mm-55 === Is there an integer polynomial in Z[x] with roots sqr 2, sqr 3?>> That depends... if you require the polynomial to have sqrt 2 and sqrt> 3 as _the_ roots, then the answer is no. Clearly one polynomial with> these roots is f(x) = (x-sqrt(2))(x-sqrt(3)) = x^2 -> (sqrt(2)+sqrt(3))x + sqrt(6). Any other polynomial with exactly> sqrt(2) and sqrt(3) as roots is a constant times this polynomial. Now> If you require the polynomial to belong to Z[x], then you can only> multiply f by integers; but since sqrt(6) is irrational, no polynomial> in Z[x] is an integer multiple of f.>> If you only require that sqrt(2) and sqrt(3) are roots of the> polynomial, then the answer is clearly yes. Try g(x) => (x-sqrt(2))(x+sqrt(2))(x-sqrt(3))(x+sqrt(3))>Indeed, however (x^2 - 2)(x^2 - 3) is reducible.Is there a irreducible poly including sqr 2, sqr 3 as roots?I think not. Now Q(sqr 6) subset Q(sqr 2, sqr 3) but equality seems most unlikely.Thus for eld F it appears, F(u,v) cant be reduced to F(w) for some w. === > Im trying to nd integer coefcients for a polynomial which has> sqrt{3} + sqrt{2} as a root. Any ideas?> See pages 5 and 6 of my notes on algebraic number theory for a method> of solving all such problems (and an example slightly harder than this).(For the OP):I ended up there while googling for a proof that the algebraicintegers form a ring. Most such proofs (as Prof. Chapman does)establish a connection between an algebraic integer and a matrix for which the algebraic number is an eigenvalue, and they show how to construct that matrix explicitly. Given algebraic integersa and b, the matrixes A and B are constructed so that Av = av and Bv = bv, with a common eigenvector v.Then (a+b) is an eigenvalue of (A+B) with eigenvector v(and ab is an eigenvalue of AB). The construction showsyou how to relate the matrix to the minimal polynomial. - Randy === > Thus for eld F it appears, F(u,v) cant be reduced to F(w) for some w.There is a theorem that if F is a eld of characteristic zero (such as the rationals) and u and v are algebraic over F then there does exist w such that F(u, v) = F(w). You just have to be a little careful choosing w - not just any old element of F(u, v) will do.-- === > Indeed, however (x^2 - 2)(x^2 - 3) is reducible.> Is there a irreducible poly including sqr 2, sqr 3 as roots?No, since Q(sqrt(2)) isnt isomorphic to Q(sqrt(3)).-- Jim Heckman === >> Thus for eld F it appears, F(u,v) cant be reduced to F(w) >There is a theorem that if F is a eld of characteristic zero >(such as the rationals) and u and v are algebraic over F then >there does exist w such that F(u, v) = F(w). You just have to be >a little careful choosing w - not just any old element of F(u, v) >will do.There is?! How careful must one be? Lets consider u = sqr 2, v = sqr 3, n = degree of w, F = RealsWhos the w with R(w) = R(u,v) ?---- === >> Indeed, however (x^2 - 2)(x^2 - 3) is reducible.> Is there a irreducible poly including sqr 2, sqr 3 as roots?>> No, since Q(sqrt(2)) isnt isomorphic to Q(sqrt(3)).>Nifty. === >> Thus for eld F it appears, F(u,v) cant be reduced to F(w)>There is a theorem that if F is a eld of characteristic zero>(such as the rationals) and u and v are algebraic over F then> >there does exist w such that F(u, v) = F(w). You just have to be>a little careful choosing w - not just any old element of F(u, v)>will do.> There is?! How careful must one be? Lets consider> u = sqr 2, v = sqr 3, n = degree of w, F = Reals> Whos the w with R(w) = R(u,v) ?> Whats wrong with w=sqrt(2)+sqrt(3) ??Then 1/2*w^3-9/2*w = sqrt(2) and 11/2*w-1/2*w^3 = sqrt(3) .Of course if you really mean F=R, then R(w) = R and R(u,v) = R so justany old element WILL do. Was that your point? === > Thus for eld F it appears, F(u,v) cant be reduced to F(w)>There is a theorem that if F is a eld of characteristic zero>(such as the rationals) and u and v are algebraic over F then>there does exist w such that F(u, v) = F(w). You just have to be> >a little careful choosing w - not just any old element of F(u, v)>will do.> There is?! How careful must one be? Lets consider> u = sqr 2, v = sqr 3, n = degree of w, F = Reals> Whos the w with R(w) = R(u,v) ?>> Whats wrong with w=sqrt(2)+sqrt(3) ??> Then 1/2*w^3-9/2*w = sqrt(2) and 11/2*w-1/2*w^3 = sqrt(3) .>ww = 5 + 2sqr 6(w/2)(2sqr6 - 4) = (sqr 6 - 2)(sqr 2 + sqr 3) = sqr 12 - 2.sqr 2 + sqr 18 - 2.sqr 3 = sqr 2(-w/2)(2.sqr 6 - 6) = -(sqr 6 - 3)(sqr 2 + sqr 3) = -(sqr 12 - 3.sqr 2 + sqr 18 - 3.sqr 3) = sqr 3Nothing.ww - 5 = 2.sqr 6(ww - 5)^2 = 24w^4 - 10w^2 - 1 = 0> Of course if you really mean F=R, then R(w) = R and R(u,v) = R so just> any old element WILL do. Was that your point?>No, F = Q.Does the general theorem have a name?How complicated is the proof? === > Thus for eld F it appears, F(u,v) cant be reduced to F(w)>There is a theorem that if F is a eld of characteristic zero>(such as the rationals) and u and v are algebraic over F then>there does exist w such that F(u, v) = F(w). You just have to be> >a little careful choosing w - not just any old element of F(u, v)>will do.>There is?! How careful must one be? Lets consider> u = sqr 2, v = sqr 3, n = degree of w, F = Reals>Whos the w with R(w) = R(u,v) ?>Whats wrong with w=sqrt(2)+sqrt(3) ??>>Then 1/2*w^3-9/2*w = sqrt(2) and 11/2*w-1/2*w^3 = sqrt(3) .>>ww = 5 + 2sqr 6>(w/2)(2sqr6 - 4) = (sqr 6 - 2)(sqr 2 + sqr 3)> = sqr 12 - 2.sqr 2 + sqr 18 - 2.sqr 3 = sqr 2>(-w/2)(2.sqr 6 - 6) = -(sqr 6 - 3)(sqr 2 + sqr 3)> = -(sqr 12 - 3.sqr 2 + sqr 18 - 3.sqr 3) = sqr 3>Nothing.>>ww - 5 = 2.sqr 6>(ww - 5)^2 = 24>w^4 - 10w^2 - 1 = 0>>Of course if you really mean F=R, then R(w) = R and R(u,v) = R so just>>any old element WILL do. Was that your point?>>No, F = Q.>>Does the general theorem have a name?>Theorem 5.3 (or whatever).>How complicated is the proof?>Its in the chapter on Galois theory. If the book is conversational, its easy. If its formal, it could be pretty hard to understand. (YMMV, some people prefer formal to conversational.) Its not complicated at all, its straightforward and usually done right where it makes sense. Its just that some books are hard to follow because of the way theyre written. (Or the way I process what I read.)I like Van der Waerden, _Algebra_ or _Modern Algebra_, depending on your version. Im suggesting it because from some other things you seem to like an old-fashioned approach to algebra, so your point of view may agree with his and make it really easy. But any algebra book that covers Galois theory will do. (Not Herstein _Topics in Algebra_, which I consider the best introduction to algebra.)Jon Miller === ... >> Thus for eld F it appears, F(u,v) cant be reduced to F(w) > >There is a theorem that if F is a eld of characteristic zero >(such as the rationals) and u and v are algebraic over F then >there does exist w such that F(u, v) = F(w). You just have to be >a little careful choosing w - not just any old element of F(u, v) >will do. > There is?! How careful must one be? Lets consider > u = sqr 2, v = sqr 3, n = degree of w, F = Reals > Whos the w with R(w) = R(u,v) ?(Lets do it over rationals, yes? u and v are elements of the reals.)If z = sqrt(2) + sqrt(3), (z^3 - 9z)/2 = sqrt(2).So w = z works.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === [.snip.]>>Does the general theorem have a name?>Theorem 5.3 (or whatever).Or the Primitive Element Theorem, since it guarantees the existence ofa primitive element for any nite separable extension in those elds. === === === =Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === = === ==Arturo Magidinmagidin@math.berkeley.edu === [in re: the Primitive Element Theorem,]> ...any algebra book that covers Galois theory will do. I think the exception here is Artins book. If I remember some other discussion correctly, Artin had something against using this theorem in an exposition of Galois Theory, and studiously avoided it.-- === > Indeed, however (x^2 - 2)(x^2 - 3) is reducible.> Is there a irreducible poly including sqr 2, sqr 3 as roots?>> > No, since Q(sqrt(2)) isnt isomorphic to Q(sqrt(3)).>> Nifty.Another approach: Every algebraic element over F is the root ofa *unique* monic irreducible polynomial over F. For sqrt(2) its(x^2 - 2), and ...-- Jim Heckman === > > Indeed, however (x^2 - 2)(x^2 - 3) is reducible.> Is there a irreducible poly including sqr 2, sqr 3 as roots?> >> No, since Q(sqrt(2)) isnt isomorphic to Q(sqrt(3)).> >> Nifty.> Another approach: Every algebraic element over F is the root of> a *unique* monic irreducible polynomial over F. For sqrt(2) its> (x^2 - 2), and ...The number z = sqrt(3) + sqrt(2), is a zero of x^4 - 10*x^2 + 1, which shows that z is an algebraic integer and the reciprocal of an algebraic integer. === >> [in re: the Primitive Element Theorem,]>> ...any algebra book that covers Galois theory will do.>> I think the exception here is Artins book. If I remember some other> discussion correctly, Artin had something against using this theorem> in an exposition of Galois Theory, and studiously avoided it.If youre talking about Michael Artins _Algebra_, he doesindeed prove the Primitive Element Theorem. In fact, he thengoes on to use it in his proof that For any nite extensionK/F, the order |G(K/F)| of the Galois group divides the degree[K:F] of the extension.One thing I nd somewhat disconcerting about Artins expositionof Galois Theory is that he proves several important theoremsout of order, i.e., he states them early on, proving them onlyin later sections. To my mind this makes following the logicaldevelopment of the ideas a little trickier than need be.-- Jim Heckman === >> [in re: the Primitive Element Theorem,]>> ...any algebra book that covers Galois theory will do.>> I think the exception here is Artins book. If I remember some other>> discussion correctly, Artin had something against using this theorem>> in an exposition of Galois Theory, and studiously avoided it.> If youre talking about Michael Artins _Algebra_,No, hes talking about Emil Artins book on Galois theory.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === > > [in re: the Primitive Element Theorem,]> ...any algebra book that covers Galois theory will do. > I think the exception here is Artins book. If I remember some other > discussion correctly, Artin had something against using this theorem > in an exposition of Galois Theory, and studiously avoided it.Yes, but Artin still discusses when a eld is generatedby a single element in his book Galois Theory. It appearson page 64 (of 82 pages) in section M called Simple Extensions.In this section, Artin gives two theorems. The rst theoremgives a necessary and sufcient condition for the existenceof primitive elements: Theorem 26. A nite extension E of F is primitive over F if and only if there are only a nite number of intermediate elds.Theorem 27 then gives the situation being discussed in this thread.To apply Theorem 27 theorem, we need to note that sqrt(2) and sqrt(3)are separable elements over Q. That is, the irreducible polynomialsfor sqrt(2) and sqrt(3) do not have repeated roots.I have not yet grasped the impact that separable elements hasin Galois theory. I just usually assume that I am working overextensions of Q, where separable follows because the characteristicof Q is 0.To nd a couterexample to having a primitive element, I thinkthat you need to have a nite extension E of F, where F is aninnite eld of characteristic p. But, I havent pursued this.-- Bill Hale === > [in re: the Primitive Element Theorem,]> ...any algebra book that covers Galois theory will do. > I think the exception here is Artins book. If I remember some other > discussion correctly, Artin had something against using this theorem > in an exposition of Galois Theory, and studiously avoided it.Yes, but Artin still discusses when a eld is generatedby a single element in his book Galois Theory. It appearson page 64 (of 82 pages) in section M called Simple Extensions.In this section, Artin gives two theorems. The rst theoremgives a necessary and sufcient condition for the existenceof primitive elements: Theorem 26. A nite extension E of F is primitive over F if and only if there are only a nite number of intermediate elds.Theorem 27 then gives the situation being discussed in this thread.To apply this theorem, we need to note that sqrt(2) and sqrt(3)are separable elements over Q. That is, the irreducible polynomialsfor sqrt(2) and sqrt(3) do not have repeated roots.I have not yet grasped the impact that separable elements hasin Galois theory. I just usually assume that I am working overextensions of Q, where separable follows because the characteristicof Q is 0.To nd a couterexample to having a primitive element, I thinkthat you need to have a non-nite extension E of Z_p, thenite eld of order p. But, I havent pursued this.-- Bill Hale === So I was doing an elementary rate of change on x with respect to time giventhe change in y w.r.t time. I also know that x and y are the two sides ofthe triangle, connected by the constant hypotenous h. Now we all know thefollowing:h^2 = x^2 + y^2so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing relations,constants are left out and...-x^2 (proportional to) y^2 #thusx (proportional to) yso since the h^2 is constant, the change in x should reect the change in ysince x^2 is proportional toy^2. However, this is not the case, meaningthat if we have a ladder standing on the oor and leaned against a wall, it begins to slide, it will not slide on x proportional to y.So the question is why? what am I missing....and perhaps I should stopreviewing math late at night?ThanxDoug === > So I was doing an elementary rate of change on x with respect to time> given the change in y w.r.t time. I also know that x and y are the> two sides of the triangle, connected by the constant hypotenous h. > Now we all know the following:> h^2 = x^2 + y^2> so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing> relations, constants are left out and...> -x^2 (proportional to) y^2 #thus> x (proportional to) y> so since the h^2 is constant, the change in x should reect the> change in y since x^2 is proportional toy^2. However, this is not the> case, meaning that if we have a ladder standing on the oor and> leaned against a wall, if it begins to slide, it will not slide on x> proportional to y. > So the question is why? what am I missing....and perhaps I should stop> reviewing math late at night?> Thanx> Doug Gday Doug,Perhaps too late at night, lets see if I can still get this right (too early in the day? )x^2 = h^2 - y^2 doesnt imply x proportional to y sincex= sqrt (h^2 - y^2)rate of change of x with respect to y:dx/dy = {d[sqrt (h^2 - y^2)]/d(h^2 - y^2)} . d(h^2 - y^2)/dy = {-1/2[sqrt (h^2 - y^2)]} . (-2y) = y/sqrt (h^2 - y^2)so x is not proportional to yJulian-- Local IT BlokeCSIRO, Forestry and Forest Products Ph: +61 8 8721 8118 === > So I was doing an elementary rate of change on x with respect to time given> the change in y w.r.t time. I also know that x and y are the two sides of> the triangle, connected by the constant hypotenous h. Now we all know the> following:> h^2 = x^2 + y^2> so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing relations,x^2 = h^2 - y^2 is neater style> constants are left out and...> -x^2 (proportional to) y^2 #thus> x (proportional to) y>I beg your pardon, but youre learning rules instead of math.y proportional to x when theres some constant c such that y = c * xThinking thus from the basics, you can see that from x^2 = h^2 - y^2x^2 is _not_ proportionally to y^2.Now if y^2 is proportional to x^2, then for some constant c y^2 = c * x^2Hence y = (sqr c) * x and thus as you wished, y is proportional to xand in this case the constant of proportionallity is sqr c. === >h^2 = x^2 + y^2>so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing relations,>constants are left out and...>-x^2 (proportional to) y^2 #thusNope. You can leave out constants in the original equation when youre describing _rates_of_change_, but not when youre describing relations of the original variables.In this case the constant h is (by denition of a right triangle) larger than either x or y. x and y are never proportional for constant h.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/You nd yourself amusing, Blackadder.I try not to y in the face of public opinion. === Julian Mattay typed:> Perhaps too late at night, lets see if I can still get this right (too > early in the day? )> x^2 = h^2 - y^2 doesnt imply x proportional to y since> x= sqrt (h^2 - y^2)> rate of change of x with respect to y:> dx/dy = {d[sqrt (h^2 - y^2)]/d(h^2 - y^2)} . d(h^2 - y^2)/dy> = {-1/2[sqrt (h^2 - y^2)]} . (-2y)> = y/sqrt (h^2 - y^2)> so x is not proportional to yJulain, if you directly differentiate the following equation,x^2 = h^2 - y^2where _h_ is a constant term, and both _x_ and _y_ variables, then you would end up with the answer,dy/dx = -x/ywhich if I have interpreted correctly denes clearly the proportional relation of _x_ and _y_. -- Ayaz Ahmed KhanYours Forever in,Cyberspace. === > Julain, if you directly differentiate the following equation,> x^2 = h^2 - y^2> where _h_ is a constant term, and both _x_ and _y_ variables, then you > would end up with the answer,> dy/dx = -x/y> which if I have interpreted correctly denes clearly the proportional > relation of _x_ and _y_. x and y would only be proportional in that equation if dy/dxwere constant. It is not and so they are not.-- Rich Carreiro rlcarr@animato.arlington.ma.us === exercise. Im still a newbie so forgive and correct me on any errors.If h is the second hand on a clock, and is 2cm long, the rate at whichy (the height) is decreasing when h is at say 2 oclock can be foundusing the following method.----------------------In a right angled triangle YHX where the opposite sides are labelledyhx, then at 2 oclock angle Y=30 at the centre, H=90 near 3 on theclock and X=60 near 2 on the clock.y/sin y = h/sin hy = (hsin y)/sin hy = 1x = sqrt(h^2 - y^2) = sqrt(3)Y is changing at the rate of (pi/30)/second sodY/dt = pi/30We know thaty = (hsin y)/sin Hand since h and sin H are constant at 2 and 1 respectivelyy = 2sin Yanddy/dY = 2cos YTo nd the rate at which y is decreasing we look for dy/dt.dy/dt = (dy/dY)(dY/dt)dy/dt = 2cos(Y)(pi/30)At 2 oclockdy/dt = 2cos(pi/6)(pi/30) = 0.1814y is decreasing at a rate of 0.1814 cm/s----------------------To nd the rate at which x is increasingdx/dt = (dx/dy)(dy/dt)We knowx = sqrt(h^2 - y^2)dx/dy = -y/[sqrt(h^2 - y^2)]So the rate at which x is increasing isdx/dt = -[y/(sqrt(h^2 - y^2))](0.1814) = -0.1047x is increasing at a rate of 0.1047 cm/s.----------------------x is not proportional to y because the fraction x/y should remainunchanged for all values, which is similar to what William said.Dave. === >h^2 = x^2 + y^2> >so : -x^2 = y^2 - h^2 #where h^2 is constant so when describingrelations,>constants are left out and...