the usual order of the reals with a > well-ordering of the reals. The former is well-defined and just what > you know it to be. > An ordered set is by definition well-ordered iff each of its nonempty > subsets has a least element. Thus, the rationals are not well-ordered > under the usual ordering. However, let f be be a bijection from the > rationals to the natural numbers. Define a relation R on the > rationals by saying q R r iff f(q) < f(r). Then the rationals are > well-ordered by the relation R. Actually, any INjection from the rationals to the naturals will induce a well-ordering of the rationals, and injections are much easier to construct. For example, each non-zero rational is uniquely some s*m/n, where s in {-1,1}, and n,m are naturals with gcd(m,n) = 1. Let f(0) = 1 and otherwise f(s*m/n) = 2^((s+1)/2)*3^m*5^n > The reals are not well-ordered by the usual order. The Axiom of Choice > implies that there exists a well-ordering of the reals. However, it is > impossible to expliclty present a well-ordering as I did with the > rationals above. This bothers some people's intuition and leads them to > be suspect of the Axiom of Choice. === Subject: Re: Points and numbers. > I'm not entirely sure I understand your quest, but this much is true: > given an affine or projective incidence geometry meeting a handful of > regularity conditions (e.g. Desargue's axiom) one can coordinatize > the geometry, i.e. show that there is a field F with the property > that the points of the geometry can be paired with the set of > ordered pairs F^2 and the lines can be paired with the solution > sets of equations of the form ax+by=c (a,b not both zero). The field > F is constructed from the geometry itself and the groups of symmetries > of the geometries; for example, the underlying set of F is constructed > first to be the set of elements in one line (any one will do -- one of > the first theorems in these geometries is that there is a one-to-one > correspondence between the elements of any two lines). Then you explain > how to add two points on the line in terms of certain geometric > constructions on the line, etc. Look up Hilbert's calculus of segments and von Staudt's algebra of throws. --Bill Dubuque === Subject: In general how to convert a system of 1st order diff equations into higher order diff equation for 1 variable and vice versa? Hi all, In general how to convert a system of 1st order diff equations into higher order diff equation for 1 variable and vice versa? For example, I am stuck with the following 1st order diff equation systems: (x1)'=2*x2+4*x3+3*exp(-t) (x2)'=x1+x2-2*x3+1 (x3)'=-2*x1+5*x3 with x1(0)=1, x2(0)=1, and x3(0)=0, solve it... where ' denotes differentiation with respect to t... the underlying variable... for x1(t), x2(t) and x3(t) ... I want to solve it by simple elimination approach by pen and pencil, not the matrix approach. Specifically, I want to see how this kind of systems of 1st order diff equations can be converted into a higher order diff equation for 1 variable and solve for it... Do you have systematic way for doing this? How many constants should there be in the final solution? I did in some ad-hoc way, I got the following: (x1)''' - 6*(x1)'' + 11* (x1)' - 6 x1 = -10 + 36*exp(-t) This is good, but it introduces three constants to x1: x1(t)=c1*exp(t)+c2*exp(2*t)+c3*exp(3*t)+5/3-3/2*exp(-t) Substituting into the third equation, I got a 1st order diff equation for x3, so x3(t)=c1/2*exp(t)+2/3*c2*exp(2*t)+c3*exp(3*t)+2/3-1/2*exp(-t)+c4*exp(5t), then substituting this into the second original equation, I got a 1st order diff equation for x2, so x2(t)=-1/3*c2*exp(2*t)-c3/2*exp(3*t)-4/3+1/4*exp(-t)-1/2*c4*exp(5*t)+c5*exp( t), which looks good but bad. Now I have 5 constants but I only have three initial values: x1(0)=1, x2(0)=1, and x3(0)=0, How can I solve for it? Please guide me some systemtic way of solving this kind of system of 1st order equations by simple elimination... but please don't matrix === Subject: Re: In general how to convert a system of 1st order diff equations into higher order diff equation for 1 variable and vice versa? > In general how to convert a system of 1st order diff equations > into higher order diff equation for 1 variable and vice versa? The only nontrivial direction is that of converting a system of first order differential equations into a linear homogeneous differential equation. For this one may employ the cyclic vector method -- see the references listed below. Variants of this algorithm are implemented in many computer algebra systems. --Bill Dubuque van der Put, M.; Singer, M. F. Differential Galois Theory, Section 2.3: Cyclic Vectors (only in first version of book) http://www4.ncsu.edu/~singer/papers/dbook.ps Kovacic, J. J.; Churchill, R.C. Cyclic Vectors http://mysite.verizon.net/jkovacic/publications/cvt.dvi Kovacic, J. J. Cyclic vectors and Picard-Vessiot extensions http://mysite.verizon.net/jkovacic/ksdalectures/cvt-ksda.dvi === Subject: Re: In general how to convert a system of 1st order diff equations into higher order diff equation for 1 variable and vice versa? Hi Bill, reading I still don't know how to apply the cyclic vector theorem onto my previous problem... They have many proofs there but they just don't have a single exam. This makes it harder. Moreover, in these papers, they are talking about homogeneous matrix equations, I think... Could you please help more on this problem? === Subject: Re: Sudoku >[...] > But, changing the topic temporarily, I did used to enjoy > the Rubik's cube, and I wonder if anyone has yet found a > workable and neat way to represent all the patterns and > operations in a group theoretic form or something similar. There is e.g. Singmaster, David: Notes on Rubik's magic cube (5th ed.) Enslow Publishers, Hillside, N.J., 1981. Marc === Subject: Re: H numbers >> Geez, you guys are nasty. > No, they're teachers. If they made their feelings about their students > known to the students' faces they'd have riots in the lecture hall; so > they take it out on sci.math posters :-) Poor things. They are not alone, look at the net, full of the same kind. > Seriously: they're very generous with their help, if you want it. If > you post something that's wrong, they'll tell you so. But that's what > you want, isn't it? You wouldn't want to go on believing a falsehood or > a muddle. Sure. But I am still confident about H numbers, more than a month after publishing. Not a confirmation, but I am encouraged by the increasing traffic on my site. === Subject: Re: H numbers >>I did not say Close in on themselves, I said close on themselves. > Which is equally meaningless. You are right, that wording is no good. I meant to say they have a circular relationship. (Quaternions) >>This is the conventional picture in line with the Fundemental Theorem of >>Algebra (FTA) as 'proven' by Gauss. >>My proposition, not yet proven, is that FTA is wrong. > Oh my gosh. Until you said that the people reading the thread > might have just assumed you had some interesting ideas but were > unaware of some well-known facts. You've just revealed yourself > to be a crackpot, congratulations. > Hint: FTA is not wrong. We know because we know how to prove it. All pots have cracks, including yours, depends on how closely you look at them. Many a proof turned out to be wrong in the past, regardless of how rigorous they seem. Your attitude is the kind of 'self confirming' one that lead to the slavish acceptance of it. Some physics training is a must for you maths people, to learn some intituion for the real stuff, the universe. === Subject: Re: H numbers >> > > I think H-numbers are commutative, quaternions not. >> Yes H numbers commute. Quaternions close on themselves, > Do you mean that the quaternions are closed, in the algebraic sense? No, I meant to say thay have a circular relationship. ij=k,jk=i,ki=j That wording is not correct. That bloody wording! >> while H numbers are >> open, in line with Godel. > And, if so, what is the connection with Godel? === Subject: What did they do with the 4? algebra. I need to learn how to expand brackets using Pascal's triangle, which I can do to a certain degree! Ask me to do (a+b)ñ, and im fine. But (a+4b)ñ has thrown me off a bit! Well not so much a bit, as in im getting to the point where I want to commit suicide after spending three hours looking at a solution trying to figure out what they did with the 4! I guess the majority of you cannot believe such a retarded person is posting on a maths group, but I was hoping someone could enlighten me! Please! What did they do with the 4? Rhi! === Subject: Re: What did they do with the 4? > Ask me to do (a+b)ñ, and > im fine. But (a+4b)ñ has thrown me off a bit! > Please! What did they do with the 4? Don't worry about the 4; just do with (4b) what previously happened with b on its own. If it helps, write out (x+y)^4, and when you're finished, replace each x by a, and each y by (4b). So, where you previously had a b^2, now you have a (4b)^2=16b^2, and so on. === Subject: Re: What did they do with the 4? > concepts of algebra. I need to learn how to expand brackets > using Pascal's triangle, which I can do to a certain degree! Ask > me to do (a+b)ñ, and im fine. But (a+4b)ñ has thrown me off a > bit! Well not so much a bit, as in im getting to the point where > I want to commit suicide after spending three hours looking at > a solution trying to figure out what they did with the 4! > I guess the majority of you cannot believe such a retarded person > is posting on a maths group, but I was hoping someone could > enlighten me! > Please! What did they do with the 4? -- First off, be careful about using non-standard characters. They might not show up properly. What I see is (a+b) followed by a superscript 3. That makes sense in the context of your question, so I'm going to assume it came through all right. The typesetting kind of features I see used most often are superscripts ( (a+b)^3 ), subscripts ( < a, b> = Sum_j(a_j*b_j)), and escape character (backslash) ( eg, x in A , where in stands for the is-an-element-of epsilon). There is a lot that you could do. Latex syntax seems to be very used a lot, but any notation you use that is even a little bit uncommon, you should probably explain, as a matter of clarity and politeness, even if your audience might be able to figure it out from context. -- Where do they put the 4? Write out the expression for (a+b)^3, then substitute in it 4b wherever b is. Eg, (a+b)^2 = a^2 + 2*a*b + b^2 -> a^2 + 2*a*(4*b) + (4*b)^2 Jim Burns === Subject: Re: What did they do with the 4? Replace b by 4b in the expansion, making sure that powers of b become powers of 4b. === Subject: Re: undefined terms > In my ongoing journey of becoming math literate, I'm finding that one > thing that textbooks don't do the greatest job of is identifying > mathematical (and logical) terms that are left undefined. This is > frustrating because a book will often mention a term, and I'm left > guessing if that term has