mm-2619 === Subject: Re: Standard Deviation of PISA >>Same for niggers and jews. >> There was a separate amendment specifically giving blacks the right to >> vote. Jews were recognized as citizens and voters from the founding. >> The 14th provided for *citizenship* and the privileges and immunities >> of *citizens* and due process for all *persons*. Nothing in the 14th >> excludes blacks or Jews (or women) from these explicit statements. >> You lose again, nincompoop. >How did our Founding Fathers feel about who can become a citizen of >America? Let's take a look: >Act of March 26, 1790 (1 Stat 103-104) >That any alien, ^^^^^ >being a free white person, who shall have resided >within the limits and under the jurisdiction of the United States for >the term of two years, may be admitted to become a citizen thereof, on >application to any common law court of record, in any one of the States >wherein he shall have resided for the term of one year at least, and >making proof to the satisfaction of such court, that he is a person of >good character... [end of quote]. That was the *naturalization* law for *aliens* who wanted to become citizens. Those who were born here weren't aliens, and thus were citizens without need of such a law. In any event, Jews were considered white under the law, and had no trouble being naturalized. But the ones who were born here were citizens even without the 1790 act. The first Jewish Congressman was elected in 1844, well before the 14th amendment. >Is a nigger a free white person? The ones who were here weren't aliens. >No, of course not. Is a jew a free white person? Yes. >Not on your life. By their own admission they're one quarter nigger, Where do you find such a silly admission? In your strange orifice, like most such things, I presume. >ACCORDING TO THE STANDARD JEWISH ENCYCLOPEDIA Which standard Jewish encyclopedia? There is no such thing. You might be referring to the 100 year old Jewish Encyclopedia, which is reasonably authoritative, but is still 100 years old and hence prone to error on matters which modern scholarship has added new information. >96% OF ALL THE JEWS KNOWN >TO THE WORLD TODAY ARE THE DESCENDANTS OF THE KHAZAR TRIBES OF RUSSIA, >EASTERN EUROPE AND WESTERN MONGOLIA http://en.wikipedia.org/wiki/Khazaria THESE ARE THE ASKNAZI JEWS, No such word. Perhaps you mean Ashkenazi, in which case you merely have multiple errors in your statement, as indicated by the above and the following: http://en.wikipedia.org/wiki/Ashkenazi THE >OTHER MAJOR SECT OF THE JEWS ARE THE SEPHARDIC JEWS, AND THEY ARE A >PERIZZITES, HIVITES, JEBUSITES, GIRGAES, KENITES, EDOMITES AND SOME >TRUE ISRAELITES. No. The Sephardic Jews are a mixing of the same Middle Eastern Jews with the indigenous people of Spain and Portugal, and with the Moorish/Arabic invaders, just as most other Spanish and Portuguese people are. There are recorded conversions from Christianity to Judaism and Judaism to Christianity, and thus lots of intermingling of peoples in terms of race. >THE JEWS HAVE NEVER BEEN ISREALITES; You haven't a clue. >THEY ARE NOT ISRAELITES NOW; AND THEY WILL NEVER BE ISRAELITES [read: Whites]. I don't accept silly nincompoop readings. >It wasn't until the beginning of the Twentieth Century that jews were >even allowed to be buried here, a mistake of monumental proportions. Pure nonsense. Here are just a few of the pre-20th century Jewish cemeteries in one city - Philadelphia. There were about 15-20 such cemeteries in Philadelphia alone: >Mikveh Israel Cemetery #1 (1740-1848) >Historical Site >Spruce St. between 8th and 9th, Philadelphia, PA 19107 Mikveh Israel Cemetery #2 (1842) >1114 Federal St., Philadelphia, PA 19146 1740 was a long time before the 20th century, nincompoop. >Why? Because God HATES their patriarch, Esau, and it's not a good idea >for Israelites to mingle with people God HATES: Tough. They did, which is why there are no longer any Israelites. >Has anything good happened on this continent since we permitted jews to >be buried here? Absolutely not. If you don't like it here, we'll be happy to see you leave. Try the all-white arctic. >Did our Founding Forefathers intend for this to apply to women? Permitting them to be buried here? Certainly. And they were. >One mistake leads to another, Your birth has led to a never-ending series of mistakes. >Do I believe No one cares what you believe. >the poll which now shows 90% want niggers to be >repatriated to Africa [and this means jews too] is accurate? It is meaningless whether it is accurate or not, since it reports only the opinions of self-selected people, most of them nincompoops like you. In any event, the blacks are still here, and it has been 11 days since December 6 with nary a single one being exiled. You lose again, nincompoop. lojbab === Subject: Re: cosh <171220051309055261%bruck@math.usc.edu> > tanh(x) = 0.5 . May be it is possible by using geometric interpretation > of hyperbolic functions demonstrated on an equilateral hypebola. > Presumably you mean to solve it? But this is obvious! Parametric equation of rectangular ( not equilateral ) hyperbola is x = cosh(t) and y = sinh (t) and slope of radius vector is tanh(t). Now t is also the area enclosed between hyperbola radius vectors at t and - t. Let line y = x/2 cut the hyperbola at some t. So the area and slope should be numerically equal at the solution point. :) === Subject: Re: proof or disproof of Riem geom + Loba geom = Eucl geom > Au contraire, it satisfies the properties of a field; see > http://mathworld.wolfram.com/Field.html > (More importantly: If I'm wrong, then why are you the only one pointing > it out? I should be swamped with replies saying that I was wrong. Also, > there are three good reasons why this construction isn't used in > practice.) > Suppose it was a Field, > Chris, then we can mapp the Earth geography globe with your invF above. > We can take every point on the globe of Earth and assign it a Real > number as per your above invF. Thus the North Pole would have a invF > and thus a Real number, ditto South Pole and ditto a city center such > as say Denver. Now, apply the associative and distributive Field law to > that of North Pole, South Pole and city center Denver. You see Chris, > your invF contrivance does not work. > Okay; let a = F(North Pole), b = F (South Pole), and c = F (Denver). > Then > North Pole plus South Pole = invF(F(North Pole) + F(South Pole)) > = invF (a + b), > South Pole plus Denver = invF(b + c), similarly, and > (North Pole plus South Pole) plus Denver > = invF (F(North Pole plus South Pole) + F(Denver)) [definition of > plus] > = invF (F(invF(a + b)) + c) [definition of plus, and c] > = invF ((a + b) + c) = invF (a + (b + c)) [associativity of real > numbers] > = invF (a + F(invF(b + c))) [F and invF are inverses of each > other] > = invF (F(North Pole) + F(South Pole plus Denver)) > = North Pole plus (South Pole plus Denver) > Therefore, the associative law does work. The distributive law works in > a similar way; convert Not really, please give the invF with actual numbers using the coordinate system of latitude and longitude as outlined below. > North Pole times (South Pole plus Denver) to > a * (b + c), then to > a * b + a * c, then to > (North Pole times South Pole) plus (North Pole times Denver). > BTW, this is what a real proof looks like, provided you're still here. > If Chris's contrivance worked or if there exists any such contrivance > that makes the points on a sphere into a Field, then we would have > replaced our global marking of latitude and longitude lines of > reference by such a contrivance. The reason we have never replaced it > is testimony to the fact that none exists. > No, that's not the reason. This construction works, but it's not > convenient, for several reasons. I will list three of them, the first > two being mathematical. > The first real reason is that there is no continuity in the function F. > The South Pole might be 0, the North Pole 1, Denver 1/2, Tempe 1/4, > Washington DC 1/8, etc., where all positive numbers represent points > north of the equator. Then invF(0) is the South Pole, but invF(1/2^n) > is a sequence of points north of the equator, which cannot converge to > the South Pole. > (See http://mathworld.wolfram.com/ConvergentSequence.html ) > The second real reason is that betweenness is violated; if city B is > between A and C, then it is not the case that F(B) is between F(A) and > F(C). > The third real reason is because the whole world does not consist of > mathematicians, and people would object to saying I lived in 1/3 until > last year, then I moved to -4. > The moment that Chris or anyone else who thinks they have a Field for > the points on a sphere by the Reals will then replace our longitude and > latitude reference system with that of Reals marking each and every > point on the globe and where you can do associative and distributive > laws on city center Denver and other points on the globe. > Do you really think that people around the world would give up names of > cities, latitude, and longitude? The USA can't even convert to metric, > for Cthulhu's sake! > --- Christopher Heckman I am sorry, I did not pick the best example to show that invF does not work. Since NorthPole, Denver and SouthPole do not form a huge triangular shape but lie on the same great circle. Even so, if Chris were to actually throw in numbers for Denver, Northpole and Southpole that it will not work. If I pick San Francisco, Moscow and Melbourne does your invF still hold up for associative and distributive and commutative. Now San Francisco is about 40 N-L by 120 W-Long Moscow is about 60 N-Lat by 40 E- Long Melbourne is about 40 S-Lat by 140 E-Long Now let us just use that grid system and say that those cities are exactly at those locations. Chris, let us further codify those latitude and longitude into a shorthand number coordinates such as perhaps 40.120 for San Francisco and -40.140 for Melbourne since it is in the southern hemisphere. And codify East and West and North and South. I am looking for a shorthand since the range is from -180 to +180 for north and south and -180 to +180 for east and west but let us not get confused and thus use 0 to 360. Perhaps use 0 to 360 for both longitude and latitude so that Moscow is 60 lat by 320 long and thus codified into 60.320 or codified into a shorthand of 60,320 where the maximum number is 360,360. Just some practical shorthand to be able to add and multiply longitude and latitude quickly. Now, would that codified system obey the Associative and Distributive and Commutative laws? The longitude and latitude coordinate system obviously encompass every point on the sphere surface. Chris, now where does the invF connect with each longitude and latitude point. What number does the invF ascribe to SanFrancisco, Moscow and Melbourne? What I need to show is that starting at San Francisco and then multiplying Moscow and adding a multiplied San Francisco to Melbourne is not the same as starting at San Francisco and then adding Moscow to Melbourne and then multiplying lead to different results. So please tell, Chris, show me where the invF corresponds one to one with the longitude latitude grid system. Basically, you have to have some numbers for letters of a, b, c and thus your above is not any sort of proof but just a I say so. Basically, Chris , if your invF really works, then it is translatable to that of the longitude and latitude grid system where you replace each point on the Earth geography such as San Francisco, Moscow, and Melbourne with actual numbers and let us see whether commutative, associative and distributive are upheld. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: proof or disproof of Riem geom + Loba geom = Eucl geom Now I suspect, Chris that the reason you feel and the reason that the authors of the MathWorld on this topic feel that invF works when it really does not work is because invF does not define multiplication or addition with strictly the surface of the sphere but as in Dik's example of a line not on the circle where it intersects a point of the circle and projected onto the Eucl plane or line. I think that a valid Field for the surface of the sphere has to be a operation that is confined strictly to the surface of the sphere and cannot be a projection from outside the sphere. Because like in Dik's example of the punctured sphere projected onto a Eucl plane or like your example of invF, both of these models fail because in essence you are dealing with more than the sphere surface itself. To be a legitimate Field for the points on a sphere surface, then the model has to be strictly defined on that surface and not some razzamatazz above or beyond the sphere surface itself. Because the inability to put in a one to one correspondence of invF with longitude and latitude coordinate system indicates that invF is more than just the points on a sphere surface. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: proof or disproof of Riem geom + Loba geom = Eucl geom Because where Chris thinks that invF satisfies the associative law for points on the surface of the sphere, is mistaken because invF is more than just the points on the surface of the sphere but a projection that is above and beyond the sphere surface and thus what appears to be associative obeyance is because of the extra baggage of the projection itself that is associative obeyance. That is why the invF is never able to be translated into the coordinates of longitude and latitude because of the extraneous projection. So invF is no longer a model of a sphere surface but a model of a sphere surface plus the appendage of its projections. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: straightening-out the ill-defined Field theory Re: proof or disproof of Riem geom + Loba geom = Eucl geom Okay, belay most of my last 3 posts tonight. I think I found the source of the trouble with Field theory and why I insist that it is only a concept restricted to Euclidean geometry and the Reals and that its application to anything outside of the Reals and Euclidean geometry is a grave mistake. Consider how we form numbers in the first place. We add 1 endlessly to form the Natural Numbers, and in this process of endlessly adding 1 we create a metric or a vector value. A vector has both distance and a direction. That is what is missing in Field and Ring theory and is the reason they are ill-defined. Do the points of Eucl geom obey the commutative and associative and distributive laws of Field theory. Of course because vectors obey those laws in Eucl geom. But for the points of a sphere (model of Riem geom) the points on a sphere do not obey the commutative law because the vectors are ambiguous. The vector on a sphere (our globe) of Moscow and Melbourne using the intersection of greenwich with the equator as point of origin then a vector of Moscow added to a vector of Melbourne can not be uniquely defined because of NonBetweenness. Does the Moscow vector go east to west or west to east. The associative law falls apart badly because multiplying vectors on a sphere surface could be circling the globe many times with different end results. So what needs changing is that Field theory needs to straighten out its rules. Its commutative and associative and distributive rules need to say that they are commutative vector rules and associative vector and distributive vector and if they obey those vector rules then a Field. That would mean that Riem geometry and the sphere model can not be a Field. It would also mean that Loba geom with its torus model cannot be a Field either. Nor a Ring. The only thing that is a Field and Ring is Euclidean geometry and its Native Numbers of Reals and Complex. Everything else in mathematics is simply not a Field nor Ring. This probably trashcanns all Finite Fields also. As I keep saying that in human history whether mathematics, economics, history, politics, when humanity finds a good thing such as the stock market in the 1990s, they over do the good thing and extend it beyond its capability and we saw the stock market meltdown in the early 2000. Field theory in mathematics was so loved and cherished that it was overextended where Field here and there and everywhere. Even phonyness of finite fields. But now we have a time where we see how overextended Field theory had become and we see how narrow is its application and its range. Field and Ring theory are simply a characteristic of Reals and Complex and Euclidean geometry but that is all. Everywhere else it is out-of-place. Commutative, associative and distributive rules do not apply to points on a sphere nor Riem geom nor Loba geom. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: straightening-out the ill-defined Field theory Re: proof or disproof of Riem geom + Loba geom = Eucl geom > Okay, belay most of my last 3 posts tonight. Hey, do you still in buckets and save it up to spread around your farm? LOL. === Subject: Proof that Field theory can never apply to Riem or Loba geom and that it is confined to only Reals and Eucl geom Re: Dik with the projection of a punctured sphere onto the Eucl plane is not a Field but a 1-1 correspondence. Chris with his invF by MathWorld is not a Field on the sphere but a 1-1 correspondence with a subset of Reals. Here is a proof that there can never be a Field or Ring made of the points on a sphere surface nor can the points of Riem or Loba geometries be a Field or a Ring. Proof: what destroys the Field and Ring is the Commutative Law, and I do not even need the Associative or Distributive. The Commutative Law rests on vectors of a distance and direction and this is missing in the Field theory in that the Commutative, Associative and Distributive are actually vectors. That makes present day Field and Ring theory ill-defined. But the proof relies on a tug of war between the fact that a sphere and Riem and Loba geometries, unlike Euclid geometry, have no unique point of origin and the fact that vectors cannot be unique to Riem and Loba geometries. On Earth's globe if we take where 0 longitude intersects with the equator and call that the origin then a unique vector for that of Melbourne or Moscow or San Francisco can not be established because you can go either east or west and thus every point has at least two vectors. Unlike Eucl geom where every point is unique because it has a unique origin and which establishes unique vectors to every point. Given any sphere surface there cannot be a unique origin and thus every vector is not unique. Thus the Commutative Law exists only in Eucl geom and not on the surface of a sphere nor Riem nor Loba geometries. Field theory and Ring theory are a characterization of Eucl geometry and is out of place anywhere else. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: inverse and derivative of a function Hello Could some one recommend some good resources with graphs to explain the Derivative of the inverse and inverse of the derivative of a function. === Subject: Re: inverse and derivative of a function Derivative of the inverse --- Mapping (x,y) to (X,Y) by interchanging the axes is easy. Look at the graph on a transparent sheet of plastic from the other side, new x-axis ( X ) running to the right. dy/dx = 1/ ( DY/ DX) For example, y = sin (x) , dy/dx = y' = cos (x) Now express cos (x) in terms of sin (x) and find its reciprocal. cos (x) = sqrt ( 1 - sin (x)^2 ). So Y' = 1/ sqrt (1-X^2) which is the derivative of arcsin(X). The applies to any other function. I dont know what is inverse of a derivative ... guess integral of the function? === Subject: Re: inverse and derivative of a function > Hello > Could some one recommend some good resources with graphs to explain the > Derivative of the inverse and inverse of the derivative of a function. Example. f(x) = x^3 is the function, defined for all real x. Its inverse is the cube root function for real numbers which we can indicate by g(x) = curt(x), sometimes denoted by x^(1/3), which is still valid for all real numbers (not all functions have global inverses, as transpires below). The derivative of the inverse would be given by g'(x) = 1/(3*g(x)*2)) for all real x except x = 0, at which it is undefined. The derivative of f(x) = x^3 is f(x) = 3*x^2, whose domain is still all reals, but since its image is only the non-negative reals, it has no global inverse, but only the inverse h(x) = sqrt(x/3), for x >= 0. All of these should be simple eough for you to graph for yourself. === Subject: Re: inverse and derivative of a function >> Hello >> Could some one recommend some good resources with graphs to explain the >> Derivative of the inverse and inverse of the derivative of a >> function. > Example. f(x) = x^3 is the function, defined for all real x. > Its inverse is the cube root function for real numbers which we can > indicate by g(x) = curt(x), sometimes denoted by x^(1/3), which is still > valid for all real numbers (not all functions have global inverses, as > transpires below). > The derivative of the inverse would be given by g'(x) = 1/(3*g(x)*2)) > for all real x except x = 0, at which it is undefined. > The derivative of f(x) = x^3 is f(x) = 3*x^2, whose domain is still all > reals, but since its image is only the non-negative reals, it has no > global inverse, but only the inverse h(x) = sqrt(x/3), for x >= 0. > All of these should be simple eough for you to graph for yourself. === Subject: Quotient Rings Hi Consider the following factoring: R= Z[X]/((x^2 + 1), (10x - 10)) where ((x^^2 + 1), (10x - 10)) is the ideal generated by these two elements. It folows that in R 20=0 So is this Z_20[i] ? === Subject: Re: Quotient Rings > Hi > Consider the following factoring: > R= Z[X]/((x^2 + 1), (10x - 10)) > where ((x^^2 + 1), (10x - 10)) is the ideal generated by these two elements. > It folows that in R 20=0 > So is this Z_20[i] ? Other posters have shown that R is not Z_20[i]. However, you can show that R is isomorphic to Z[i]/(10i - 10). I'm not sure how clear this is, so I'll go through the argument. First I'm going to state a couple of easy to prove but useful lemmas for these kinds of problems. Lemma 1: If R and S are rings with f: R -> S an isomorphism, and I is an ideal of R, then R/I is isomorphic to S/f(I). (This is very easy to show) Lemma 2: Let R be a ring with x and y elements of R. Then (x + yR) R/yR is isomorphic to (x, y) R/yR. This notation may look funny, but I use it to distinguish where ideals are being generated. So x + yR is just the image of x in R/yR, and (x + yR) is being generated as an ideal in the ring R/yR. Similarly, the ideal (x, y) is the ideal generated in R/yR. (This lemma is also easy to prove) For your ring R, by the third isomorphism theorem we have: R = Z[x]/(x^2 + 1) / (x^2 + 1, 10x - 10)/(x^2 + 1) By Lemma 2, the right hand side of the above isomorphism becomes: RHS = Z[x]/(x^2 + 1) / ((10x - 10) + (x^2 + 1)Z[x]) By evaluating at i, Z[x]/(x^2 + 1) = Z[i]. So by Lemma 1, the RHS of the above ismorphism becomes: RHS = Z[i]/(10i - 10 + (i^2 + 1)Z[x]) = Z[i]/(10i - 10) Mike === Subject: Re: Quotient Rings >> Consider the following factoring: factor-ring (or maybe factorring), or quotient ring. >> R= Z[X]/((x^2 + 1), (10x - 10)) >> where ((x^^2 + 1), (10x - 10)) is the ideal generated by these two elements. >> It folows that in R 20=0 >> So is this Z_20[i] ? >Other posters have shown that R is not Z_20[i]. However, you can show >that R is isomorphic to Z[i]/(10i - 10). Note that 10i-10 = 5 (i-1)^3 up to units, so by the Chinese Remainder Theorem, Z[i]/(10i-10) = Z[i]/5 + Z[i]/(i-1)^3 . Since Z/5 already HAS a square root of -1, Z[i]/5 = (Z/5)+(Z/5). The other factor is a quotient of Z[i]/(i-1)^4 = Z[i]/4 = (Z/4)[i] so you can work out this factor as (Z/4)[i]/(2i-2) = (Z/4)[X]/(X^2=-1, 2X=2) so additively this ring is isomorphic to (Z/4) + (i-1) (Z/2) (i.e., to (Z/4) x (Z/2) ) . In particular the ring has 200 elements, while (Z/20)[i] (which is additively isomorphic to (Z/20)^2 ) has 400 elements. dave === Subject: Re: Quotient Rings >Consider the following factoring: >R= Z[X]/((x^2 + 1), (10x - 10)) >where ((x^^2 + 1), (10x - 10)) is the ideal generated by these two elements. >It folows that in R 20=0 >So is this Z_20[i] ? No, R is not Z_20[i]. In Z[x], let I be the ideal (x^2+1,10x-10) and let J be the ideal (x^2+1,20). Since 20 is in I, it follows that I contains J. Claim 10x-10 is not in the ideal J. Suppose instead 10x-10 is in J. Then a*(x^2+1) + b*20 = 10x-10 for some a,b in Z[x]. Write a=a1+a2 where a1 is odd and a2 is even. Similarly, write b=b1+b2 where b1 is odd and b2 is even. then a2*(x^2+1)+20*b2 = -10 It follows that a2 is a multiple of 10 in Z[x]. Let c2=a2/10. Then c2*(x^2+1)+2*b2* = -1. This would imply that 2 is a unit in Z[i], which is false. Thus, 10*x-10 is not in J, as claimed. It follows that I strictly contains J. But Z[x]/J is clearly a finite ring, hence, since I is a larger ideal than J, Z[x]/I is a smaller ring than Z[x]/J. Therefore the rings Z[x]/J and Z[x]/I are not isomorphic. quasi === Subject: Re: Quotient Rings > Consider the following factoring: > R= Z[X]/((x^2 + 1), (10x - 10)) > where ((x^^2 + 1), (10x - 10)) is the ideal generated by these two elements. Call that I. > It folows that in R 20=0 It does? A comma after R would help. 20 = 10(x^2 + 1) - (10x - 10)(x + 1) in I > So is this Z_20[i] ? Does 10i - 10 = 0 ? === Subject: Number theory Just check if this result is true or not. If p is a prime of the form 4k+3, then 4^p-3 is also a prime. Please reply soon, Sudhanva === Subject: Re: Number theory > Just check if this result is true or not. > If p is a prime of the form 4k+3, then 4^p-3 is also a prime. true for only 3, 7, 11, 47, 347 and perhaps a few more not true for 4^19+3 = 274877906947 = 7^2 * 5609753203 and 4^23+3 = 70368744177667 = 13^2 * 416383101643 and other primes (31,43,...) you can use GP-Pari to check for other values. so the conjecture is obviously false === Subject: Re: Number theory > Just check if this result is true or not. > If p is a prime of the form 4k+3, then 4^p-3 is also a prime. > Please reply soon, > Sudhanva Try p = 19. === Subject: Re: Number theory > Just check if this result is true or not. > If p is a prime of the form 4k+3, then 4^p-3 is also a prime. > Please reply soon, > Sudhanva Counterexample: p=19. Then 11| 4^19 - 3. === Subject: is it true? plz help if integral from -oo to+oo of h(x) is equal to the integral from -oo to +oo of h(x), can I say h(x)=g(x) ? === Subject: Re: is it true? plz help 12/17/2005 at 11:05 PM, VijaKhara@gmail.com said: >if integral from -oo to+oo of h(x) is equal to the integral from -oo >to +oo of h(x), can I say h(x)=g(x) ? What's to stop you? Sure, you can say it, but it won't be true, not even if you fix the typo in your question. Is that a homework question? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: is it true? plz help > if integral from -oo to+oo of h(x) is equal to the integral from -oo to > +oo of h(x), can I say h(x)=g(x) ? No. As an exercise, try to find two different functions whose integrals from -infinity to infinity are both zero. Here's a hint: think about odd functions. === Subject: Re: is it true? plz help >if integral from -oo to+oo of h(x) is equal to the integral from -oo to >+oo of h(x), can I say h(x)=g(x) ? Of course not -- draw a picture. === Subject: Re: is it true? plz help > if integral from -oo to+oo of h(x) is equal to the > integral from -oo to +oo of h(x), can I say h(x)=g(x) ? No, not even if the second ingrand is g(x). Consider h(x) = 1/x^3, g(x) = -h(x). === Subject: Re: is it true? plz help >>if integral from -oo to+oo of h(x) is equal to the >>integral from -oo to +oo of h(x), can I say h(x)=g(x) ? > No, not even if the second ingrand is g(x). > Consider h(x) = 1/x^3, g(x) = -h(x). How about sticking with integrable functions? 1/x^3 is not integrable from -infinity to infinity. === Subject: Re: is it true? plz help <40kmnmF19o69iU3@individual.net> >>if integral from -oo to+oo of h(x) is equal to the >>integral from -oo to +oo of h(x), can I say h(x)=g(x) ? > No, not even if the second ingrand is g(x). > Consider h(x) = 1/x^3, g(x) = -h(x). > How about sticking with integrable functions? 1/x^3 > is not integrable from -infinity to infinity. f_n(x) = 1/x^(2n+1), x not in (-1,1) = 0 otherwise For all n in N, integral(-oo,oo) f_n(x) = 0 = integral(-oo,oo) -f_n(x) === Subject: Re: is it true? plz help >>if integral from -oo to+oo of h(x) is equal to the >>integral from -oo to +oo of h(x), can I say h(x)=g(x) ? >No, not even if the second ingrand is g(x). >Consider h(x) = 1/x^3, g(x) = -h(x). >>How about sticking with integrable functions? 1/x^3 >>is not integrable from -infinity to infinity. > f_n(x) = 1/x^(2n+1), x not in (-1,1) > = 0 otherwise Or (a bit simpler!) f(x) = 1 if 01 === Subject: Re: is it true? plz help <40kmnmF19o69iU3@individual.net> <40l81nF1aljqiU1@individual.net> > >>if integral from -oo to+oo of h(x) is equal to the >>integral from -oo to +oo of h(x), can I say h(x)=g(x) ? > >No, not even if the second ingrand is g(x). >>How about sticking with integrable functions? > f_n(x) = 1/x^(2n+1), x not in (-1,1) > = 0 otherwise > Or (a bit simpler!) > f(x) = 1 if 0 -1 if -1 0 if |x|>1 f_n(x) = nx, x in [-1,1] = 0 otherwise integral(-oo,oo) a(f_n(x))^k dx = 0 for all a in R, odd k in N === Subject: Re: Zuhair's Circles > Hi Zuhair, > Please think about a point and a segment, where the segment is a non-composed element (no sub-elements exist within its domain). > There is no problem to define the accurate location of a point, and I am not talking about metrical space, but about the Belonging concept itself where a point is in {.} XOR out {}. of some set. > But a non-composed segment can be simultaneously in AND out of some set _{_}; therefore its exact membership does not exist. > Now please think about a point and a segment in terms of a metric-space. > In that case only a point has an accurate location (it is a local-element) where a segment does not have an accurate location (it is a non-local element). > Be aware of the fact that in order to define the location of a non-composed segment, you have no choice but to use a point in order to do the job, but also be aware that a point (which is a local-element) is not an inherent part of a non-composed segment, because of at least two more reasons (in addition to the fundamental difference between locality and non-locality): > 1) A non-composed segment has no sub-elements within its domain. > 2) A Direction is an inherent property of a non-composed segment, but it is not an inherent property of a point. > In other words, a point and a segment cannot be defined in terms of each other, because each one of them is a different atomic building-block of Mathematics, or in more general definition, there are at least two fundamental and different concepts in the basis of the mathematical research that cannot be ignored anymore, which are: The Local and the Non-local. > The fundamental mistake of the mathematical research of the past 2500 years since Euclid was to ignore the independent existence of the Non-local from the Local, by define it in terms of the Local. > Please look how Modern Mathematics defines R set and actually defines the Set concept itself only in terms of the Local. > At the moment that you understand that a segment is not made of infinitely many local elements, but it is a Non-local element as well, then and only then you will start to understand my arguments. Fair enough, and what is potential infinity and actual infinity and why do you call them like that , You said actual infinity is nothing less than a non composed line segment , as if you want to say that actual infinity is SOLID (atomic)continous infinity while potential infinity as you say is a non finite collection , and I don't know what do you mean by non finite collection I should guess that the set of all sets is a non finite collection , Also Zuhair's circles is a non finite collection , now these non-finite collections ( if my interpretation of your ideas is true) are not actually infinite because they are formed of parts or elements within its domain, ie they are segregate non finite sets that's why you call them potentially infinite , while solid continous non finite sets are actual infinite. Now let me ask you something? let me take a line segment and consider it a one whole that cannot be described in terms of having parts within it ( what I call a point line ) now suppose that line is 1 cm long , is that line an actual infinity. Now back to Zuhair's circles , since they cannot be defined on the basis of potential infinity , Do the WHOLE of Zuhair's circles represent a defined actual infinity ? Zuhair === Subject: Re: Zuhair's Circles > I should guess that the set of all sets is a non finite collection The term *all* cannot be related to a non-finite collection because a non-finite collection is incomplete (it is no more than a potential infinity) by definition because it cannot reach the completeness of a non-local singleton that cannot be understood as a one of many abstract mathematical element. Therefore the set of *all* sets does not exist. The potential-infinity is based on THE PERMANENT EXISTENCE OF THE NEXT. > ...now suppose that line is 1 cm long , is that line an > actual infinity? Yes, because a non-local singleton is not a one of many element , but at the moment that you look at it as a one of many element (e.g. the line is 1 cm long) you are no longer look at it as an actual-infinity, but you look at it as a member of some collection, and a non-finite collection/sequence is no more than a potential-infinity. Do you get the difference? > Now back to Zuhair's circles , since they cannot be defined on the > basis of potential infinity , Do the WHOLE of Zuhair's > circles represent a defined actual infinity ? You still miss it, because Zuhair's circles are no more than a potential-infinity and they get their identity because of THE PERMANENT EXISTENCE OF THE NEXT, which is an inherent property of potential-infinity. Do you start to get my point of view of the Non-finite? === Subject: Re: Zuhair's Circles <6384527.1134896117197.JavaMail.jakarta@nitrogen.mathforum.org> > I should guess that the set of all sets > is a non finite collection > The term *all* cannot be related to a non-finite collection because a non-finite collection is incomplete (it is no more than a potential infinity) by definition because it cannot reach the completeness of a non-local singleton that cannot be understood as a one of many abstract mathematical element. > Therefore the set of *all* sets does not exist. > The potential-infinity is based on THE PERMANENT EXISTENCE OF THE NEXT. > ...now suppose that line is 1 cm long , is that line an > actual infinity? > Yes, because a non-local singleton is not a one of many element , but at the moment that you look at it as a one of many element (e.g. the line is 1 cm long) you are no longer look at it as an actual-infinity, but you look at it as a member of some collection, and a non-finite collection/sequence is no more than a potential-infinity. > Do you get the difference? > Now back to Zuhair's circles , since they cannot be defined on the > basis of potential infinity , Do the WHOLE of Zuhair's > circles represent a defined actual infinity ? > You still miss it, because Zuhair's circles are no more than a potential-infinity and they get their identity because of THE PERMANENT EXISTENCE OF THE NEXT, which is an inherent property of potential-infinity. > Do you start to get my point of view of the Non-finite? Yes === Subject: Re: Zuhair's Circles In that case dear Zuhair I wish to know YOUR point of view about these ideas, because I think that you have an open mind that can find new things about this subject. Doron === Subject: Re: Zuhair's Circles <28561938.1134903381933.JavaMail.jakarta@nitrogen.mathforum.org> > In that case dear Zuhair I wish to know YOUR point of view about these ideas, because I think that you have an open mind that can find new things about this subject. > Doron I need to read all of your posts before making any opinion, besides I am not a mathematician , and my opinion is not important really. You say I am open minded , I think that this is true since for example I like the jews while I am in reality arab . jews are good people. Zuhair === Subject: Logic question The statement if A then B is considered true when A is impossible. Is this just a convenient definition? I can think of cases where it can be proved that B would not be true were A to be true, but A is never true so .. How does math/logic handle this? -- Peter Fairbrother === Subject: Re: Logic question > The statement if A then B is considered true when A is impossible. Is this > just a convenient definition? One of the policies of my credit union means that If I bounce a check, I will have to pay a penalty. I don't bounce checks. I will never bounce a check. I have never bounced a check. So, do you believe that what I said about the policy is true? -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Logic question >One of the policies of my credit union means that If I bounce a check, >I will have to pay a penalty. For your sake, I hope the condition is iff. === Subject: Re: Logic question >>One of the policies of my credit union means that If I bounce a check, >>I will have to pay a penalty. > For your sake, I hope the condition is iff. It isn't, at least for my credit union. They also issue credit cards, and they charge a penalty for late payments. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Logic question > The statement if A then B is considered true when A is impossible. Is this > just a convenient definition? > I can think of cases where it can be proved that B would not be true were A > to be true, but A is never true so .. > How does math/logic handle this? > -- > Peter Fairbrother Thats a very good question .You might like to look at the work :Logic,Language,and Meaning ,vol 1,Introduction to Logic (and vol 2 :Intensional Logic and Logical Grammar ),University of Chicago Press (paperback) written by Dutch experts under the collective name L.T.F.Gamut.These people are interested in logic for the study of natural language in contrast to mathematical logic and many questions of this type are discussed (leading sometimes to 3 an 4 valued logic).The statement form you have;if A then B, which is to be regarded as true except when A is true and B is false ,is called material implication .By making this most convenient choice (certainly for science and mathematics) is called making it truth functional (the truth or falsity is determined by the truth or falsity of the constituent statements A and B. Gamut introduces material implication on page 33 of volume 1.He says In everyday language ,(if....,then...) can usually not be considered truth functional. They discuss the sentence: If John has just bumped his head then he is now crying . They say in part-but what if John has not bumped his head? One would not wish to say that the sentence must always be false in that case ,but it also does not seem attractive to say that it must be true .Since we have agreed that indicative sentences are either true or false ,let us choose the least unatractive alternative and say that material conditionals(implication) are true if their antecedent (i.e. A) is untrue. They give an illustration from mathematics where the statement really involves quantifiers and is sometimes called formal implication . If a number is larger then 5,then it is larger then 3. (In predicate calculus one would say: for very x (if x is larger than 5 ,then x is larger than 3).This is clearly a true statement.The assertion of the truth of a universal statement( that just means ;for every ...) implies the truth of all of its instatiations( choices of x).This is an indisputable principle of reasoning (the implicit meaning of :for every) For example let x be 2 ,we get -if 2 is larger then 5 (which it isn't) then 2 is larger then 3 (which is isn't either) So the truth of this material implication is forced on us by the needs of formal implication. This is a prime example of Vulcan philosophy :the needs of the many outweigh the needs of the few.(thats the part Spock put in) === Subject: Re: Logic question > So the truth of this material implication is > forced on us by the needs of formal implication. I don't see that. I don't think it would be any different if it was considered to be false - but it's a convention, I guess. What I'd like to see is it being considered meaningless, and thus not a wff (in FOL terms). > This is a prime example of Vulcan philosophy :the needs of the many > outweigh the needs of the few.