>-x^2 (proportional to) y^2 #thus>> Nope. You can leave out constants in the original equation when> youre describing _rates_of_change_, but not when youre describing> relations of the original variables.What do you mean by this? Any rate of change equations is also one thatlinks the twovariables whose proportionality were trying to discern. For example:The Volume of a sphere is: V = 4/3(pi)r^3so V is related to r by the above equation, and V is proportional to r^3.>> In this case the constant h is (by denition of a right triangle)> larger than either x or y. x and y are never proportional for> constant h.I must say that I dont understand this statement as well. Of course x andy will not be proportionalto constant h as they change, since its a constant and it doesnt change asthe other variables change.>> --> Stan Brown, Oak Road Systems, Cortland County, New York, USA> http://OakRoadSystems.com/> You nd yourself amusing, Blackadder.> I try not to y in the face of public opinion.Doug === This actually makes the most sense, but can you come up with a quick proof ?You would thinkthat as the rate of change of x or y increases to a signicant amount, theconstant added on would startto become negligable, and thus allow for what Im trying to do, no?Doug>> So I was doing an elementary rate of change on x with respect to timegiven> the change in y w.r.t time. I also know that x and y are the two sidesof> the triangle, connected by the constant hypotenous h. Now we all knowthe> following:> h^2 = x^2 + y^2> so : -x^2 = y^2 - h^2 #where h^2 is constant so when describingrelations,>> x^2 = h^2 - y^2 is neater style>> constants are left out and...> -x^2 (proportional to) y^2 #thus> x (proportional to) y>> I beg your pardon, but youre learning rules instead of math.> y proportional to x when theres some constant c such that> y = c * x>> Thinking thus from the basics, you can see that from> x^2 = h^2 - y^2> x^2 is _not_ proportionally to y^2.>> Now if y^2 is proportional to x^2, then for some constant c> y^2 = c * x^2> Hence y = (sqr c) * x and thus as you wished, y is proportional to x> and in this case the constant of proportionallity is sqr c.> === So seeing how you were waking up when me was hitting the sack, (and by theGDay), Ill assumeyoure from down-under...so thanx for replying AND for bringing us up herein the north theCrocodile Hunter - one crazy bloke ...contradict with a counter example.Im guessing that youre assuming x will only be proportional to y if dy/dx= 1? for example as in:x^2=y^2 #where we all agree that x is proportional to y.x=yd/dy(x) = d/dy(y)x = 1But now consider adding a constant to one side of the equation:x^2=3y^2 #so thatx = (3y^2)^(1/2)and if you nd the derivative of d/dy...this time youll get...(and Illlet you do the math)dx/dy = (3y)/((3y^2)^(1/2))which certainly no longer appears to be proportional> So I was doing an elementary rate of change on x with respect to time> given the change in y w.r.t time. I also know that x and y are the> > two sides of the triangle, connected by the constant hypotenous h.> Now we all know the following:> h^2 = x^2 + y^2> so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing> relations, constants are left out and...> -x^2 (proportional to) y^2 #thus> x (proportional to) y>> so since the h^2 is constant, the change in x should reect the> > change in y since x^2 is proportional toy^2. However, this is not the> case, meaning that if we have a ladder standing on the oor and> leaned against a wall, if it begins to slide, it will not slide on x> proportional to y.>> So the question is why? what am I missing....and perhaps I should stop> reviewing math late at night?>> Thanx> Doug>> Gday Doug,>> Perhaps too late at night, lets see if I can still get this right (too> early in the day? )>> x^2 = h^2 - y^2 doesnt imply x proportional to y since> x= sqrt (h^2 - y^2)> rate of change of x with respect to y:> dx/dy = {d[sqrt (h^2 - y^2)]/d(h^2 - y^2)} . d(h^2 - y^2)/dy> = {-1/2[sqrt (h^2 - y^2)]} . (-2y)> = y/sqrt (h^2 - y^2)> so x is not proportional to y>> Julian> --> Local IT Bloke> CSIRO, Forestry and Forest Products Ph: +61 8 8721 8118 === Yeah, I dont think this holds either...go over the example I just put up toJulian Mattays commentand tell me if Im still lost in the woods on this?Thanxdoug>> Julain, if you directly differentiate the following equation,>> x^2 = h^2 - y^2>> where _h_ is a constant term, and both _x_ and _y_ variables, then you> would end up with the answer,>> dy/dx = -x/y>> which if I have interpreted correctly denes clearly the proportional> relation of _x_ and _y_.>> x and y would only be proportional in that equation if dy/dx> were constant. It is not and so they are not.>> --> Rich Carreiro rlcarr@animato.arlington.ma.us === Pardon, fergot to simplify, which would give us:dx/dy = (3y)/((3y^2)^(1/2)) = 3/(3^(1/2)) = constantand now that I think about it, what Rich said makes complete and utter sensewith respectto proportionality equations having constants when ones change is comparedto another.And I know that taking the derivative of d/dy or d/dx will not yield aconstant fory^2 + x^2 = h^2, but why? Simply because we have a constant that is beingadded to both?Doug> So seeing how you were waking up when me was hitting the sack, (and by the> GDay), Ill assume> youre from down-under...so thanx for replying AND for bringing us up here> in the north the> Crocodile Hunter - one crazy bloke ...>> contradict with a counter example.> Im guessing that youre assuming x will only be proportional to y ifdy/dx> = 1? for example as in:> x^2=y^2 #where we all agree that x is proportional to y.> x=y> d/dy(x) = d/dy(y)> x = 1>> But now consider adding a constant to one side of the equation:> x^2=3y^2 #so that> x = (3y^2)^(1/2)> and if you nd the derivative of d/dy...this time youll get...(and Ill> let you do the math)> dx/dy = (3y)/((3y^2)^(1/2))> which certainly no longer appears to be proportional>> So I was doing an elementary rate of change on x with respect to time> given the change in y w.r.t time. I also know that x and y are the> two sides of the triangle, connected by the constant hypotenous h.> Now we all know the following:> h^2 = x^2 + y^2> so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing> > relations, constants are left out and...> -x^2 (proportional to) y^2 #thus> x (proportional to) y>> > so since the h^2 is constant, the change in x should reect the> change in y since x^2 is proportional toy^2. However, this is not the> case, meaning that if we have a ladder standing on the oor and> leaned against a wall, if it begins to slide, it will not slide on x> proportional to y.>> So the question is why? what am I missing....and perhaps I should stop> reviewing math late at night?>> Thanx> Doug>> Gday Doug,> >> Perhaps too late at night, lets see if I can still get this right (too> early in the day? )>> x^2 = h^2 - y^2 doesnt imply x proportional to y since> x= sqrt (h^2 - y^2)> rate of change of x with respect to y:> dx/dy = {d[sqrt (h^2 - y^2)]/d(h^2 - y^2)} . d(h^2 - y^2)/dy> = {-1/2[sqrt (h^2 - y^2)]} . (-2y)> = y/sqrt (h^2 - y^2)> so x is not proportional to y>> Julian> --> Local IT Bloke> CSIRO, Forestry and Forest Products Ph: +61 8 8721 8118>> === >> So I was doing an elementary rate of change on x with respect to time> given the change in y w.r.t time. I also know that x and y are the> two sides of the triangle, connected by the constant hypotenous h.> Now we all know the following:> h^2 = x^2 + y^2> so : -x^2 = y^2 - h^2 #where h^2 is constant so when describingSince you know the change in y with respect to time andwant the change in x with respect to time, differentiatethe entire equation with respect to time: x^2 + y^2 = h^2 d/dt (x^2 + y^2 = h^2) 2x dx/dt + 2 y dy/dt = 0 x dx/dy + y dy/dt = 0You said you were given dy/dt, so solve for dx/dt interms of everything else: x dx/dt = - y dy/dt dx/dt = -(y/x) dy/dtor dx/dt = -(sqrt(h^2 - x^2))/x dy/dtor dy/dt = -y/sqrt(h^2 - y^2) dy/dtAssuming you know either of x or y along with dy/dt, youthus know dx/dt.But to tie into some earlier statements, nothing inany of those equations implies x and y are proportional.Heck, even in something as simple as x + by = h (b, h are non-zero constants)x and y are not proportional.Remember, dy/dx = constant is a necessary *but notsufcient* condition for x and y to be proporational.-- Rich Carreiro rlcarr@animato.arlington.ma.us === I got the problem right, and I know that x and y in pythagoras are notworking out to beproportional, whats bothering me is that I cant see why the wouldnt beproportional. The answerseems to be simply because theres a constant that is added on to one side(the h), and this seems tothrough everything off balance. With respect to the simple example you gavebelow (y = mx +b)you see, y and x ARE proportional if b = 0. Meaning, if we transform thecartesian plane by making surethat our segment goes through the origin, well be ne, and able tocalculate the proportionality of x w.r.t y.And this further cleries (and perhaps nally makes me see) why in thepythagoras eqtn x and y are notproportional....it seems, you cant simply eliminate constants that arentrelated via a multiplication to the variableDoug>> So I was doing an elementary rate of change on x with respect totime> given the change in y w.r.t time. I also know that x and y are the> two sides of the triangle, connected by the constant hypotenous h.> Now we all know the following:> > h^2 = x^2 + y^2> so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing>> Since you know the change in y with respect to time and> want the change in x with respect to time, differentiate> the entire equation with respect to time:> x^2 + y^2 = h^2> d/dt (x^2 + y^2 = h^2)> 2x dx/dt + 2 y dy/dt = 0> x dx/dy + y dy/dt = 0>> You said you were given dy/dt, so solve for dx/dt in> terms of everything else:> x dx/dt = - y dy/dt> dx/dt = -(y/x) dy/dt> or> dx/dt = -(sqrt(h^2 - x^2))/x dy/dt> or> dy/dt = -y/sqrt(h^2 - y^2) dy/dt>> Assuming you know either of x or y along with dy/dt, you> thus know dx/dt.>> But to tie into some earlier statements, nothing in> any of those equations implies x and y are proportional.>> Heck, even in something as simple as> x + by = h (b, h are non-zero constants)> x and y are not proportional.>> Remember, dy/dx = constant is a necessary *but not> sufcient* condition for x and y to be proporational.>> --> Rich Carreiro rlcarr@animato.arlington.ma.us === >h^2 = x^2 + y^2>>so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing>relations,>>constants are left out and...>> >-x^2 (proportional to) y^2 #thus>> Nope. You can leave out constants in the original equation when>> youre describing _rates_of_change_, but not when youre describing>> relations of the original variables.>>What do you mean by this? Any rate of change equations is also one that>links the two>variables whose proportionality were trying to discern. He was talking about an _additive_ constant. Since the derivative of a constant is 0, you can ignore additive constants when taking the derivative.>The Volume of a sphere is: V = 4/3(pi)r^3>so V is related to r by the above equation, and V is proportional to r^3.Sure, but its not proportional to r.>> In this case the constant h is (by denition of a right triangle)>> larger than either x or y. x and y are never proportional for>> constant h.>>I must say that I dont understand this statement as well. Of course x and>y will not be proportional to constant h as they change, since its a constant >and it doesnt change as the other variables change.You misread the preposition. I did not say x and y were non-proportional _to_ constant h, but _for_ (i.e., given or under the circumstances of) constant h.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/You nd yourself amusing, Blackadder.I try not to y in the face of public opinion. === Following your style, I too will top post.A proof of what? If y = cx, then when x = 0, y = 0.Tho y isnt proportional to x, its of the same order of magnitude as x.> This actually makes the most sense, but can you come up with a quick proof ?> You would think> that as the rate of change of x or y increases to a signicant amount, the> constant added on would start> to become negligable, and thus allow for what Im trying to do, no?>> Doug>> So I was doing an elementary rate of change on x with respect to time> given> > the change in y w.r.t time. I also know that x and y are the two sides> of> the triangle, connected by the constant hypotenous h. Now we all know> the> following:> h^2 = x^2 + y^2> so : -x^2 = y^2 - h^2 #where h^2 is constant so when describing> relations,>> x^2 = h^2 - y^2 is neater style>> constants are left out and...> -x^2 (proportional to) y^2 #thus> x (proportional to) y>> > I beg your pardon, but youre learning rules instead of math.> y proportional to x when theres some constant c such that> y = c * x>> Thinking thus from the basics, you can see that from> x^2 = h^2 - y^2> x^2 is _not_ proportionally to y^2.>> Now if y^2 is proportional to x^2, then for some constant c> y^2 = c * x^2> Hence y = (sqr c) * x and thus as you wished, y is proportional to x> and in this case the constant of proportionallity is sqr c. === Rich Carreiro typed:> x and y would only be proportional in that equation if dy/dx> were constant. It is not and so they are not.Upon feeding arbitrary values of _x_ in the equation, and nding with the help of the given equation the values of _y_, I nd that _x_ and _y_ and indeed proportional, but only inversely. -- Ayaz Ahmed KhanYours Forever in,Cyberspace. === >Upon feeding arbitrary values of _x_ in the equation, and nding with >the help of the given equation the values of _y_, I nd that _x_ and >_y_ and indeed proportional, but only inversely. Sorry, but no. If the original equation is x^2 + y^2 = h^2, which I believe it was, then y = +/- sqrt(h^2 - x^2), and theres no way x and y are inversely proportional.x and y would be inversely proportional if and only if xy=const.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/You nd yourself amusing, Blackadder.I try not to y in the face of public opinion. === Im trying to solve an algebra problem and I came up with fourequations (and some side conditions) that I think can be condensed.Here goes:Let G be an abelian group, nite, generated by x and y. Let H be acyclic subgroup of G. Also, suppose it is known that X = c(X-Y), Y =(c-1)(X-Y), and G/H = , where X is the coset of X, Y is thecoset of Y, etc. Ive shown thatblahblahblahblahblahis true if and only if there exist relatively prime positive integersm_1, n_1 such that0 <= [G/H:]*n_1 <= |G/H|-1 andm_1*order(X) + n_1*([G/H:]-e_1) = |G/H|-1,where e_1 is the unique integer satisfying0 <= e_1 <= order(X) and[G/H:]*Y = e_1*X_and_ there exist relatively prime positive integers m_2, n_2 suchthat0 <= [G/H:]*n_2 <= |G/H|-1 andm_2*order(Y) + n_2*([G/H:]-e_2) = |G/H|-1,where e_2 is the unique integer satisfying0 <= e_2 <= order(Y) and[G/H:]*X = e_2*Y.I suspect I can (using relationships between X and Y) reduce thenumber of equations (and conditions) needed to express thisinformation. Heres what Ive tried so far:Since order(X)=|G/H|/gcd(c,|G/H|) and order(Y)=|G/H|/gcd(c-1,|G/H|),we must have order(X)/order(Y) = gcd(c-1,|G/H|)/gcd(c,|G/H|). Callthis quotient 1/D. Then we haveblahblahblahblah is true iff there exist relatively prime positive integers m_1, n_1 such that0 <= |G/H|/order(X) * n_1 <= |G/H|-1 andm_1*order(X) + n_1*(|G/H|/order(X) - e_1) = |G/H|-1,where e_1 is the unique integer satisfying0 <= e_1 <= order(X) and|G/H|/order(X) * Y = e_1*X_and_ there exist relatively prime positive integers m_2, n_2 suchthat0 <= |G/H|/(D*order(X)) * n_2 <= |G/H|-1 andm_2*order(X)*D + n_2*(|G/H|/(D*order(X)) - e_2) = |G/H|-1,where e_2 is the unique integer satisfying0 <= e_2 <= order(X)*D and|G/H|/(D*order(X)) * X = e_2*Y.Even if these equations cant be expressed as a set of fewerequations, is there anything we can say number-theoretically about|G/H|, |H|, or order(X) and order(Y) (besides the _obvious_inequalities obtained from simply combining these equations)??? Anyoneprocient with Mathematica want to give this one a whirl and post anysimplications you discover? (I dont have access to it or any othersymbolic algebra package from where I currently work). === >> Let G be an abelian group, nite, generated by x and y. Let H be a> cyclic subgroup of G. Also, suppose it is known that X = c(X-Y), Y => (c-1)(X-Y), and G/H = , where X is the coset of X, Y is the> coset of Y, etc. Ive shown that>Ill presume G is additive as its Abelian group.H is subgroup. Whats c? Some integer?Whats meant X is coset of X ?Do you mean for x,y in G, let X = x+H, Y = y+H ?Then -Y = -y+H, X-Y = x-y + H, c(X-Y) = c(x-y) + cH ?> blahblahblahblahblah>Whats that? Bushs latest public speach?> is true if and only if there exist relatively prime positive integers> m_1, n_1 such that>> 0 <= [G/H:]*n_1 <= |G/H|-1 and> m_1*order(X) + n_1*([G/H:]-e_1) = |G/H|-1,>> where e_1 is the unique integer satisfying>> 0 <= e_1 <= order(X) and> [G/H:]*Y = e_1*X>> _and_ there exist relatively prime positive integers m_2, n_2 such> that>> 0 <= [G/H:]*n_2 <= |G/H|-1 and> m_2*order(Y) + n_2*([G/H:]-e_2) = |G/H|-1,>> where e_2 is the unique integer satisfying>> 0 <= e_2 <= order(Y) and> [G/H:]*X = e_2*Y.>> I suspect I can (using relationships between X and Y) reduce the> number of equations (and conditions) needed to express this> information. Heres what Ive tried so far:>> Since order(X)=|G/H|/gcd(c,|G/H|) and order(Y)=|G/H|/gcd(c-1,|G/H|),> we must have order(X)/order(Y) = gcd(c-1,|G/H|)/gcd(c,|G/H|). Call> this quotient 1/D. Then we have>> blahblahblahblah>> is true iff>> there exist relatively prime positive integers m_1, n_1 such that>> 0 <= |G/H|/order(X) * n_1 <= |G/H|-1 and> m_1*order(X) + n_1*(|G/H|/order(X) - e_1) = |G/H|-1,>> where e_1 is the unique integer satisfying>> 0 <= e_1 <= order(X) and> |G/H|/order(X) * Y = e_1*X>> _and_ there exist relatively prime positive integers m_2, n_2 such> that>> 0 <= |G/H|/(D*order(X)) * n_2 <= |G/H|-1 and> m_2*order(X)*D + n_2*(|G/H|/(D*order(X)) - e_2) = |G/H|-1,>> where e_2 is the unique integer satisfying>> 0 <= e_2 <= order(X)*D and> |G/H|/(D*order(X)) * X = e_2*Y.>> Even if these equations cant be expressed as a set of fewer> equations, is there anything we can say number-theoretically about> |G/H|, |H|, or order(X) and order(Y) (besides the _obvious_> inequalities obtained from simply combining these equations)??? Anyone> procient with Mathematica want to give this one a whirl and post any> simplications you discover? (I dont have access to it or any other> symbolic algebra package from where I currently work).