a definition of its own (and that I just > can't find it), or whether that term is really undefined. I guess a > famous example would be 'set'. Another good example would be > 'variable'. And probably another good example would be 'form' (as in, > a linear equation is of the form y = mx + b) Maybe also 'symbol', > 'substitution'??? > I'm sure this request isn't quite as simple as it sounds, since there > are probably terms that have been attempted to be defined by some, but > maybe aren't accepted by others. Also, I'm not talking about > intuitive definitions that are sometimes given to give the learner an > idea of what the term is; for the purposes of my request that doesn't > count as a definition. I'm taking only about rigorous definitions that > can be used in formal proofs. Well, I think that you left quite a few words undefined in your posting. What is the definition of in? What is the definition of my? What is the definition of ongoing? What is the definition of journey? What is the definition of of? What is the definition of becoming? ..... Well, you get the idea. More seriously, when we communicate we have to assume a certain amount of universal knowledge. I mean, you could completely reduce all of your math proofs to first order language, but even those books that define first order language start out by using regular language and undefined terms. === Subject: Re: undefined terms >I'm reminded of Halmos Applied mathematics is bad mathematics. >Something else I've not read. Have you, and did he mean what his title >seems to suggest? Haven't read it. The title seems like a needlessly divisive stance. You prefer abstraction to application? Fine -- it does have more elegance, and usually more depth. On the other hand, the applications pay the bills. I mean really, universities don't have a Department of Model Railroading, much as some people find that to be fun. The only reason mathematics thrives on the scale that it does is because it's arguably useful. For me, well, I get a kick out of seeing occasions when some moderately powerful mathematics gives a slick answer to an applied problem. dave === Subject: Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF See how Your last developments are working for n=3 and may be Your final Q^3 + P^3 arguments could be completed in all off correct developments of any of Yours parameters Success Ro-bin === Subject: Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF However the argument sketched above is well known, and is very old -- I'm sure it goes back at least as far as Fermat, so the places where the proof is likely to fail lie ahead. quasi First You say that I do not understand my Arguments then you say: However the argument sketched above is well known, and is very old -- I'm sure it goes back at least as far as Fermat, so the places where the proof is likely to fail lie ahead. quasi SO YOU HOPE!Well ,Discover It.Good Luck. If you do notundersdand Some argument since is unreadeble for you ask me to make it lucid for you. As I see you start to catch _up with my unredeable statements as I Began to explain them. But You still can not take the fact that George Might be right and thousands of mathematicienns missed to find this elementary proof while you have to face this possibility. Dig and fight with a strong Math. Argument this FLT Proof. George === Subject: Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF Lets say that W=X+Y and R have as a common divisor a prime number ,for exemple number 5 R=X^2-X*Y+Y^2R has another two expressions: 1.R=(X+Y)^2-3*(X+Y)*Y +3*Y^2 2.R=(X+Y)^2-3*(X+Y)*X+3*X^2 If (X+Y)=W is divisible by 5 then R is divisible by 5 if 3*Y^2 or 3*X^2 is divisible by 5 .That means Y^2 or X^2 is divisible by 5. Therfore (X+Y) and Y^2 or (X+Y)=W and X^2 have as common divisor 5(a prime is allways chosen to prove this) That means that X is divisible by 5. Therefore W-X=(X+Y)-X is divisible by 5. Therfore Y is divisible by 5 but X and Y are relative prime . Therefore R is not divisible by 5 Therefore R and W do not have as a common divisor number 5 or any other prime number chosen by us george === Subject: Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF Lets say that (X+Y) =W and R have as a common divisor the prime number 5 R has another two expressions: 1.R=(X+Y)^2-3*(X+Y)*Y +3*Y^2 2.R=(X+Y)^2-3*(X+Y)*X+3*X^2 If (X+Y)=W is divisible by 5 then R is divisible by 5 if 3*Y^2 or 3*X^2 is divisible by 5 .That means Y^2 or X^2 is divisible by 5. Therfore (X+Y) and Y^2 or (X+Y)=W and X^2 have as common divisor 5(a prime is allways chosen to prove this) That means that X is divisible by 5. Therefore W-X=(X+Y)-X is divisible by 5. Therfore Y is divisible by 5 but X and Y are relative prime . Therefore R is not divisible by 5 Therefore W=X+Y and R are Relative prime with respect to prime number 5 In this way we can use any prime number to show that (W)=X+Y and R are relative prime ( no common divisor) george === Subject: Re: how to solve this equation by paper and pencil? > > Use the cublic formula. http://mathworld.wolfram.com/CubicFormula.html > > > > > > I don't think the roots are rational or real on this one. > > > > Every cubic has at least one real root. (Look at the left- and > > > right-hand behavior, and use the Intermediate Value Theorem.) > > Tell it to x^3 - i. > HAHA! Nice one. > But, Proginoskes is prob meant if the coefficients are real. Yes. An mr0x was specificially referring to x^3 + x^2 - 4, whose coefficients _are_ all real. But it was a fair cop. --- Christopher Heckman > And, I meant > none of the roots are rational and some of them are not even real! > > -- > > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: INFINITY Revisited > If at noon there are no balls remaining, does this have implications > about the naturals having a one to one correspondence with the real > decimalic numbers? No. > By noon an infinite number of balls (with printing on the surface) has > been added to the vase and removed. Correct: there are infinitely many balls with finite labels, and each of those was added, and later removed before noon. > This seems to me to imply that by noon all balls with infinite > decimalic representations such as 1/3 = 0.333..., Pi/10 = > 0.31415..., and sqrt (2)/2 = 0.70710678..., as well as many that > can't be denumerated in real life, must also have been placed in the > vase and removed. At what time was the ball labelled with 1/3 placed in the vase? How long did it stay there? - Tim === Subject: Re: INFINITY Revisited By noon an infinite number of balls (with printing on the surface) has > > been added to the vase and removed. > Correct: there are infinitely many balls with finite labels, and each > of those was added, and later removed before noon. think finally understood what you were saying and avoiding. While I agree that every ball with a finite label will be eventually be removed before noon, the number of balls in the vase at any time before noon has increased and will continue to increase at each time interval (# balls = 10^n - n). At any time before noon the number of balls is larger than the previous time. At no time before noon are there zero balls in the vase. Yet by noon the miracle has happened. > > This seems to me to imply that by noon all balls with infinite > > decimalic representations such as 1/3 = 0.333..., Pi/10 = > > 0.31415..., and sqrt (2)/2 = 0.70710678..., as well as many that > > can't be denumerated in real life, must also have been placed in the > > vase and removed. > At what time was the ball labelled with 1/3 placed in the vase? How > long did it stay there? I don't know when, but I am sure it was the same time the vase became empty. My other response to this would be at noon the time of of the miraculous coming of the infinite. At noon (the time when our limits are measured) the infinite set of naturals rose for the measuring, the infinite decimalic balls arrived and disappeared instantly. > - Don === Subject: Re: INFINITY Revisited >> > By noon an infinite number of balls (with printing on the surface) has >> > been added to the vase and removed. This seems to me to imply that by >> > noon all balls with infinite decimalic representations such as 1/3 = >> > 0.333..., Pi/10 = 0.31415..., and sqrt (2)/2 = 0.70710678..., as well >> > as many that can't be denumerated in real life, must also have been >> > placed in the vase and removed. >> No. According to the scheme you have described, only the balls with >> terminating decimal expressions have been added and removed. > I clearly agree that at any time before noon only the balls with > terminating decimal expressions have been added or removed and at any > time before noon only a finite number of balls have been added, > representing a finite subset of the infinite natural numbers. However, > by noon things have changed: the finite became infinite - the naturals > became an infinite set. What about the sequence of balls represented > by 0.3, 0.33, 0.333, ...? At noon don't there become an infinite > number of digits associated with the limit point of the sequence formed > from balls with only the numerals 3 printed on the surface? If not, > at noon how many digits are there printed on the surface of the ball > containing the largest number of numeral threes and which has only 3s > as numerals? What makes you think there is such a thing as a largest number of numeral threes and which has only 3s as numerals? Do you also think there is a largest integer? Each ball that is marked with a terminating decimal string is added at a specific time before noon. There is no time when any ball with a nonterminating string is added. > This thought experiment seems like a method prone to generate > misunderstanding. > What is the proper mathematical term for a decimalic digit dj for the > jth place to the right of the decimal point in a number represented in > decimalic form? For example: 0.333...dj. Knowing proper terms would > be useful when trying to precisely specify and discuss issues on these > matters. I hope the language I use in this post will not be > misunderstood. The language you are using is fine. > The sequence of rational numbers 0.3, 0.33, 0.333, ... converges to a > real limit point (as dj the number of decimalic digits approaches > infinity) of 0.333...; where 0.333... is the repeating decimalic > representation of the rational number 1/3. > Does the list of natuarl numbers corresponding to the infinite set of > natural numbers enumerate all of the digits dj where j = 1 -> oo of > 1/3 = 0.333... ? For each natural number n, the terminating decimal of the form sum_{k=1}^n 3/10^k is on the label of a ball that is added before noon. The nonterminating string sum_{k=1}^oo 3/10^k = 1/3 is not on the label of any ball that is added to the vase according to your description. > 0, 1, 2, 3, ... > 0 . 