(thats the part Spock put in) I think Spock changed his mind about that as an absolute after he was re-thingied - born? incarnated? married? :) -- Peter Fairbrother === Subject: Re: Logic question > I don't see that. I don't think it would be any different if it was > considered to be false - but it's a convention, I guess. > What I'd like to see is it being considered meaningless, and thus not a wff > (in FOL terms). A bit like dividing by zero in algebra. In fact, quite a lot like that. -- Peter Fairbrother Find me uh cave 'n talk the bears In t' takin' me in Don van Vleit === Subject: Re: Logic question Discussion, linux) >> I don't see that. I don't think it would be any different if it was >> considered to be false - but it's a convention, I guess. >> What I'd like to see is it being considered meaningless, and thus not a wff >> (in FOL terms). > A bit like dividing by zero in algebra. In fact, quite a lot like > that. Well-formedness is supposed to be a purely syntactic property. It shouldn't depend on the meaning (or truth values) of the components. -- I am the barbarian at the gates, raw creative force, willpower, and the will to fight for the truth no matter what, no matter who stands against me, no matter how many of you band [...] together in your weakness to fight against the math. -- James S. Harris === Subject: Re: Logic question > The statement if A then B is considered true when A is impossible. Is this > just a convenient definition? > I can think of cases where it can be proved that B would not be true were A > to be true, but A is never true so .. That translates to ~A => ~B which is equivalent to B => A. If A is impossible then A is false hence A => B. However one cannot infer B from A in this instance since A cannot be asserted therefore one cannot apply modus pones to infer B. Ex falsi quodlibet. Bob Kolker === Subject: Re: Logic question >> The statement if A then B is considered true when A is impossible. Is this >> just a convenient definition? >> I can think of cases where it can be proved that B would not be true were A >> to be true, but A is never true so .. > That translates to ~A => ~B which is equivalent to B => A. > If A is impossible then A is false hence A => B. Why? How does that follow? > However one cannot > infer B from A in this instance since A cannot be asserted therefore one > cannot apply modus pones to infer B. > Ex falsi quodlibet. I know that's said, (well, with falso) but it's not true - you cannot infer or entail anything from a falsehood because, as you say, you cannot use modus ponens to do the inferring or entailing. -- Peter Fairbrother When I see you floating down the gutter I'll buy you a bottle of wine. === Subject: Re: Logic question >The statement if A then B is considered true when A is impossible. Is this >just a convenient definition? Suppose if A, then B. This means if A is true, then B is true. It says nothing about what B should be if A is false. This means, that the statement is valid if A is false, no matter what B is. Consider the statement A or B which means either A is true or B is true. Another way to say this is if A is false, then B is true; that is, if not A, then B. So A or B is the same as if not A, then B. Now A or B is true when A is true no matter what B is. Therefore, if not A, then B is true when not A is false, no matter what B is. >I can think of cases where it can be proved that B would not be true were A >to be true, but A is never true so .. >How does math/logic handle this? Your first statement says if A, then not B. If A cannot be true, the statement says nothing about the truth of B. What's to handle? Rob Johnson take out the trash before replying === Subject: Re: Logic question >> The statement if A then B is considered true when A is impossible. Is this >> just a convenient definition? > Suppose if A, then B. This means if A is true, then B is true. It > says nothing about what B should be if A is false. This means, that > the statement is valid if A is false, no matter what B is. I didn't follow that last step - how is it valid? > Consider the statement A or B which means either A is true or B is > true. Another way to say this is if A is false, then B is true; Itym A xor B > that is, if not A, then B. So A or B is the same as if not A, > then B. Now A or B is true when A is true no matter what B is. Nope - you have already used or in the xor sense. If A and B are true then the statement A xor B is not true. -- Peter Fairbrother === Subject: Re: Logic question > The statement if A then B is considered true when A is impossible. Is this > just a convenient definition? >> Suppose if A, then B. This means if A is true, then B is true. It >> says nothing about what B should be if A is false. This means, that >> the statement is valid if A is false, no matter what B is. >I didn't follow that last step - how is it valid? The statement if A, then B says that if A is true, so is B. If A is false, nothing is to be concluded, so the statement is valid; that is, it will not lead to an incorrect conclusion as it says nothing in that case. >> Consider the statement A or B which means either A is true or B is >> true. Another way to say this is if A is false, then B is true; >Itym A xor B Typo, I had meant to delete the either, but forgot. A or B means A is true or B is true, of course. >> that is, if not A, then B. So A or B is the same as if not A, >> then B. Now A or B is true when A is true no matter what B is. >Nope - you have already used or in the xor sense. If A and B are true then >the statement A xor B is not true. and only if B. Rob Johnson take out the trash before replying === Subject: Re: Logic question >> The statement if A then B is considered true when A is impossible. Is >> this >> just a convenient definition? > > Suppose if A, then B. This means if A is true, then B is true. It > says nothing about what B should be if A is false. This means, that > the statement is valid if A is false, no matter what B is. >> I didn't follow that last step - how is it valid? > The statement if A, then B says that if A is true, so is B. If A is > false, nothing is to be concluded, so the statement is valid; I don't follow that last - how can a statement that says nothing be true (or false for that matter). -- Peter Fairbrother === Subject: Re: Logic question > > The statement if A then B is considered true when A is impossible. Is > this > just a convenient definition? >> >> Suppose if A, then B. This means if A is true, then B is true. It >> says nothing about what B should be if A is false. This means, that >> the statement is valid if A is false, no matter what B is. > > I didn't follow that last step - how is it valid? >> The statement if A, then B says that if A is true, so is B. If A is >> false, nothing is to be concluded, so the statement is valid; > I don't follow that last - how can a statement that says nothing be true (or > false for that matter). The statement if A, then B means that A cannot happen without B. If A cannot happen at all, then it is certainly true that A cannot happen without B, and therefore the statement is satisfied. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Logic question > The statement if A, then B means that A cannot happen without B. It actually says if A is true then B is true as a consequence, ie B will be true because A is true. > If A cannot happen at all, then There are no consequences. > it is certainly true that A cannot happen > without B, and therefore the statement is satisfied. Nice try though!!! -- Peter Fairbrother On Earth the lords of the Instrumentality met. The presiding chairman looked about and said, Well, gentlemen, all of us have been bribed by Raumsog. We have all been paid off individually. I myself received six ounces of stroon in pure form. Will the rest of you show better bargains? Around the room, the councilors announced the amount of their bribes. The chairman turned to the secretary. Enter the bribes in the record and then mark the record off-the-record. The others nodded gravely. Golden The Ship Was - Oh! Oh! Oh!:1-1 Paul Linebarger === Subject: Re: Logic question >> The statement if A, then B means that A cannot happen without B. > It actually says if A is true then B is true as a consequence, ie B will be > true because A is true. No, the statement if A, then B does not mention anything at all about consequences. It simply says that A can't happen without B. That's all. Consider the statement, If it's Tuesday, it must be Belgium. Surely Tuesday does not cause Belgium. >> If A cannot happen at all, then > There are no consequences. No consequences are required for the statement to be satisfied. >> it is certainly true that A cannot happen >> without B, and therefore the statement is satisfied. > Nice try though!!! Likewise. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Logic question Discussion, linux) <20051218.003943@whim.org> >> The statement if A, then B means that A cannot happen without B. > It actually says if A is true then B is true as a consequence, ie B will be > true because A is true. That's strange. If A then B means if A is true then B is true as a consequence which means if A is true then (B is true as a consequence) as a consequence which means if A is true then ((B is true as a consequence) as a consequence) as a consequence.... What an interesting explanation. It's a bit unusual when the meaning of a phrase Phi involves a longer phrase involving Phi, but never mind. That might be what you think if A then B means, but it is not the meaning of the propositional connective as used in mathematics and elsewhere. -- Jesse F. Hughes In theory there is no difference between theory and practice. In practice there is. -- Yogi Berra === Subject: Re: Logic question > > The statement if A then B is considered true when A is impossible. Is > this > just a convenient definition? >> >> Suppose if A, then B. This means if A is true, then B is true. It >> says nothing about what B should be if A is false. This means, that >> the statement is valid if A is false, no matter what B is. > > I didn't follow that last step - how is it valid? >> The statement if A, then B says that if A is true, so is B. If A is >> false, nothing is to be concluded, so the statement is valid; >I don't follow that last - how can a statement that says nothing be true (or >false for that matter). This is what is called vacuous truth. If a statement says nothing, then it is valid to use in a proof and will not lead to an incorrect conclusion. See . This is pretty much what the next part of the quote that was not copied said: that is, it will not lead to an incorrect conclusion as it says nothing in that case. Rob Johnson take out the trash before replying === Subject: Re: Logic question > The statement if A then B is considered true when A is impossible. In Propositional Logic, there are no impossible statements. Each statement is either true or false. > Is this just a convenient definition? Well, that can be said about any definition. Do you want to propose another definition? > I can think of cases where it can be proved that B would not be true were A > to be true, but A is never true so .. The word never is strange here. Either A is true or false. To say that A is never true is like saying that 2 is never equal to 3. It's correct, but also rather weird, right?! > How does math/logic handle this? By regarding implication like a kind of contract: If you don't return my car until midnight, you'll have to pay me $100. There will be a contract breach only if you don't return the car until midnight *and* you don't pay me $100. But if you do return the car before midnight then, whatever happens next (that is, whether or not you pay me $100) there is no contract breach. Jose Carlos Santos === Subject: Re: Logic question >> The statement if A then B is considered true when A is impossible. > In Propositional Logic, there are no impossible statements. Each > statement is either true or false. Okay, replace impossible by untrue. I am not up on math terminology. >> Is this just a convenient definition? > Well, that can be said about any definition. Do you want to propose > another definition? >> I can think of cases where it can be proved that B would not be true were A >> to be true, but A is never true so .. > The word never is strange here. Either A is true or false. To say that > A is never true is like saying that 2 is never equal to 3. It's correct, > but also rather weird, right?! I don't see anything weird about it. But let's rephrase my sentence as: I can think of cases where it can be proved that B would not be true were A to be true, but A is false so .. >> How does math/logic handle this? > By regarding implication like a kind of contract: If you don't return > my car until midnight, you'll have to pay me $100. There will be a > contract breach only if you don't return the car until midnight *and* > you don't pay me $100. But if you do return the car before midnight > then, whatever happens next (that is, whether or not you pay me $100) > there is no contract breach. Interesting. More? Implication is not truth? Is the contract completed? -- Peter Fairbrother (btw that's not strictly a contract, though it might be a clause or consideration in one - in law a contract is where person A says I will do this if you, person B, do that in exchange and person B agrees to do that in exchange for person A doing this) === Subject: Re: Logic question > The statement if A then B is considered true when A is impossible. > Is this> just a convenient definition? It can be proved. > I can think of cases where it can be proved that B would not be true were A > to be true, but A is never true so .. > How does math/logic handle this? Assume A is false. Now I want to prove A implies B. First assume A. Now for reducio ad absurdum, assume B false. We now have contradiction, namely A and not A. Thus the assumption not B leads to contradiction and by reduci ad absurdum, B is true. Thus A implies B has been established. Hence you see that not A implies (A implies B) === Subject: Re: Logic question >>The statement if A then B is considered true when A is impossible. >>Is this> just a convenient definition? > It can be proved. How do you proof a definition, since if A then B is the abbreviation for the function NON(A AND NON(B)) in mathematical logics? J. === Subject: Re: Logic question <43A526E9.5020408@web.de> >>The statement if A then B is considered true when A is impossible. >>Is this> just a convenient definition? > It can be proved. > How do you proof a definition, since if A then B is the abbreviation > for the function NON(A AND NON(B)) in mathematical logics? Depends upon your logic. Not all logics use that as definition. Some do just the opposite, defining a & b as ~(a -> ~b) Here in natural deduction without giving reasons and using tab to indicate assumption, proves the equivalence of two expressions. a -> b a & ~b a a -> b b ~b b & ~b ~(a & ~b) (a -> b) -> ~(a & ~b) ~(a & ~b) a ~b a ~b a & ~b ~(a & ~b) ~~b ~~b -> b b a -> b ~(a & ~b) -> (a -> b) (a -> b) <-> ~(a & ~b) === Subject: Re: Logic question >>The statement if A then B is considered true when A is impossible. >>Is this> just a convenient definition? >It can be proved. >>How do you proof a definition, since if A then B is the abbreviation >>for the function NON(A AND NON(B)) in mathematical logics? > Depends upon your logic. Not all logics use that as definition. > Some do just the opposite, defining a & b as ~(a -> ~b) > Here in natural deduction without giving reasons > and using tab to indicate assumption, proves the > equivalence of two expressions. > a -> b > a & ~b > a > a -> b > b > ~b > b & ~b > ~(a & ~b) > (a -> b) -> ~(a & ~b) > ~(a & ~b) > a > ~b > a > ~b > a & ~b > ~(a & ~b) > ~~b > ~~b -> b > b > a -> b > ~(a & ~b) -> (a -> b) > (a -> b) <-> ~(a & ~b) Looks funny to me. I am not sure what you have proven here. As you said, everything depends on the basis you choose. I choose Hilbert's. If I understand you right, then you might deduce Modus Ponens from other axioms. Still looks funny to me! :-) De gustibus non est disputandum. J. === Subject: Re: Logic question <43A526E9.5020408@web.de> <43A55450.5070602@web.de> >> >>How do you proof a definition, since if A then B is the abbreviation >>for the function NON(A AND NON(B)) in mathematical logics? >> > Depends upon your logic. Not all logics use that as definition. > Some do just the opposite, defining a & b as ~(a -> ~b) > Here in natural deduction without giving reasons > and using tab to indicate assumption, proves the > equivalence of two expressions. > a -> b > a & ~b > a > a -> b > b > ~b > b & ~b > ~(a & ~b) > (a -> b) -> ~(a & ~b) > ~(a & ~b) > a > ~b > a > ~b > a & ~b > ~(a & ~b) > ~~b > ~~b -> b > b > a -> b > ~(a & ~b) -> (a -> b) > (a -> b) <-> ~(a & ~b) > Looks funny to me. I am not sure what you have proven here. Consider it to be a tabular form of iterated uses of the deduction > As you said, everything depends on the basis you choose. I choose > Hilbert's. If I understand you right, then you might deduce Modus > Ponens from other axioms. Still looks funny to me! :-) Modus Ponens is a inference rule in most logics. > De gustibus non est disputandum. === Subject: Re: Logic question >The statement if A then B is considered true when A is impossible. >Is this> just a convenient definition? >>It can be proved. > How do you proof a definition, since if A then B is the abbreviation Sorry: How do you *prove* ... > for the function NON(A AND NON(B)) in mathematical logics? > J. === Subject: Re: Logic question >> The statement if A then B is considered true when A is impossible. >> Is this> just a convenient definition? > It can be proved. >> I can think of cases where it can be proved that B would not be true were A >> to be true, but A is never true so .. >> How does math/logic handle this? > Assume A is false. > Now I want to prove A implies B. > First assume A. You can't. It's impossible. If you want to use it's impossibility later in the proof then you cannot assume A here. -- Peter Fairbrother === Subject: Re: Logic question > The statement if A then B is considered true when A is impossible. > Is this> just a convenient definition? > >> It can be proved. > I can think of cases where it can be proved that B would not be true were A > to be true, but A is never true so .. > > How does math/logic handle this? > >> Assume A is false. >> Now I want to prove A implies B. >> First assume A. >You can't. It's impossible. If you want to use it's impossibility later in >the proof then you cannot assume A here. To prove A implies B, you must first show that if A is true, then B is true. This means you get to assume A is true. As you point out, this introduces a contradiction. However, this contradiction is just what is used to prove B is true: assume B is false, get contradiction, thus B is true. Of course, by the same method, we could have proven that A implies not B, but this is okay; since A is false, we don't arrive at a contradiction in the end. Of course, I can see how you might find this method of proof, proof by contradiction, unsettling since it uses if A and not A, then B, which is the very syllogism about which you uncertain. Rob Johnson take out the trash before replying === Subject: Re: Logic question > >> The statement if A then B is considered true when A is impossible. >> Is this> just a convenient definition? >> > It can be proved. > >> I can think of cases where it can be proved that B would not be true were A >> to be true, but A is never true so .. >> >> How does math/logic handle this? >> > Assume A is false. > Now I want to prove A implies B. > > First assume A. >> You can't. It's impossible. If you want to use it's impossibility later in >> the proof then you cannot assume A here. > To prove A implies B, you must first show that if A is true, then B is > true. This means you get to assume A is true. As you point out, this > introduces a contradiction. So you can't assume A is true without introducing a contradiction. As you say, it's the same contradiction you use to prove the truth of the statement by contradiction.. So you have assumed a contradiction, and then used it's existence to prove something by contradiction. That doesn't sound like a valid proof to me. In fact I think it even has a name as a logical fallacy, petitio principii. -- Peter Fairbrother === Subject: Re: Logic question > >> > The statement if A then B is considered true when A is impossible. > Is this> just a convenient definition? > >> It can be proved. >> > I can think of cases where it can be proved that B would not be true were A > to be true, but A is never true so .. > > How does math/logic handle this? > >> Assume A is false. >> Now I want to prove A implies B. >> >> First assume A. > > You can't. It's impossible. If you want to use it's impossibility later in > the proof then you cannot assume A here. >> To prove A implies B, you must first show that if A is true, then B is >> true. This means you get to assume A is true. As you point out, this >> introduces