> === Basic algebra problem in my pre-calc refresher course that I am taking onaudit this summer.Been such a long time (12 years) since I did this that I need someadditional help to help refresh my memory. I used to know this but nowneed to know this again :-) One of our review questions asks to writethe slope-intercept forms of equations of lines that are (a) parallel and(b) perpendicular to the given line (x = 4)...The given point is (2,5)The answer in the back of the book (which does not show how the problem isworked out, just the answer is shown) is(a) = x = 2(b) = y = 5I do not believe that for (a) that the slope-intercept form y = mx + b canbe written since theres no slope as the slope would be undened. Myteacher wont show the steps to this as he says hes not giving it for thetest but he told me that m is undened and should be treated as 0 inthis type of problem. Thats ok that he did not want to work this out onthe board but Im doing this to re-learn and understand this. He said forhis test, he will make sure both X and Y have a value so that m will notbe undened. This is why he did not want to work this out on the board.I understand how to do these problems when theres both a Y and X in theequation but I am confused when either Y or X is missing (as in the caseabove).Can someone show me if my steps are correct -or- if it was overkill in mypart to write out the steps like this for cases where m is undened?For (a) parallel m = m2 (m2 is undened)1. y - y1 = m(x - x1) where m is undened...2. y - y1 / m = x - x1 ==> (y - 5) / m = x - 2 where m isundened -or- innity.3. Since anything over innity = 0 (according to my instructor) , then(y - 5)/{innity} = 04. So rewrite equation as 0 = x - 25. x = 2For (b) parallel m = -1 / m2 (m2 is undened)1. y - y1 = m(x - x1) where m is undened therefore m(x - x1)(according to my instructor) would be rewritten as 0 in this case so go tostep 22. y - y1 = 03. y - 5 = 04. y = 5thanks in advance... === >> Basic algebra problem in my pre-calc refresher course that I am taking on> audit this summer.> Been such a long time (12 years) since I did this that I need some> additional help to help refresh my memory. I used to know this but now> need to know this again :-) One of our review questions asks to write> the slope-intercept forms of equations of lines that are (a) parallel and> (b) perpendicular to the given line (x = 4)...>> The given point is (2,5)>> The answer in the back of the book (which does not show how the problem is> worked out, just the answer is shown) is> (a) = x = 2> (b) = y = 5>> I do not believe that for (a) that the slope-intercept form y = mx + bcan> be written since theres no slope as the slope would be undened.Correct.> My> teacher wont show the steps to this as he says hes not giving it for the> test but he told me that m is undened and should be treated as 0 in> this type of problem._Absolutely_ not. A slope of 0 and an undened slope are two VERYdifferent things. The two lines x=2 and y=5 are perpendicular. The rsthas undened slope while the second has 0 slope. Looks like this teacheris the one who should be sitting in on this class.> Thats ok that he did not want to work this out on> the board but Im doing this to re-learn and understand this. He said for> his test, he will make sure both X and Y have a value so that m will not> be undened.Well, OK, but that completely avoids your question to him.> This is why he did not want to work this out on the board.> I understand how to do these problems when theres both a Y and X in the> equation but I am confused when either Y or X is missing (as in the case> above).>> Can someone show me if my steps are correct -or- if it was overkill inmy> part to write out the steps like this for cases where m is undened?Slope intercept form y=mx+b applies only when you *have* a slope. Verticallines dont have a slope, so stay away from trying to do anything with thisform y=mx+b on equations of the form x=, with no y term. I mean, how wouldyou go about solving for y in x=5? Its just not there...> For (a) parallel m = m2 (m2 is undened)>> 1. y - y1 = m(x - x1) where m is undened...The right side is therefore undened as a result of m being undened....and if the left side is equal to the right, then the lhs is alsoundened. So no, you better stay away from this approach. Theslope-intercept form only applies when you really do have an m, that is, adened slope.Heres what your teacher should have put on the board when you asked, or atthe very least explained after class. The slope of the line x=5, if itexists, is dened as delta(y)/delta(x) for any two distinct points on theline. Pick two. How about (5,0) and (5,1):delta(y) / delta(x)= (0-1) / (5-5)= -1/0, which is an undened expression, so this line doesnt even have aslope, nor does any vertical line. You will get a similar undenedexpression regardless of which two points you select, for any vertical line.Similarly, it can be shown that any horizontal line has a slope of 0 sinceyou will get a delta(y) of 0 and a delta(x) something nonzero.Next, he could have taken that same denition of slope:delta(y) / delta(x) = m, where delta(x)<>0and demonstrated that...(y - y1) / (x - x1) = m ...for some two distinct points (x,y) and (x1,y1)...then multiplied both sides by x-x1, again so long as x-x1 is NOT 0,getting:y - y1 = m(x - x1) ...this is the point-slope formThe y-intercept is dened to be where x=0, right? Call the y-coordinate ofthis point b, that is, let the cordinates of this point be (0,b). Thenlet (x1,y1) *be* the y-intercept, ie let x1 be 0 and y1 be b:y - b = m(x - 0)y - b = mxthen add b to both sides:y = mx + bAs you see, *all* of this is based on m actually existing, ie m is*dened*. With m being undened, as in the case of the line x=5, yousimply dont *get* to the form y=mx+b.Either your teacher is being extremely nonhelpful on purpose (perhaps hedoesnt have time to go over such basics in a precalc class) or he atout doesnt know what hes doing. Normaly, I would give the benet of thedoubt and assume he is just being arrogant and will not explain to yousomething so basic. After all, you really should know this going in toa precalculus course. Its basic algebra. Or do they call basic algebraprecalculus? dunno...However, if he really told you to consider the slope of the line x=5 to be0, then at worst he at out doesnt know what he is doing and shouldnt beallowed to teach mathematics, or at best he likes to lie alot. Its one orthe other. this is not one of those things that you can just have a slip ofthe tonue and inadvertently say consider an undened slope to be a 0slope.> 2. y - y1 / m = x - x1 ==> (y - 5) / m = x - 2 where m is> undened -or- innity.> 3. Since anything over innity = 0 (according to my instructor) ,then> (y - 5)/{innity} = 0You say this is a precalc course? In such a context, innity doesnt even_exist_, much less has a recipricol of 0. Anything over innity is not adened expression, at least not in this context (which is real numbers).Its only when you study alternative number systems (as opposed to thestandard real and imaginary numbers) that you get into things like1/innity.> 4. So rewrite equation as 0 = x - 2> 5. x = 2>> For (b) parallel m = -1 / m2 (m2 is undened)The result that parallel lines have slopes that are negative reciprocols,again, only applies when they *have* slope. A horizontal line has a slopeof 0. 0 has no recipricol, much less a negative recipricol.The above explanation I offered is more appropriate for your context,surely. Doesnt your book have a similar derivation of these forms? Justabout any elementary algebra text should. Even college algebra texts (orprecalculus texts).My advice: Get out of his class as fast as you can and audit someone elsesclass instead, assuming the act of audit is for your benet. If theaudit truly is an audit, as in lets see if anythings wrong here thenreport your ndings.-- Darrell === ><...>> For (b) parallel m = -1 / m2 (m2 is undened)>> The result that parallel lines have slopes that are negative reciprocols,> again, only applies when they *have* slope. A horizontal line has a slope> of 0. 0 has no recipricol, much less a negative recipricol.Seems we both made the same typo here. Replace both occurances ofparallel with perpendicular.-- Darrell === [ccd to previous poster; follow-ups in newsgroup suggested]>One of our review questions asks to write the slope-intercept forms of equations >of lines that are (a) parallel and (b) perpendicular to the given line (x = 4)>The given point is (2,5)>>The answer in the back of the book (which does not show how the problem is>worked out, just the answer is shown) is>(a) = x = 2>(b) = y = 5>>I do not believe that for (a) that the slope-intercept form y = mx + b can>be written since theres no slope as the slope would be undened. My>teacher wont show the steps to this as he says hes not giving it for the>test but he told me that m is undened and should be treated as 0 in>this type of problem.I _hope_ you misheard your teacher. The line x = 2 has a slope of 0; the line y = 5 has undened slope and that can _not_ be treated as 0.Heres how they were worked out.The line x = 4 has no slope (meaning that its slope is undened, not that its slope is 0). Why? slope is rise over run or change in y over change in x. Between any two points on that line, such as (4, 0) and (4, 12),the run or change in x is 0, so you have division by zero (forbidden) when trying to nd slope.The general rule is lines are parallel if and only if their slopes are equal but more precisely it should read ... if their slopes are equal or both slopes are undened. So for the parallel line you need another line with undened slope. The parallel line must go through given point (2, 5). Other points with a run of 0 from that rst point will all have the same x coordinate: (2, 0), (2, -11), (2, 1887.625)and so forth. If all of them must have the same x coordinate, the equation must be x = 2.Perpendicular lines have slopes that are negative reciprocals, you have learned. Again, that should be either their slopes are negative reciprocals, or one has undened slope and the other has slope = 0.So any line perpendicular to x = 4 must have slope of 0 (since x=4 has undened slope). Since you have the slope (0) and a point (2,5) on the desired line, you can start with the point-slope form and then convert to slope-intercept: y - y1 = m(x - x1) y - 5 = 0(x - 2) y - 5 = 0 y = 5But really, I tell my students its easier not to try to treat horizontal and vertical lines as general lines, rather to learn them as special cases: ** The x axis is horizontal, and it has all different x coordinates but the ys are all 0; therefore the equation of the x axis is y = 0. Any other horizontal line must also have equation y = const. ** The y axis is vertical, and the xs are all 0 but it has all different y coordinates; therefore the equation of the y axis is x = 0. Any other vertical line must also have equation x = const.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/Walrus meat as a diet is less repulsive than seal. === Your derogation of the instructor may not be justied. See below.> One of our review questions asks to> write the slope-intercept forms of equations of lines that are (a)> parallel and (b) perpendicular to the given line (x = 4)...>> The given point is (2,5)>> The answer in the back of the book (which does not show how the problem> is worked out, just the answer is shown) is> (a) = x = 2> (b) = y = 5If your text really said to write the slope-intercept forms, then thetext is at fault for part (a). There is no such thing as slope-interceptform for the equation of a vertical line.> I do not believe that for (a) that the slope-intercept form> y = mx + b can> be written since theres no slope as the slope would be undened.>> Correct.>> My> teacher wont show the steps to this as he says hes not giving it for> the test but he told me that m is undened and should be treated as> 0 in this type of problem.>> _Absolutely_ not. A slope of 0 and an undened slope are two VERY> different things. The two lines x=2 and y=5 are perpendicular. The> rst has undened slope while the second has 0 slope. Looks like this> teacher is the one who should be sitting in on this class.Youd be right, Darrell, _if_ thats what the teacher actually said. Butjudging from what music said below (items 2. and 3.), it seems far morelikely that the instructor said something like If m is undened, treatits _reciprocal_ as being 0.> Thats ok that he did not want to work this out on> the board but Im doing this to re-learn and understand this. He said> for his test, he will make sure both X and Y have a value so that m> will not be undened.>> Well, OK, but that completely avoids your question to him.I agree that that avoidance is unfortunate.> > This is why he did not want to work this out on the board.> > I understand how to do these problems when theres both a Y and X in> the equation but I am confused when either Y or X is missing (as in the> case above).>> Can someone show me if my steps are correct -or- if it was overkill> in my part to write out the steps like this for cases where m is> undened?>> Slope intercept form y=mx+b applies only when you *have* a slope.Specically, a _nite_ slope.> Vertical lines dont have a slope,At least, not a nite one.> so stay away from trying to do> anything with this form y=mx+b on equations of the form x=, with no y> term. I mean, how would you go about solving for y in x=5? Its just> not there...Right.> For (a) parallel m = m2 (m2 is undened)> >> 1. y - y1 = m(x - x1) where m is undened...>> The right side is therefore undened as a result of m being undened.> ...and if the left side is equal to the right, then the lhs is also> undened. So no, you better stay away from this approach. The> slope-intercept form only applies when you really do have an m, that> is, a dened slope.Rather, one could say the form applies only when m is nite.> Heres what your teacher should have put on the board when you asked, or> at the very least explained after class. The slope of the line x=5, if> it exists, is dened as delta(y)/delta(x) for any two distinct points on> the line. Pick two. How about (5,0) and (5,1):>> delta(y) / delta(x)>> = (0-1) / (5-5)>> = -1/0, which is an undened expression, so this line doesnt even have> a slope, nor does any vertical line. You will get a similar undened> expression regardless of which two points you select, for any vertical> line.Of course, thats undened if you refuse to deal with an innite value.But it seems that the teacher is willing to deal with an innite value, inwhich case it is dened.BTW, IMO, the best answer to musics question might be to use initiallythe form* (x3 - x2)(y - y1) = (y3 -y2)(x - x1)since it is valid in _all_ cases. To get an equation of the line through(x1,y1) = (2,5) which is parallel to the vertical line x = 4, get any twopoints on that line, say, (x2,y2) = (4,0) and (x3,y3) = (4,7). Thensubstituting in my recommended form * we get (4 - 4)(y - 5) = (7 - 0)(x - 2) 0 = 7(x - 2) 0 = x - 2and so x = 2.[Of course, its much faster just to realize that a vertical linesequation can always be written as x = constant. But if you dont want tohave to deal with a special case, then I think the form * is the bestchoice.]> Similarly, it can be shown that any horizontal line has a slope of 0> since you will get a delta(y) of 0 and a delta(x) something nonzero.>> Next, he could have taken that same denition of slope:>> delta(y) / delta(x) = m, where delta(x)<>0>> and demonstrated that...>> (y - y1) / (x - x1) = m ...for some two distinct points> (x,y) and (x1,y1)>> ...then multiplied both sides by x-x1, again so long as x-x1 is NOT 0,> getting:>> y - y1 = m(x - x1) ...this is the point-slope form>> The y-intercept is dened to be where x=0, right? Call the y-coordinate> of this point b, that is, let the cordinates of this point be (0,b).> Then let (x1,y1) *be* the y-intercept, ie let x1 be 0 and y1 be b:>> y - b = m(x - 0)>> y - b = mx>> then add b to both sides:>> y = mx + b>> As you see, *all* of this is based on m actually existing, ie m is> *dened*. With m being undened, as in the case of the line x=5, you> simply dont *get* to the form y=mx+b.>> Either your teacher is being extremely nonhelpful on purpose (perhaps he> doesnt have time to go over such basics in a precalc class) or he at> out doesnt know what hes doing. Normaly, I would give the benet of> the doubt and assume he is just being arrogant and will not explain to> you something so basic. After all, you really should know this going> in to a precalculus course. Its basic algebra. Or do they call basic> algebra precalculus? dunno...>> However, if he really told you to consider the slope of the line x=5 to> be 0, then at worst he at out doesnt know what he is doing and> shouldnt be allowed to teach mathematics, or at best he likes to lie> alot. Its one or the other. this is not one of those things that you> can just have a slip of the tonue and inadvertently say consider an> undened slope to be a 0 slope.>> 2. y - y1 / m = x - x1 ==> (y - 5) / m = x - 2 where m> is undened -or- innity.??? Undened and innity are, in general, very different! However, inthis context of slopes, what youve written is understandable. If we onlyallow real numbers, then a vertical line cannot have a dened slope. OTOH,if we allow an unsigned innity in our system, then we do have a denedslope for vertical lines.> 3. Since anything over innity = 0 (according to my instructor) ,> then (y - 5)/{innity} = 0>> You say this is a precalc course? In such a context, innity doesnt> even _exist_,If the instructor chooses to consider it, it does exist. [But of course,that does not address the question of whether the instructors choice wouldthen be pedagogically sound for students at that level.]> much less has a recipricol of 0. Anything over innity> is not a dened expression, at least not in this context (which is real> numbers).If you _have_ something over innity, then innity is obviouslyinvolved, which means that the context _cannot_ be just the reals. And,yes, any nonzero quantity divided by innity yields 0.> Its only when you study alternative number systems (as opposed> to the standard real and imaginary numbers) that you get into things like> 1/innity.>> 4. So rewrite equation as 0 = x - 2> 5. x = 2>> For (b) parallel m = -1 / m2 (m2 is undened)As Darrell noted in a later post, parallel above and below here shouldbe perpendicular.> The result that parallel lines have slopes that are negative reciprocols,> again, only applies when they *have* slope. A horizontal line has a> slope of 0. 0 has no recipricol, much less a negative recipricol.The previous sentence is correct if we restrict ourselves to real values.But if we take 1/0 = unsigned innity and 1/(unsigned innity) = 0, thenwe can state neatly that reciprocals.And, BTW, we can also state neatly thatDavid Cantrell> The above explanation I offered is more appropriate for your context,> surely. Doesnt your book have a similar derivation of these forms?> Just about any elementary algebra text should. Even college algebra> texts (or precalculus texts).>> My advice: Get out of his class as fast as you can and audit someone> elses class instead, assuming the act of audit is for your benet.> If the audit truly is an audit, as in lets see if anythings wrong> here then report your ndings. === >> Your derogation of the instructor may not be justied. See below.Well we may disagree which is ne, but if he is telling them to treat avertical line with slope 0 then theres a serious rpblem there. And itaint just a slip of the toungue.