3 3 3 ... > 0,1,2,3, ..., n, ... > 0.3 3 3 ..., dj, ... > Can the set of natural numbers be put in a one to one correspondence > with all of the digits of any specific decimalic number (including any > nonterminating rational or irrational number)? Yes. > Can there ever be any decimalic digits dj of a real number that such a > one to one correspondence (between the digits of the real number and > the infinite set of natural numbers) miss? Cantors diagonal proof > seems to depend upon this. No, there are no digit positions that are missed, and no, the Cantor diagonal argument does not depend on any such thing. How many 3-digit strings can you make from the ten decimal digits? 1000, right? Isn't 1000 larger than 3? The number of digits in a nonterminating decimal digit string is aleph_0 (that's the name we use for the infinity of the natural numbers). The number of different digit strings of length aleph_0 is 10^aleph_0 = c, the cardinality of the continuum. Just as 10^3 is bigger than 3, 10^aleph_0 is bigger than aleph_0. That's what the diagonal argument shows. > Does such a rule serve as an alternative definition of a real number > eliminating numbers such as 0.000...1 ( the termination of the sequence > of numbers 0.1, 0.01, 0.001, ... that apparently lacks a real limit > point) from consideration as real numbers? What rule? A real number is not a digit string. A real number is a collection of rational numbers that is nonempty, bounded above, downward closed, and has no maximum element. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: INFINITY Revisited > This thought experiment seems like a method prone to generate > misunderstanding. > What is the proper mathematical term for a decimalic digit dj for the > jth place to the right of the decimal point in a number represented in > decimalic form? For example: 0.333...dj. Knowing proper terms would > be useful when trying to precisely specify and discuss issues on these > matters. I hope the language I use in this post will not be > misunderstood. Its normally written as the jth digit d_j but you are fine with dj. > The sequence of rational numbers 0.3, 0.33, 0.333, ... converges to a > real limit point (as dj the number of decimalic digits approaches > infinity) of 0.333...; where 0.333... is the repeating decimalic > representation of the rational number 1/3. Yes > Does the list of natuarl numbers corresponding to the infinite set of > natural numbers enumerate all of the digits dj where j = 1 -> oo of > 1/3 = 0.333... ? No. 1/3 doesn't appear in the listas you have written it. It appears in others, such as the standard mapping of p/q to Natural numbers. > 0, 1, 2, 3, ... > 0 . 3 3 3 ... > 0,1,2,3, ..., n, ... > 0.3 3 3 ..., dj, ... > Can the set of natural numbers be put in a one to one correspondence > with all of the digits of any specific decimalic number (including any > nonterminating rational or irrational number)? Yes, trivially. 1 -> 3 2 -> 1 3 -> 4 4 -> 1 5 -> 5 If you want a mapping between N and approximations to pi, pick 1 -> 3 2 -> 3.1 3 -> 3.14 If you want pi on the list, 1 -> pi 2 -> 3 3 -> 3.1 4 -> 3.14 > Can there ever be any decimalic digits dj of a real number that such a > one to one correspondence (between the digits of the real number and > the infinite set of natural numbers) miss? Cantors diagonal proof > seems to depend upon this. Sort of. We need to know that if you count up from 1 you eventually get to all numbers. This is provable from the axioms of arithmetic. > Does such a rule serve as an alternative definition of a real number > eliminating numbers such as 0.000...1 ( the termination of the sequence > of numbers 0.1, 0.01, 0.001, ... that apparently lacks a real limit > point) from consideration as real numbers? > Yes, the rule says that the limit point of 0.1, 0.01, 0.001 is 0, and tough titty if that is not part of the sequence. You can construct things that look like 0.000..1, infintismals, but they don't help with Cantor. > Don Whitehurst === Subject: Re: INFINITY Revisited <4310a75c$0$19001$afc38c87@news.optusnet.com.au> Does the list of natuarl numbers corresponding to the infinite set of > > natural numbers enumerate all of the digits dj where j = 1 -> oo of > > 1/3 = 0.333... ? > No. 1/3 doesn't appear in the listas you have written it. It appears in > others, such as the standard mapping of p/q to Natural numbers. > > Can the set of natural numbers be put in a one to one correspondence > > with all of the digits of any specific decimalic number (including any > > nonterminating rational or irrational number)? > Yes, trivially. > 1 -> 3 > 2 -> 1 > 3 -> 4 > 4 -> 1 > 5 -> 5 > If you want a mapping between N and approximations to pi, pick > 1 -> 3 > 2 -> 3.1 > 3 -> 3.14 > If you want pi on the list, > 1 -> pi > 2 -> 3 > 3 -> 3.1 > 4 -> 3.14 Here is where I get lost. Above in essence you said that the infinite set naturals can trivially be placed in a one to one correspondence with all of the digits of pi; and yet you now seem to be suggesting there are not enough natural numbers in the infinite set of natural numbers for a mapping between N the approximations of pi and pi, unless pi is placed as an indivdual element corresponding to some finite natural (in other words pi cannot be the last element). Why not if the set of naturals is infinite? > Don Whitehurst === Subject: Re: INFINITY Revisited > > Does the list of natuarl numbers corresponding to the infinite set of >> > natural numbers enumerate all of the digits dj where j = 1 -> oo of >> > 1/3 = 0.333... ? >> > No. 1/3 doesn't appear in the listas you have written it. It appears in >> others, such as the standard mapping of p/q to Natural numbers. >> > > > > Can the set of natural numbers be put in a one to one correspondence >> > with all of the digits of any specific decimalic number (including any >> > nonterminating rational or irrational number)? >> > Yes, trivially. >> 1 -> 3 >> 2 -> 1 >> 3 -> 4 >> 4 -> 1 >> 5 -> 5 >> If you want a mapping between N and approximations to pi, pick >> 1 -> 3 >> 2 -> 3.1 >> 3 -> 3.14 >> If you want pi on the list, >> 1 -> pi >> 2 -> 3 >> 3 -> 3.1 >> 4 -> 3.14 > Here is where I get lost. Above in essence you said that the infinite > set naturals can trivially be placed in a one to one correspondence > with all of the digits of pi; and yet you now seem to be suggesting > there are not enough natural numbers in the infinite set of natural > numbers for a mapping between N the approximations of pi and pi, unless > pi is placed as an indivdual element corresponding > to some finite natural (in other words pi cannot be the last element). > Why not if the set of naturals is infinite? pi cannot be the last element because there is no last element. The set of naturals are infinite and so there is no last natural number to map to pi. Stephen === Subject: GRAVITY AND RADIATION MECHANICS === Subject: GRAVITY IS NOT A FORCE PLANETS ORBIT THE SUN TO CONSERVE TOTAL ENERGY GRAVITATION IS NOT A FORCE BUT AN ILLUSION Copyright 1984-2005 Allen C. Goodrich A planet or any mass such as the earth orbits the sun simply because it would require the gain or loss of a tremendous amount of energy to make it travel in any other orbit or path.This is the only path where its kinetic and potential energies,relative to the rest of the universe, are equal in magnitude, and their sum is a constant. But,why do we seem to be attracted to the earth by a force of gravity? SUMMARY OF PAST HISTORY: The precise measurements of planetary motion by Tycho Brahe (1546-1601) and observations by Galileo Galilei (1564-1642) were plotted by Johann Kepler (1571-1630 ) resulting in Kepler's Three laws: 1. The planets move about the sun in elliptical orbits with the sun as one focus of the ellipse. 2. The straight line joining the sun and a given planet sweeps out equal areas in equal intervals of time. 3. The square of the period of revolution of the planet about the sun is proportional to the cube of the mean distance from the sun. t^2 = K L^3 Sir Isaac Newton (1642-1721 ) concluded that it was a force F = mL/t^2 = k m_1 x m_2 /L^2 that caused the orbital motion. Allen C. Goodrich defined the cause as a conservation of total energy. The concentration of the Kinetic Energy of mass increases as the Potential Energy of the universe decreases with the expansion of the universe at constant total energy. Planets orbit the sun in a state of equiliurium,where no change to total energy occurs. At Equilibrium the sum of kinetic and potential energies is a constant. A positive change of kinetic energy equals a negative change of potential energy. + delta m (2 pi L)^2/t^2 = - delta G (M-m)m / L . or Delta e (2 pi L)^2/t^2 = - Delta K e^2 / 4 pi E_o L. if a charge is present. The mass of the human body, on the earrth's surface, is not in an equilibrium orbit. If a force ,such as the surface of the earth , was not present, the body would not stay where it is. IT TRIES TO MOVE TO AN EQUILIBRIUM ORBIT.( No change of total energy) This force is what is felt to rqual gravitational force. A gravitational force is not needed in a state of orbital equilibrium. Galileo demonstrated the effect of gravitational force. Newton assumed that a gravitational force between all masses pulled them together. Was this a correct assumption? Einstein and many other scientists felt that there must be more to gravitation than an attraction at a distance. Action at a distance was considered to be impossible in the absence of a transfer of energy at the speed of light. A change of kinetic energy is not always the result of a force. In an equilibrium system at constant total energy, kinetic energy can increase as potential energy decreases, with the total energy remaining constant.. Hubble then showed that the distant Galaxies were moving away from the earth and that the universe was expanding in all directions. If this is true , What else must be true? 1. The potential energy of the rest of the universe must be decreasing relative to the mass of the earth. It has long been assumed that the first law of thermodynamics, which says that the total energy of the universe is a constant, was a fact of nature. If this is true what then? 2. The kinetic energy of the universe must be increasing at the same rate that the potential energy is decreasing as the universe expands. How is this possible? Masses must be accelerating, because, kinetic energy is the result of an acceleration. 