a contradiction. >So you can't assume A is true without introducing a contradiction. As you >say, it's the same contradiction you use to prove the truth of the statement >by contradiction.. >So you have assumed a contradiction, and then used it's existence to prove >something by contradiction. That doesn't sound like a valid proof to me. In >fact I think it even has a name as a logical fallacy, petitio principii. In any proof by contradiction, something is assumed that leads to a contradiction. That is how proof by contradiction works. If you are not comfortable with that, then you are not alone. There is a group of people, called intuitionists, who question proofs by contradiction. See . Rob Johnson take out the trash before replying === Subject: Re: Logic question >> > >> The statement if A then B is considered true when A is impossible. >> Is this> just a convenient definition? >> > It can be proved. > >> I can think of cases where it can be proved that B would not be true were >> A >> to be true, but A is never true so .. >> >> How does math/logic handle this? >> > Assume A is false. > Now I want to prove A implies B. > > First assume A. >> >> You can't. It's impossible. If you want to use it's impossibility later in >> the proof then you cannot assume A here. > > To prove A implies B, you must first show that if A is true, then B is > true. This means you get to assume A is true. As you point out, this > introduces a contradiction. >> So you can't assume A is true without introducing a contradiction. As you >> say, it's the same contradiction you use to prove the truth of the statement >> by contradiction.. >> So you have assumed a contradiction, and then used it's existence to prove >> something by contradiction. That doesn't sound like a valid proof to me. In >> fact I think it even has a name as a logical fallacy, petitio principii. > In any proof by contradiction, something is assumed that leads to a > contradiction. That is how proof by contradiction works. If you are > not comfortable with that, then you are not alone. There is a group of > people, called intuitionists, who question proofs by contradiction. I have no real problem with proof by contradiction - just with that particular proof. I didn't quite realise it before, but what it actually proves is that A is false (the assumption is A is true, and the contradiction proves the assumption is false) - when that is a given anyway - but it says nothing about the truth of if A then B when A is false. To say you can use it in a proof and not get an incorect result is also wrong or misleading - you can't _usefully_ use it in a proof, as it says nothing at all. Take it out of the proof and it changes nothing. -- Peter Fairbrother I hear that the emperor of china used to wear iron shoes with ease Incredible String Band === Subject: Re: Logic question > To say you can use it in a proof and not get an incorect result is also > wrong or misleading - you can't _usefully_ use it in a proof, as it says > nothing at all. Take it out of the proof and it changes nothing. It's unclear to me what you have in mind here, but the inference from not-A to if A then B is used in proofs all the time. Disallowing this inference, apart from having no obvious basis, is a problematic matter, since we would then have to somehow restrict one of three other very common forms of inference: Given that B follows from P1,..,Pn,A, conclude that if A then B follows from P1,..,Pn. === Subject: Re: Logic question >> To say you can use it in a proof and not get an incorect result is also >> wrong or misleading - you can't _usefully_ use it in a proof, as it says >> nothing at all. Take it out of the proof and it changes nothing. > It's unclear to me what you have in mind here, but the inference from > not-A to if A then B is used in proofs all the time. It is?? Can you give a real example or three please? Because I don't believe you are correct in saying that. > Disallowing this inference, apart from having no obvious basis, is a > problematic matter, since we would then have to somehow restrict one > of three other very common forms of inference: Eh? Why? > Given that B follows from P1,..,Pn,A, conclude that if A then B > follows from P1,..,Pn. I'm not a logician, and I don't understand that. First, the given is meaningless (if I understand you correctly to mean that the P_i's are statements or conditions all of which must be true for the follows to be true - if A is false, then it doesn't matter whether the follows is a true consequence or not, the inference will never be made). Second, - well, should I bother? That's just plain irrelevant. A or B has little or nothing to do with if A then B. The first implies the second is untrue, the second says nothing about the first. Ditto. -- Peter Fairbrother Corrupt, wise, weary old Earth fought with masked weapons, since only hidden weapons could maintain so ancient a sovereignty - sovereignty which had long since lapsed into a titular paramountcy among the communities of mankind. Earth won and the others lost, because the leaders of Earth never put other considerations ahead of survival. And this time, they thought, they were finally and really threatened. The Raumsog war was never known to the general public except for the revival of wild old legends about golden ships. Golden The Ship Was - Oh! Oh! Oh!:0-2 Paul Linebarger === Subject: Re: Logic question > It is?? Can you give a real example or three please? Because I don't believe > you are correct in saying that. Take an introductory mathematics text and read through the proofs, keeping in mind that a statement like every zero of p(x) is positive is understood, and proved, as for every x, if p(x)=0, x is positive. As for the two rules they are highly relevant, as you can easily demonstrate for yourself. === Subject: Re: Logic question > That's just plain irrelevant. A or B has little or nothing to do with if > A then B. The first implies the second is untrue, Got xor on the brain. Sorry. -- Peter Fairbrother But I think you misunderstand. I am not here to keep darkness out. I'm here to keep it in. Call me ... the Guarding Dark. Imagine how strong I must be. from Thud!, by Terry Pratchett === Subject: Re: Logic question Discussion, linux) <20051218.002031@whim.org> <20051218.061121@whim.org> >>So you can't assume A is true without introducing a contradiction. As you >>say, it's the same contradiction you use to prove the truth of the statement >>by contradiction.. >>So you have assumed a contradiction, and then used it's existence to prove >>something by contradiction. That doesn't sound like a valid proof to me. In >>fact I think it even has a name as a logical fallacy, petitio principii. > In any proof by contradiction, something is assumed that leads to a > contradiction. That is how proof by contradiction works. If you are > not comfortable with that, then you are not alone. There is a group of > people, called intuitionists, who question proofs by contradiction. > See . Far as I can tell, intuitionists are not opposed to all proofs by contradiction. Just those that result in removing (rather than adding) a negation. In other words, the following proof form is intuitionistically valid: Assume P. Derive contradiction. Conclude ~P. (*) This proof form is not intuitionistically valid: Assume ~P. Derive contradiction. Conclude P. (**) Note: *This* form *is* valid. Assume ~P. Derive contradiction. Conclude ~~P. (***) Clearly, (***) is just a special case of (*). Since intuitionists believe it is not the case that ~~P -> P in general, (***) does not imply that (**) is valid. -- Jesse F. Hughes I'm ruler, said Yertle, of all that I see. But I don't see enough. That's the trouble with me. -- Yertle the Turtle, by Dr. Suess === Subject: Re: Logic question > In any proof by contradiction, something is assumed that leads to a > contradiction. That is how proof by contradiction works. If you are > not comfortable with that, then you are not alone. There is a group of > people, called intuitionists, who question proofs by contradiction. Proofs by contradiction are perfectly OK intuitionistically, but not indirect proofs. === Subject: Re: Logic question >>In any proof by contradiction, something is assumed that leads to a >>contradiction. That is how proof by contradiction works. If you are >>not comfortable with that, then you are not alone. There is a group of >>people, called intuitionists, who question proofs by contradiction. > Proofs by contradiction are perfectly OK intuitionistically, but not > indirect proofs. Torkel, what is the difference. If something is assumed hypothetically and it implies a contradiction, doesn't that mean the hypothesis is false? Bob Kolker === Subject: Re: Logic question > what is the difference. There is no generally agreed upon terminology, so proof by contradiction and even indirect proof are widely used both for proofs that derive a contradiction from the assumption A and conclude not-A (constructive reductio) and proofs that derive a contradiction from the assumption not-A and conclude A (non-constructive reductio). === Subject: Re: Logic question >> The statement if A then B is considered true when A is impossible. >> Is this> just a convenient definition? >> > It can be proved. >> I can think of cases where it can be proved that B would not be true were A >> to be true, but A is never true so .. >> >> How does math/logic handle this? >> > Assume A is false. > Now I want to prove A implies B. > First assume A. > You can't. It's impossible. If you want to use it's impossibility later in > the proof then you cannot assume A here. The hell I can't. I can assume anything I want. Like Bush is the best president in all of US history. However you make important point about modal logic where necessary, possible, true and false are four different attributes. In modal logic, A may be false and at the same time possible. For example, even tho it's possible Bush could be intelligent, he isn't. However, as you stated A is impossible, which implies A is false, what I proved is still correct, even within modal logic. For example, since it's not possible for a brain dead president to be intelligent, our brain dead president isn't intelligent. However he did discover life after brain death. === Subject: Re: Logic question > The statement if A then B is considered true when A is impossible. Is > this > just a convenient definition? If A then B is true if A is false. Yes, it is a convenient definition, because it works. > I can think of cases where it can be proved that B would not be true were > to be true, but A is never true so .. You mean like: If I am a unicorn, then you are a zebra. > How does math/logic handle this? Just fine. That is why the definition is set up how it is - to make sure that logic handles all cases in a consistent manner. === Subject: Re: Logic question >> The statement if A then B is considered true when A is impossible. Is >> this >> just a convenient definition? > If A then B is true if A is false. Yes, it is a convenient definition, > because it works. >> I can think of cases where it can be proved that B would not be true were >> A >> to be true, but A is never true so .. > You mean like: > If I am a unicorn, then you are a zebra. Nope, I mean more like If Sheila Monaghan from down the road kissed me then I would have been kissed But there is no Sheila Monaghan from down the road (I sincerely hope!). And how about when B is (also) impossible, like: If she would let me I would kiss Sheila Monaghan from down the road? -- Peter Fairbrother === Subject: Re: I think we don't need the 0 !!! >> The 0 is fat, ugly and useless, cause it creates a >> division by zero >> error in computers all the time. >> >> I think we should abandon the 0 ! > Besides, the nasty old thing was invented by Arabs as a way > confounding western civilization. It screwed up the > calender and Roman Numerals and was rumored to have caused > insanity in several Bohemian prostitutes (they were the > girls who just couldn't say 0). Actually Babylonians, Mayans and Hindus are credited with inventing zero with Babylonians perhaps being the first. 1. http://www-history.mcs.st-andrews.ac.uk/HistTopics/Zero.html 2. http://www.mediatinker.com/blog/archives/008821.html The first rules of arithmetic including the division by zero came from Brahmagupta: http://www-history.mcs.st-andrews.ac.uk/Mathematicians/Brahmagupta.html though he was wrong about the results from division by zero. Jitendra === Subject: Find a Ring Hello to all. I need help with such question: Find a ring R without unity 1, but which has a subring S with unity 1. If I take R=Z*Zwith addition by components (just as for groups) and multiplikation (a,b)(c,d)=(ac,2bd) it will be ring ( Why?). Then there is no identity in the whole ring. If I take S=Z, defined with the elements of type (a,0) is subring of R ( Why? ) And it has one unity (1,0). Why? Can someone explain this to me. Tanja === Subject: Re: Find a Ring > Hello to all. > I need help with such question: > Find a ring R without unity 1, but which has a subring S with unity > 1. > If I take R=Z*Zwith addition by components (just as for groups) and > multiplikation (a,b)(c,d)=(ac,2bd) it will be ring ( Why?). > Then there is no identity in the whole ring. > If I take S=Z, defined with the elements of type (a,0) is subring of R > ( Why? ) > And it has one unity (1,0). Why? > Can someone explain this to me. > Tanja It would help if you said what was giving you trouble because as it is it just seems like you want someone to do your homework for you. Anyway, most of this question is very straightforward. To show R is a ring, just show that it satisfies every part of the definition for being a ring; there's nothing tricky here. Possibly showing that it has no identity is giving you trouble? Suppose that R has an identity (a, b). Then for any (c, d) in R, we have (a, b) * (c, d) = (c, d). But (a, b) * (c, d) = (ac, 2bd). These two facts show that 2bd = d. Can you see the contradiction here? As for S, it is again a very straightforward exercise in checking definitions to show that it is a subring, and it is also clear that (1, 0) is an identity for S. Maybe some of the things I'm saying are obvious aren't obvious to you. If that's the case, you need to be more specific about what's giving you trouble. Mike === Subject: Re: Find a Ring I am only beginner in Abstrakt Algebra and it was very nice of You to help me. Tanja === Subject: Re: Find a Ring days. My association with the Department is that of an alumnus. >Hello to all. >I need help with such question: >Find a ring R without unity 1, but which has a subring S with unity >1. >If I take R=Z*Zwith addition by components (just as for groups) and >multiplikation (a,b)(c,d)=(ac,2bd) it will be ring ( Why?). Presumably, because it satisfies all the axioms. Check them. >Then there is no identity in the whole ring. >If I take S=Z, defined with the elements of type (a,0) is subring of R >( Why? ) Presumably, because it satisfies all the axioms. Check them. >And it has one unity (1,0). Why? Because (1,0)*(a,0) = (a,0), so (i) (1,0) is in S; and (ii) (1,0)*s = s*(1,0)=s for all s in S. Hence, it is a unity of S (by definition of unity). >Can someone explain this to me. Nothing to explain. Check that the things that are supposed to be rings are actually rings, check that the things that are supposed to be unities are unities; and check that no element is a unity of the big ring. HINT: is there an element (x,y) such that (x,y)*(0,1)=(0,1)? If not, then how can R have a unity? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Ring of polynomial I confused , but I need more help. Let R be a ring R[x] of polynomial with real coefficients. Look at Principal ideal in R.. Show that every element in factor ring R/ has one unambiguous define representative of that subset E of R, which consist of those polynomials of R which has degree maximum 1. This cause of bijective function f: E --> R/. Explaine why? I cannot Tanja === Subject: Re: Ring of polynomial > I confused , but I need more help. > Let R be a ring R[x] of polynomial with real coefficients. > Look at Principal ideal in R.. > Show that every element in factor ring R/ has one unambiguous > define representative of that subset E of R, which consist of those > polynomials of R which has degree maximum 1. > This cause of bijective function f: E --> R/. > Explaine why? > I cannot > Tanja I'm not sure how many algebraic results you're familiar with, but R is a field, so R[x] is a PID and therefore has a division algorithm So for any f in R[x], there exists g and r in R[x] such that f = (x^2 + 1)g + r where deg(r) < deg(x^2 + 1) = 2. That is, r is a polynomial of degree 0 or 1. Now, if we look at the image of f in R[x]/(x^2 + 1), it is represented by r since the (x^2 + 1)g terms is zero in this ring. Since the division algorthim gives unique remainders, each element of this ring has a unique representative of degree at most 1. This essentially proves there is a bijective function f: E -> R[x]/(x^2 + 1). Let me point out that there is a more standard way of viewing this ring that may make what's happening in this problem more transparent. For any polynomial f(x) in R[x], we can evaluate f(x) at the imaginary unit i to get f(i) in C, the complex numbers. It's easy to show that evaluation at i gives a surjective homomorphism from R[x] to C. Note that the kernel of this homomorphism is the ideal generated by (x^2 + 1) since R[x] is a PID and x^2 + 1 is the smallest degree polynomial satisfied by i. So R[x]/(x^2 + 1) is just isomorphic to C, and each of those degree at most one representatives above can just be viewed as complex numbers a + bi. Mike === Subject: Re: Ring of polynomial Tanja === Subject: Show that Z_m^* is group Can someone help me? Z m^*={1,2,...,m-1} with bin.8ar operation multiply modulo m. Show that Z m^* is group if and only if m is prime number. Tanja === Subject: Re: Show that Z_m^* is group >Can someone help me? >Z_m^*={1,2,...,m-1} with binar operation multiply modulo m. >Show that Z_m^* is group if and only if m is prime number. Well, Z_m^*, the units modulo m, form a group no matter what integer m is. Z_m, the integers modulo m, form a group if and only if m is a prime number. What rule(s) do the integers modulo 4 not satisfy so that they are not a group under multiplication? For what m does this not occur? Hint: consider the difference between Z_m^* and Z_m. Rob Johnson take out the trash before replying === Subject: Re: Show that Z_m^* is group <20051218.012746@whim.org> Tanja === Subject: Algebraic over field We know that root of 2 in R is algebraic of degree 2 over Q and of degree 1 over R. Can vi tell that 3/2 is algebraic over Z (integers)? I know that (Z,+,*) is a ring, but not a field,... Can someone explain this to me very-very simple? Tanja === Subject: Re: Algebraic over field > We know that root of 2 in R is algebraic of degree 2 over Q and of > degree 1 over R. > Can vi tell that 3/2 is algebraic over Z (integers)? > I know that (Z,+,*) is a ring, but not a field,... > Can someone explain this to me very-very simple? 