>> One of our review questions asks to> write the slope-intercept forms of equations of lines that are (a)> > parallel and (b) perpendicular to the given line (x = 4)...>> The given point is (2,5)>> The answer in the back of the book (which does not show how theproblem> > is worked out, just the answer is shown) is> (a) = x = 2> (b) = y = 5>> If your text really said to write the slope-intercept forms, then the> text is at fault for part (a). There is no such thing as slope-intercept> form for the equation of a vertical line.>> I do not believe that for (a) that the slope-intercept form> y = mx + b can> be written since theres no slope as the slope would be undened.>> Correct.>> My> teacher wont show the steps to this as he says hes not giving it for> the test but he told me that m is undened and should be treated as> 0 in this type of problem.>> _Absolutely_ not. A slope of 0 and an undened slope are two VERY> different things. The two lines x=2 and y=5 are perpendicular. The> rst has undened slope while the second has 0 slope. Looks like this> teacher is the one who should be sitting in on this class.>> Youd be right, Darrell, _if_ thats what the teacher actually said.Well, it goes without saying, David. Of course, it thats not what was saidthen all bets are off. I stand by my remarks as written, since theyobviously apply only when thats what the teacher actually *said.* if heactually said what was written, that m being undened should be treated asthe slope is 0, then something aint right with the teacher. That simple.It happens. But> judging from what music said below (items 2. and 3.), it seems far more> likely that the instructor said something like If m is undened, treat> its _reciprocal_ as being 0.Well, perhaps, being that he talks about 0 being the reciprocol of innity.But rather than play guessing games as to what was actually said, I amsimply responding to what was written. At any rate, I certainly do notagree that telling people in a precalculus course, who are reviewing theslope of a dad gum line, that 1/innity is 0 is at all appropriate.>> > Thats ok that he did not want to work this out on> the board but Im doing this to re-learn and understand this. He said> for his test, he will make sure both X and Y have a value so that m> will not be undened.>> Well, OK, but that completely avoids your question to him.>> I agree that that avoidance is unfortunate.Which is another reason why the teacher is at fault, IMO. He seemed not tohave time to actually address this persons question directly in somestandard fashion, which is a VERY legitimate concern BTW, but somehow hefound the time to enter into discussions of innity and 1/innity=0 andsuch. and we wonder why so many people, when they take a calculus course,actually *believe* 1/innity=0.>> This is why he did not want to work this out on the board.> I understand how to do these problems when theres both a Y and X in> the equation but I am confused when either Y or X is missing (as inthe> > case above).>> Can someone show me if my steps are correct -or- if it was overkill> in my part to write out the steps like this for cases where m is> undened?>> Slope intercept form y=mx+b applies only when you *have* a slope.>> Specically, a _nite_ slope.>> Vertical lines dont have a slope,>> At least, not a nite one.Note to OP-- these distinctions are valid, but not that important to yourissue. A dened slope would fall into Davids category of nite slope.Theres just more than one way to say it, thats all.>> so stay away from trying to do> > anything with this form y=mx+b on equations of the form x=, with no y> term. I mean, how would you go about solving for y in x=5? Its just> not there...>> Right.>> For (a) parallel m = m2 (m2 is undened)>> 1. y - y1 = m(x - x1) where m is undened...>> The right side is therefore undened as a result of m being undened.> ...and if the left side is equal to the right, then the lhs is also> undened. So no, you better stay away from this approach. The> slope-intercept form only applies when you really do have an m, that> is, a dened slope.>> Rather, one could say the form applies only when m is nite.Of course.>> Heres what your teacher should have put on the board when you asked, or> at the very least explained after class. The slope of the line x=5, if> it exists, is dened as delta(y)/delta(x) for any two distinct pointson> the line. Pick two. How about (5,0) and (5,1):>> delta(y) / delta(x)>> = (0-1) / (5-5)>> = -1/0, which is an undened expression, so this line doesnt even have> a slope, nor does any vertical line. You will get a similar undened> expression regardless of which two points you select, for any vertical> line.>> Of course, thats undened if you refuse to deal with an innite value.> But it seems that the teacher is willing to deal with an innite value,in> which case it is dened.My point exactly. You may disagree, but I really dont think he should.This is a precalculus class, where they are reviewing basic algebraconcerning lines and slope.>> BTW, IMO, the best answer to musics question might be to use initially> the form>> * (x3 - x2)(y - y1) = (y3 -y2)(x - x1)>> since it is valid in _all_ cases. To get an equation of the line through> (x1,y1) = (2,5) which is parallel to the vertical line x = 4, get any two> points on that line, say, (x2,y2) = (4,0) and (x3,y3) = (4,7). Then> substituting in my recommended form * we get>> (4 - 4)(y - 5) = (7 - 0)(x - 2)>> 0 = 7(x - 2)>> 0 = x - 2>> and so x = 2.>> [Of course, its much faster just to realize that a vertical lines> equation can always be written as x = constant. But if you dont want to> have to deal with a special case, then I think the form * is the best> choice.]Yes, I agree.>> Similarly, it can be shown that any horizontal line has a slope of 0> since you will get a delta(y) of 0 and a delta(x) something nonzero.>> Next, he could have taken that same denition of slope:>> delta(y) / delta(x) = m, where delta(x)<>0>> and demonstrated that...>> (y - y1) / (x - x1) = m ...for some two distinct points> (x,y) and (x1,y1)>> ...then multiplied both sides by x-x1, again so long as x-x1 is NOT 0,> getting:>> y - y1 = m(x - x1) ...this is the point-slope form>> The y-intercept is dened to be where x=0, right? Call they-coordinate> of this point b, that is, let the cordinates of this point be (0,b).> Then let (x1,y1) *be* the y-intercept, ie let x1 be 0 and y1 be b:> >> y - b = m(x - 0)>> y - b = mx>> then add b to both sides:>> y = mx + b>> As you see, *all* of this is based on m actually existing, ie m is> *dened*. With m being undened, as in the case of the line x=5, you> simply dont *get* to the form y=mx+b.>> Either your teacher is being extremely nonhelpful on purpose (perhaps he> doesnt have time to go over such basics in a precalc class) or heat> out doesnt know what hes doing. Normaly, I would give the benet of> the doubt and assume he is just being arrogant and will not explain to> you something so basic. After all, you really should know this going> in to a precalculus course. Its basic algebra. Or do they call basic> algebra precalculus? dunno...>> However, if he really told you to consider the slope of the line x=5to> be 0, then at worst he at out doesnt know what he is doing and> shouldnt be allowed to teach mathematics, or at best he likes to lie> alot. Its one or the other. this is not one of those things that you> can just have a slip of the tonue and inadvertently say consider an> undened slope to be a 0 slope.Again, just to clarify my position on your earler remarks, note the secondword in this paragraph is if.>> 2. y - y1 / m = x - x1 ==> (y - 5) / m = x - 2 wherem> is undened -or- innity.>> ??? Undened and innity are, in general, very different! However, in> this context of slopes, what youve written is understandable. If we only> allow real numbers,...which of course, they do> then a vertical line cannot have a dened slope. OTOH,> if we allow an unsigned innity in our system, then we do have a dened> slope for vertical lines....which is a nonstandard system, at least n the sense that they do not useonly the standard real numbers. Are you sure what you infer may be theactual context of this particular persons class he is taking? I haveserious doubts...>> 3. Since anything over innity = 0 (according to my instructor),> then (y - 5)/{innity} = 0> >> You say this is a precalc course? In such a context, innity doesnt> even _exist_,>> If the instructor chooses to consider it, it does exist.I see. Theres no way in hell he could *possibly* be wrong.>[But of course,> that does not address the question of whether the instructors choicewould> then be pedagogically sound for students at that level.]Translate: He was wrong.>> much less has a recipricol of 0. Anything over innity> is not a dened expression, at least not in this context (which is real> numbers).>> If you _have_ something over innity, then innity is obviously> involved, which means that the context _cannot_ be just the reals. And,> yes, any nonzero quantity divided by innity yields 0.Well, no, any nonzero quantity divided by innity does not exist, _in thecontext of reals_, just as stated. I fail to see why we need to delve intosuch nonstandard numbre systems at this point. The simple and obviousconclusion is, the teacher simply failed to adequately (meaning appropriatefor the scope of the class) address theis persons question. Instead, hechose to tell him things that were simply invalid _in such a contect_.> Its only when you study alternative number systems (as opposed> to the standard real and imaginary numbers) that you get into thingslike> 1/innity.>> 4. So rewrite equation as 0 = x - 2> 5. x = 2>> For (b) parallel m = -1 / m2 (m2 is undened)>> As Darrell noted in a later post, parallel above and below here should> be perpendicular.>> The result that parallel lines have slopes that are negativereciprocols,> again, only applies when they *have* slope. A horizontal line has a> slope of 0. 0 has no recipricol, much less a negative recipricol.>> The previous sentence is correct if we restrict ourselves to real values.> But if we take 1/0 = unsigned innity and 1/(unsigned innity) = 0, then> we can state neatly that>> reciprocals.>> And, BTW, we can also state neatly that>Of course, in the context you imply. But in this context, again, I stand bymy claim that these statements are simply wrong. Of course, I myself may bewrong but I am far from convicned of that as of yet.-- Darrell === [snip]> At any rate, I> certainly do not agree that telling people in a precalculus course, who> are reviewing the slope of a dad gum line, that 1/innity is 0 is at all> appropriate.Youre certainly entitled to your opinion as to what is pedagogicallyappropriate -- and so is the instructor.[snip]> and we wonder why so many people, when they take a calculus> course, actually *believe* 1/innity=0.Well, its _very good_ they believe it; its true. I dont think I getyour point.[snip]> Vertical lines dont have a slope,>> At least, not a nite one.>> Note to OP-- these distinctions are valid, but not that important to your> issue. A dened slope would fall into Davids category of nite> slope. Theres just more than one way to say it, thats all.Not quite. With the slope of a vertical line being unsigned innity, theline has a dened slope. A dened slope need not be nite.[snip]> Are you sure what you infer may be> the actual context of this particular persons class he is taking? I> have serious doubts...Id be almost certain that the text doesnt mention any such thing as aninnite slope, but the instructor apparently has mentioned such, therebybroadening the context.> 3. Since anything over innity = 0 (according to my> instructor), then (y - 5)/{innity} = 0>> You say this is a precalc course? In such a context, innity> doesnt even _exist_,>> If the instructor chooses to consider it, it does exist.>> I see. Theres no way in hell he could *possibly* be wrong.Of course he could be wrong _pedagogically_, but theres certainly nothingwrong mathematically with, for example, saying 1/innity = 0.> [But of course, that does not address the question of whether the> instructors choice would> then be pedagogically sound for students at that level.]>> Translate: He was wrong.Hmm. You must be translating from a language other than mine!I certainly dont think the instructor was wrong.> If you _have_ something over innity, then innity is obviously> involved, which means that the context _cannot_ be just the reals. And,> yes, any nonzero quantity divided by innity yields 0.>> Well, no, any nonzero quantity divided by innity does not exist, _in> the context of reals_, just as stated. I fail to see why we need to> delve into such nonstandard numbre systems at this point.We dont _need_ to, but IMO (and probably in that of the instructor) itsvery nice to. Isnt it nice for _all_ lines to have slopes?But of course youre entitled to your opinion.> The simple and> obvious conclusion is, the teacher simply failed to adequately (meaning> appropriate for the scope of the class) address theis persons question.> Instead, he chose to tell him things that were simply invalid _in such a> contect_.The instructor may have chosen to _broaden_ the context so that what youreobjecting to is actually valid. David> Its only when you study alternative number systems (as opposed> to the standard real and imaginary numbers) that you get into things> like 1/innity.>> 4. So rewrite equation as 0 = x - 2> 5. x = 2>> For (b) parallel m = -1 / m2 (m2 is undened)>> As Darrell noted in a later post, parallel above and below here> should be perpendicular.>> The result that parallel lines have slopes that are negative> reciprocols,> again, only applies when they *have* slope. A horizontal line has a> slope of 0. 0 has no recipricol, much less a negative recipricol.>> The previous sentence is correct if we restrict ourselves to real> values. But if we take 1/0 = unsigned innity and 1/(unsigned> innity) = 0, then we can state neatly that>> reciprocals.>> And, BTW, we can also state neatly that> Of course, in the context you imply. But in this context, again, I stand> by my claim that these statements are simply wrong. Of course, I myself> may be wrong but I am far from convicned of that as of yet. === > > Note to OP-- these distinctions are valid, but not that important toyour> issue. A dened slope would fall into Davids category of nite> slope. Theres just more than one way to say it, thats all.>> Not quite. With the slope of a vertical line being unsigned innity, the> line has a dened slope. A dened slope need not be nite.David, I dont understand why you are having so much trouble undrstandingthat I have no problem with anything you are saying given an appropriatecontext. Now, we can disagree till the coes come home over what is or isnot appropriate context for such a discussion. let me make myself perfectlyclear: I do not believe it to be wise to refer to an undened slope as adened slope when he is reviewing basic algebra. Its either deed, orits not. Because it may be dened as unsigned innity in certaincontexts (usually much later than a precalculus course) does not make itdened in basic algebra. In the REAL NUMBERS, which for some reason youinsist we should not limit ourselves to when introducing algebra students tothe various forms of a line n the plane, 1/innity simply is notencountered. Division by 0 as results when trying to plug two points intothe slope formula for a vertical line, results in an UNDEFINED expression,period.OK, its undened FOR NOW, allright? Do you delve into complex numbers THEMOMENT a beginiing algebra student is introduced to the notion of a squareroot? No... Why do the same for slope?>> [snip]> Are you sure what you infer may be> the actual context of this particular persons class he is taking? I> have serious doubts...>> Id be almost certain that the text doesnt mention any such thing as an> innite slope, but the instructor apparently has mentioned such, thereby> broadening the context.everything Ive read, not the least of which is the OPs expressedmisunderstanding of what to do with these lines when x and/or y is notpresent, it is most certainly inappropriate to delve into such discussions.thats not to say that a student approaching his instructor seekingadditional insight should not be entitled to such insight, even it it isnormally outside the scope of such a course, *assuming* that he is alreadycomfortable with the *standard* methods of doing such things. In this case,the OP clearly did not understand even the standard methods concerning theexplanation of these forms of lines, how slope factos into the equation,etc. The instructor needs to *stoop* to the level of who hes dealing withhere, else he aint an effective instructor. its that simple.<...>-- Darrell === We need to end our dialog. Ill make just two brief comments below.>> Note to OP-- these distinctions are valid, but not that important to> your issue. A dened slope would fall into Davids category of> nite slope. Theres just more than one way to say it, thats all.> >> Not quite. With the slope of a vertical line being unsigned innity,> the line has a dened slope. A dened slope need not be nite.>> David, I dont understand why you are having so much trouble undrstanding> that I have no problem with anything you are saying given an appropriate> context. Now, we can disagree till the coes come home over what is or is> not appropriate context for such a discussion. let me make myself> perfectly clear: I do not believe it to be wise to refer to an> undened slope as a> dened slope when he is reviewing basic algebra. Its either deed, or> its not. Because it may be dened as unsigned innity in certain> contexts (usually much later than a precalculus course) does not make it> dened in basic algebra. In the REAL NUMBERS, which for some reason you> insist we should not limit ourselves to when introducing algebra students> to the various forms of a line n the plane,Neither in this thread nor elsewhere have I ever _insisted_ on any suchthing!> 1/innity simply is not> encountered. Division by 0 as results when trying to plug two points> into the slope formula for a vertical line, results in an UNDEFINED> expression, period.>> OK, its undened FOR NOW, allright? Do you delve into complex numbers> THE MOMENT a beginiing algebra student is introduced to the notion of a> square root? No... Why do the same for slope?> [snip]> Are you sure what you infer may be> the actual context of this particular persons class he is taking? I> have serious doubts...>> Id be almost certain that the text doesnt mention any such thing as> an innite slope, but the instructor apparently has mentioned such,> thereby broadening the context.>> We know, David, but it doesnt make it *appropriate* necessarily.And of course I never said it was appropriate, from a pedagogicalstandpoint, for students at that level.David> everything Ive read, not the least of which is the OPs expressed> misunderstanding of what to do with these lines when x and/or y is not> present, it is most certainly inappropriate to delve into such> discussions. thats not to say that a student approaching his instructor> seeking additional insight should not be entitled to such insight, even> it it is normally outside the scope of such a course, *assuming* that he> is already comfortable with the *standard* methods of doing such things.