3. Orbital motion could then be the result of the expansion of the universe. The Gravitational illusion could be the result. Based on the first law of thermodynamics The total mass energy of the universe is a constant. ((total kinetic (mass) energy plus total potential energy is a constant)). m is any mass say that of the earth. Planets, moons, and electrons are normally in equilibrium orbits where the total energy is constant. m(2 pi L)^2/t^2 + G(M-m)m/L+ X e(2 pi L)^2/t^2 + Z e^2/4 pi E_o L = a constant. In the absence of a charge, from this equation the equation Delta m (2 pi L)^2 / t^2 = - Delta G (M-m)m/L follows mathematically. The earth orbit is a result of an energy equilibrium, ( the absence of a change of total energy ) and not the result of a force of gravity between masses. Force of gravity is the resulting illusion assumed by Newton to be a force. If a planet (say earth) moved away from the sun its potential energy would decrease as L increased. Its kinetic energy would decrease because it is no longer accelerating toward the sun in orbital motion. Total energy would have to decrease. A very great change of total energy would have to take place. POTENTIAL ENERGY = G(M-m)m/L KINETIC ENERGY = m(2 pi L)^2/t^2 m(2 pi L)^2/t^2 + G(M-m)m/L = A constant = M G= Gravitational constant; M = total energy of the universe (or effective universe) ; m = mass in question. t = time ; L = radial distance. No mechanism exists for this to occur rapidly. So it could not happen. The magnitudes of kinetic and potential energies of planets and moons travelling in orbital motion are nearly equal and any increase or decrease of orbital distance L results in an equal change in magnitude of both.This is the only value of L where no change of total energy will occur if the value of L changes. At any other distance L, an increase of kinetic energy will be at a different rate than potential energy decreases. Orbital motion conserves total energy. Force of gravity isn't needed to explain orbital motion or any other motion at a distance. GRAVITY MECHANICS AND RESEARCH ON ASTRONOMICAL OCEAN TIDES Copyright 1984 to 2002 Allen C. Goodrich An examination of United States Coast and Geodetic Survey Tidal Data, which was gathered by extensive measurements over long periods of time,was compared with astronomical data showing the phases of the moon at corresponding times for many years. This correlation of the two sets of data revealed a very interesting fact, in a manner that had never before been mentioned in the literature. It is invariably and exactly the lowest tide that exists directly under the full and new moons at deep ocean ports. TABULATED co-op.nos.noaa.gov and space.jpl.nasa.gov DATA: OCEAN TIDES AND PHASES OF THE MOON AT DEEP OCEAN PORT- MYRTLE BEACH LOWEST TIDE (YEARS 1992 AND 1993) 1992 FULL MOON---1992 NEW MOON (at moons highest point in the sky) DATE---TIME(std)-DATE---TIME(std) Mar.18--12:00Mid-Mar.3---12:00Noon Apr.17--12:00Mid-Apr.2---12:00Noon May.17--12:00Mid-May.2---12:00Noon Jun.15--12:00Mid-Jun.29--12:00Noon July.13-12:00Mid-July.29-12:00Noon Aug.12--12:00Mid-Aug.27--12:00Noon Sept.11-12:00Mid-Sept.26-12:00Noon Oct.11--12:00Mid-Oct.26--12:00Noon Nov.10--12:00Mid-Mov.25--12:00noon Dec.10--12:00Mid-Dec.25--12:00noon 1993 FULL MOON---1993 NEW MOON (at moons highest point in the sky) DATE---TIME(sdt)-DATE---TIME(sdt) Jan.8--12:00Mid--Jan.24-12:00Noon Feb.6--12:00Mid--Feb.21-12:00Noon Mar.8--12:00Mid--Mar.23-12:00Noon Apr.6--12:00Mid--Apr.21-12:00Noon May.6--12:00Mid--May.20-12:00Noon Jun.4--12:00Mid--Jun.19-12:00Noon July.3-12:00Mid--Juy.18-12:00Noon Aug.2--12:00Mid--Aug.17-12:00Noon Sep.1--12:00Mid--Sep.16-12:00Noon Sep.30-12:00MId--Oct.15-12:00Noon Oct.30-12:00Mid--Nov.14-12:00Noon Nov.29-12:00Mid--Dec.13-12:00Noon Dec.28-12:00Mid--Jan.12-12:00Noon This was a very interesting discovery because current physics,based on the gravitational theory, discussed in the following U.S.Gov. documents: PREDICT THE OCEAN TIDES http://co-ops.nos.noaa.gov/restles1.html SEE PHASES OF THE MOON FROM EARTH http://space.jpl.nasa.gov/ ,would lead one to believe that,except for many possible reasons, the highest tides tend to be under the full and new moons. The dictionary and encyclopedia as well as physics texts predict this with pictures of the earth and oceans bulging on the side facing the full moon. Of course it never happens as the gravitational theory predicts, and many reasons are given for the discrepancies. CONCLUSION: No discrepancies were found in the occurence of exactly the lowest tide directly under the full and new moons, at deep ocean ports. A lowest tide also occurs on the earth's ocean directly opposite to the new and full moons. SIGNIFICANCE: One must admit that this is beyond question one of the most important discoveries of modern physics research. It indicates that a change must be made in the theory of gravitation. One can no longer assume that a force between the moon and the water of the earth's oceans, is causing the ocean tides. The force of gravity must be an illusion caused by some other, more basic, reason. What would this be? If the total energy ( kinetic and potential ) of the universe is assumed to be a constant,from this fundamental equation, many interesting things follow. If the rest of the universe is expanding ( potential energy decreasing) relative to masses, the masses must be shrinking ( increasing in kinetic energy ) (gravitation) relative to the rest of the universe. THE FIRST LAW OF MOTION-(GOODRICH) Copyright 1984 to 2002 ALLEN C. GOODRICH A body (m) continues in a state of rest (equilibrium) or motion in a straight or curved line (equilibrium) as long as no change occurs in its total (kinetic and potential) energy, relative to the rest of the effective universe (M-m), Delta m(2 pi L)^2/t^2 = - Delta K(M-m)m/L equilibrium = no change in the total energy relative to the rest of the effective universe (M-m). ^ = to the power of. Orbital motion complies with this equation. This equation is derived from the fundamental equation of the universe which states that the total energy of the universe is a constant. The sum of kinetic and potential energies is a constant. m(2 pi L)^2/t^2 + K(M-m)m/L = A constant. INERTIA AND MOMENTUM are the properties of a mass that evidence its reluctance to change its total energy, or it is its need to maintain a constant total energy. If it could more easily obtain or lose energy, it would have less inertia or momentum. SEE THE UNIVERSE- A GRAND UNIFIED THEORY OF MASS ENERGY SPECTRUM OF THE BUFFALO ASTRONOMICAL ASSOCIATION INC. NOV.1996 TO FEB.1997 :( CLICK BLACK AND BLUE PAGES BELOW ) http://ourworld.cs.com/gravitymechanic2/myhomepage/business.html http://ourworld.cs.com/gravitymechanic2/myhomepage/profile.html TIDES AND GRAVITY MECHANICS http://ourworld.cs.com/gravitymechanic2/myhomepage/resume.html A new theory of gravitation is given, which predicted, stimulated the above research,and is consistent with, the new findings. The universe has been found to be expanding at an accelerating rate as predicted in 1984 by this new theory. ELECTROMAGNETIC ,PHOTON AND CHARGE EFFECTS. ARE DEFINED IN THE FOLLOWING BOOK.-- THE UNIVERSE:--Allen C. Goodrich Copyright 1984 to 2005 Allen C. Goodrich FORCE OF GRAVITATION DOES NOT EXIST. If One calculates the kinetic and potential energies of the planets relative to the rest of the effective universe, using the formulas kinetic energy = m(2 pi L )^2/t^2 and potential energy = -G(M-m) m/L, M is the gm mass of the sun and all planets; m ,L,and t are the gm mass, mean radial cm. distance, and orbital time in sec, of one of the planets. ( THIS IS THE ONLY CORRECT METHOD, it explains the T.R.Young-two slit interference pattern which involves the rest of the universe ). One will find that they are of nearly equal magnitude but opposite in sign. One will also find that their sum is a constant, the equilibrium energy for the particular planet.This is the energy that remains constant as the universe expands. its potintial energy continually decreasing and its kinetic energy continually increasing. Only at the orbital distance will a small change of kinetic energy equal an opposite change of potential energy.This is the total energy that requires no force , with its necessary acceleration and change of total energy, to maintain it as a constant.No force of gravity is necessary to explain the motion of the planets in the expanding universe. The planets motion around the center of the rest of the universe at the specific distance L is the equilibrium condition for constant total energy of the orbiting planet in the expanding universe. THE SOLAR SAIL Copyright 1984 to 2005 Allen C. Goodrich The Solar Sail, which is being tested by Russia and the United States, for possible propulsion in interstellar space travel, is additional evidence that no change of potential energy to kinetic energy of the photon takes place unless the potential energy is absorbed .The photon does not have mass ( kinetic energy). A change of direction of the photon's potential energy can occur at the reflective surface but no potential to kinetic energy change takes place there. A change of potential to kinetic energy takes place at the black absorption surface.which has the correct frequency response as well as direction and density (time ) in the expanding universe.This is evidence that the photon is potential not kinetic energy.The light photon does not have mass or kinetic energy.until the photon is absorbed by a mass of the correct frequency response as well as direction and density (time ), no potential to kinetic energy change can take place.in the expanding universe, in the absence of a mass.. THE VELOCITY OF LIGHT IS AN ILLUSION Copyright 1984 to 2005 Allen C. Goodrich A negative kinetic energy change of a mass, is a positive potential energy change of the rest of the effective universe relative to a mass of the proper frequency, direction ,distance L and time change t (density), in the expanding universe. This is consistent with the first law of thermodynamics, whch conserves total energy.. The L/t is currently falsely assumed to be a velocity of light. This explains the T.R. Young two slit interference pattern. change of the entire universe, that can become a positive kinetic energy change of a mass such as the electron if the frequency, direction, distance L , and time change t (density) are correct.. === Subject: Another Coxeter group question Another question re Grove's and Benson's _Finite Reflection Groups_: Exercise 6.8(e) asks the reader to show that an irreducible Coxeter group G whose graph has an edge with an even mark -- i.e., the groups Bn for all n, F4 and what Grove and Benson call H2^n, the finite dihedral subgroups of O(2,R) of order 2n for n even -- has |G/G'| = 4. It's straightforward to do this for Bn since |Bn/Dn| = 2 and we already know that |Dn/Dn'| = 2 (see below), and of course Bn has a rotation subgroup of index 2, so |Bn/Dn'| = 4 is abelian. And of course it's easy to do for H2^n, n even, from elementary group theory. But I don't see how to do it for F4, and in any case I suspect that these approaches aren't what the authors had in mind, given the previous parts of the exercise (all of which I could do): Part (a) asks the reader to show that the rotation subgroup H of a Coxeter group G consists of all elements of G that are the product of an even number of fundamental reflections; part (b) that if G is irreducible, then every element of G can be written as a product of adjacent fundamental reflections; part (c) that if two adjacent fundamental reflections are joined by an edge with an odd mark, then their product is a commutator; and part (d) that if G is irreducible and every mark on its graph is odd, then G' = H and so |G/G'| = 