3/2 is the root of 2x - 3 in Z[x]. So it's algebraic of degree 1 over Z. By adjuncting 1/2 to Z, one finds 3/2 in Z[1/2]. Of course Z[1/2] = Z[3/2] and it's not a field. -- To Google and MathForum users: Reply only if adequate context is included _within_ the reply. Otherwise all contexts are removed from my view, the flow of thought disrupted and chaos reigns. In particular for Google users: Instead of simply hitting the prominent Reply link, which doesn't include a copy of the post to which one is replying, click the Show Options link (toward the top of an item in the thread), which causes a shaded area of links to appear next to the top of the item, including Reply (first) that does introduce a copy of the previous text (offset by > signs in the usual fashion). ---- === Subject: Re: Algebraic over field >>We know that root of 2 in R is algebraic of degree 2 over Q and of >>degree 1 over R. >>Can vi tell that 3/2 is algebraic over Z (integers)? >>I know that (Z,+,*) is a ring, but not a field,... >>Can someone explain this to me very-very simple? > 3/2 is the root of 2x - 3 in Z[x]. > So it's algebraic of degree 1 over Z. What about googling the expression 'algebraic' first? I doubt your result, since 3/2 is not integral over Z. > By adjuncting 1/2 to Z, one finds 3/2 in Z[1/2]. > Of course Z[1/2] = Z[3/2] and it's not a field. > -- To Google and MathForum users: > Reply only if adequate context is included _within_ the reply. > Otherwise all contexts are removed from my view, > the flow of thought disrupted and chaos reigns. > In particular for Google users: > Instead of simply hitting the prominent Reply link, which doesn't > include a copy of the post to which one is replying, click the Show > Options link (toward the top of an item in the thread), which causes > a shaded area of links to appear next to the top of the item, including > Reply (first) that does introduce a copy of the previous text (offset > by > signs in the usual fashion). > ---- J. === Subject: Re: Algebraic over field <43A5554C.1070208@web.de> >>We know that root of 2 in R is algebraic of degree 2 over Q and of >>degree 1 over R. >>Can vi tell that 3/2 is algebraic over Z (integers)? >>I know that (Z,+,*) is a ring, but not a field,... >>Can someone explain this to me very-very simple? > 3/2 is the root of 2x - 3 in Z[x]. > So it's algebraic of degree 1 over Z. > What about googling the expression 'algebraic' first? I doubt your Google sucks ever since it went commercial. > result, since 3/2 is not integral over Z. Let F be an extension field of K and u and element of F. If there's some non-zero polynomial p(x) in K[x], then u is said to be algebraic over K, Since the question was about Z, which isn't a field, the above definition needs be extended to rings to give meaning to OP's question. For fields, it's moot question whether p is monic or not. > By adjuncting 1/2 to Z, one finds 3/2 in Z[1/2]. > Of course Z[1/2] = Z[3/2] and it's not a field. === Subject: Re: Algebraic over field >>We know that root of 2 in R is algebraic of degree 2 over Q and of >>degree 1 over R. >>Can vi tell that 3/2 is algebraic over Z (integers)? >>I know that (Z,+,*) is a ring, but not a field,... >>Can someone explain this to me very-very simple? >3/2 is the root of 2x - 3 in Z[x]. >So it's algebraic of degree 1 over Z. >>What about googling the expression 'algebraic' first? I doubt your > Google sucks ever since it went commercial. >>result, since 3/2 is not integral over Z. > Let F be an extension field of K and u and element of F. > If there's some non-zero polynomial p(x) in K[x], then > u is said to be algebraic over K, > Since the question was about Z, which isn't a field, > the above definition needs be extended to rings to give > meaning to OP's question. Exactly. The clear notion established in algebra is 'integral'. E.g. in number theory. > For fields, it's moot question > whether p is monic or not. Over rings it is one of *the* questions. >By adjuncting 1/2 to Z, one finds 3/2 in Z[1/2]. >Of course Z[1/2] = Z[3/2] and it's not a field. J. === Subject: Re: Algebraic over field <43A5554C.1070208@web.de> <43A559D5.6030003@web.de> > >>We know that root of 2 in R is algebraic of degree 2 over Q and of >>degree 1 over R. >>Can vi tell that 3/2 is algebraic over Z (integers)? >>I know that (Z,+,*) is a ring, but not a field,... >>Can someone explain this to me very-very simple? > >3/2 is the root of 2x - 3 in Z[x]. >So it's algebraic of degree 1 over Z. >>result, since 3/2 is not integral over Z. > Let F be an extension field of K and u and element of F. > If there's some non-zero polynomial p(x) in K[x], then > u is said to be algebraic over K, > Since the question was about Z, which isn't a field, > the above definition needs be extended to rings to give > meaning to OP's question. > Exactly. The clear notion established in algebra is 'integral'. E.g. in > number theory. That right, for 1/2 to be an algebraic integer of Z, it has to be a solution of a monic polynomial. However OP isn't asking about the abstruse notion of algebraic integers, he's asking about algebraic. > For fields, it's moot question > whether p is monic or not. > Over rings it is one of *the* questions. >By adjuncting 1/2 to Z, one finds 3/2 in Z[1/2]. >Of course Z[1/2] = Z[3/2] and it's not a field. > > J. === Subject: Re: Algebraic over field William Elliot Tanja === Subject: Re: Algebraic over field > We know that root of 2 in R is algebraic of degree 2 over Q and of > degree 1 over R. > Can vi tell that 3/2 is algebraic over Z (integers)? > I know that (Z,+,*) is a ring, but not a field,... > Can someone explain this to me very-very simple? > Tanja It's hard to tell exactly what your question is. I think that you're asking if there's a notion similar to algebraic for general rings rather than fields. Let R and S be rings with R a subring of S. Then an element x of S is said to be integral over R is x satisfies a monic polynomial with coefficients in R. So if R and S are fields, you get your familiar definition of x being algebraic over R. However, I should point out that 3/2 is not integral over Z because it does not satisfies a monic polynomial in Z. So maybe I'm not interpreting your question correctly? Mike === Subject: Re: Algebraic over field I am only beginner in Abstrakt Algebra and IK think that You interpreting my question correctly. Tanja === Subject: $10 000 prise the new conjecture > [...] > I noticed he was quoting Wikipedia on unsolved conjectures. > And just the other day someone (certainly not I) added a link > to one of my Collatz web pages to the Wikipedia entry on the > Collatz Conjecture. > In that Wikipedia entry at: > http://en.wikipedia.org/wiki/Collatz_conjecture > the real/complex iteration can obviously be expressed as: > 4.f(z) = (4z + 1) - (2z + 1).cos(PI.z) [*] > from where: > f(z) = z/2 if cos(PI.z) = 1, so that z = 2Z > = (3z + 1)/2 if cos(PI.z) = -1, so that z = 2Z + 1 > For integers z, [*] is equivalent to: > | 1 -1 -1 | > | | > 4.f(z) = 2z + | 1 z 2z | > | | > | cos(PI) cos(-z.PI) cos(-2z.PI) | If you squint hard enough, the RHS looks vaguely similar to the form of Hamiltonians that arise in scattering theory. See http://www.du.edu/~jcalvert/phys/smatrix.htm It's a very long shot, but if anyone is seriously interested, even absolutely determined, to solve the Collatz conjecture, maybe that might be worth at least a cursory investigation and lead to a novel and possibly hitherto unexplored approach. John R Ramsden === Subject: Re: $10 000 prise the new conjecture > > [...] > > I noticed he was quoting Wikipedia on unsolved conjectures. > And just the other day someone (certainly not I) added a link > to one of my Collatz web pages to the Wikipedia entry on the > Collatz Conjecture. > In that Wikipedia entry at: > http://en.wikipedia.org/wiki/Collatz_conjecture > the real/complex iteration can obviously be expressed as: > 4.f(z) = (4z + 1) - (2z + 1).cos(PI.z) [*] > from where: > f(z) = z/2 if cos(PI.z) = 1, so that z = 2Z > = (3z + 1)/2 if cos(PI.z) = -1, so that z = 2Z + 1 > For integers z, [*] is equivalent to: > | 1 -1 -1 | > | | > 4.f(z) = 2z + | 1 z 2z | > | | > | cos(PI) cos(-z.PI) cos(-2z.PI) | > If you squint hard enough, the RHS looks vaguely similar to > the form of Hamiltonians that arise in scattering theory. > See http://www.du.edu/~jcalvert/phys/smatrix.htm > It's a very long shot, but if anyone is seriously interested, > even absolutely determined, to solve the Collatz conjecture, > maybe that might be worth at least a cursory investigation > and lead to a novel and possibly hitherto unexplored approach. Drat, I certainly won't be the one to solve it. I looked at that link but it was all Greek to me. > John R Ramsden === Subject: Re: $10 000 prise the new conjecture > > > [...] > > I noticed he was quoting Wikipedia on unsolved conjectures. > And just the other day someone (certainly not I) added a link > to one of my Collatz web pages to the Wikipedia entry on the > Collatz Conjecture. > > In that Wikipedia entry at: > > http://en.wikipedia.org/wiki/Collatz_conjecture > > the real/complex iteration can obviously be expressed as: > > 4.f(z) = (4z + 1) - (2z + 1).cos(PI.z) [*] > > from where: > > f(z) = z/2 if cos(PI.z) = 1, so that z = 2Z > > = (3z + 1)/2 if cos(PI.z) = -1, so that z = 2Z + 1 > > For integers z, [*] is equivalent to: > > | 1 -1 -1 | > | | > 4.f(z) = 2z + | 1 z 2z | > | | > | cos(PI) cos(-z.PI) cos(-2z.PI) | > If you squint hard enough, the RHS looks vaguely similar to > the form of Hamiltonians that arise in scattering theory. > See http://www.du.edu/~jcalvert/phys/smatrix.htm > It's a very long shot, but if anyone is seriously interested, > even absolutely determined, to solve the Collatz conjecture, > maybe that might be worth at least a cursory investigation > and lead to a novel and possibly hitherto unexplored approach. > Drat, I certainly won't be the one to solve it. > I looked at that link but it was all Greek to me. Me too, and to be honest glancing at it again I'm not sure if that was the exact link where the Hamiltonian similar to the above appeared. But I wouldn't worry too much, as it may very well all be a wild goose chase anyway. === Subject: Re: Well Ordering the Reals > Hi Ross, > I must say, Ross, congratulations a nicely analyzed post. (I take back > accusations about sloppiness...it looks like you've really thought this > through.) > The only area I am not sure about is your mapping of f(o) to the reals. > I will have to think about that more. However your point of extending > Cantor's first to the countable ordinals is not lost on me. > A couple of things you want to keep in mind moving forward... Con(ZFC) > <-> Con(ZFC+CH), so you could decouple to analysis to two cases: ZFC+CH > and ZFC+~CH. Secondly, nonstandard ordinals and cardinals are > different from standard ones. In fact, I believe it's straightforward > to prove that nonstandard omega is of size standard |2^N|. It is > curious that |*N| > |N|, whereas |*R| = |R|. Thus we have |N| < |*N| = > |R| = |*R|. Thus, nonstandard countable is the size of the reals. > However, larger sets of course remain uncountable, such as the set of > all functions f:R->R (which is of size |2^R|). > Good Luck! > Well that's all fine and good, but my purpose is to find a > contradiction with the reals in ZFC so I can justify to all comers > re-evaluating the definition(s) of the real numbers. > Where we've been talking about ZFC, the Continuum Hypothesis, and > Cantor's nested intervals in well-ordering the reals, basically all > we've seen is that if CH holds then for each ordinal o in the > well-ordering f that there is simply constructible a sequence that > converges to f(o), and, that the intersection of those nested intervals > is a degenerate interval containing only one point, [f(o)]. > I'm looking for the previous point that was removed from that interval > to leave a single point, that is, in consideration of what a degenerate > interval is. Basically that involves the consideration of a maximal > element, for example in a model where the maximal element is less than, > or equal, to o. > I find that distasteful because I eschew models in theory and classes > in set theory. While that miay be so, it is my goal to show these > features in the context of ZFC, thus that they are palatable to those > weaning on the pablum of non-naive, regular set theory. Now, that > might be seen as plainly antagonistic, and to some extent it is, > because there is no universe in ZFC yet for a predicate that always > evalutes to true, every set would satisfy the membership requirement. > In returning to the notion of a non-constant a in the progression and > generation of a set containing nested intervals a, b, there is again > the question of a mechanism to alternate between a and b, and to store > an intermediate value. That is where if nested intervals are to be > constructed without having one of the endpoints be constant, then > there is the notion of having an auxiliary set that would store a > half-constructed interval. That may be simpler by constructing a > superset of AB, containing each element (a,b), and also (a,x) for some > x not E R, in that way providing a subset of AB that is AX that > contains a ordered set of points, er, the point being to selet the > largest in value of those points which leads to problems, because, some > people will not accept there being a largest value in an open interval. > With the non-constant intervals endpoints, again f(0) and f(1) and > endpoints a, b or an interval (a, b). If f(1) < f(0) then a notion is > to compose f through g(x)= -f(x) with the point about <{ reversing, so > f(1) > f(0) can be presumed. So anyways, then AB contains an interval > (a_0, b_0), that's an ordered pair that represents interval endpoints, > not an open interval, AB contains (a_0, b_0) which represents an > interval [a_0, b_0]. Then, in infinite induction, as o increases from > 1 then the first value f(o) that is between a and b for each (a,b) in > AB is then a half-constructed interval a. It can be put in a the set > as (a,o), because it's easier to compare ordinals in the manner of > getting the greatest from the set. Then, continuing from o, f(o') is > evaluated, and if it is between each (a,b), then each (a,o) is checked > to see if there exists any (a,b), if not, then the for the maximum > value of o from the pairs (a, o), a pair (a,b) is added to the set. > It's easy to see that the construction rules as above need to be > re-ordered, but the point is that that AB contains for each ordinal o a > set constructed of nested interval between o and o+1 that is not > constantly (f(0), b) or (a, f(o+1)) unless there were adjacent points > in the reals. > Then, if CH, each o is equal to a c, a countable ordinal. Then, for > each o E O', it's countable, and thus the degenerate interval or > intersection of the nested intervals in AB, well dang I want to say > that's it not there but then there's the question of it being f(o') for > o' < o for AB_o. To avoid contradiction it must be, aha, but there is > no o' < 0. > Thus, if CH, do you see how thus Cantor's first would apply to that > well-ordering? In the variation of Cantor's first proof of the uncountability of the reals, the second being the powerset rresult and the third the antidiagonal result, the nested intervals of Cantor's first are applied not just to a bijection from the natural numbers to the real numbers but to a bijection between any ordinal and the real numbers. A set AB_o is is constructed that contains nested intervals in the form (a, b), with a < b. Those values converge towards each other as f is a well-ordering of the reals or interval of the reals O in a bijection from ordinal O' <-> R, and then their intersection would be a single point or degenerate interval, but it can't be said to be that point because that point in the construction of the set of nested intervals would have been within more than one of the nested intervals via induction, per the density of the reals in themselves as complete ordered field. So, Cantor's first applies to well-ordering the reals in the same way as it applies to bijection of natural numbers to the reals. In the set theory ZFC, Zermelo-Fraenkel set theory with the Axiom of Choice, ZF+C or ZFC, the axiom of choice axiomatizes the well-ordering principle, so in said theory there must exist a well-ordering of the reals as it is a direct consequence of the axiomatization. Similarly, if the reals have a cardinal, a cardinal is an ordinal, and an ordinal is well-ordered. If nested intervals thus implies that the well-ordering of the reals as a complete ordered field can not exist, yet the well-ordering principle that it must exist, besides dilemmas of contradiction there is the notion that the structure of the real numbers can be represented in a different way than the standard representation of the real numbers and unless they are there is a contradiction within ZFC of the standard notion of the real numbers. One possible notion then that might be reasonable in consideration is that of the contiguous sequence of adjacent points that form the continuum of the real numbers. Were the reals expressible in that form, then a well-ordering would simply match a sequence of those elements and thus not find a contradiction in the variation of the nested intervals result. Ross === Subject: Aha! Found out about manifold stripings - Foliations [continuation of month-old thread to which Google won't let me reply :-< > A month or so ago I posted a question in a sci.math > thread titled What manifolds allow an everywhere > parallel vector field? > I recently found that the idea I was trying to express > is called Foliation Theory, and is currently an active > geometry and 3-manifolds by James Otterson at: > http://math.berkeley.edu/~alanw/242papers02/otterson.pdf > on symplectic geometry. > I then owe you an apology. It seemed clear to me you were talking > about foliations, but I assumed you were familiar with the term, > only holding back from speaking too technically. While I'm no > expert by any stretch of the imagination, I have heard a bit of the > language used by native speakers. Well, it was years ago, but I've > done my penance by googling up a bunch of relevant things. > [..] > If I recall any other relevant terminology, you'll hear from me again. Seymour, if you see this). I still feel like I'm trying to run before I can walk, but hopefully the the references and search term hints you provided will be very useful. John R Ramsden === Subject: Re: A Recursive Game >>Consider the example whare one box has $5 and the other none. This >>has an expected value to the player of about $5.00063426. >Thus, at this point, the player might as well settle for $5 rather >than play on for an expected additional gain of less than a penny. In >fact, if settlement was allowed, the player should already have >settled at state ($0,$4). >>With optimum play, the $0 box is unlocked with probability about >>0.83342646 and is chosen with probability about 0.00063421. >No, this contradicts the math. >With this problem, it's very easy to get the intuition wrong, and it >looks like you accidentally got things reversed. So it seems >By p[n], we mean the probability that the house _locks_ the empty box. >By calculation, p[5]=0.83342646, hence the $0 box is _locked_ with >probability 0.83342646. >However, you have the correct value for the probability that the $0 >box is chosen by the player. With $5 at risk, the player should almost >never choose the empty box. >But the error you made above leads to some errors below. >>So the outcomes are: >>1. Game ends with player getting nothing with prob about 0.00052856 >>2. Game ends with player getting $5 with prob about 0.16646790 >>3. Game moves to $5 and $1 (worth about an expected $4.00380771+$1 = >> $5.00380771) with prob about 0.83289790 >>4. Game moves to $6 and $0 (worth about an expected $6.00009060) >> with prob about 0.00010564 >The probabilities for statements 2 and 3 should be swapped. The probabilities for 1 and 4 also need to be swapped. I did try to check this, but didn't notice that 0*0.00052856 + 5*0.1664679 + 5.00380771*0.8328979 + 6.0000906*0.00010564 = 0*0.00010564 + 5*0.8328979 + 5.00380771*0.1664679 + 6.0000906*0.00052856 This happens with minimax problems, where getting one player's strategy correct will produce the correct values even if you get the other player's strategy wrong >The corrected statements should be: >2. Game ends with player getting $5 with prob about 0.83289790 >3. Game moves to $5 and $1 with prob about 0.16646790 >>means that >> >>1) vast majority of dollars go into one box, >Unfortunately not. The player doesn't have the luxury of building up >the big box. Instead, the player must select the big box most of the >time, so as not to risk ending up with the smaller amount. >>So usually new dollars go into the box with fewer already >The house also wants the game to end, not continue, so given that the >player won't risk selecting the small box very often, the house can >lock it. >>I an not sure about that: the house usually locks the big box (though >>less often than the player chooses it). >No, same error as before. In fact, the house usually locks the small >box (except when they're equal in which case the boxes are chosen >equally likely). >Thus, with optimal play, the game ends quickly. >>It depends what you mean by quickly. Starting at 0, the expected >>number of games seems to me (i.e. unchecked) to be about 2.0986 >Yes, approximately 2 rounds on average, although I get a little less >than 2. OK - I will try about 1.90237 >>(slightly more than in the St Petersburg paradox: of course there is a >>50% chance it finishes after 1 go). The expected number of remaining >>rounds seems to go up as the difference between the two boxes >>increases, as far as I can tell tending towards n/2 + 1. >No -- same error. The expected number of remaining rounds goes down as >n increases. The higher the value of n (where n=big-small), the more >likely