> In this case, the OP clearly did not understand even the standard methods> concerning the explanation of these forms of lines, how slope factos into> the equation, etc. The instructor needs to *stoop* to the level of who> hes dealing with here, else he aint an effective instructor. its that> simple. === >> We need to end our dialog. Ill make just two brief comments below.? Either we need to end it or we dont. If you feel we need to end it,then you should feel no urge to include more comments. If you include morecomments, you are implicitly asking for a response to those comments....but instead you seem to feel you must have the last word.I get it now, I think. End the conversation means continue theconversation. At least youre consistent, since an undened slope is,well, dened.>> Note to OP-- these distinctions are valid, but not that important to> your issue. A dened slope would fall into Davids category of> nite slope. Theres just more than one way to say it, thatsall.>> > Not quite. With the slope of a vertical line being unsigned innity,> the line has a dened slope. A dened slope need not be nite.>> David, I dont understand why you are having so much troubleundrstanding> that I have no problem with anything you are saying given an appropriate> context. Now, we can disagree till the coes come home over what is oris> not appropriate context for such a discussion. let me make myself> perfectly clear: I do not believe it to be wise to refer to an> undened slope as a> dened slope when he is reviewing basic algebra. Its either deed,or> its not. Because it may be dened as unsigned innity in certain> contexts (usually much later than a precalculus course) does not make it> dened in basic algebra. In the REAL NUMBERS, which for some reasonyou> > insist we should not limit ourselves to when introducing algebrastudents> to the various forms of a line n the plane,>> Neither in this thread nor elsewhere have I ever _insisted_ on any such> thing!Huh? So you were _not_ the one that corrected my explanation that avertical line has an undened slope? So you were _not_ the one that kepton saying that if we accept unsigned innity as the slope of a verticalline? David, this isnt James Harris youre talking to. I am not searchingfor nor do I enjoy a fruitless argument. You seem to completely understandthat within the context of my remarks, my remarks are correct. Likewise, Ibelieve you understand (you should, at least) that I feel your remarks arecorrect in the context in which you present them.Some way, can we get back to the context of the OPs question? Remember,just because his instructor delved into 1/innity and the like, does notimply that he answered the OPs question in the context the OP presented it.Very clearly, the OP felt that a vertical line does not have a denedslope, therefore the equation of such line does not posses a slope-interceptform, despite the instructions in his book. The proper response to hisquestion CONSIDERING THE CONTEXT IN WHICH IT WAS PRESENTED is not to somehowdance around the issue and present a context in which a vertical line HASslope, but rather to simply explain to the OP what it is he is specicallyasking for. That was done, and may I say in a matter consistent with mostprecalculus texts. You yourself even acknowledged that in all probabilityhis book does not approach the issue the way you and his instructor has.Apparently you are of the opinion that whatever an instructor says, isappropriate BY DEFINITION, regardless of the cpontext in which the questionwas asked of him. he can know all the math in the world but unless he canget it across, he is worth nothing as a teacher. Undened slope means zeroslope? Cmon. Undened slope means dened slope? Cmon. If hisapproach is so effective, then why is the OP asking a newsgroup the samequestion he asked his instructor? did the OP not later acknowledge that theresponses he received from myself and another were helpful? Of course. Isay that not looking for any glory, but simply to point out which was themore effective (and APPROPRIATE) answer to his question.>> 1/innity simply is not> encountered. Division by 0 as results when trying to plug two points> into the slope formula for a vertical line, results in an UNDEFINED> expression, period.>> OK, its undened FOR NOW, allright? Do you delve into complex numbers> THE MOMENT a beginiing algebra student is introduced to the notion of a> square root? No... Why do the same for slope?> > [snip]> Are you sure what you infer may be> > the actual context of this particular persons class he is taking?I> have serious doubts...>> Id be almost certain that the text doesnt mention any such thing as> an innite slope, but the instructor apparently has mentioned such,> thereby broadening the context.>> We know, David, but it doesnt make it *appropriate* necessarily.>> And of course I never said it was appropriate, from a pedagogical> standpoint, for students at that level.Therefore your explanation, which is along the same lines, is also notnecessarily appropriate. I go so far as to say it was actuallyunappropriate and I have clearly justied the reasons why.-- Darrell === > Therefore your explanation, which is along the same lines, is also not> necessarily appropriate. I go so far as to say it was actually> unappropriate and I have clearly justied the reasons why.Wow, did I really say unappropriate? Youre right, it must be time to endthe conversation ;-). === >>We need to end our dialog. Ill make just two brief comments below.>>Note to OP-- these distinctions are valid, but not that important to>>your issue. A dened slope would fall into Davids category of>>nite slope. Theres just more than one way to say it, thats all.>Not quite. With the slope of a vertical line being unsigned innity,>the line has a dened slope. A dened slope need not be nite.>David, I dont understand why you are having so much trouble undrstanding>>that I have no problem with anything you are saying given an appropriate>>context. Now, we can disagree till the coes come home over what is or is>>not appropriate context for such a discussion. let me make myself>>perfectly clear: I do not believe it to be wise to refer to an>> undened slope as a>>dened slope when he is reviewing basic algebra. Its either deed, or>>its not. Because it may be dened as unsigned innity in certain>>contexts (usually much later than a precalculus course) does not make it>>dened in basic algebra. In the REAL NUMBERS, which for some reason you>>insist we should not limit ourselves to when introducing algebra students>>to the various forms of a line n the plane,>>Neither in this thread nor elsewhere have I ever _insisted_ on any such>thing!>But I _insist_ that students at any level who can appreciate it be exposed to it, and that instructors be allowed to teach it if they want. The important thing is to get the students to _think_, and not to blindly manipulate.>>1/innity simply is not>>encountered. Division by 0 as results when trying to plug two points>>into the slope formula for a vertical line, results in an UNDEFINED>>expression, period.>>OK, its undened FOR NOW, allright? Do you delve into complex numbers>>THE MOMENT a beginiing algebra student is introduced to the notion of a>>square root? No... Why do the same for slope?>[snip]>Are you sure what you infer may be>>the actual context of this particular persons class he is taking? I>>have serious doubts...>Id be almost certain that the text doesnt mention any such thing as>an innite slope, but the instructor apparently has mentioned such,>thereby broadening the context.>We know, David, but it doesnt make it *appropriate* necessarily.>>And of course I never said it was appropriate, from a pedagogical>standpoint, for students at that level.>Well, it depends on what that level is. Theres no reason to hide these ideas from children. On the other hand, if this is a college pre-algebra course, these students have already painted themselves into a corner of frustration and/or indifference, and introducing one more unnecessary concept isnt any help at all.But when theyre still young and curious (or even older and have merely had their curiosity aroused), then they ask questions like why?. And why not? And the solution there is to explore different things.People always ask about innity. It seems to be a magnet to the human imagination. Theres no reason for us to shy away from it, including the fact that we cant actually *grasp* innity, so we have to make some assumptions about it. But as long as we make those assumptions explicit, then theres no harm in exploring it. Except if we only have 15 weeks to get ready for a calculus course, in which case come see me in my ofce, we cant take up class time for this.Jon Miller === > The line x = 2 has a slope of 0; >the line y = 5 has undened slope and that can _not_ be treated as >0.I did not read the rest of your message.The quoted piece should be rearranged. Slope is vertical change divided byhorizontal change. The line x=2 has any vertical change, but zero horizontalchange. The slope for x=2 is (anyRealNumber/0) but this is undened; thehorizontal change of zero gives a useless(?) divisor, so the slope isundened. The other line, y=5, has no horizontal change. As you pick any x, the changein x (horizontal change is almost any real number that you want), the verticalchange, change in y, is zero. Here, slope is (0/anyRealNumber), which isequal to zero. In the simple, crudest language, no slope means horizontally at, no up anddown as you move accross. As I say, I did not read the rest of your message; I just comment on the stuffquoted above.G C === > The line x = 2 has a slope of 0;>the line y = 5 has undened slope and that can _not_ be treated as>0.Well, Stan simply misspoke. That sort of thing can happen to anyone.> I did not read the rest of your message.> The quoted piece should be rearranged. Slope is vertical change divided> by horizontal change. The line x=2 has any vertical change, but zero> horizontal change. The slope for x=2 is (anyRealNumber/0) but this> is undened; the horizontal change of zero gives a useless(?) divisor,> so the slope is undened.Yes indeed, assuming that were restricting ourselves to real numbers.> The other line, y=5, has no horizontal change. As you pick any x, the> change in x (horizontal change is almost any real number that you want),By almost any real number that you want, I suppose you mean any nonzeroreal number that you want.> the vertical change, change in y, is zero. Here, slope is> (0/anyRealNumber), which is equal to zero. ^ nonzero> In the simple, crudest language, no slope means horizontally at, no up> and down as you move accross.That is so crude that it is simply incorrect.I assume that youre dealing with only real numbers for slopes. If so,then no slope means vertical, rather than horizontal.By contrast, slope of 0 means horizontal.Theres a huge difference between having no slope and having a slope whichhappens to be 0.(But I suppose you already knew that and, like Stan, simply misspoke.)David> As I say, I did not read the rest of your message; I just comment on the> stuff quoted above.>> G C === >I _hope_ you misheard your teacher. The line x = 2 has a slope of 0; >the line y = 5 has undened slope and that can _not_ be treated as >0.This was backwards, but it was a writeo: honestly I do know The line x = const is vertical and its slope is undened (as far as the real numbers are concerned). The run is 0 for any rise, so any attempt to calculate slope gives you a division by 0.The line y = const is horizontal and its slope is 0 because the rise is 0 for any run.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/Walrus meat as a diet is less repulsive than seal. === With some hesitation, Ill add my pittance.First, in regard to OPs question. Suppose we have the line 2*x + 3*y = 6. So, y = (-2/3)*x + 2 and x = (-3/2)*y + 3. Which is*the* slope-intercept form?For the line through (2, 5) parallel to the y - axis, x = 2 seems likea perfectly valid slope-(x-)intercept form.Likewise, for the line through (2, 5) parallel to the x-axis, y = 5seems ne for the slope-(y-)intercept form.Now, for the oo (= innity sysmbol). Topologically, there is no problem with R / { oo }. It is the onepoint compactication of R, is well understood, not particularlyinteresting and is (topologically) the same as a circle.So far, so good.Algebraically, however, there are big time problems. For example: oo * oo = oo? If so, then oo = 1 if you still want mutiplicativeinverses. oo - oo = ? oo + a = oo + b => a = b?Now I know some want to restrict to very special cases: 1/oo = 0 or 1/0= oo. But then (1/0)*0 = ?In fact, I cant think of _any_ algebra that works as one would like itto if oo is included. That forces locutions like If a + b = a + c then b = c if a =/= oo. I know, I know, we say Ifa*b = a*c then b = c if a =/= 0 but with oo the qualier would haveto be used virtually _all_ the time.There is a way around all this, Non-Standard Analysis, but that is alot more than just tossing oo into the pot.For the same reason that division by zero is not dened, oo is notconsidered as any kind of number: it screws things up.Terminology like lim(f(x), x -> oo) or lim(f(x), x -> a) = oo or oo/oo(for LHopitals rule) should, IMHO, be viewed as macros; the oo byitself is meaningless.-- Paul SperryColumbia, SC (USA) === > With some hesitation, Ill add my pittance.Ill add a few comments -- without any hesitation.(Fools rush in where angels fear to tread.)> First, in regard to OPs question. Suppose we have the line> 2*x + 3*y = 6. So, y = (-2/3)*x + 2 and x = (-3/2)*y + 3. Which is> *the* slope-intercept form?An excellent comment. Of course, when his text says slope-intercept form,it presumably means slope-(y-intercept) form, to be more precise.> For the line through (2, 5) parallel to the y - axis, x = 2 seems like> a perfectly valid slope-(x-)intercept form.Hmm. Perhaps whether it would be perfectly valid depends on onesperspective. I would expect that some people would balk, claiming that Aline without a slope couldnt possibly have a valid slope-anything form.I, OTOH, would suggest that, taking the slope of vertical lines to beunsigned innity, we havey = m*x + b, slope-(y-intercept) form, valid for all nonvertical linesand symmetricallyx = y/m + a, slope-(x-intercept) form, valid for all nonhorizontal lines.Doing that, x = 2 would indeed be a perfectly valid slope-(x-intercept)form.> Likewise, for the line through (2, 5) parallel to the x-axis, y = 5> seems ne for the slope-(y-)intercept form.>> Now, for the oo (= innity sysmbol).>> Topologically, there is no problem with R / { oo }. It is the one> point compactication of R, is well understood, not particularly> interesting and is (topologically) the same as a circle.>> So far, so good.>> Algebraically, however, there are big time problems. For example:> oo * oo = oo?Of course.> If so, then oo = 1 if you still want mutiplicative inverses.So what? The introduction of 0 into the number systems produced similarbig time problems. Merely replacing oo with 0 in what you said abovegives:For example: 0 * 0 = 0? If so, then 0 = 1 if you still want mutiplicativeinverses.> oo - oo = ?Right.And similarly 0/0 = ?[Actually, in the computer algebra system Derive, thats precisely whatyou get: 0/0 simplies to ? .]> oo + a = oo + b => a = b?No more than 0 * a = 0 * b => a = b.> Now I know some want to restrict to very special cases: 1/oo = 0 or 1/0> = oo.Not quite _that_ special! Let R* denote the one-point compactication ofR. Then certainly x/oo = 0 for all nite x, and x/0 = oo for all nonzero x.> But then (1/0)*0 = ?Right. Thats normally taken to be undened.> In fact, I cant think of _any_ algebra that works as one would like it> to if oo is included.And I cant think of _any_ algebra that works as one would like it to if0 is included either. (But then weve grow accustomed to dealing withzeros eccentricities.)> That forces locutions like If a + b = a + c then b = c if a =/= oo.> I know, I know, we say If a*b = a*c then b = c if a =/= 0 but with oo> the qualier would have to be used virtually _all_ the time.Hmm. virtually _all_ the time? I dont think I agree.> There is a way around all this, Non-Standard Analysis, but that is a> lot more than just tossing oo into the pot.>> For the same reason that division by zero is not dened, oo is not> considered as any kind of number: it screws things up.Some mathematicians (and computer scientists) do consider it to be anumber. Zero screws things up too, and for a long time was not regardedas a number either. But whether something is called a number or not isnot really important, at least IMO.> Terminology like lim(f(x), x -> oo) or lim(f(x), x -> a) = oo or oo/oo> (for LHopitals rule) should, IMHO, be viewed as macros; the oo by> itself is meaningless.??? It _can_ be given precise meaning. If youre familiar with thedevelopment of real numbers as equivalence classes of Cauchy sequences ofrationals, then all you need to do to obtain R* from R is to adjoin anotherequivalence class which consists of all rational sequences which increasewithout bound in absolute value.David Cantrell === One more comment, and this fool is done :-)<...>> Terminology like lim(f(x), x -> oo) or lim(f(x), x -> a) = oo or oo/oo> (for LHopitals rule) should, IMHO, be viewed as macros; the oo by> itself is meaningless.>> ??? It _can_ be given precise meaning.????????????????????Regardless if one *can* give it precise meaning or not, at least at thisjuncture (precalculus, and for that matter calI,II,III and more,) it *isnt*given precise meaning, OK? I wish for one moment you would pretend--justoo was given precise meaning.Please put things into proper context. Clearly, in the proper context,Pauls above statement should be as deserving of a ??? response about thesame as a claim of a vertical line having no dened slope should bedeserving of the same response. ...but yet, you seem to insist withouthesitation that we should just go ahead and use innity as though it was awell dened mathematical object, saying things like 1/oo=0 and the like.Ask 100 precalulus teachers which response they would give to a studentinquiring about vertical/horizontal lines, where they do not believe that avertical line has a dened slope:a. The slope of a vertical line is innite (which *is* dened), the slopeof a horizontal line is 0, and since 1/oo=0=(-)0 their slopes are indeednegative recipricols, which is the same result for two perpendicular linesthat are not vertical/horizontal.b. The slope of a vertical line is undened, the slope of a horizontalline is 0, therefore the rule that perpendicular lines have negativerecipricol slopes doesnt apply for vertical/horizontal lines.My moneys on b.