2. -- Jim Heckman === Subject: Re: What should a mod() function do then? >> You haven't specified the allowed range of r. Judging by the physical >> problem, it sounds like for the case of d = 2 pi, you want >> -pi < r <= pi. > -2 pi < r <= 2 pi Oh. Then r has two allowed values for every value of v? Fair enough, but ... unusual. You've got a periodic function, I would have expected that a 2 pi change in phase should make no difference to the output, because it's physically indistinguishable. - Tim === Subject: Re: What should a mod() function do then? >I think of the mod operation as the remainder of integer division: > > >>q d + r = v > > You haven't specified the allowed range of r. Judging by the physical > > problem, it sounds like for the case of d = 2 pi, you want > > -pi < r <= pi. > -2 pi < r <= 2 pi If you're doing calculations modulo 2 pi, this will not give you a unique value of r, but rather two possible values: pi = 1 * 2pi + -pi pi = 0 * 2pi + pi So do you want MOD(pi, 2 pi) to be pi or -pi? (Continued below.) > >>Then the remainder is: > > >>r = v - int(q d), > > I think what you mean is r = v - d int(v / d). > Yes. The next line made that clear. > > However, this doesn't give you what you want for your definition of > > int(x) = { floor(x), x>=0 ; ceiling(x), x<0 }. > > For example, let v = 3.9 pi: you get r = 1.9 pi, when I think you want > > r = -0.1 pi. Likewise for v = -3.9 pi you will get r = -1.9 pi when > > you more likely want r = 0.1 pi. > No. Should be like this: > T mod(T, 2 pi) > -4.1 pi -0.1 pi > -3.9 pi -1.9 pi > -2.1 pi -0.1 pi > -1.9 pi -1.9 pi > 0 0 > 1.9 pi 1.9 pi > 2.1 pi 0.1 pi > 3.9 pi 3.9 pi > 4.1 pi 0.1 pi (Continued from above.) Based on your table, you probably want to take MOD(|v|, 2pi) (using the standard formula) and multiply by the sign of v. Thus, you'd have the condition on r being 0 <= r < 2 pi if r is nonnegative, and -2 pi < r <= 0 if r is negative, which _does_ result in a unique value for r. Your function will then be MOD( abs(v), 2 pi) * sgn(v) = (abs(v) - 2 pi * floor(abs(v) / 2 pi)) * sgn (v). --- Christopher Heckman === Subject: Re: Han's startling new set theory. <87hddhrc3s.fsf@phiwumbda.org> <87oe7nvzfk.fsf@phiwumbda.org> <51d26$430ee238$82a1e3ad$24520@news2.tudelft.nl > Note that mathematics in general, does make testable predictions. > >> In what sense is the square root of 2 is an irrational number a > testable prediction? > > >>The decimal expansion of sqrt(2) does not repeat. You can check that > >>by computing. > > Really?? I mean, this is finite computing, with your microscope, > > right? > > How long, very roughly, does it take to check that sqrt(2) doesn't > > repeat? > > (Please answer in real, microscopable, units, like years, or kalpas.) > You can't check if it doesn't repeat, but you can check some if it does: > 1/7 = 0.14285714285714285714285714285714285714285714285714285714285 > sqrt(2) = 1.41421356237309504880168872420969807856967187537694807317668 A few years ago, an error was found in the floating point section of the Pentium processor, which caused it to give incorrect results to certain functions: http://www.cs.earlham.edu/~dusko/cs63/fdiv.html How do you know that your calculations above are correct (i.e., that whatever computer you perform them on is actually correctly modelling the conclusions which come from the axioms of arithemtic)? === Subject: Re: Han's startling new set theory. > There's a reason why mathematicians use the word proof and > scientists do not. sqrt(2) is provably irrationaly. Just because something is provable doesn't make it true. I mean, I fully expect it to be true, but I don't absolutely *believe* it. - Tim === Subject: Re: Han's startling new set theory. <87hddhrc3s.fsf@phiwumbda.org> <87oe7nvzfk.fsf@phiwumbda.org> There's a reason why mathematicians use the word proof and > > scientists do not. sqrt(2) is provably irrationaly. > Just because something is provable doesn't make it true. > I mean, I fully expect it to be true, but I don't absolutely *believe* > it. Interesting stance. So you believe that it might be possible that there are co-prime natural numbers p and q, such that p^2 = 2*q^2? Do you also believe it might be possible that a natural number m can be divisible evenly by 2 (i.e., is even) and simulateously not be divisible evenly by 2 (i.e., is odd)? === Subject: Re: Han's startling new set theory. > So you believe that it might be possible that there are co-prime > natural numbers p and q, such that p^2 = 2*q^2? It depends what you mean by natural numbers, co-prime and so forth. Usually these terms are defined in terms of some particular system, and these systems aren't necessarily all equivalent. For example, I'm not certain that we can't derive a suitably formalized version of there exists p,q in N such that p^2 = 2 q^2 from the axioms of, say, ZFC. I'm more confident (very very confident, but still not absolutely certain) that we can't derive such a theorem from PA. I'm even more close to certain that we can't find an actual example with decimal numerals. I'm absolutely certain that I haven't such an example in mind. - Tim === Subject: Re: Han's startling new set theory. <87hddhrc3s.fsf@phiwumbda.org> <87oe7nvzfk.fsf@phiwumbda.org> So you believe that it might be possible that there are co-prime > > natural numbers p and q, such that p^2 = 2*q^2? > It depends what you mean by natural numbers, co-prime and so forth. > Usually these terms are defined in terms of some particular system, > and these systems aren't necessarily all equivalent. > For example, I'm not certain that we can't derive a suitably > formalized version of there exists p,q in N such that p^2 = 2 q^2 > from the axioms of, say, ZFC. > I'm more confident (very very confident, but still not absolutely > certain) that we can't derive such a theorem from PA. Let's work in the axioms of PA. Suppose we take a more constructivist approach, and demand instead that an existence proof alone is insufficient; you must present me with a construction for naturals p and q such that 2*q^2 = p^2. How is it not certain in your mind that this is not possible? === Subject: Re: Han's startling new set theory. <87hddhrc3s.fsf@phiwumbda.org> <87oe7nvzfk.fsf@phiwumbda.org> > So you believe that it might be possible that there are co-prime > > > natural numbers p and q, such that p^2 = 2*q^2? > > It depends what you mean by natural numbers, co-prime and so forth. > > Usually these terms are defined in terms of some particular system, > > and these systems aren't necessarily all equivalent. > > For example, I'm not certain that we can't derive a suitably > > formalized version of there exists p,q in N such that p^2 = 2 q^2 > > from the axioms of, say, ZFC. > > I'm more confident (very very confident, but still not absolutely > > certain) that we can't derive such a theorem from PA. > Let's work in the axioms of PA. Suppose we take a more constructivist > approach, and demand instead that an existence proof alone is > insufficient; you must present me with a construction for naturals p > and q such that 2*q^2 = p^2. > How is it not certain in your mind that this is not possible? Perhaps it is a rationalization of self-determination. Tim here is rational, he's pretty smart, he's just acknowledging the possibility of someone turning out the lights, a higher power or his own assertion of objective existence, not Rayndian objectivism, or something along those lines. I think he's not saying 2 + 2 = 5 is not always wrong. That's incorrect, it's not true, it's not a truism, the entire chain of inference to arrive at 2 + 2 = 5 contains a flaw, and he might agree with that. About 2 q^2 = p^2, (sqrt 2) q = p, so one or the other of p and q is irrational if either one is a natural finite integer. If we get off into some nonstandard model of the natural integers for the consideration of, say, induction in the limit and what that means, then it might imply the compactification or existence of the infinite element and thus the cumulative hierarchy and thus the construction of the number system, but that machinery is not necessary to illustrate that sqrt 2 and its reciprocal are algebraic irrationals, and the product of a finite natural integer, a natural integer for most intents and purposes, and an algebraic irrational is an algebraic irrational. The mathematical question is answerable. Anyways, I don't speak for Tim, he'll say what he wants. Part of his point is that he does. Ross -- Ross A. Finlayson === Subject: Re: Han's startling new set theory. > The fact that sqrt(2) is irrational is a mathematical statement, and > doing an empirical investigation to attempt to falsify it is just > silly - like doing a search for a four-sided triangle. It's not an attempt to falsify the mathematical statement: it's an attempt to falsify the *applicability* of that statement to the real world. For example, that you will never find an object modelled by a rational that yields an object modelled by 2 when an operation modelled by squaring is applied to it. You only need to look at the history of mathematics to find systems that did not model what they were supposed to model. Nor do you need to search hard for systems that turned out to be inconsistent and hence modelled nothing at all. To believe that today's mathematics fails in neither way would be a tenet of faith. > Han has a point - some bits of mathematics have engineering > consequences in the real world, some don't at present. But > predicting what branches of maths will never turn out to have real > world consequences is a tricky business There is no need for prediction. Some do have consequences at present, and they suffice as objects of this discussion. - Tim === Subject: Re: Algebraic closure of transcendental extensions and roots > given a simple transcendental extension L(x)/L, how do we construct the > algebraic closure of L(x)? How do the roots behave? In the complex > numbers, any nonzero element has n distinct n-th roots. Does the same > hold for the algebraic closure of L(x)? In particular, does anyone know > of an effective algorithm for computing the distinct n-th roots of the > transcendental element x? > DMW Either I haven't understood your question or you don't understand what is going on. Is L a subfield of the complex field C and x some element of C that is transcendental over C? In that case, you can compute an nth root of x by writing x in polar form as r(cos t + i sin t) and then an nth root is r^{1/n}(cos t/n + i sin t/n) and an nth root of r using, say logarithms. Multiplying by nth roots of 1 give all the others. On the other hand, if L is an abstract field and x a transcendental, then it is meaningless to ask for an effective algorithm, since x is not given in any effective sense. As for the algebraic closure, well, you just adjoin roots of irreducible polynomials and then more such roots, until there are no more. It is inevitably an infinite process. There is one case in which there is an effective procedure for describing all the