it is that the game will end immediately on the next round, >with the player selecting the big box and the house obliging by >leaving it empty. Fair enough - now it seems to decrease towards something like 1 + 1/n >The typical ending is that the player selects the big box which the >house had obligingly left open >In fact, about only about 6% of all games will end with the player >unhappily choosing a box which is strictly least. === Subject: sort of difference equation Since I'm able to solve first order difference equations only, I ask for some help in solving these (I'm not even sure they are difference equations...). For the number of combinations in a game I'm trying to solve, I found: p^k_0 = 4 p^{k-1}_0 + 3 p^{k-1}_1 p^k_1 = 3 p^{k-1}_0 + 2 p^{k-1}_1 with initial condition p^2_0 = 7 and p^2_1 = 5 I know that if y_k = 4 y_{k-1} the solution I would be looking for is something like y_k = 4^k y_0, but the problem (for me) is that two equations are combined here. Can somebody help me with that? ps > Some example values I computed (p^k = p_0^k + p_1^k) k p_0^k p_1^k p^k [1,] 2 7 5 12 [2,] 3 43 31 74 [3,] 4 265 191 456 [4,] 5 1633 1177 2810 [5,] 6 10063 7253 17316 [6,] 7 62011 44695 106706 [7,] 8 382129 275423 657552 [8,] 9 2354785 1697233 4052018 [9,] 10 14510839 10458821 24969660 [10,] 11 89419819 64450159 153869978 [11,] 12 551029753 397159775 948189528 [12,] 13 3395598337 2447408809 5843007146 [13,] 14 20924619775 15081612629 36006232404 [14,] 15 128943316987 92937084583 221880401570 === Subject: Re: sort of difference equation > Since I'm able to solve first order difference equations only, > I ask for some help in solving these (I'm not even sure they are > difference equations...). > For the number of combinations in a game I'm trying to solve, I found: > p^k_0 = 4 p^{k-1}_0 + 3 p^{k-1}_1 > p^k_1 = 3 p^{k-1}_0 + 2 p^{k-1}_1 > with initial condition p^2_0 = 7 and p^2_1 = 5 Your notation is inconsistent and confusing. Is this a correct and accurate translation? Find solutions to the recursive equations p_k = 4p_(k-1) + 3q_(k-1) q_k = 3p_(k-1) + 2q_(k-1) with initial conditions p_0 = 7, q_0 = 5. Use monospace font to view matrix equations. p_k = 4 3 * p_(k-1) q_k 3 2 q_(k-1) p_k = (4 3)^k * p_0 q_k (3 2) q_0 So the problem becomes to find a useful expression for (4 3)^k (3 2) > I know that if y_k = 4 y_{k-1} the solution I would be looking for > is something like y_k = 4^k y_0, but the problem (for me) is that > two equations are combined here. Can somebody help me with that? > ps > Some example values I computed (p^k = p_0^k + p_1^k) > k p_0^k p_1^k p^k > [1,] 2 7 5 12 > [2,] 3 43 31 74 > [3,] 4 265 191 456 === Subject: Re: sort of difference equation > Since I'm able to solve first order difference equations only, > I ask for some help in solving these (I'm not even sure they are > difference equations...). > For the number of combinations in a game I'm trying to solve, I found: > p^k_0 = 4 p^{k-1}_0 + 3 p^{k-1}_1 > p^k_1 = 3 p^{k-1}_0 + 2 p^{k-1}_1 > with initial condition p^2_0 = 7 and p^2_1 = 5 > Your notation is inconsistent and confusing. > Is this a correct and accurate translation? > Find solutions to the recursive equations > p_k = 4p_(k-1) + 3q_(k-1) > q_k = 3p_(k-1) + 2q_(k-1) > with initial conditions p_0 = 7, q_0 = 5. > Use monospace font to view matrix equations. > p_k = 4 3 * p_(k-1) > q_k 3 2 q_(k-1) > p_k = (4 3)^k * p_0 > q_k (3 2) q_0 > So the problem becomes to find a useful expression for > (4 3)^k > (3 2) Which can be done by diagonalizing the matrix, i.e. finding a matrix P and diagonal matrix D such that: (4 3) (3 2) = P D P^-1 Making: (P D P^-1)^k = P D^k P^-1 D is easy to exponentiate since all its terms are along the diagonal. > I know that if y_k = 4 y_{k-1} the solution I would be looking for > is something like y_k = 4^k y_0, but the problem (for me) is that > two equations are combined here. Can somebody help me with that? In this case, the solution is of the form: p_k = a r^k + c s^k q_k = b r^k + d s^k For certain initial conditions, c and d would be zero, and the form p'_k = a r^k q'_k = b r^k would work. (Likewise, there are initial conditions which make a and b zero.) For such a solution we have: p'_k = r p'_(k-1) q'_k = r q'_(k-1) p'_k = r 0 * p'_(k-1) = 4 3 * p'_(k-1) q'_k 0 r q'_(k-1) 3 2 q'_(k-1) 4-r 3 * p'_(k-1) = 0 3 2-r q'_(k-1) 0 |4-r 3 | | 3 2-r| = 0 (4-r)(2-r) - (3)(3) = 0 We can solve this equation to find the values of r and s: r = 3 + sqrt(10) s = 3 - sqrt(10) Relating this back to what I said about diagonalizing the matrix, r and s are the elements of D: D = r 0 0 s The numbers r and s are called the eigenvalues of (4 3) (3 2). A set of initial conditions p'_0 q'_0 for which c = d = 0 is called an eigenvector associated with the eigenvalue r. The solution will be dominated by the terms proportional to r^k. The part proportional to s^k will approach zero in this case since |s| < 1. > ps > Some example values I computed (p^k = p_0^k + p_1^k) > k p_0^k p_1^k p^k > [1,] 2 7 5 12 > [2,] 3 43 31 74 > [3,] 4 265 191 456 === Subject: Re: sort of difference equation > Since I'm able to solve first order difference equations only, > I ask for some help in solving these (I'm not even sure they are > difference equations...). > > For the number of combinations in a game I'm trying to solve, I found: > p^k_0 = 4 p^{k-1}_0 + 3 p^{k-1}_1 > p^k_1 = 3 p^{k-1}_0 + 2 p^{k-1}_1 > with initial condition p^2_0 = 7 and p^2_1 = 5 > > Your notation is inconsistent and confusing. > Is this a correct and accurate translation? > Find solutions to the recursive equations > p_k = 4p_(k-1) + 3q_(k-1) > q_k = 3p_(k-1) + 2q_(k-1) > with initial conditions p_0 = 7, q_0 = 5. > Use monospace font to view matrix equations. > p_k = 4 3 * p_(k-1) > q_k 3 2 q_(k-1) > p_k = (4 3)^k * p_0 > q_k (3 2) q_0 > So the problem becomes to find a useful expression for > (4 3)^k > (3 2) > Which can be done by diagonalizing the matrix, i.e. finding a matrix P > and diagonal matrix D such that: > (4 3) > (3 2) = P D P^-1 > Making: > (P D P^-1)^k = P D^k P^-1 > D is easy to exponentiate since all its terms are along the diagonal. For a real symmetric matrix, as you have here, one can even do it in the form M = (P^t D P), where P^t is the transpose of P, rather than the inverse === Subject: Re: sort of difference equation You were totally right about the notation. JeeBee. === Subject: A Segment's Edge and a Point A point is based only on the Length concept, which means, even if its length equal to zero we still use the Length concept. For example, in the case of Emptiness, the Length concept itself is not used. We can ask: What is the difference between a segment's edge and a point? The answer is: Since a segment's edge is an inseparable thing of the segment itself, it has the properties of a segment, which means: It is defined by at least Length and Direction, where Direction is the property of being in more than one state wihin the same element, for example: To exist in and out of some set _{_}. But in the case of a point, the Direction concept does not exist at all (a point cannot be but in {.} XOR {}. of some set). My length concept is not a geometrical concept but based on the Function concept as the 1-1 mapping. Length 0 is the 1-1 mapping of an element to itself, where Length NOT= 0 is the 1-1 mapping between at least two distinguished elements. The minimal information form that we get in the case of 0 Length, is a Point {.} . The minimal information form that we get in the case of Non_0 Length, is a Segment {._.} . If it is still do not understood, then please consider that No_Length_at_all has Rank 0 where a Length has Rank 1 (http://mathworld.wolfram.com/Rank.html), for example: {} (No_Length_at_all) {.} 0 Length (Point) {} .81©Length.81¬ {{}} is a 1-1 mapping between at least two distinguished elements (Segment) {__} Non-measurable Length (Fullness) Each building-block {}, {.}, {._.}, {__} is a free element (not composed by the other building-blocks). Some example: Let us say that any positive number is based on the Length concept. It means that if the Length concept itself is omitted, then our set is Empty and we get {}, which is the Empty set. By saying .81g0 length.81h we are using the Length concept and in this case we get number 0, which is equivalent to a Point. Any other number, which is not a point, is at least some Segment, which is always longer than a Point (and length > 0 has a direction). We can take any arbitrary segment and call it 1. Now we have two building-blocks that can help use to define any given positive R member that we wish. So our positive R members are based on the Length concept, and each one of these numbers has a unique length, that can be represented by a point {.} or by a segment {._.} . Now let us check what is infinity according to this approach. We know that {} means that the Length concept itself is omitted from our framework. Can we define a state, which is the opposite of {}? The answer is: Yes, and we can call this state Fullness which is the totality of the length concept itself, or in other words it is an infinitely long element that can be represented by us as {__} or {.__}. It means that we cannot define its length and use it to get a particular number, or in other words, the Length concept itself is too strong to be used by us. Now let us examine Point*Fullness, which is equivalent to: Zero*Infinity According to the Length-concept framework, we can ask: What is the result of Point*Emptiness? What is the result of Point*Fullness? In both cases we can clearly see that we cannot get any meaningful result within this framework, because neither Fullness nor Emptiness can be used as input (or legitimate participators) in these multiplication operations. (By the way, if we add the Direction concept to the above framework, we can represent the entire R members, by using x_0, 0, 0_x forms). The answer to these questiones is very simple from this point of view. === Subject: Re: Tim's golden Coordinates (polysigned news) > I did discover the cone in the cartesian coordinate system too. I > okay, we need four points for space, just the origin in cartesian > coordinates with a point at equal distance from it on each positive > axis. Connecting these gives us a tetrahedron, but off centre and with > > unequal distances between the points. > But this tetrahedron 0 > x=1 > y=1 > z =1 ist a quarter of a bigger > tetrahedron, which is nice on center with 0 and with equal sides and > angles. Let a ray start at the origin with equal distance from all three > negative arms with length 1 and connect to x=1 > y=1 > z =1. A tetra > with side length sqrt ( 8 / 3 ). This was wrong ! One can have a tetra with equal distances from the center of 1 unit and three of the rays are on the x-, y-, and z-axes, so with a right angle between two of these. And the fourth ray of length 1 is on a ray starting at the origin with equal distances from all three negative arms. But this tetra will have three edges of length sqrt 2, and three edges of a different length. > Now, if You rotate round the axis of equal distance to all three axes, > You get a cone with tip at zero and an angle of arcus tan sqrt 8, > between the rotation axis and the surface This is of course for the cartesian tetra with unequal side length, the surface of the cone mentioned and the angle is here created by the rotation of the x-, y-, and z-axes. An equal edged tetra rotated around one ray (from center to corner) gives a cone surface created from the other three rays with an angle between axis of this cone and it's surface at the tip of the cone ( which is here the center of the tetra ) and this angle is arc cos 1/4. A true cartesian system inside an equal edged tetra i gave in my last post. The cartesian three axes are six rays from the center. And the tetra has six edges, so connect the middles of each edge with the center and voil.88. center to each square side and to the oppsite is one cartesian axis. And on each square one diagonal, conected in the right way will connect four of the eight corners to a perfect tetra. And the side length of the tetra is related to it's distance from the center by sqrt 8. You see, i too have some beginner's difficulties to visualize and i apologize for my fault's. Hero === Subject: Re: about Rudin >(Preceding material elided) >> I find this hard to follow as well. You have this book >> by Rudin, maybe it's Principles of Mathematical Analysis >> or maybe it's the book that the OP and I have been >> talking about. You see that you can't do any of >> the exercises, and that's discouraging. So far I'm >> with you. But if you'd had a solutions manual that >> then the fact that you couldn't do the exercises >> would have been less discouraging because you could >> look up the answers? I don't get that. >Then there is the obverse side of the coin: I remember >working out some of the problems in Principles of Mathematical >Analysis (aka Baby Rudin) as an undergraduate (especially >some of the series problems) and being totally elated. >That elation helped to cement the conviction that THIS was what >I wanted to do the rest of my life. Precisely. I woulda thought that that's how everybody thought about this sort of thing. >What use would a solutions manual have been to me? The question >was, COULD I DO THEM? If I'd given up and used a solutions manual, >the answer would have been, NO, and I can hardly imagine anything >more damaging to my self-esteem. >And if I couldn't have done them? A shrug of the shoulders when I >heard how the problem was done, and no damage to a 19-year-old's ego, >which is usually pretty insufferable anyway. >I've seen answers to some of the more interesting questions on the web. >I don't know if they're still out there, but a really determined search >will find something. And a really determined student will look for >other books, like Knopp's. >--Ron Bruck ************************ David C. Ullrich === Subject: The Helsinki Code, Part 3 After writing the computing urban legend The Helsinki Code, I spent several nights thinking up how in the world Gustav Larsson, the Finnish PDP-8 computer programmer, could have managed to receive such a miraculous message from God. Surely, for a 1-byte computer program such as @ to compile successfully (in RTPS FORTRAN), a miracle compression algorithm would be necessary. Then it dawned on me. Gustav had accidentally stumbled upon a compiled Fortran program compressed by the Impossibly Efficient Zeus Compression Algorithm (ZCA). That is, a compression algo- rithm so efficient that pages upon pages of text can be compressed into a single byte. The ZCA, in theory, can compress the entire contents of the Holy Bible into a single byte. We presently know of no technology which can accomplish this, but the ZCA work. In future years, it will happen. BTW, the story of the Helsinki Code can still be found online (just search for it on Usenet). Paul === Subject: Re: The Helsinki Code, Part 3 > Then it dawned on me. Gustav had accidentally stumbled upon > a compiled Fortran program compressed by the Impossibly Efficient > Zeus Compression Algorithm (ZCA). That is, a compression algo- > rithm so efficient that pages upon pages of text can be compressed > into a single byte. The ZCA, in theory, can compress the entire > contents of the Holy Bible into a single byte. Huh? I can easily write you a program that can compress the Holy Bible into one bit, and decrompress back again, if you like. It can also compress other things, but not as effeciently.... Soren === Subject: Re: The Helsinki Code, Part 3 >> Then it dawned on me. Gustav had accidentally stumbled upon >> a compiled Fortran program compressed by the Impossibly Efficient >> Zeus Compression Algorithm (ZCA). That is, a compression algo- >> rithm so efficient that pages upon pages of text can be compressed >> into a single byte. The ZCA, in theory, can compress the entire >> contents of the Holy Bible into a single byte. > Huh? I can easily write you a program that can compress the Holy Bible > into one bit, and decrompress back again, if you like. It can also > compress other things, but not as effeciently.... Well, define easily. Unless it's a lossy compression algorithm, you would have to compare the input to the Bible in order to verify that a Bible-like document doesn't get accidentally compressed to the same value as the bible itself. Probably this would involve having a copy of the Bible in the compression/decompression algorithm itself. Since any other person later decompressing the compressed text would need the program, and the program contains an uncompressed Bible (or, one compressed by some other non-lossy technique), in what way have you saved any space? -- J. Giles I conclude that there are two ways of constructing a software design: One way is to make it so simple that there are obviously no deficiencies and the other way is to make it so complicated that there are no obvious deficiencies. -- C. A. R. Hoare === Subject: Re: The Helsinki Code, Part 3 > Well, define easily. Unless it's a lossy compression > algorithm, you would have to compare the input to the > Bible in order to verify that a Bible-like document > doesn't get accidentally compressed to the same > value as the bible itself. Probably this would involve > having a copy of the Bible in the compression/decompression > algorithm itself. Since any other person later decompressing > the compressed text would need the program, and the program > contains an uncompressed Bible (or, one compressed by > some other non-lossy technique), in what way have you saved > any space? If it's the Bible, output a 0. Stop. If not, output a 1, followed by the input (possibly compressed in some other way, doesn't matter). Soren :) === Subject: Packinon 0.04 Released I have just released Packinon 0.04 http://packinon.sourceforge.net/download/ This release contains the set of programs I currently use. I haven't yet documented the changes to off2pov that allow custom drawing. Otherwise, it works as before but will now accept multiple OFF files for input and can also output objects in a self-contained POV file suitable for including in another POV scene file. There are some undocumented (beyond using option -h for help) unfinished programs in the release: off_report gives some basic information about the geometry in an OFF file; twist is the program I use for twisting regular and quasi-regular polyhedra; slice4d gives slices and projections of hypercubes, 24-cells and their packs. Adrian Rossiter Email: adrian_r@... Web Site: http://www.terra.es/personal/adrian_r === Subject: Re: order of magnitute I have one more question. I have such a serie; sum(Ai)=O(n) where i is from 1 to n. Then what is order of magnitude(big Oh) of sum(Ai^2) where i is from 1 to n. I tested the serie in matlab it is O(n). How? === Subject: Re: order of magnitute > I have one more question. I have such a serie; > sum(Ai)=O(n) where i is from 1 to n. > Then what is order of magnitude(big Oh) of > sum(Ai^2) where i is from 1 to n. > I tested the serie in matlab it is O(n). How? It's not clear what you meant by asking How?. A simple case in which what you state as the orders of magnitude are sharp is the case of A_i being some constant, as also (A_i)^2 will be constant (with respect to increasing n). Thus SUM A_i FOR i = 1 to n and SUM (A_i)^2 FOR i = 1 to n would be constant multiples of n. More generally if A_i is bounded by a constant independent of i, then so is (A_i)^2 bounded by a constant independent of i, and the estimates of O(n) on both sums (over n terms) are true, but not necessarily sharp. === Subject: Re: order of magnitute Chip, you are right.But I thought about this, I think we need more information. Knowing sum(Ai)=O(n) is not enough( i is from 1 to n). In your case it is again O(n), right. However ,look at that serie, for n=2, A1=1, A2=1 for n=3, A1=1, A2=1, A=2 for n=4, A1=1, A2=1, A3=1, A4=3 for n=4, A1=1, A2=1, A3=1, A4=1, A5=4 ..... then the serie is 2, 4,6,8.....it is O(n) , but let's look at to sum(Ai^2), then the serie is 0,2,6,12,.... which is not O(n). So it seems as we need more infromation about Ai. But what === Subject: Re: Representation of analytic functions > Conceptually speaking, how do you mentally visualize > analytic functions? When dealing with functions from > R^k to R^n, everything is about the same as in the > simpler cases (say R->R^2 for instance), meaning the > notions of continuity and differentiability give no > big problem. > For analytic functions I think it's different. First > I could not grasp any global intuition about their > nature (ok, I can pick paths from R^2 and see that the > function must be continuous and complex-differentiable > along this path. But this is not enough to grasp the > meaning of analytic because reasoning on a path is not > enough ( I should pick elementary discs instead ). > So, after going through a dozen problems, I realize I > tend to apply the theorems mechanically rather than > develop an intuition (for instance when dealing with a > more or less bounded function you know you'll have > to use the maximum modulus principle; or when the > function has a few zeros, the argument principle, > etc ...). However when comes a problem I cannot tackle, > I am left with no opinion about it. > How shall I understand all of this ? > -- > Julien Santini As Stuart said f:C --> C can be thought of as a mapping f:R^2 --> R^2. It should be hard to visualize such a function since it's four dimensional. A friend of mine gave me a tip on visualizing objects in four dimensions. It doesn't help a great deal, and I suppose it would take some practice, but it is interesting. Consider what we do when we visualize a function from R^2 --> R. We normally let R^2 be the xy-plane and have the function send those points vertically up or down parallel with the z-axis. Now to visualize a function R^2 --> R^2 attach a color to each point. Let green being a point close to zero, red & orange be points less than zero, and blue & violet be points greater than zero. My professor also mentioned a similar technique; just assign integers to each point. Either way, it's challenging to do all of this mentally without aid of paper, pencil, or computer. Kyle === Subject: Re: Representation of analytic functions <15899538.1134919437846.JavaMail.jakarta@nitrogen.mathforum.org> > Conceptually speaking, how do you mentally visualize > analytic functions? When dealing with functions from > R^k to R^n, everything is about the same as in the > simpler cases (say R->R^2 for instance), meaning the > notions of continuity and differentiability give no > big problem. > For analytic functions I think it's different. > .......... > As Stuart said f:C --> C can be thought of as a mapping > f:R^2 --> R^2. It should be hard to visualize such a > function since it's four dimensional. Another way is a vectorarrowfield, the domain are the points on a plane and the function attaches a fieldvectorarrow to each point. But first of all, what is C ? By strict definition the set is Rī, and an addition and a multiplication makes it a algebraic field ( something like having a group for both compositions ). Nothing more! By this it is the usual vectorspace Rī with addition and multiplication with a scalar. and the scalars are embedded in the first component of the ordered pairs. The multiplication into Rī makes C complete and this is more powerfull than just the multiplication with a scalar. But it is not as powerfull as introducing the dot-product, which gives a metric and so with this Rī is called euclidian space. and rotations with scaling by multiplication - but no reflection as in the euclidian. So in the first lesson in university the prof introduces the conjugate into C, changing (3, 4) into ( 3, - 4). Now C with the extra conjugate will give a metric and than one has the dot product, an euclidian space, but without it stays beeing C. Inside Rī with addition, muliplication with scalars and the dot product the multiplication of C can be constructed without introducing something new from outside. As on a plane one cannot change the orientation of rotation clockwise or reverse, actually there are two C's, a map from leftoriented C to leftoriented C is called holomorphic, but from leftoriented C to rightoriented C is called antiholomorphic - if i'm not mistaken here, You will find out what the correct relationshipt between left and antiholomorphic is ( and may be send me a mail). Intro of the conjugate changes the situation and one is in the euclidian Rī. An interesting point is for analytic functions (without conjugate) one cannot seperate the components of the vectors, so functions are like f : C ---> C : ( x, y ) ------> f ( (x,y) ), whereas in the euclidian Rī one can have f : C ---> C : ( x, y ) ------> f ( x, y ) = ( f1 ( x, y) , f2 ( x, y )) You can find out somethings and somelinks about i not longer beeing imaginary on my website http://1iz. de and the story: Algebraic fairy tale : C is euclidian And the best tool for visualization is the programm f(z) from Martin Lapidus on his: http://lascauxsoftware.com And i always want to get to the Cauchy theorem understanding but always i have to look at some simpler things like the i and just these days about reflections and transitions so i hope i'll get there some day also. Have fun Hero === Subject: Re: Representation of analytic functions On 17 Dec 2005 09:35:59 -0800, Julien Santini >Conceptually speaking, how do you mentally visualize analytic functions >? When dealing with functions from R^k to R^n, everything is about >the same as in the simpler cases (say R->R^2 for instance), meaning the >notions of continuity and differentiability give no big problem. >For analytic functions I think it's different. First I could not grasp >any global intuition about their nature (ok, I can pick paths from >R^2 and see that the function must be continuous and >complex-differentiable along this path. But this is not enough to grasp >the meaning of analytic because reasoning on a path is not enough ( I >should pick elementary discs instead ). >So, after going through a dozen problems, I realize I tend to apply the >theorems mechanically rather than develop an intuition (for instance >when dealing with a more or less bounded function you know you'll have >to use the maximum modulus principle; or when the function has a few >zeros, the argument principle, etc ...). However when comes a problem I >cannot tackle, I am left with no opinion about it. >How shall I understand all of this ? I'll be eager to see any constructive answers you may get to all this. I have none, but I can offer some sympathy: it's not just you. The last few times I taught the course I made up my own notes instead of using a book. That was partly because I'd become frustrated with all the books I'd tried, and partly because I felt I didn't understand Cauchy's Theorem and thought it might help if I worked it all out myself. Didn't work. The notes might become a book someday, which would be nice, and I know the details of a lot of proofs a lot better, which is very helpful in figuring out other things. But I still don't feel like I understand why Cauchy's Theorem is true. ************************ David C. Ullrich === Subject: Re: Representation of analytic functions An analytic function is a function from a part of the complex plane C (this is the cartesian plane R^2 except that the real numbers ( ( (as the real number line ) has been substituted for the x-axis ( an element of C is an ordered pair of real numbers (x,y) except that this pair has the additional property (x,0)=x , i =(0,1) and (x,y)=x+iy )and added to the vector space structure of vector addition and scalar multiplication is a vector multiplication which makes it a field (division is possible-thats unusual))) .This function also takes values in this field which you could think of as a 1 complex dimensional vector space C and thus also a 2 dimensional real vector space R^2. So think of an analytic function as a vector valued function on part of the plane,a special case of vector valued function on part of R^2 which is one interpretation of a function from R^2 into R^2 ,except that these vector values have a division ,so there somewhat like quantities (real numbers ,only there is no useful ordering) and they are 2 real component objects. Another way to think about functions from R^2 into R^2 is as a mapping which takes sets into sets and curves into curves The one to one analytic functions are the conformal maps meaning that he angle between the tangent vectors to two intersecting curves is the same at the point of intersection as the angle between the image curves at the === Subject: convergence question this was a question on my final and i have handed it in but did not follow the hint that the prof gave so am wondering if i'm right without using the hint: The question is: suppose sum(k=1 to inf ) ak converges. prove that the tail end of the series goes to zero. that is given epsilon > 0, there is an N such that for n>=N we have abs( sum(k=n to inf) ak ) < epsilon. The hint was use Cauchy criteria. I just said that sum(k=n to inf ) = sum( k = 1 to inf ) - sum( k = 1 to n-1 ) and since then that sum( k = 1 to inf ) converges then for every epsilon> 0 there exists an N st for all n >= N abs( sum( k = 1 to inf ) - sum( k = 1 to N ) ) < epsilon and we are done. === Subject: L^1(R) problem L^1(R) and log | f | in L^1(R) ? === Subject: Re: L^1(R) problem > L^1(R) and > log | f | in L^1(R) ? If |f| is small, what happens to log(|f|)? === Subject: Re: L^1(R) problem Hint: for any 1>eps>0, the set { |f(x)| >=eps } has measure < +oo by integrability of f. === Subject: Re: L^1(R) problem Hint: for any 1>eps>0, the set { |f(x)| >=eps } < +oo by integrability of f. === Subject: Re: Newton and the best neighborhood > Hello > trying to understand Newton method used to find a solution of a non-linear > system, so > I am trying to deduce one of the forms below from the other. > Picking a best guess to start = x_0 > scalar valued function. > x_n | f'(x_(n-2) ) = 0 > and the other sequence > vector-valued function. > x_n = x_(n-1) - f(x_(n-1) ) / Jf(x_(n-1) ) Be careful: f is a vector and J is a matrix, so you need to get the order of operation right. It should be x_n = x_(n-1) - [Jf(x_(n-1))]^(-1) f(x_(n-1)). You might also consider using a true Newton's Method, which incorporates a line-search. That is, a better method is often to use x_(n+1) = x_ - t [Jf(x_n)]^(-1) f(x_n), where t > 0 is determined by a one-dimensional search, for example by trying to minimize F = f(x(t))^T f(x(t)) along the line x(t) = x_n - t J^(-1) f(x_n) > where x_1, x_2, ... , x_n is the sequence which is expected to converge to > the solution of the non-linear system. > the second point I am trying to understand is how do you pick up the best > guess I don't mean Lipschitz or Kantrovic methods but rather, do you use > the inverse function theorem to pick your neighborhood? Most applications I have seen either guess a good starting point, perhaps using engineering experience (if it is an applied problem similar to others already known), or else just use some convenient point---trying to stay away from bad points if these can be identified. Sometimes people use several randomly-generated starting points in an attempt to generate several possible solutions if the problem may not have a unique solution (or if this is unknown).. R.G. Vickson === Subject: Re: Newton and the best neighborhood > Hello > trying to understand Newton method used to find a solution of a non-linear > system, so > I am trying to deduce one of the forms below from the other. > Picking a best guess to start = x_0 > scalar valued function. > x_n | f'(x_(n-2) ) = 0 > and the other sequence > vector-valued function. > x_n = x_(n-1) - f(x_(n-1) ) / Jf(x_(n-1) ) > where x_1, x_2, ... , x_n is the sequence which is expected to converge to > the solution of the non-linear system. You may have been having trouble deriving one case from the other because the scalar form appears garbled. The scalar-valued form is the special case of the vector-valued form in which the vector has a single component. It's a bit misleading to divide by the Jacobian. If the Jacobian is non-singular and the number of independent variables equals the number of components in vector f, then multiplying by the inverse of the Jacobian makes more sense. If it is thoroughly clear to you what's going on and you want to abuse the notation in this fashion, I'd accept your artistic license to do so. The correct statement of a Newton iteration for scalar valued f(x) is: x_n = x_{n-1} - f(x_{n-1})/f '(x_{n-1}) > the second point I am trying to understand is how do you pick up the best > guess I don't mean Lipschitz or Kantrovic methods but rather, do you use > the inverse function theorem to pick your neighborhood? Picking a good starting guess is a critical part of any successful application of Newtons method. Some problems have global convergence, but typically in the neighborhood of a simple root/solution, there is a basin of attraction within which any starting point leads to a convergent sequence of Newton iterates. A common strategy is to use a more robust root-finding method for a few iterations with hope of getting into such a basin of attraction for the desired root. Some knowledge of the specific problem or experimentation is usually needed to develop a robust approach. === Subject: Singleton codomain -- how to write I want to describe a little function I have cooked. It returns either the smpty set, or a singleton set of an element from a universe Sigma. How do I write that nicely? Has it returned any subset of Sigma, I would have written f: blahblah --> 2^Sigma, but what's the normal notation for singletons sets, or for singleton-or-empty sets? Soren === Subject: Re: Singleton codomain -- how to write > I want to describe a little function I have cooked. It returns either > the smpty set, or a singleton set of an element from a universe Sigma. > How do I write that nicely? Has it returned any subset of Sigma, I would > have written f: blahblah --> 2^Sigma, but what's the normal notation for > singletons sets, or for singleton-or-empty sets? f: D -> { {x} in P(X) : x in X} for clarity, or f: D -> { {x} : x in X} for brevity? How's that? === Subject: Re: Equations I see in Economics > On 16 Dec 2005 20:28:04 -0800, Hunter > >> >> >> >>It really has been pretty mysterious. How's the average Joe >>to account >>for the astounding loss in value of the dollar in a period >>when >>inflation seems to have been vigorously controlled? Are any >>of the >>experts pointing fingers at the particular policies, >>relevant policy >>makers and describing the problems in English? I don't see >>much of it >>in the financial press--to the extent I read it. > 1. Huge deficits in federal spending > 2. Zero or negative personal savings rates > 3. Huge trade deficits with other countries. > Thats all happening. And its the driving force for the > actual mechanism of the dollar going worthless, as wages > stagnate. > The Fed is 'fixing' the problems you mention by issuing funny > money, that is covertly giving or loaning money to the banks, > and instructing them to give low interest loans to non credit > worthy people... a way to float the economy and mask the > problems you have listed...it is also using the funny money to > make massive stock purchases ever time the stock market takes > a dip... The Fed itself is using our funny money to invest in stocks!? Tell me it ain't so. > Investors are thus encouraged to keep investing in the mostly > bogus stock market... its hot air. > Will it collapse soon? Not if the hyper inflation > continues, the stock market will continue 'up'... but when you > sell the stock and spend the money, it wont buy much... right > now the DOW supposedly in the 10,000 dollar range is actually > around 6,000 in 1999 dollars. So it has actually *lost 40% > not returned to its 1999 peak. The man on the street > sometimes figures that out when he needs the money to buy a > house, medical care or gas. > The same stock that if sold in 1999 would have bought him a > house, will only buy half of the house today. How's the guy invested in real estate going to do? Is there any way to win this game? Or perhaps more correctly, who are the winners? === Subject: TURN $6 INTO $60,000!!!! HOW TO TURN $6 INTO $60,000!!!!!! READING THIS COULD CHANGE YOUR LIFE! All 2 messages in topic - view as tree webstream1 Jul 12, 11:15 pm author Local: Tues, Jul 12 2005 11:15 pm === Subject: HOW TO TURN $6 INTO $60,000!!!!!! READING THIS COULD CHANGE YOUR LIFE! Reply to Author | Forward | Print | Individual Message | Show original | Report Abuse I found this on a bulletin board and decided to try it. 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This is now my fourth week and I have made a total of just over $42,000.00 and it's still coming in rapidly. It's certainly worth $6.00, and 6 stamps, I have spent more than that on the lottery!! Let me tell you how this works and most importantly, why it works.... Also, make sure you print a copy of this Article NOW, so you can get the information off of it as you need it. I promise you that if you follow the directions exactly, that you will start making more money than you thought possible by doing something so easy! Suggestion:Read this entire message carefully! (print it out or download it.) Follow the simple directions and watch the money come in! It's easy. It's legal. And, your investment is only $6.00 (Plus postage) IMPORTANT: This is not a rip-off; it is not indecent; it is not phoney; and it is virtually no risk - it really works!!!! If all of the following instructions are adhered to, you will receive extraordinary dividends. PLEASE NOTE: Please follow these directions EXACTLY, and $50,000 or more can be yours in 20 to 60 days. This program remains successful because of the honesty and integrity of the participants. Please continue its success by carefully adhering to the instructions. You will now become part of the Mail Order business. In this business your product is not solid and tangible, it's a service. You are in the business of developing Mailing Lists. Many large corporations are happy to pay big bucks for quality lists. However, the money made from the mailing lists is secondary to the income, which is made from people like you and me asking to be included in that list. Here are the 4 easy steps to success: STEP 1: Get 6 separate pieces of paper and write the following on each piece of paper; PLEASE PUT ME ON YOUR MAILING LIST. Now get 6 US $1.00 bills and place ONE inside EACH of the 6 pieces of paper so the bill will not be seen through the envelope (to prevent thievery).Next, place one paper in each of the 6 envelopes and seal them. You should now have 6 sealed envelopes, each with a piece of paper stating the above phrase, your name and address, and a $1.00 bill. What you are doing is creating a service. THIS IS ABSOLUTELY real! You are requesting a legitimate service and you are paying for it! Like most of us I was a little skeptical and a little worried about the legal aspects of it all. So I checked it out with the U.S. Post Office (1-800-725-2161) and they confirmed that it is indeed legal! Mail the 6 envelopes to the following addresses; (#1) Brian Kuhns 15 Chestnut Dr. Doylestown, PA 18901. (#2) I. Linderoth 1060 Township Line Rd. Phoenixville, PA 19460. (#3) E. Butler 5844 Cobble Trail Columbus, GA 31907. (#4) CHRISTIAN GEENER 381 Casa Linda Plaza #193 Dallas, TX 75218. (#5) D. Nelson P.O. Box 459 Midway (#6) B. Alexander 2800 Reisterstown Road Baltimore, MD. 21215 City, CA 92655. STEP 2: Now take the #1 name off the list that you see above, move the other names up (6 becomes 5, 5 becomes 4, etc...) and add YOUR Name as number 6 on the list. STEP 3: Change (I think there are close to 24,000 groups) All you need is 200, but remember, the more you post, the more money you make! This is perfectly real! If you have any doubts, refer to Title 18 Sec. 1302 & 1341 of the Postal lottery laws. Keep a copy of these steps for yourself and, whenever you need money, you can use it again, and again. PLEASE REMEMBER that this program remains successful because of the honesty and integrity of the participants and by their carefully adhering to the directions. Look at it this way. If you are of integrity, the program will continue and the money that so many others have received will come your way. NOTE:You may want to retain every name and addresses sent to you, either on a computer or hard copy and keep the notes people send you. This VERIFIES that you are truly providing a service. (Also, it might be a good idea to wrap the $1 bill in dark paper to reduce the risk of mail theft.) So, as each post is downloaded and the directions carefully followed, six members will be reimbursed for their participation as a List Developer with one dollar each. Your name will move up the list geometrically so that when your name reaches the #1 position you will be receiving thousands of dollars in CASH!!! What an opportunity for only $6.00 ($1.00 for each of the first six people listed above) Send it now, add your own name to the list and you're in business! ---DIRECTIONS ----- FOR HOW TO POST TO NEWSGROUPS------------ Step 1) You do not need to re-type this entire letter to do your own posting. Simply put your cursor at the beginning of this letter and drag your cursor to the end of this paper, and select 'copy' from the edit menu. This will copy the entire letter into the computer's memory. Step 2) Open a blank 'notepad' file and place 'paste'. This will paste a copy of the letter into notepad so that you can add your name to the list. Step 3) Save your new notepad file as a .txt file. If you want to do your postings in different settings, you'll always have this file to go back to. Step 4) Use Netscape or Internet explorer and try searching for various newsgroups (on-line forums, message boards, chat sites, discussions.) You can search on millionaire message board, money making message board, employment message board, money making discussions, money making forums, and business message boards. Step 5) Visit these messages boards and post and selecting paste from the edit menu. Fill in the Subject, this will be the header that everyone sees as they scroll through the list of postings in a particular group, click the post message tab. You're done with your first one! Congratulations...THAT'S IT! All you have to do is jump to different newsgroups and post away, after you get the hang of it, it will take about 30 seconds for each newsgroup! * *REMEMBER, THE MORE NEWSGROUPS YOU POST IN, THE MORE MONEY YOU WILL MAKE!! BUT YOU HAVE TO POST A MINIMUM OF 200** That's it! You will begin receiving money from around the world within days! You may eventually want to rent a P.O.Box due to the large amount of mail you will receive. If you wish to stay anonymous, you can invent a name to use, as long as the postman will deliver it. **JUST MAKE SURE ALL THE ADDRESSES ARE CORRECT.