-- Darrell === If anyone else wants the last word, they are welcome to it.[...]> First, in regard to OPs question. Suppose we have the line> 2*x + 3*y = 6. So, y = (-2/3)*x + 2 and x = (-3/2)*y + 3. Which is> *the* slope-intercept form?> An excellent comment. Of course, when his text says slope-intercept form,> it presumably means slope-(y-intercept) form, to be more precise.I expect you are right.> For the line through (2, 5) parallel to the y - axis, x = 2 seems like> a perfectly valid slope-(x-)intercept form.> Hmm. Perhaps whether it would be perfectly valid depends on ones> perspective. I would expect that some people would balk, claiming that A> line without a slope couldnt possibly have a valid slope-anything form.Who says x = 2 doesnt have a slope? Maybe I choose to draw the y-axishorizontally and the x-axis vertically and use rise/run.[...]> Now I know some want to restrict to very special cases: 1/oo = 0 or 1/0> = oo.> Not quite _that_ special! Let R* denote the one-point compactication of> R. Then certainlyCertainly? Sez who? Not the topologists AFIK.> x/oo = 0 for all nite x, and> x/0 = oo for all nonzero x.> > But then (1/0)*0 = ?> Right. Thats normally taken to be undened.> In fact, I cant think of _any_ algebra that works as one would like it> to if oo is included.> And I cant think of _any_ algebra that works as one would like it to if> 0 is included either. Wow! It is really, really nice that the reals are a eld.> (But then weve grow accustomed to dealing with> zeros eccentricities.)> That forces locutions like If a + b = a + c then b = c if a =/= oo.> I know, I know, we say If a*b = a*c then b = c if a =/= 0 but with oo> the qualier would have to be used virtually _all_ the time.> Hmm. virtually _all_ the time? I dont think I agree.I guess we could (but wont) argue about virtually. How about solutions to x^2 + 1 = 2*x? Are there two of them: 1 and oo? What do you propose for the old negative reciprocal rule? A horizontalline has slope 0 so a vertical line has slope -1/0 = -oo? I guess one could say oo = -oo but then what could be said about x if x = -x? Or x + (-x) for that matter.The slope intercept form for a vertical line is y = oo*x + b? (So oo isthe only point on the line?)Adding fractions is really easy: 1/2 + 1/3 = (1*oo)/(2*oo) + (1*oo)/(3*oo) = undened?Do you recommend that algebra refer to the Real Circle rather thanthe Real Line?[...]> Some mathematicians (and computer scientists) do consider it [ oo ] to be a> number. Sure, topologists and theres no accounting for computer scientists :-)[...]> Terminology like lim(f(x), x -> oo) or lim(f(x), x -> a) = oo or oo/oo> (for LHopitals rule) should, IMHO, be viewed as macros; the oo by> itself is meaningless.> ??? It _can_ be given precise meaning. If youre familiar with the> development of real numbers as equivalence classes of Cauchy sequences of> rationals, then all you need to do to obtain R* from R is to adjoin another> equivalence class which consists of all rational sequences which increase> without bound in absolute value.And how does that get handled algebraically?> > David Cantrell-- Paul SperryColumbia, SC (USA) === >> If anyone else wants the last word, they are welcome to it.Ditto.[...]> Now I know some want to restrict to very special cases: 1/oo = 0 or> 1/0 = oo.>> Not quite _that_ special! Let R* denote the one-point compactication> > of R. Then certainly>> Certainly? Sez who? Not the topologists AFIK.Surely anyone whos ever considered arithmetic on R*. The two statementsbelow are also valid, of course, in C*, and you should be able to nd themin any complex analysis text which discusses C* well. And R* is, after all,just the real slice through the Riemann sphere.> > x/oo = 0 for all nite x, and> x/0 = oo for all nonzero x.[...]> In fact, I cant think of _any_ algebra that works as one would like> it to if oo is included.>> And I cant think of _any_ algebra that works as one would like it to> if 0 is included either.>> Wow! It is really, really nice that the reals are a eld.True, of course. But then it depends on what your priorities are. IMO,wheels are pretty nice too. [Of course, Im referring to the algebraicstructure called a wheel, as opposed to something else.][...]> How about solutions to x^2 + 1 = 2*x? Are there two of them: 1 and oo?In R, theres just one solution of course. But in R* (or C*), there aretwo.> What do you propose for the old negative reciprocal rule?I gave it ealier in this thread.> A horizontal line has slope 0 so a vertical line has slope -1/0 = -oo?Sure. But of course, in R*, oo is unsigned, like 0.Thus, just as we have -(0) = 0, we also have -(oo) = oo.> I guess one could say oo = -oo but then what could be said about x if> x = -x?That x is either 0 or oo.> Or x + (-x) for that matter.x + (-x) would normally be considered undened if x = oo, much asx/x would normally be considered undened if x = 0 or oo. [But in a wheel,these would all be dened. But thats another story...]> The slope intercept form for a vertical line is y = oo*x + b? (So oo is> the only point on the line?)Certainly not. I thought Id covered that matter earlier:y = m*x + b, slope-(y-intercept) form, valid for all nonvertical linesx = x/m + a, slope-(x-intercept) form, valid for all nonhorizontal lines> Adding fractions is really easy: 1/2 + 1/3 => (1*oo)/(2*oo) + (1*oo)/(3*oo) = undened?Adding fractions is easy, of course. But the example is just as wrong assaying that 1/2 = (1*0)/(2*0).> Do you recommend that algebra refer to the Real Circle rather than> the Real Line?For R*, real circle might be a good name. Ive also seen it calledthe Riemann circle.> Terminology like lim(f(x), x -> oo) or lim(f(x), x -> a) = oo or> oo/oo (for LHopitals rule) should, IMHO, be viewed as macros; the> oo by itself is meaningless.>> ??? It _can_ be given precise meaning. If youre familiar with the> development of real numbers as equivalence classes of Cauchy sequences> of rationals, then all you need to do to obtain R* from R is to adjoin> another equivalence class which consists of all rational sequences> which increase without bound in absolute value.>> And how does that get handled algebraically?With much the same sort of care which must be used when handling zero.David Cantrell === > One more comment, and this fool is done :-)> <...> Terminology like lim(f(x), x -> oo) or lim(f(x), x -> a) = oo or> oo/oo (for LHopitals rule) should, IMHO, be viewed as macros; the> oo by itself is meaningless.>> ??? It _can_ be given precise meaning.>> ????????????????????>> Regardless if one *can* give it precise meaning or not, at least at this> juncture (precalculus, and for that matter calI,II,III and more,) it> *isnt* given precise meaning, OK?I certainly agree with that. Indeed, even once students have nished thebasic calculus sequence, they dont know how to give precise meaning tosomething as simple as 1/2 + 1/3. [Related to this, you might enjoy thePreface to the Student in Landaus classic text _Foundations of Analysis_.At one point he says I speak only of such numbers as you have alreadydealt with in high school. ... Please forget what you have learned inschool; you havent learned it.]> I wish for one moment you would> paper, where oo was given precise meaning.OK, ne. I just did as you asked. That moment has now passed. (And I mustnow wonder what the purpose of that little exercise was.)> Please put things into proper context. Clearly, in the proper context,> Pauls above statement should be as deserving of a ??? response about> the same as a claim of a vertical line having no dened slope should be> deserving of the same response. ...but yet, you seem to insist without> hesitation that we should just go ahead and use innity as though it was> a well dened mathematical object, saying things like 1/oo=0 and the> like.I have never _insisted_ that one _should_ do such.Your misrepresentations (inadvertent though Im sure they are) of what Ihave said are very annoying. Thats why I had tried to stop our dialogearlier in this thread. I hope that readers will read what Ive said,rather than what youve said that Ive said!Please feel free to continue to respond, Darrell. But if you do, as faras I am concerned, it will be your monologue henceforth.> Ask 100 precalulus teachers which response they would give to a student> inquiring about vertical/horizontal lines, where they do not believe that> a vertical line has a dened slope:>> a. The slope of a vertical line is innite (which *is* dened), the> slope of a horizontal line is 0, and since 1/oo=0=(-)0 their slopes are> indeed negative recipricols, which is the same result for two> perpendicular lines that are not vertical/horizontal.>> b. The slope of a vertical line is undened, the slope of a horizontal> line is 0, therefore the rule that perpendicular lines have negative> recipricol slopes doesnt apply for vertical/horizontal lines.>> My moneys on b.Of course. Id wager that more than 90 of them would choose b. I wouldntbe surprised if _all_ of them chose b.David Cantrell === > experiences in the military, where I actually had the honor of giving> a lecture on the physics of lasers to the medical personnel at Madigan> Army Medical Center, including the surgeons, other doctors and nurses,> for their medical continuing education credits, I feel like I can> speak condently on the subject.Radiation Protection Ofce, Madigan Army Medical Center who posted(which is archived at Vanderbilt). Now the question is: Are you the sameJames Harris who later posted to RADSAFE as James.Harris@rfets.gov, K-HManager, 771 Radiological Safety in February 2001? If so, Id *love* tohear all about the safety violations in Building 771 (including workersbeing ned $385,000 a few months after you posted that message.-- Wayne Brown | When your tails in a crack, you improvisefwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === > Could somebody clarify the connection between the denition of stochastic> independence for a pair of events [Pr(A*B)=Pr(A)Pr(B)] and the notion of> independent sequences of trials (eg: Bernoulli trials)? Mucho appreciado for> any and all helps and hints.> Peace,> EJ> I know that you ip a coin several times, each time you ip it is a trial, and each trial is independent of the other trials. If you were to ip two coins, the probability of getting two heads would be equal to the probability of getting a head on the rst coin times the probability of getting a head on the second coin.I suppose that a pair of events could occur with a deck of cards, where one event could be getting an ace, and another event could getting a spade. These two events, and the probability of getting an ace of spades is equal to the probability of getting an ace times the probability a spade.Bernoulli trials usually deal with success or failure ( they only have two possible outcomes ). And each trial has the same probability of success.I hope this helps.- Stephenhttp://pages.prodigy.net/stephenmorais === Okay, I will make the assumption the Mr. Harris truly wants to learnabout this eld. I understand the history, here, but will try tomake no assumptions this once.Wiles proof can be seen to begin from a very interestingrepresentation of the problem (found independently by Gerhart Frey andYves Hellegouarch) as a form of elliptic curve. In particular,beginning with the Fermat equation a^p + b^p +c^p = 0, they looked at:E: y^2 = x (x - a^p) (x - b^p)Now, for this curve, it can be shown that there are relationshipsbetween the diophantine properties of the Fermat equation and thearithmetic properties of E. In particular, and this is important, Ecan be shown to be semistable.The next step can be taken in reference to Galois representations. Inparticular, there is a particular Galois representation associatedwith E, and one can prove some important properties of this particularrepresentation. In particular, it is absolutely irreducible, odd,unramied outside 2p, and at at p. These are four very uniqueproperties.Now one looks at newforms. To these modular forms, one may associatea Galois representation through the theory of Eichler-Shimura. Onecan prove that such representations obey certain equations on theirtrace and determinant and also that they are unramied at allsufciently large primes. And this can be generalized, so that wemay call Galois representations modular if they are unramiedoutside the multiples of p and obey extensions of the trace anddeterminant equations obeyed by the representations related tonewforms with the generalization being on setting the equalities inrelation to certain homomorphisms from Hecke operators to the eldthat is the base of the representation.The modular Galois representations were shown by Ribet to haveproperties in contradiction to the four properties listed above forthe representation associated to E. Thus E does not have a modularrepresentation.prove the hard part. He proved that every semistable elliptic curveover Q is associated to a modular representation (is modular). Since E was semistable, it should be modular, but we have already seenit is not. Thus, a contradiction, and there is therefore no solutionto the equation a^p + b^p + c^p = 0 over Q.So, it has to be seen that Wiles proof does not follow the logic ofMr. Harris post. It is a classic proof by contradiction. Of course,there is much more going on here. In particular, there are thedetails of the proofs mentioned. Its a beautiful theory, and if oneis generally interested in expanding their expertise, I would suggeststudying the necessary elds and, in particular, the theory requiredto understand the proofs (elliptic curves, Galois representations, andmodular forms).I do hope this was not a waste of my time... === > Virgil says...>>If Wiles is correct, someone needs to come forward and say what both>>elliptic curves and modular forms are, and that thing will have the>>desired requirement, such that you can talk about either without>>mentioning the other.>>Someone has, or no one wold be speaking about either >>elliptic curves and modular forms. >>Originally they were spoken of quite separately. Much of the >>importance of Wiles proof, and subsequent developments along >>the same lines, is that they can meaningfully be spoken of >>together.>>So Harris desire to isolate them from each other is an >>attempt to step backwards in time and knowledge.> Im not sure that I understood the point of the original post on this thread,> but what I *think* Harris was trying to say was this: If there is a proof of> FLT that goes by way of real numbers and elliptic curves, then there should be a> way to rewrite it so that the proof only refers to natural numbers.> So whether or not thats what he meant, it is a legitimate question: Given Wiles> proof, is it possible (in principle) to construct a proof purely within PA? (Or> maybe PA augmented with stronger induction principles?) If such a pure-PA proof> is possible, would it necessarily be much more complex than Wiles proof? In> other words (which I think it Harris main question) is Wiles detour into reals> and elliptic curves necessary?Heres a way to turn Wiles proof into a proof in PA. Take a sufciently strong, but crippled system (say an ad hoc subset of ZFC chosen so that only the things used in the proof are included) that sufces to carry out the proof, and prove in PA that this fragment is sound with respect to Pi_1 sentences about numbers. Presumably Wiles proof can be recast in a form A --> Ax(x is a four-tuple and x_1 > 2 --> x_2^x_1 + x_3^x_1 != x_4^x_1, and thus the Pi_1 sentence expressing Fermats last theorem should be true. Of course, if Fermats last theorem is independent of PA, then PA will be unable to prove the soundness for Pi_1 sentences about numbers for any theory which proves the theorem.However, the question whether there are genuinely arithmetical (whatever that means) propositions undecidable in PA is subject to much debate. D. Isaacson has argued, for example, that PA is the correct codication of arithmetic and that all undecidable propositions are really higher order statements suitably coded.-- Aatu Koskensilta (aatu.koskensilta@xortec.)Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === >I do hope this was not a waste of my time...Im sure that regardless of how James Harris responds, there will be otherswho appreciate your effort.It might be worth pointing out that there is a somewhat more elementary wayto state the modularity conjecture, that doesnt require understanding allthat jazz about Galois representations. For every positive integer N, letGamma_0(N) be the multiplicative group of 2x2 integer matrices (a b | c d)such that c is divisible by N and ad-bc = 1. An element (a b | c d)in Gamma_0(N) acts on the upper half of the complex plane by sendingthe complex number z to (az+b)/(cz+d). If we now take the quotient ofthe upper half plane by this group action and then compactify suitably(glossing over some details here ...not difcult, but a bit tedious),we get a compact Riemann surface that goes by the name of X_0(N).X_0(N) is an algebraic curve over C, the complex numbers. It turns outthat there is a natural way to give it the structure of an algebraiccurve over Q, the rationals (glossing over more details here...this timea little more substantial). Then the modularity conjecture (now theorem)says that given any elliptic curve E over Q, there exists an integerN such that there exists a nonconstant morphism (of algebraic curves,dened over Q) from X_0(N) to E.Modular forms live naturally on things like X_0(N). They have a lotof structure that lets you compute the coefcients of their Fourierexpansions. Roughly speaking, the modularity conjecture carries thisstructure over to E and among other things tells you that these Fouriercoefcients count solutions to certain polynomial equations that maynot be easy to analyze otherwise.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === let a billion owers bloom!(I dont think that it should become a requirementfor graduating from LAUSD, yet .-) > Modular forms live naturally on things like X_0(N). They have a lot> of structure that lets you compute the coefcients of their Fourier> expansions. Roughly speaking, the modularity conjecture carries this> structure over to E and among other things tells you that these Fourier> coefcients count solutions to certain polynomial equations that may> not be easy to analyze otherwise.