elements of an algebraic closure. Begin with an ordered field that has has a decidable order relation. The rationals do, the reals don't, for example. The field of algebraic numbers do, although that is far from obvious. Add a transcendental along with an effective procedure for deciding where it goes in the order. This can be done with pi and e, for example. At any rate, there is an effective procedure for describing the real closure of a decidably ordered field. The argument is long and tedious, but not truly difficult. Now adjoin a square root of -1 and you get an algebraic closure of the original field. Incidentally, the real closure is decidably ordered as well. === Subject: Re: 0.999... = 1? (I know, a beaten dead horse) > > This is hard for me to explain because I don't know whether I am using > > the wrong terminology or the wrong concept or both. But what I'm > > getting at is that I can see, for example, that the limit of 1/x as x > > approaches 0 is infinity (although I also recall reading that 1/0 is > > undefined), > It is true that > the limit of 1/x as x approaches 0 is infinity . . . . (*) > so long as x is greater than 0. Perhaps unfortunately, you didn't mention compactifications of the reals until later, in the last paragraph of yours which I quoted. So it seems that your comments here were intended to be taken in the context of the reals, rather than in some extension of them. In that context, your comments here are correct. But since you did later mention the extensions, it might be worthwhile to re-examine your earlier comments in terms of the extensions. The one-point extension of the reals is R U {oo}, where oo is unsigned infinity. The two-point extension of the reals is [-oo, +oo], where -oo and +oo are signed infinities. (BTW, the infinities in these extensions are not merely fa.8dons de parler. They are actual elements of these systems.) In [-oo, +oo]: The limit of 1/x as x approaches 0 from the right is +oo, and from the left is -oo. Thus the bilateral limit does not exist. Also note that 1/0 is undefined. In R U {oo}: The limit of 1/x as x approaches 0 (bilaterally) is oo. Also note that 1/0 is defined: 1/0 = oo. The function 1/x is continuous at x = 0. David W. Cantrell > (If x tends to 0 through negative > values the limit is negative infinity. It might be better to write > positive infinity in (*)) But when (*) is written out in full using > the official definitions, the word infinity (or the symbol one is > using for it) disappears. That is to say, there is no _number_ infinity > that (*) is referring to--it is a fa.8don de parler. (My apologies if the > wotsit under the c doesn't come out right.) Not only isn't there a > _number_ infinity that (*) is referring to, there isn't any other kind > of entity infinity that it is referring to either. > It is also true that 1/0 is undefined. There is no contradiction here > because (*) doesn't involve x being 0, so it has nothing to do with > 1/0. Generally limit of thing as x tends to A... doesn't involve x > being equal to A in thing. For example > limit of sin(x)/x as x tends to 0 is 1 > is true and useful but sin(0)/0 is meaningless. > > but -- unless I am wrong -- infinity cannot be used in a > > direct calculation because it is not a number. > Just to confuse you, the real numbers can be compactified in two ways; > one involving a number infinity, and one involving two numbers: positive > infinity and negative infinity. Calculating with them can be handy, for > example in the theory of integration and measure. Probably best to > forget about that until the need arises. Those (three) infinities are > not real numbers, they are additions which can enter into some limited > arithmetic. === Subject: Re: 0.999... = 1? (I know, a beaten dead horse) > I know that the typical rule is not to respond to abusive people such > as yourself, but as they say, sometimes it's necessary to break the > rules. Don't take it personally. Every newsgroup has its resident asshole(s) and you've just run afoul of one. -js === Subject: Re: 0.999... = 1? (I know, a beaten dead horse) Jim, Shepherdmoon > > > > If he means key into a calculator > > the entire sequence of 0.999... because it is infinite, so the actual > > number of 9s used is always finite. > True, but there is a standard way of writing 0.999..., it is to write > 0.9 with a dot ever the 9, and it is read Nought point nine > recurring. With something like 0.8427427427..., one would write 0.8427 > with a line over the 427. So notations get invented so that you can > write down what you can't write down, if you see what I mean. > > This is hard for me to explain because I don't know whether I am using > > the wrong terminology or the wrong concept or both. But what I'm > > getting at is that I can see, for example, that the limit of 1/x as x > > approaches 0 is infinity (although I also recall reading that 1/0 is > > undefined), > It is true that > the limit of 1/x as x approaches 0 is infinity . . . . (*) > so long as x is greater than 0. (If x tends to 0 through negative > values the limit is negative infinity. It might be better to write > positive infinity in (*)) But when (*) is written out in full using > the official definitions, the word infinity (or the symbol one is > using for it) disappears. That is to say, there is no number infinity > that (*) is referring to--it is a fa.8don de parler. (My apologies if the > wotsit under the c doesn't come out right.) Not only isn't there a > number infinity that (*) is referring to, there isn't any other kind > of entity infinity that it is referring to either. > It is also true that 1/0 is undefined. There is no contradiction here > because (*) doesn't involve x being 0, so it has nothing to do with > 1/0. Generally limit of thing as x tends to A... doesn't involve x > being equal to A in thing. For example > limit of sin(x)/x as x tends to 0 is 1 > is true and useful but sin(0)/0 is meaningless. > > but -- unless I am wrong -- infinity cannot be used in a > > direct calculation because it is not a number. > Just to confuse you, the real numbers can be compactified in two ways; > one involving a number infinity, and one involving two numbers: positive > infinity and negative infinity. Calculating with them can be handy, for > example in the theory of integration and measure. Probably best to > forget about that until the need arises. Those (three) infinities are > not real numbers, they are additions which can enter into some limited > arithmetic. > > Similarly, I was > > wondering if that was the same concept as approaching the number one > > from the 0 side of the number line, since in both cases something > > infinite is involved. > There's no approaching where 0.999... (infinitely many digits) is > concerned. There is (in a sense that can be made precise) where 0.9, > 0.99, 0.999, ... (each number having a finite number of digits) is > concerned. > > But I guess that's wrong, and it is not the case that 0.999... > > potentially reaches 1 but doesn't ever do so; > One might say that 0.9, 0.99, 0.999, ... (each number having a finite > number of digits) gets close to, but doesn't reach 1. But even that's a > bit unclear because it hints at some temporal process, and there is > none. It can be made clear with limits. > > rather, if I understand > > correctly, it is the case that 0.999... actually IS 1 and both are in > > fact the same number shown two different ways (and not two different > > real numbers). And the infinity of the 9s doesn't affect the fact that > > 0.999... represent a number and not any incalculable infinity. Thus > > both 0.999... and 1 would be marked at the same point on the number > > line, the same way 0.5 and 1/2 would be. > Correct. > > I guess another question I have regarding 0.999... = 1 is, one of the > > proofs of 0.999... = 1 is that there is no number 0.000...001 (a 1 > > after an infinity of 0s), > It doesn't sound a very good proof to me. Best to say that > 0.999... = 9/10 + 9/100 + 9/1000 + ... > (that's true by definition of 0.999...) and then observe that > 9/10 + 9/100 + 9/1000 + ... > is a geometric series that you probably learned to sum at school. > > so what is the smallest number greater than > > 0? I guess there is no smallest positive number, which I think is what > > I learned in school. > That's right, you did. > > That's as far as I understand so far, but I wanted to clarify what I > > was thinking about when I said calculate. > If you want to learn about limits, I'd recommend J C Burkill's A First > Course in Mathematical Analysis Cambridge, which has the advantage of > brevity (as well as clarity, etc, etc). Also, Courant and Robbins's > What is Mathematics? might be helpful for a more chatty discussion. > -- > I don't know who you are Sir, or where you come from, > but you've done me a power of good. === Subject: Re: 0.999... = 1? (I know, a beaten dead horse) > > I know that the typical rule is not to respond to abusive people such > > as yourself, but as they say, sometimes it's necessary to break the > > rules. > > I began my comment with the following: > > --- > > I'm not a math expert but an interested layman, so please keep that in > > mind. > Allow me to quote Robert A. Heinlein > Anyone who can not cope with mathematics is not fully human. > At best he is a tolerable subhuman who has learned to wear shoes, > bathe, and not make messes in the house > You came here claiming to understand math up to the college freshman > level, while it is clear that you do not, despite your initial > disclaimer. I call that arrogance. > One does not need to be a math expert. This is **BASIC** JR. HIGH > SCHOOL > mathematics. > Yet despite your disclaimer, you presumed to be argumentative. > If someone does not understand something they either: > (1) Accept what experts tell them > or > (2) ASK experts to explain. > You did neither. You came here spouting gibberish and being > argumentative. You're lying. I did do (2). See question 4 below. What you perceived as insolent arguments were regarding things I thought I did understand, and even then I was tentative about it. ( And I also don't have any clever objections that I think proves the opposite case. ) If you had read the post with an open mind, you wouldn't have responded the way you did. I refuse to accept what math experts tell me without understanding it. Maybe *you* expect everyone to bow and scrape at your feet when you deliver your pronouncements, but there is a difference between being argumentative and respectfully and tentatively presenting one's current understanding (even if it turns out to be wrong). It is *your* personal problem that you interpreted my post as some kind of affront to your expertise. Go deal with that problem and leave me alone. Others are being a lot more helpful and considerate to me than you are. > > --- > > Again, because I have seen people arguing this point, and because I am > > a layman, I was pointing out that while I don't have any expertise on > > this