** Now the WHY part: Out of 200 postings, say I receive only 5 replies (a very low example). So then I made $5.00 with my name at #6 on the letter. Now, each of the 5 persons who just sent me $1.00 make the MINIMUM 200 postings, each with my name at #5 and only 5 persons respond to each of the original 5, that is another $25.00 for me, now those 25 each make 200 MINIMUM posts with my name at #4 and only 5 replies each, I will bring in an additional $125.00! Now, those 125 persons turn around and post the MINIMUM 200 with my name at #3 and only receive 5 replies each, I will make an additional $626.00! OK, now here is the fun part, each of those 625 persons post a MINIMUM 200 letters with my name at #2 and they each only receive 5 replies, that just made me $3,125.00!!! Those 3,125 persons will all deliver this message to 200 newsgroups with my name at #1 and if still 5 persons per 200 newsgroups react I will receive $15,625,00! With an original list, you just take the latest posting in the newsgroups, and send out another $6.00 to names on the list, putting your name at number 6 again. And start posting again. The thing to remember is: do you realize that thousands of people all over the world are joining the So, can you afford $6.00 and see if it really works?? I think so... People have said, & quote; what if the plan is played out and no one sends you the money? So what! What are the chances of that happening when there are tons of new honest users and new honest people who are joining the Internet and newsgroups everyday and are willing to give it a try? Estimates are at 20,000 to 50,000 new users, every day, with thousands of those joining the actual Internet. Remember: play FAIRLY and HONESTLY and this will really work!! Much SUCCESS TO ALL OF YOU! InternetZone Jul 16, 4:57 pm this author === Subject: Re: HOW TO TURN $6 INTO $60,000!!!!!! READING THIS COULD CHANGE YOUR LIFE! Reply to Author | Forward | Print | Individual Message | Show original | Report Abuse http://www.safelists-4-u.com/cgi-bin/Blessed/cl.cgi?230 === Subject: Re: How to cheat in exams using mobile phones and calculators >> No. No should about it. The issue of top or bottom posting is >> an issue of personal preference. >> No, it is a matter of common courtesy and sense. If you want to end up in >> many killfiles, post how you like. If however, you want to communicate, >> then >> do so in a legible and conventional manner. Posting upside down is just >> showing your contempt for the readership at large. > As to legibility - the ASCII text presenting my post is the same ASCII > that presents yours. As to convention - the sentence structure and > words that > I use are from the same language that you use. What causes > you difficulty in reasing them? Your spelling, for one. -- Andy. === Subject: Completeness of C_o(X) In http://www-math.mit.edu/~rbm/18.155-F04-notes/Lecture-notes.pdf space C o(X), where X is a metric space, is defined as C o(X) = { f in C(X) | for all E > 0 there is a compact subset K of X so that sup {x not in K} |f(x)| <= E } Is the space C o(X) complete? Do we have to assume that X is complete and/or locally compact for the completeness of C o(X)? I'm particularly interested in the case X=R^n. BTW, does C(R) mean the set of all continous real functions or the set of all continuous and bounded real functions? Suppose f is a function from R^n to C. Is lim {|x|->infty} f(x) = a equivalent to for all E > 0 there is a r > 0 so that for all x in R^n, |x| > r : | f(x) - a | < E ? Can you recommend a good book for this kind of subjects in functional analysis? Particularly for the commonly encountered function spaces. - Tommi H.9ayn.8al.8anmaa - === Subject: Re: Abstract question: sample size vs. lot size OpenPGP: id=BC5E1282; url=random.sks.keyserver.penguin.de -----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 You may reduce the problem to the folowing Question: You habe a Bag with 200 Balls that coud be white or black. You extract a number of x Balls and they are all white Waht is the probablilti that all 200 Balls are white. or viceversa: How many white Balls (y) do you have to extract, so that you may say that all the rest of 200 Balls are with a probability of x% white. -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.1 (GNU/Linux) Comment: Using GnuPG with Thunderbird - http://enigmail.mozdev.org iD8DBQFDpcdrjRWhlrxeEoIRAot9AJ97LRKSPDQCYFCNJkbtjrCzR8aJZQCfVlft n034N17qwPBcMl4RRJ7Ill8= =digf -----END PGP SIGNATURE----- === Subject: Re: Abstract question: sample size vs. lot size >-----BEGIN PGP SIGNED MESSAGE----- >Hash: SHA1 >You may reduce the problem to the folowing Question: >You habe a Bag with 200 Balls that coud be white or black. >You extract a number of x Balls and they are all white >Waht is the probablilti that all 200 Balls are white. >or viceversa: >How many white Balls (y) do you have to extract, so that you may say >that all the rest of 200 Balls are with a probability of x% white. No, those are the wrong questions. You can't say anything about the probability that all 200 are white. You could do a Bayesian analysis if you had a prior distribution that the 200 balls were taken from, but you don't. The questions statisticians usually answer are: assuming there are n black balls among the 200, 1) what would be the probability of getting at least one black ball in a sample of y balls? 2) how large must y be in order to make that probability at least x? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Abstract question: sample size vs. lot size > In our manufacturing process, we produce a product that gets shipped to > the customer in lots of 200. The wizards in the QA department have > determined that we need a sample size of 13 to pass a destructive test > with zero failures. So we actualy produce 213 units per shipment. > One day one of the wizards showed me the computer program that they use > to calculate such things. Essentially, he keys in the lot size of 200 > and gets out a sample size of 13. The 13 represent the 200 we ship. > My question is: since we actually produce 213, doesn't that mean that > any process variability would be spread over 213 products rather than > 200.....and so the lot size should really be 213? I have not reviewed the math. It is my impression that typically the sample is considered part of the lot in such calculations. So I think you are correct in thinking the lot size is 213. > And, of course, if he keyed in 213 as the lot size, then some *other* > number would get spit out for the sample size......and we'd end up > producing even more product....which would mean the actual lot was even > larger.....so on and so forth..... > What am I missing??? I don't see why there couldn't be a lot size, say 215, where the sample size was 15. As I recall these sorts of calculations, there is something akin to rounding up the sample size. So it could be that a lot size of 214 requires a sample size slightly greater than 214, while a lot size of 215 requires a sample size of 15. By the way, I hope you are not rely solely on testing to achieve reliability. Do you have a program that includes design and the monitoring of processes? These programs might have names like Total Quality Management or Six Sigma. If you are in the U.S. and have further questions, you might find it helpful to contact the people running this site: -- Mostly economics: r c v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Abstract question: sample size vs. lot size > In our manufacturing process, we produce a product that gets shipped to > the customer in lots of 200. The wizards in the QA department have > determined that we need a sample size of 13 to pass a destructive test > with zero failures. So we actualy produce 213 units per shipment. > One day one of the wizards showed me the computer program that they use > to calculate such things. Essentially, he keys in the lot size of 200 > and gets out a sample size of 13. The 13 represent the 200 we ship. > My question is: since we actually produce 213, doesn't that mean that > any process variability would be spread over 213 products rather than > 200.....and so the lot size should really be 213? > And, of course, if he keyed in 213 as the lot size, then some *other* > number would get spit out for the sample size......and we'd end up > producing even more product....which would mean the actual lot was even > larger.....so on and so forth..... > What am I missing??? the sample size only indicates that the probability of part failure is less than 98% or so. larger sample size greater confidence, ie if all passed using 26 parts, failure rate could be less than 99.9%. your lot sizes could be in the thousands and 13 samples would still be good. Small Lot size of 200 indicate you probably have a high failure rate as the part may be complex to produce, or your orders are small, or unit cost is high. equn. They also have the # required for testing if 1 failes or if 2 failes (besides no failures), to meet a predicted failure rate. === Subject: Re: Abstract question: sample size vs. lot size <43a4e9f0$0$61683$892e7fe2@authen.yellow.readfreenews.net> question. Should the lot size be considered as the number of items shipped or the total number of items produced? Thanx! === Subject: Re: Abstract question: sample size vs. lot size > question. > Should the lot size be considered as the number of items shipped or the > total number of items produced? > Thanx! not realy cases you have to test 13 The Curves are pretty close, the test rate aproxamtes C*ln(batch size) -- The world is flat it's pi that's round! There is only one number. === Subject: A first-order PDE I was trying to help someone the following exercise for an advanced calculus course. All derivatives are partials. Consider the partial-differential equation y dz/dx + x dz/dy = 0. Use the change of coordinates u = x^2 - y^2, v = 2 x y to convert this to into equation only in u. I could not solve this. By the usual application of the chain rule, the best I could come up with was (x^2 + y^2) dz/dv = 0, and x^2 + y^2 is not a nice function of u and v. Am I missing something? Or did the guy write copy the problem wrong? (He only had a written copy of the problem.) -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Dave Touretzky, the net-lifter the alleged free speech activist David Touretzky goes after publications that he doesn't like. Why doesn't he say that the only free speech he stands for is just his own free speech? BTW, his claim that I am obsessed with him is completely ridiculous. If I would have not been harassed with a porn letter about his sex toy purchases, which he ordered from his CMU office, I would probably never have researched him. All I want is that his wrongful activities are being corrected, and I gladly will forget that he exists. Barbara Schwarz -- http://www.thunderstar.net/~Schwarz/ (About Dave Touretzky) He also has bomb instructions on the net: http://www.religiousfreedomwatch.org/extremists/touretzky00.html How David Touretzky censored my website My name is Barbara Schwarz and for the past few years I have been attempting to expose racist statements made by a rat brain researcher at Carnegie Mellon University, David Touretzky. I have also been attempting to expose the fact this individual has purchased sexual implements using the university's phone number. Touretzky has systematically used his position as research scientist to intimidate every single ISP or web hosting to remove my website. Touretzky threatens with libel suits, but he never provided any evidence as to what the libelous statement was. Touretzky has never threatened to sue me for what I have made available on the web. Instead he goes behind my back and lies to ISPs and threatens them with a law suit. Why do you think he never sued me? Because he knows I have the documentation to prove what I write in my website. He knows that in discovery I will subpoena his telephone records, the places where he purchased his sex toys, using Carnegie Mellon facilities. Touretzky has consistently attacked my right to express my opinion. He has censored my website many times and is trying to silence me for telling the truth about him. Touretzky claims to be a free speech advocate, but not when someone is critical of him. For example, Touretzky is mirroring bomb information and promotes that website from his Carnegie Mellon University site. He uses a public chat room to make racist statements against African Americans, Hispanics, and other ethnic groups. Touretzky was confronted with his racist statements in a radio show and he did not deny them. Here are some of the racist statements David Touretzky made in a public chat forum: She [Diane Watson] is the former ambassador to Micronesia! and she's black. I should have known. What are all the really st00000pid congresswomen black? The black underclass is a very, very sick culture, with terrible internal problems such as high rates of black-on-black murder, teenage pregnancy, educational failure, drug abuse, and so on. AA programs can't help the black underclass, because those people are so screwed up they're virtually unemployable. ... ... they republished a long rant of mine about how women should have sole responsibility for conception ... a rant which was very well worded, and which I'm rather proud of now that they reminded me of it. There's also a long rant about the sickness of black underclass culture, which again, I completely stand behind. ... the Salvation Army ... buncha heads. Man, Hispanics are ed up, which is why they're still working class. Dips. the Muslims. .... check out the bloody Muslims .. jeez what loons. Hell, I'm for post-partum abortion! Retro-active abortion! Up until, say, the age of six months. Here is the site that Touretzky doesn't want you to see: PUBLIC SERVICE ANNOUNCEMENT To parents of students at Carnegie Mellon University If you are the parent of a student who attends Carnegie Mellon University, this PSA is to help you. David Touretzky is a research scientist at Carnegie Mellon University, and he teaches some of your children. David Touretzky owns a website that contains instructions on how to build bombs. One of these instructions encourages people to throw the bomb into police cars. Touretzky mirrored a site that was removed by the FBI after its original creator was arrested and prosecuted. David Touretzky claims he created the mirror as a matter of free speech. In my view this is not an innocent lack of judgment by David Touretzky for, as they say, where there is smoke there is fire, and Touretzky is involved in other activities. For example, during his work time he associates with individuals on the internet who harass religions and ethnic groups. Some of them posted threats against people they dislike. Touretzky has perversions that parents of CMU students should be informed about. He is a customer of sex shops and purchases sexual implements. The invoice posted on this website is the evidence of such activity, and it also shows that Touretzky used his University office phone number for the sex shop to contact him. You can compare the phone number listed in the invoice, 412 268-7561 (www.csd.cs.cmu.edu/people/faculty_s-z.html), with the phone number listed in the university directory which goes to David Touretzky's office. Touretzky has pornographic photographs on his CMU website, which he covered in part and removed some after I exposed him. There are some questions you should ask yourself when it comes to David Touretzky: Why would Touretzky provide instructions on how to build explosives? What if a student, like it happened in Columbine, downloads Touretzky's information, builds it and sets it off against people, who would be responsible? Why would the University allow Touretzky to use his office phone to order sex toys? Who is paying for Touretzky's activities when he does it from the University? Is it your tax dollar or tuition fees? There are deranged people and wannabe terrorists out there who want to know how to build explosive devices and use them against innocent victims. Even if Touretzky believes he has no responsibility for what he puts up on his website, he will eventually be accountable for the action that someone will undertake after having downloaded his bomb instructions and set it off. Let Carnegie Mellon University know about your views of David Touretzky. Barbara Schwarz === Subject: Re: Continuum hypothesis <0ainf.3227$j7.77401@news.indigo.ie> > Two things to note: > 1. You did not provide an example of a way to provide a uniform > distribution on the naturals. Not on the _set_ of the naturals, indeed, but on the _sequence_? > 2. The result in the example you gave was also a function of the > ordering you placed on the naturals (the usual ordering). If you had > chosen a different ordering, you would have gotten a different result. [ ... example OK but deleted here ... ] > This is why you must be careful about what you mean when you say it is > equally probable to choose an even number or an odd number from the > /set/ of all naturals. Not just the set, but also the ordering of the > set, is relevant in this case. Yes.That has been pointed out to me in the past, by Dik T. Winter: Han de Bruijn === Subject: Re: Probability problem. > Let S be a random subset of size k from {1, 2, 3,..., n} (0 < k > <= n). Let Y = min S. What is EY? > The discussion in this thread suggest the following calculation: > Let X be uniformly distributed on {1, 2., 3,...., n}, with X > independent ot Y. > P{X <= Y} = EP(X<=Y | Y) = E[Y/n] = EY / n. > P{X <= Y} = P(X<=Y | X in S) P{X in S} + P(X<=Y | X not in S) P{X > not in S} > = 1/k k/n + 1/(k+1) (n-k)/n > = (n+1) / [n(k+1)] > Setting the two equal, we get EY = (n+1) / (k+1). More generally, for any nonnegative random variable X: If 0 <= X <= a almost surely, then EX = a P{X >= U}, where U has Uniform(0,a) distribution and X and U are independent. If the range of X is contaned in {0, 1, 2,..., n}, then EX = n P{X >= U}, where U has uniform distribution on {1, 2,..., n} and X and U are independent. These follow for the formulae of the expectaton as the integral/sum of the tail. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: The Truth <83471$43a14141$82a1e3ad$11965@news1.tudelft.nl> > > > What is the point? > > Again. What would be the good and the bad for the following function: > > f(x) = 1/x ? > > f(x) has no goal here [ ... snip ... ] > Not true. See below. > Define the Goal of f(x) and will tell you the answer? > No doubt, the goal of f(x) is to calculate 1/x from x . > Now tell me the answer, as you promised. > Han de Bruijn > If x=0 this is bad for f(x) > if x<>0 this is good for f(x) Exactly! Han de Bruijn