--A church-school McCrusade (Blairs ideals?):Harry-the-Mad-Potter wants US to kill Iraqis?...For a 1000-year anglo-american hegemony?HEY, JIMMY; LETS US and SU FIGHT -then-PM of England & Zbiggyhttp://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/les/curriculum/Cosmo.PCX== ==> Okay, I will make the assumption the Mr. Harris truly wants to learn> about this eld. I understand the history, here, but will try to> make no assumptions this once.Ive read expositions about Wiless work before, and even glanced atthe actual papers. What readers should notice is the vericationbelow of my saying that 4 descriptors are used that are the linkbetween elliptic curves and modular forms.The logical error the mathematicians are falling prey to is actuallyan easy one to explain as I did in my original post.Fighting a claim of logical error would necessarily involve attackingthat claim, not simply repeating the outlines of the argument with theerror.It turns out though that Wiless mistake is so clear that theressimply no correcting for it.The problem is that you cannot prove that a set of objects of onething are limited, based on another set of objects, unless you provean inclusion set, which denes how they are in some sense both thesame thing.Thats the identity relationship.For instance, cars have wheels. So you can talk of wheels and Mustangconvertibles and Volkswagen bugs, but you can also just talk of cars.That cars have wheels puts them in a superset of objects with wheels.Wiless work requires a superset, and as given would need to apply toobjects with the 4 descriptors. Um, thats an innite number ofobjects other than elliptic curves and modular forms, and there is noeffort made from what Ive seen to show a more dened set.For instance, many of you have 4 ngers on your hand. Write certainnumbers on each nger, and your ngers are included in the set thatWiless work covers. > Wiles proof can be seen to begin from a very interesting> representation of the problem (found independently by Gerhart Frey and> Yves Hellegouarch) as a form of elliptic curve. In particular,> beginning with the Fermat equation a^p + b^p +c^p = 0, they looked at:> E: y^2 = x (x - a^p) (x - b^p)> > Now, for this curve, it can be shown that there are relationships> between the diophantine properties of the Fermat equation and the> arithmetic properties of E. In particular, and this is important, E> can be shown to be semistable.What Ive seen from mathematicians is a love of big words. They seemto not only like big words, they seem to like big, complicated wordsthat few people understand.Id appreciate it if they were more logical.Now I like big words as well, at times, but I like logic more.> The next step can be taken in reference to Galois representations. In> particular, there is a particular Galois representation associated> with E, and one can prove some important properties of this particular> representation. In particular, it is absolutely irreducible, odd,> unramied outside 2p, and at at p. These are four very unique> properties.There they go. Notice that short shrift is given to the 4 descriptorsthat supposedly do such wondeful things, like supposedly provide a wayto prove Fermats Last Theorem.> Now one looks at newforms. To these modular forms, one may associate> a Galois representation through the theory of Eichler-Shimura. One> can prove that such representations obey certain equations on their> trace and determinant and also that they are unramied at all> sufciently large primes. And this can be generalized, so that we> may call Galois representations modular if they are unramied> outside the multiples of p and obey extensions of the trace and> determinant equations obeyed by the representations related to> newforms with the generalization being on setting the equalities in> relation to certain homomorphisms from Hecke operators to the eld> that is the base of the representation.What makes that fascinating is that its possible that internallyWiless work could be correct, but still be wrong.Logic is just that way.GIGO. Garbage In, Garbage Out.> The modular Galois representations were shown by Ribet to have> properties in contradiction to the four properties listed above for> the representation associated to E. Thus E does not have a modular> representation.And there go the 4 descriptors I mentioned ying past you.Some of you labor under the misapprehension that knowledge provestruth. But say I know what the length of the Nile river is?So what? Mathematicians can rattle off all kinds of knowledge which in contextis as useless as the length of the Nile river generally is.I mean, its neat to know the answer, like for the TV show Jeopardy,but a lot of the time, it doesnt matter a lot.> prove the hard part. He proved that every semistable elliptic curve> over Q is associated to a modular representation (is modular). > Since E was semistable, it should be modular, but we have already seen> it is not. Thus, a contradiction, and there is therefore no solution> to the equation a^p + b^p + c^p = 0 over Q.And once again, it doesnt matter if he succeeded because the approachfails no matter what.Linking one set of objects to another set of objects based on 4descriptors just isnt logically correct if you wish to prove that alimitation on the one set must apply to the other.> So, it has to be seen that Wiles proof does not follow the logic of> Mr. Harris post. It is a classic proof by contradiction. Of course,> there is much more going on here. In particular, there are the> details of the proofs mentioned. Its a beautiful theory, and if one> is generally interested in expanding their expertise, I would suggest> studying the necessary elds and, in particular, the theory required> to understand the proofs (elliptic curves, Galois representations, and> modular forms).The other fascinating thing I nd from mathematicians is that theymake bold statements that contradict the facts. Nothing statedrefutes my assertions, and in fact you could see those 4 descriptorsbeing mentioned more than once, which supports it.Some of you may get a little sick to your stomachs if someone givesthose 4 descriptors with an example so you can see some actual values,as suddenly you may fear that the basis of Western civilization isbogus crap.Dont worry, unlike the work of Wiles, we know a lot of things workbecause they connect to the real world, like thermodynamic theory andcars.Now a lot of pure mathematics may be total crap, but for most of youit probably wont affect your lives. It affects mine though.> I do hope this was not a waste of my time...Well I keep hoping that day after day as I continue to explain thetruth as simply as I can and face people who get away with statingthings that are false using really complicated stuff.The mathematical truth doesnt change day-to-day people.Wiless argument was wrong when he presented it, and its wrong today.Itll be wrong tomorrow, as well.And mathematicians can write reams of technical jargon which wontchange the truth.And yes, I do feel a certain sense of satisfaction at being someonewho actually *believes* in logic.James Harris === >Wiless work requires a superset, and as given would need to apply to>objects with the 4 descriptors.The conjecture that Wiles (partially) proved can be restated as follows. Every elliptic curve is a modular curve.Both elliptic curves and modular curves are curves. A parabola is a curve.A circle is a curve. An elliptic curve is a curve. A modular curve is acurve.I recognize that I have not referred to the 4 descriptors or used theexact term modular form. Thats because neither of these is neededin order to address the specic question you are asking.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === >Wiless work requires a superset, and as given would need to apply to> >objects with the 4 descriptors.> The conjecture that Wiles (partially) proved can be restated as follows.> Every elliptic curve is a modular curve.No. Thats the result he *wanted* if Im to accept that your claim iscorrect.But its meaningless unless a modular form is a modular curve.The reader should note that this poster did not dene modular curve. Im assuming that he means the result that every elliptic curve ismodular.Unfortunately Ive noticed a tendency within the math community toreact to correct challenges by replying with unsupported statements.My assessment is that when faced with the truth, as they cant refutethe truth, they just make something up. > Both elliptic curves and modular curves are curves. A parabola is a curve.> A circle is a curve. An elliptic curve is a curve. A modular curve is a> curve.But what is a modular form then?Remember Ribet purportedly proved that if FLT were false an ellipticcurve would exist thats not modular, but Wiles worked to link thoseelliptic curves to modular forms by noting 4 descriptors between themto supposedly prove that an elliptic curve had to be modular.Like I said in the post to which youre replying, if you have 4ngers, you can put your ngers in the SAME set by writing certainnumbers on them.And like I said in my post to which youre replying I think somereaders who may think the mathematicians *must* have something wouldget sick if they actually saw some of those numbers.Why doesnt some mathematician post an example giving the 4descriptors?> I recognize that I have not referred to the 4 descriptors or used the> exact term modular form. Thats because neither of these is needed> in order to address the specic question you are asking.If Wiles had a proof then he could prove something about ellipticcurves without reference to modular forms by referring to thesuperset.So, *if* he had a proof all mention of modular forms could be deleted.Debates wont change logic.Note: Part of the reason I included the newsgroup sci.logic is toemphasize that point as Ive noticed posters on math newsgroups to soeasily switch to non-logic statements and get general *agreement* frommath newsgroups like sci.math that it made *me* sick.James Harris === >> Every elliptic curve is a modular curve.>>But its meaningless unless a modular form is a modular curve.No. If youve gotten the impression that Wiles claimed that ellipticcurves and modular forms are the same type of object, then either youor the account youve read is mistaken. The simplest denition of athe upper half of the complex plane satisfying certain symmetry properties.So a modular form is a function, whereas an elliptic curve is a curve.As you are pointing out, it therefore doesnt make sense to say that theyare the same thing.But the fault lies not with Wiles or other establishment mathematicians,since they havent claimed that elliptic curves and modular forms are thesame thing.>The reader should note that this poster did not dene modular curve. >Im assuming that he means the result that every elliptic curve is>modular.Thats right.>Unfortunately Ive noticed a tendency within the math community to>react to correct challenges by replying with unsupported statements.Are you saying that I made an unsupported statement? Which one?>But what is a modular form then?Explained above. I glossed over the exact symmetry properties, but I canprovide them if necessary. I dont think its relevant to the argument.>Remember Ribet purportedly proved that if FLT were false an elliptic>curve would exist thats not modular, but Wiles worked to link those>elliptic curves to modular forms by noting 4 descriptors between them>to supposedly prove that an elliptic curve had to be modular.Thats all correct, except that the part about 4 descriptors betweenthem is not precise. The 4 descriptors are properties of *Galoisrepresentations*, not of elliptic curves or modular forms per se. Toan elliptic curve one can associate a Galois representation, and to amodular form one can associate a Galois representation. Then we cantalk about comparing Galois representations to each other. One doesnot directly identify an elliptic curve with a modular form anywhere.Notice that even in the way youve stated the facts, there is no statementthat says that Wiles claimed to prove that modular forms and ellipticcurves are the same type of thing.>Why doesnt some mathematician post an example giving the 4 descriptors?Example of what? Of a Galois representation?>If Wiles had a proof then he could prove something about elliptic>curves without reference to modular forms by referring to the>superset.>>So, *if* he had a proof all mention of modular forms could be deleted.I dont follow you here. If I want to prove that every integer isan algebraic integer, how do I do this without mentioning algebraicintegers?-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === nevermind. lets start another item! > But the fault lies not with Wiles or other establishment mathematicians,> since they havent claimed that elliptic curves and modular forms are the> same thing. > Thats all correct, except that the part about 4 descriptors between> them is not precise. The 4 descriptors are properties of *Galois> representations*, not of elliptic curves or modular forms per se. To> an elliptic curve one can associate a Galois representation, and to a> modular form one can associate a Galois representation. Then we can> talk about comparing Galois representations to each other. One does> not directly identify an elliptic curve with a modular form anywhere. > I dont follow you here. If I want to prove that every integer is> an algebraic integer, how do I do this without mentioning algebraic> integers?--les ducs dEnron! http://quincy4board.homestead.com/ Funny.html (schoolboard stufn) === > Okay, I will make the assumption the Mr. Harris truly wants to learn> about this eld. I understand the history, here, but will try to> make no assumptions this once.>> Ive read expositions about Wiless work before, and even glanced at> the actual papers. What readers should notice is the verication> below of my saying that 4 descriptors are used that are the link> between elliptic curves and modular forms.The fact that you understand the quantity (4) of properties used in theproof, does not mean you understand what they mean or even what they applyto (the Galois representations). And this is more historical result toWiles proof, which follows from work by Ribet.> The logical error the mathematicians are falling prey to is actually> an easy one to explain as I did in my original post.>> Fighting a claim of logical error would necessarily involve attacking> that claim, not simply repeating the outlines of the argument with the> error.>> It turns out though that Wiless mistake is so clear that theres> simply no correcting for it.Wiles proved (with the help of the previous results) thatA has property xA does not have property xTherefore not AI see only one logical attack on this form of proof:do not assume tertium non datur (re: Heyting algebras)However such an assumption invalidates your own proof as well.(excepting the fact that you attempt certain factorings which you areattempting to prove and certain other well known mistakes mentioned on otherthreads)> The problem is that you cannot prove that a set of objects of one> thing are limited, based on another set of objects, unless you prove> an inclusion set, which denes how they are in some sense both the> same thing.>> Thats the identity relationship.Do you mean you cannot say that a particular algebraic number over a eldcannot be mapped to an ideal of the ring of polynomials over that eld?You do that in your proof.> For instance, cars have wheels. So you can talk of wheels and Mustang> convertibles and Volkswagen bugs, but you can also just talk of cars.>> That cars have wheels puts them in a superset of objects with wheels.>> Wiless work requires a superset, and as given would need to apply to> objects with the 4 descriptors. Um, thats an innite number of> objects other than elliptic curves and modular forms, and there is no> effort made from what Ive seen to show a more dened set.There is a functor from the category of elliptic curves to that of Galoisrepresentations constructed thus:One constructs the Tate module of the groups of torsion points on theelliptic curve at a given prime (in the limit over the powers).The p-adic Galois representation is given by the mapping from the Galoisgroup over Q to the general linear group over the Tate module on E.> For instance, many of you have 4 ngers on your hand. Write certain> numbers on each nger, and your ngers are included in the set that> Wiless work covers.???Wiles work covers semistable elliptic curves. No matter how many ngers Ihold up or write numbers on, I do not see semistable elliptic curves (maybeif I squint harder?).> Wiles proof can be seen to begin from a very interesting> representation of the problem (found independently by Gerhart Frey and> Yves Hellegouarch) as a form of elliptic curve. In particular,> beginning with the Fermat equation a^p + b^p +c^p = 0, they looked at:>> E: y^2 = x (x - a^p) (x - b^p)>> Now, for this curve, it can be shown that there are relationships> > between the diophantine properties of the Fermat equation and the> arithmetic properties of E. In particular, and this is important, E> can be shown to be semistable.>> What Ive seen from mathematicians is a love of big words. They seem> to not only like big words, they seem to like big, complicated words> that few people understand.>> Id appreciate it if they were more logical.>> Now I like big words as well, at times, but I like logic more.I explained in a nearby leaf of this thread what semistable means. But youshould take some initiative yourself, as well.> The next step can be taken in reference to Galois representations. In> > particular, there is a particular Galois representation associated> with E, and one can prove some important properties of this particular> representation. In particular, it is absolutely irreducible, odd,> unramied outside 2p, and at at p. These are four very unique> properties.>> There they go. Notice that short shrift is given to the 4 descriptors> that supposedly do such wondeful things, like supposedly provide a way> to prove Fermats Last Theorem.I could explain absolutely irreducible, odd, unramied at an ideal,and at, but I would suggest spending some time with it yourself andcoming back when you have difculties. If you are questioning big wordsnow, I have no idea how far I would have to go back to stop the ad hominemand argument from ignorance fallacies.[...]> prove the hard part. He proved that every semistable elliptic curve> over Q is associated to a modular representation (is modular).> Since E was semistable, it should be modular, but we have already seen> it is not. Thus, a contradiction, and there is therefore no solution> to the equation a^p + b^p + c^p = 0 over Q.>> And once again, it doesnt matter if he succeeded because the approach> fails no matter what.>> Linking one set of objects to another set of objects based on 4> descriptors just isnt logically correct if you wish to prove that a> limitation on the one set must apply to the other.How many properties of an object must be known before one may proveconsequences of those properties, then? 10? 100? Or is it impossible toprove properties of an object? The linking between object categories isdone by a constructed functor, so we are talking only about properties onone category and its restriction to that functor.> So, it has to be seen that Wiles proof does not follow the logic of> > Mr. Harris post. It is a classic proof by contradiction. Of course,> there is much more going on here. In particular, there are the> details of the proofs mentioned. Its a beautiful theory, and if one> is generally interested in expanding their expertise, I would suggest> studying the necessary elds and, in particular, the theory required> to understand the proofs (elliptic curves, Galois representations, and> modular forms).