argument, I DO see the logic of most of the explanations for > > 0.999... = 1, but I still have questions. > But you didn't ASK questions. Instead, you gave arguments. You're lying. I asked these questions: Question 1: --- Wouldn't that violate the definition that any given real number is either greater than or less than any other? --- Question 2: --- But that makes me think, well, is there a discontinuity in what we think of as the real number line? --- Question 3: --- If 0.999... is equal to 1, then what value is given to the point just to the left of 1 on the real number line? --- Question 4: --- Could someone help me understand this? --- So, again, it seems you distort my post to serve whatever your personal purposes are. > > Furthermore, I am not stupid (nor a clod nor a lout), I simply have > > gaps in my knowledge. Unless you are perfect, you have gaps in your > > knowledge too, so we are equal on that score. > But I do not presume to be argumentative about those things I do > not understand. I either accept what experts tell me or I *ask*. Since I wasn't argumentative -- that was your mistaken interpretation -- then this point is moot. > i.e. Please explain why .99999... = 1. > not: > it seems to be that there are discontinuities in the real line > I do NOT argue with someone whose knowledge is superior to mine. > It gets TIRESOME to see the same stupid drivel over and over again > about this subject. This is a *simple* bit of mathematics. It is > *elementary*. One does not need to be a mathematician to understand > this. > It just requires basic intelligence. Yet, the more experts try to > explain > this, the *more* argumentative people who do not understand seem to > become. I DID ask someone to explain it at the end, but I also included things I thought I understood. Honest people do that all the time; you know, including the phrase as I understand it or I thought. It's the chip on your shoulder that caused you to overreact in such an abusive manner and think that I was somehow subverting your mighty math authority. You seem to do it so reflexively that you completely misinterpreted the tone of my post. So I wasn't argumentative to being with, and I didn't become more argumentative (my defending myself against your abusive tone has nothing to do with math arguments). As I said before, please leave me alone. I will discuss this issue with others. Shepherdmoon === Subject: Re: 0.999... = 1? (I know, a beaten dead horse) Iam looking for walter reffick. Put us in touch. He can explain so much. Your doubts are not sensacional. but his explanations gonna shock the world. i invite you to see menorah group mano soler. === Subject: Math help Forum We have started a new Math Help Comunity on the web. To help you with Homework and projects. Please visit www.mathhelpforum.com the membership is free. === Subject: Re: Math help Forum > We have started a new Math Help Comunity on the web. To > help you with Homework and projects. Why not use THIS forum? What feature or improvement do you offer? What level of help are you planning to aim at (we have everything from elementary to phd)? > Please visit www.mathhelpforum.com the membership is free. Killing oneself is free to. Still, not a reason to do it, hehe. :) -- V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sense to be lazy --------------------------------------------------- === Subject: Re: Set Theoretical approach to compression > > Example: suppose L = {1, 2, 3, 4,...} and P(i) = 2^-i. In the > > classical model one could assign codes 0, 10, 110, 1110,... which is > > optimal. How would you choose L'? > Just rember that L is the set of strings over an alphabet SIGMA which > in your case seems to be {0,1,2,3,4,5,6,7,8,9}. Since you are encoding > whole strings of L and not the symbols of SIGMA you might as well > assign codes 1,10,11,100 etc. or to stay with the same alphabet There is a 1-1 mapping between L and the set of all integers, so you could ask the same question about any L with the probability distribution I described. > I will try to give an example of how this approach may open new paths. > Suppose you have some string p which is a substring of PI. If you try > to compress it with the classic approach, you will fail. However, if > you know that p is the substring of PI starting at the x-th position > and having the length l, you might be able to compress p. Of course > this will not work for every such substring, but for those in which it > does, we need to have an explanation. And that explanation is that you > can characterize the subset that contains such a p with a string of > length lesser than the length of p. Every bit string is a substring of the binary encoding of PI (we think). However, on average an encoding of the offset to the first match is going to be longer than the string you are encoding. -- Matt Mahoney === Subject: Re: The First to Present a Complete Mathematical Formulation of Spacetime Finding the invariants of the poincare group is a standard exercise >> If you believe that all the invariants of the Poincar.8e group >> have been found, how many Poincar.8e invariants are there >> and who found the very last one? > 10 and your second question makes no sense. The 10 you > choose depend upon the represention you choose. I don't understand how physicists can be so knowledgeable about the exact processes that brought the universe into existence out of nothingness yet know so little of mathematics as to think that the number of invariants of the Poincar.8e group somehow equals the number of parameters of that group. === Subject: Re: Standard Deviation of PISA <0001HW.BF31E2BA003AA11FF0284550@news.giganews.com> Nobody needs to document how much jews engage in scatology and niggers engage in profane language, because your own mouths are testament enough to your limited intellect. Your jew minds can't engage in any thought other than scatology, like that which fills up the Talmud, because your minds are situated lower than your butt which requires you to look up to and respect it. http://christianparty.net/talmud.htm Your miserable PISA score of 422 and your miserable TIMSS score of 466 simply document something we always suspected--jews are not the brightest candles in the box. http://christianparty.net/pisa.htm http://christianparty.net/timssjews.htm John Knight === Subject: Re: Standard Deviation of PISA <1ldvg1hsebqa4ev01uhju2ggahavbjqmdq@4ax.com> Precisely the point. IF the scoring were linear, and IF there weren't so many noise level questions which even Koko the Gorilla could answer, and IF each racial and sex group were tested separately, then the standard deviation would be MUCH smaller and the vast differences between intellect would be MUCH more apparent. Whether or not you believe it's valid to claim that Koko had an IQ of 85, the simple fact that sub-Saharan niggers were tested by the same criteria and ended up with an IQ of 60 proves that there is a vast difference in intellect between a sub-Saharan nigger who earns $20 per month and a Jap who earns $20,000 per month. It's the use of scaled scores which conceals the REAL intellectual differences. The ONLY way to explain why so many on this forum don't seem to realize these huge differences between the races is that they don't understand how scaled scoring has been used to conceal them. John Knight ps--using just the above two data points, and not giving any credibility whatsoever to jew Wechsler and his IQ test [whose fellow jews in Israel proved with PISA to have room temperature IQs], each one point increase in IQ is equivalent to an extra $400 in earnings per month. === Subject: Re: Standard Deviation of PISA >Precisely the point. You are pointless, except at the top of what passes for a head in the nincompoop species. >IF the scoring were linear, then you still wouldn't know what you were talking about. >and IF there weren't so many noise level >questions which even Koko the Gorilla could answer, then you still wouldn't know what you were talking about. (Let's see some examples of these questions. Then prove that YOU know how to answer them - unless you are dumber than Koko, which is plausible). >and IF each racial and sex group were tested separately, then you still wouldn't know what you were talking about. >then the standard deviation would be MUCH smaller No. It would likely have been larger. >and the vast differences between intellect would be MUCH more apparent. The only vast difference in intellect is between you and everyone else in the discussion. >Whether or not you believe it's valid to claim that Koko had an IQ >of 85, Why would I believe that? Unless it is because your IQ is 84. >It's the use of scaled scores which conceals the REAL intellectual differences. Nonsense. >The ONLY way to explain why so many on this forum don't seem to realize >these huge differences between the races is that they don't understand >how scaled scoring has been used to conceal them. Actually, I suspect that the rest of us understand scaled scoring far better than you do. For one thing, we don't go making nonsensical statements about scores. >ps--using just the above two data points, What two? The one on your head, and some other one? >and not giving any >credibility whatsoever to jew Wechsler and his IQ test [whose fellow >jews in Israel proved with PISA to have room temperature IQs], PISA is not an IQ test, so it proved no such thing. >each one point increase in IQ is equivalent to an extra $400 in earnings >per month. So Robert Johnson, the black billionaire, and his wife Sheila Crump Johnson, also a black billionaire, have IQs of around 200,000? And your IQ must be negative, since you are a net.loser. lojbab -- lojbab lojbab@lojban.org Bob LeChevalier, Founder, The Logical Language Group (Opinions are my own; I do not speak for the organization.) Artificial language Loglan/Lojban: http://www.lojban.org === Subject: Re: Standard Deviation of PISA To claim that the the only thing which 1) makes a difference in the ability to play basketball is height is a provably false thesis which the Olympics PROVED to be false. Why were our 7 foot plus giant billion dollar niggers BEAT so soundly by relative squat butts from Italy and Venezuela and Puerto Rico?: 1) They were too tall? 2) They were too short? 3) Lack of funds? 4) Lack of media coverage? 5) Lack of fan support? 6) Entropy? 7) Discrimination? 8) Simply being niggers? John Knight === Subject: Re: Standard Deviation of PISA >To claim that the the only thing which 1) makes a difference in the >ability to play basketball is height is a provably false thesis which >the Olympics PROVED to be false. Who made such a claim? >Why were our 7 foot plus Which ones were those? No player on the US team was 7 feet tall. Iverson is only 6' tall. >giant billion dollar niggers BEAT so soundly >by relative squat butts from Italy and Venezuela and Puerto Rico?: >1) They were too tall? >2) They were too short? >3) Lack of funds? >4) Lack of media coverage? >5) Lack of fan support? >6) Entropy? >7) Discrimination? >8) Simply being niggers? 9) They had a bad day? Even the best teams have an off day. 10) The team did not have the best possible American players because several top choices turned down slots on the team for a variety of reasons. 11) The other teams had just as competent players as the US did, and probably more motivation. In the case of Puerto Rico, they 3 NBA players on their team as well, including two who WERE more than 7 feet tall (7'3 and 7'1), so your silliness about relative squat butts is just more noisome material from your strange orifice. Their people were taller than ours. Argentina had NBA stars as well, and they weren't short. Manu Ginobili was their top player. He's 6'6. Their second best player in the Olympics was Luis Scola. He's 6'9, but not yet in the NBA. The other two NBA players are 6'6 and 6'7. The other team we lost against was Lithuania in the preliminary round, but we beat them a few days later in the bronze medal round, by a larger margin. Lithuania only had one NBA player at the time. He's 6'9. Another of their players has since joined the NBA. We didn't play against Italy, and Venezuela was not even in the Olympics. So you seem to have pulled more noisome material from your strange orifice. You seem to do that a lot. lojbab -- lojbab lojbab@lojban.org Bob LeChevalier, Founder, The Logical Language Group (Opinions are my own; I do not speak for the organization.) Artificial language Loglan/Lojban: http://www.lojban.org === Subject: Re: Standard Deviation of PISA To claim that the the only thing which 1) makes a difference in the > >ability to play basketball is height is a provably false thesis which > >the Olympics PROVED to be false. > Who made such a claim? > >Why were our 7 foot plus > Which ones were those? No player on the US team was 7 feet tall. > Iverson is only 6' tall. > >giant billion dollar niggers BEAT so soundly > >by relative squat butts from Italy and Venezuela and Puerto Rico?: > >1) They were too tall? > >2) They were too short? > >3) Lack of funds? > >4) Lack of media coverage? > >5) Lack of fan support? > >6) Entropy? > >7) Discrimination? > >8) Simply being niggers? > 9) They had a bad day? Even the best teams have an off day. > 10) The team did not have the best possible American players because > several top choices turned down slots on the team for a variety of > reasons. > 11) The other teams had just as competent players as the US did, and > probably more motivation. > In the case of Puerto Rico, they 3 NBA players on their team as well, > including two who WERE more than 7 feet tall (7'3 and 7'1), so your > silliness about relative squat butts is just more noisome material > from your strange orifice. Their people were taller than ours. > Argentina had NBA stars as well, and they weren't short. Manu > Ginobili was their top player. He's 6'6. Their second best player in > the Olympics was Luis Scola. He's 6'9, but not yet in the NBA. The > other two NBA players are 6'6 and 6'7. > The other team we lost against was Lithuania in the preliminary round, > but we beat them a few days later in the bronze medal round, by a > larger margin. Lithuania only had one NBA player at the time. He's > 6'9. Another of their players has since joined the NBA. > We didn't play against Italy, and Venezuela was not even in the > Olympics. So you seem to have pulled more noisome material from your > strange orifice. You seem to do that a lot. > lojbab > -- > lojbab lojbab@lojban.org > Bob LeChevalier, Founder, The Logical Language Group > (Opinions are my own; I do not speak for the organization.) > Artificial language Loglan/Lojban: http://www.lojban.org Really? http://christianparty.net/basketball.htm http://www.newsitaliapress.it/interna.asp?sez=240&info=91966 Italian papers celebrate shocking win over U.S. Olympic basketball team celebrated a shock victory over the U.S. men's Olympic basketball team, claiming the Dream Team mantle for Italy's players. We are the Dream Team, read the front-page headline of Wednesday's sports daily Gazzetta dello Sport. A magnificent Italy humiliates the USA, the paper said. The Dream Team is blue, said Corriere dello Sport on its front page, referring to the Italian national team's colours. The Martians from the USA are overwhelmed. The defending Olympic champion Americans were routed 95-78 by Italy in Cologne, Germany, on Tuesday - the most one-sided loss ever for a U.S. squad fielding NBA players. The Italians were clear underdogs despite a good record in international play. In a soccer-obsessed country, their performances usually win little attention. Corriere held back from predicting Olympic glory for the Italian team. No, it's not that we will win the Olympics because of this, the paper commented. But Italian basketball has never lived a http://sports.yahoo.com/oly/basketball The so-called Dream Team suffered through a nightmare game in Athens, losing to Puerto Rico by 19 points, the first loss since the U.S started sending NBA stars to the Olympics. The American squad trailed by 22 at halftime and could never make up the gap. === Subject: Re: Standard Deviation of PISA ability to play basketball is height is a provably false thesis which > the Olympics PROVED to be false. I see you ignored the second part, namely that it's the only thing which makes a difference which also has been conclusively related to race. === Subject: Retractions and topological spaces Definition: Let X and Y be topological spaces and Y is a subset of X. We say that Y is a neighbourhood retract of X iff there exists a retraction r: U -> Y for some open set U in X , Y subset of U such that r(y)=y for all y in Y. Theorem: Let A be a subset of R^n. If A is a neighbourhood retract of R^n then A is locally compact. Proof: Suppose A is a neighbourhood retract of R^n then by definition there exists an open set U of R^n such that A is a subset of U and a retraction r: U-> A. It can be shown that A is closed in U. Therefore A = F intersection U for some closed set F in R^n. Therefore A is the intersection of a closed set and an open set in R^n. Ok so far I follow the above argument. The part I don't get is the following: Given a point p in A then let K subset U be a compact neighbourhood of p in R^n. Then K intersection A = K intersection F is a compact neighbourhood of p in A. My question: How do you know that you can always find a compact neighbourhood K of p such that K is contained in U ? === Subject: Re: Retractions and topological spaces >Theorem: Let A be a subset of R^n. If A is a neighbourhood >retract of R^n then A is locally compact. >Proof: >Suppose A is a neighbourhood retract of R^n then by definition there >exists an open set U of R^n such that [...] >Given a point p in A then let K subset U be a compact neighbourhood >of p in R^n. [...] >My question: How do you know that you can always find a compact >neighbourhood K of p such that K is contained in U ? In a locally compact Hausdorff space, if U is a neighborhood of p, then there is compact subset K of U whose interior contains p. The proof of this is a liitle involved. However, in Euclidean space, this is easy. If B(p, a) is a subset ot U, then the closure of B(p, a/2) is a compact subset of U. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Retractions and topological spaces <43111D17.3090005@netscape.net> === Subject: ISO algorithm to list all divisors of a number Hi! I'm looking for an algorithm to list all the divisors (not necessarily prime) of an arbitrary positive integer. E.g. for 24, this algorithm should return 1 2 3 4 6 8 12 24 I thought this would be easy to find with Google, but to my surprise I have not been able to find anything. I suppose that my Googling has been exceptionally inept. If anyone can point me in the right direction, I'd be most grateful. kj P.S. Yes I can code a grossly inefficient monstrosity to generate all the divisors, but I'm looking for an algorithm with a clue. I imagine there must be a standard for doing this in the general case. -- NOTE: In my address everything before the first period is backwards; and the last period, and everything after it, should be discarded. === Subject: Re: ISO algorithm to list all divisors of a number >Hi! I'm looking for an algorithm to list all the divisors (not >necessarily prime) of an arbitrary positive integer. E.g. for 24, >this algorithm should return > 1 2 3 4 6 8 12 24 >I thought this would be easy to find with Google, but to my surprise >I have not been able to find anything. I suppose that my Googling >has been exceptionally inept. If anyone can point me in the right >direction, I'd be most grateful. >P.S. Yes I can code a grossly inefficient monstrosity to generate >all the divisors, but I'm looking for an algorithm with a clue. >I imagine there must be a standard for doing this in the general >case. First find the prime factorization of the number. Then use the same prime factorization but allow the exponents to vary (independently) from 0 to the value of the exponent in the prime factorization. For example, to find the divisors of 24, first obtain the prime factorization: 24 = 2^3*3 = 2^3*3^1 Then the positive integer divisors of 24 are precisely the numbers of the form: x=2^a*3^b where 0<=a<=3 and 0<=b<=1. To see this explicitly: 1=2^0*3^0 2=2^1*3^0 3=2^0*3^1 4=2^2*3^0 6=2^1*3^1 8=2^3*3^0 12=2^2*3^1 24=2^2*3^1 So once you have the prime factorization, generating the list of divisors is easy. quasi === Subject: Retractions and closed sets Let X and Y be topological spaces. Let Y be a retract of X. i.e Y is a subset of X and there is a continuous map r: X-> Y such that r(y)=y. The author claims that if Y is a retract of X then Y is closed in X. Proof: Y = {x in X : r(x) = x } = { x in X : r(x)=I_d_X (x) } subset of X Where Id(X) denotes the identity map : X-> X. But don't you need to assume Y is Hausdorff in this case? === Subject: Re: Retractions and closed sets > Let X and Y be topological spaces. Let Y be a retract of X. > i.e Y is a subset of X and there is a continuous map r: X-> Y > such that r(y)=y. > The author claims that if Y is a retract of X then Y is closed in > X. > Proof: > Y = {x in X : r(x) = x } = { x in X : r(x)=I_d_X (x) } subset of X > Where Id(X) denotes the identity map : X-> X. > But don't you need to assume Y is Hausdorff in this case? Clearly some sort of separation axiom is needed. Consider X = {a,b} with the trivial topology, Y = {b}. Now the constant map r(x) = b is continuous, as the pre-image of the only open set in X is indeed open, but Y is not closed. I don't know if you need X Hausdorff. Seems like points are closed would be enough, but my brain is not focusing properly... send coffee !! === Subject: Re: Retractions and closed sets Sorry, I meant X Hausdorff. === Subject: Re: Retractions and closed sets > Sorry, I meant X Hausdorff. That's the way Munkres gives this, as an exercise: TOPOLOGY, A First Course Sec. 4-3 Exer. 7(a) Show that is Z is Hausdorff and Y is a retract of Z, the Y is closed in Z.