>> The other fascinating thing I nd from mathematicians is that they> make bold statements that contradict the facts. Nothing stated> refutes my assertions, and in fact you could see those 4 descriptors> being mentioned more than once, which supports it.4> Some of you may get a little sick to your stomachs if someone gives> those 4 descriptors with an example so you can see some actual values,> as suddenly you may fear that the basis of Western civilization is> bogus crap.4...[...]I see it was a waste of my time for you, Mr. Harris. I gave you thebenet. Instead, you explored nothing. You are indeed either a troll (abad one at that, since you have to keep posting to keep a thread alive) orseriously psychologically avoidant. Which is a shame, since looking at yourwebsite I see that you do appear to enjoy manipulating equations, and if youcould just focus yourself on seeing where you were making mistakes (whichothers have pointed out in other threads), I am sure you could deriveresults on your own that are truly unique, as many do every day. I hopethat you are a troll, and that you do real math somewhere behind yourscreen, and that this is just your way of letting out steam. Thealternative is rather sad.Unfortunately, I wont be responding to any more of your threads. === >> Every elliptic curve is a modular curve.>>But its meaningless unless a modular form is a modular curve.> No. If youve gotten the impression that Wiles claimed that elliptic> curves and modular forms are the same type of object, then either you> or the account youve read is mistaken. The simplest denition of a> the upper half of the complex plane satisfying certain symmetry properties.> So a modular form is a function, whereas an elliptic curve is a curve.> As you are pointing out, it therefore doesnt make sense to say that they> are the same thing.I didnt say that as I noted that your statement was meaninglessunless a modular form is a modular curve.What Wiles wants is to prove something about one set based on another.In mathematics that requires an inclusion set, by which I mean a setwhere you include both of the other sets.For instance, Mustang convertibles are cars, and Volkswagen Bugs arecars.Cars are the inclusion set, or the identity set, or the superset, orthe superclass, as interestingly enough there are any number of waysto describe.Cars as a set is the inclusion set because it includes both Mustangconvertibles and Volkswagen Bugs.Cars as a set is the identity set because Mustang convertibles arecars, and Volkswagen Bugs are cars.It can be considered the superset because its a higher order set, andsimilarly from object oriented thinking its a superclass because itsa higher order class.My point is that Wiles needs a superset.> But the fault lies not with Wiles or other establishment mathematicians,> since they havent claimed that elliptic curves and modular forms are the> same thing.I didnt say they had. But what mathematicians *have* to have is asuperset.So far they have modular forms and elliptic curves, and having readfurther down I can now add Galois Representations.However, I still dont see mention of a superset. >The reader should note that this poster did not dene modular curve. >Im assuming that he means the result that every elliptic curve is>modular.> Thats right.I was making certain that modular curve isnt a phrase that mightmean something other than curve that is modular, as mathematics canget quirky.The verication is that the obvious meaning is correct.I conclude then that modular curve does not necessarily mean modularelliptic curve.>Unfortunately Ive noticed a tendency within the math community to>react to correct challenges by replying with unsupported statements.> Are you saying that I made an unsupported statement? Which one?I copy the following from your post to which I replied and made thatcomment:>Wiless work requires a superset, and as given would need to apply to>objects with the 4 descriptors.The conjecture that Wiles (partially) proved can be restated asfollows. Every elliptic curve is a modular curve.My post was challenging that Wiles had a proof because he doesntidentify a superset, in reply you simply restated the claim, and addedin the modular curve which had to be further dened, though I couldhave *assumed* what you meant.Lets say that I have a proof that the sun orbits the earth. Alsolets say that world opinion is that it does, so I have that on myside, with supposed proofs of that claim.You reply with observations that invalidate my proof.I reply that the sun moves in such a way that it tends to keep theearth at a loci in its orbit.You reply that my statement is unsupported.The context is that given your disproof of the original statement, Icant simply restate the same claim in a different way, as that is anunsupported claim in the context of the discussion.>But what is a modular form then?> Explained above. I glossed over the exact symmetry properties, but I can> provide them if necessary. I dont think its relevant to the argument.My point is, what superclass includes elliptic curves *and* modularforms.Just like theres a superclass that includes Mustang convertibles*and* Volkswagen bugs.The problem is that if youre not dealing with the superclass, if younotice something about one it doesnt necessarily apply to the other.For instance, lets say Volkswagen Bugs have single-side mirrors, andI notice Mustang convertibles with single-side mirrors.Now I imagine a Mustang convertible with dual-side mirrors, look atall the Volkswagen Bugs and nd they are all single-side, andproclaim Ive proven that this particular Mustang convertible doesntexist!!!Thats what Wiless and Ribets work basically is.The logical error is not complicated.In mathematics you need an identity or inclusion set. >Remember Ribet purportedly proved that if FLT were false an elliptic>curve would exist thats not modular, but Wiles worked to link those>elliptic curves to modular forms by noting 4 descriptors between them>to supposedly prove that an elliptic curve had to be modular.> Thats all correct, except that the part about 4 descriptors between> them is not precise. The 4 descriptors are properties of *Galois> representations*, not of elliptic curves or modular forms per se. To> an elliptic curve one can associate a Galois representation, and to a> modular form one can associate a Galois representation. Then we can> talk about comparing Galois representations to each other. One does> not directly identify an elliptic curve with a modular form anywhere.Which is just a substitution of Galois representation for 4descriptors, and you have the same problem.The logic is rather direct and absolutely, as always, rigid.> Notice that even in the way youve stated the facts, there is no statement> that says that Wiles claimed to prove that modular forms and elliptic> curves are the same type of thing.And THATS THE problem!!!>Why doesnt some mathematician post an example giving the 4 descriptors?> Example of what? Of a Galois representation?Just give the 4 numbers, isnt L something in there somewhere?>If Wiles had a proof then he could prove something about elliptic> >curves without reference to modular forms by referring to the>superset.>>So, *if* he had a proof all mention of modular forms could be deleted.> I dont follow you here. If I want to prove that every integer is> an algebraic integer, how do I do this without mentioning algebraic> integers?But you see, algebraic integers is the superset.So *of course* you can use the superset.Integers are included in that set.Its all quite logical.Wiles needs to show a superset. Without it, he cant have a proof.How about this? Ask yourself, wheres the identity relationship inWiless work?James Harris === >I was making certain that modular curve isnt a phrase that might>mean something other than curve that is modular, as mathematics can>get quirky.Fair enough. In fact, perhaps I should have said Weil curve ratherthan modular curve since thats the more common term, and sometimespeople use modular curve for something else. Let me use Weil curvefrom now on. So the (allegedly unsupported) claim is, Every ellipticcurve is a Weil curve.>I conclude then that modular curve does not necessarily mean modular>elliptic curve.Huh? Actually, it does mean a modular elliptic curve.>>Why doesnt some mathematician post an example giving the 4 descriptors?>> Example of what? Of a Galois representation?>Just give the 4 numbers, isnt L something in there somewhere?Four numbers? What numbers? Do you mean listing the four properties?Galathea did that elsewhere in this thread.>But you see, algebraic integers is the superset.>So *of course* you can use the superset.>Integers are included in that set.Ah, in that case, the answer to your question is:Weil curves comprise the superset.Elliptic curves are included in that set.Every elliptic curve is a Weil curve is analogous to every integer is analgebraic integer.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === >That brings me to Wiless proof >Wiles produced a proof (solid), not a proof (suspect). That proofhas been vetted to an extent comparable to, and probably greater than,notions such as electrons, AIDS, relativity, etc. Of course, youhave given no reason to disbelieve in anything Wiles did.>Wiles purportedly proves >Wiles actually, not purportedly, proved Fermats Last Theorem.If you have any reason to disbelieve that which is not also a reasonto disbelieve in photons, DNA, and the Jurassic era, nothing preventsit from being posted. === >Wiles never nds a superset but instead tries to map two sets of>objects, where one set contains objects called elliptic curves, while>the other contains objects called modular forms.No, thats not what he tried to do. You were misinformed. He tried toshow that every elliptic curve is a Weil curve. The set of Weil curvesis the superset.>Mathematicians noticed that for objects they checked 4 descriptors>could be matched between objects that were members of these setsDoes this sentence even make grammatical sense?>Wiles set out to map innity against innity and claimed proof that>a limitation on modular forms was a limitation on elliptic curves.Nope, thats not what he set out to do. You were misinformed.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === ------------------------------------------------- -------------------->Wiles produced a proof (solid), not a proof (suspect). That proof>has been vetted to an extent comparable to, and probably greater than,notions such as electrons, AIDS, relativity, etc.>Of course, you>>have given no reason to disbelieve in anything Wiles did.>>To the contrary, Ive made a specic assertion based on logical>principles.>Your assertion was no reason to disbelieve in anything Wiles did, as it (yourassertion, that is) was illogical and mistaken.In terms of your analogies, Wiles proved a statement of the formfor every man there is a suit that ts him. Nowhere does he claimthat the set of Men is equal (or equivalent, isomorphic, etc) to theset of Suits. It would be preposterous to demand that for his proof tobe valid, he nd some superset(as opposed to discussing tailors,clothing stores and so on). All he needs to prove, logically speaking, is that for every item of the rstkind (a man) there is at least one item of the second kind (a suit) bearing acertain relationship (tting) to the rst item. Replace the words [man, suit, ts]by [semistable elliptic curve, modular curve, modular parametrization] and you getexactly the statement proved by Wiles.> Wiles purportedly proves > Wiles actually, not purportedly, proved Fermats Last Theorem.>> If you have any reason to disbelieve that which is not also a reason>> to disbelieve in photons, DNA, and the Jurassic era, nothing prevents>> it from being posted.>Yet Wiless work is whats called pure math and I think it telling>that given that mathematicians have *bragged* about pure math not>being relevant to the real world that your attempt now is to link>Wiless work to the real world.>of thereal-world relevance or irrelevance of Wiles work or any other mathematics.What I actually stated is that there is nothing to differentiate your complaintsabout Wiles from complaints about photons, DNA, the Jurassic era, evolution,relativity, and the rest. === >Wiles never nds a superset but instead tries to map two sets of>objects, where one set contains objects called elliptic curves, while> >the other contains objects called modular forms.> No, thats not what he tried to do. You were misinformed. He tried to> show that every elliptic curve is a Weil curve. The set of Weil curves> is the superset.My reference for this discussion ishttp://mathworld.wolfram.com/ Taniyama-ShimuraConjecture.html and fromthat page I quote:In effect, the conjecture says that every rational elliptic curve isa modular form in disguise.A little further down its stated:Equivalently, for every elliptic curve, there is a modular form withthe same Dirichlet L-series.Note for readers: The Dirichlet L-series goes back to the 4descriptors Ive repeatedly mentioned.Besides even if what you claim is correct, itd mean that modularforms are Weil curves. Is that your assertion?A good example for those who feel confused about the logical aw IveKnowles with the context of having children or capable of havingchildren.The superset in that context is the set of women, as women are the sexcapable of having children.Notice that with the superset identied I dont have to talktalk about women and having children.For Wiles to have a proof, he needs a superset, and if he had asuperset, you wouldnt have to talk about elliptic curves or modularforms, as you could talk about the superset.You could dene the limitation on the superset, and itdautomatically apply to both.>Mathematicians noticed that for objects they checked 4 descriptors>could be matched between objects that were members of these sets> Does this sentence even make grammatical sense?Possibly you need more punctuation.Mathematicians noticed, that for objects they checked, 4 descriptorscould be matched between objects that were members of these sets.Technically (or maybe grammarians can correct me) the commas dontbelong for a subordinate clause beginning with that.>Wiles set out to map innity against innity and claimed proof that>a limitation on modular forms was a limitation on elliptic curves.> Nope, thats not what he set out to do. You were misinformed.Well that goes back to your earlier claim, where you mention Weilcurve.Continuing on, some readers may think that theres no way thatmathematicians would go on for years making a claim that I can soeasily show to be false as it is based on a logically awed approach,but in my experience mathematicians live in a society that is veryconforming as well as being strictly hierarchical.The following is speculation, but I think it outlines what may havehappened.Wiles decided to prove Fermats Last Theorem. Isolating himself heset out to nd a way to show that the Taniyama-Shimura Conjecture waswhile letting his colleagues believe he was working on something else. Thats a deception thats important in context.Assume that he really wanted to prove Fermats Last Theorem, andconsider that he was spending a lot of time by himself. Lets supposethat he had an approach and decided that hed take it, and convincedhimself that itd work.My understanding is that Wiles was a well established and respectedmathematician: well within the ranks of the mathematical elite. Whenhis fellow colleagues within his circle heard what he was working on,and that he felt that hed succeeded, they probably felt elation.Now lets suppose that slowly it sinks in that his path is logicallyawed.To tell him would mean forcing him to realize that hed thrown awayall those years, and would probably humiliate him. Hed be like somecrank, like so many others whod thought theyd proven the greatproblem.So instead they look carefully and nd a gap, which possibly wouldhave let him out with some dignity. But Wiles gets a former student,and they work for a year to nd a workaround.Its easier after so much time for mathematicians to give in, as theirsociety is so against confrontation on such matters, as their socialorder would not handle one of their elite being so humbled.Now I have personal experience to add, as I contacted Ken Ribet a fewyears back when I had an unwise bet about one of my earlier awedattempts at proving FLT (yeah, I lost the bet). He not only repliedbut offered to have one of his graduate students look over my work.That should surprise you. He said he was intrigued by the idea of thebet.I suggest to you that there may have been more to it.Advanced Polynomial Factorization and he replied back withencouragement and some pointed questions. I replied back of courseincluding the nal version, but havent heard from him since.What Ive presented is speculation and circumstantial evidence basedon limited contact that Ive had with people most of you probably onlyread about, if you know about FLT, where you can read about KenRibets paper, or Barry Mazurs role in the FLT saga.But the people you just read about, Ive contacted, and in some cases,as Ive mentioned, theyve answered back.Who knows, maybe they *want* the truth to be known, but are trappedsomewhat by circumstance and training. What might have began in atime of heavy emotion has continued now to a point where they may feelincapable of stopping it.I suggest to you the possibility that they were trapped in the wilesof Wiles.But that of course is speculation, and you may consider it wildspeculation, which I understand. Im merely trying to nd somerational reason for a situation which is rather bizarre.For what there is no doubt about is that without a superset, Wilesdoes not have a proof of the Taniyam-Shimura Conjecture, andtherefore, has not proven Fermats Last Theorem.The logic is clear.James Harris === >http://mathworld.wolfram.com/ Taniyama-ShimuraConjecture.html>In effect, the conjecture says that every rational elliptic curve is>a modular form in disguise.The words in disguise should tip you off to the fact that this sentence isnot intended to be a strictly accurate mathematical statement, but is aninformal statement in which the writer has taken some poetic license.>Equivalently, for every elliptic curve, there is a modular form with>the same Dirichlet L-series.Ah, well, this quotation helps explain where you got your impression ofwhat Wiles was claiming. You are perfectly correct in analyzing thissentence by pointing out that on the one hand we have elliptic curves,and on the other hand we have modular forms, and there is no supersetof objects of which elliptic curves and modular forms are both members.So just because you get this L-series thingy from an elliptic curve,and you can get the same L-series thingies from modular forms, how canthis possibly imply anything like all elliptic curves are modular?The answer is that you are right, it *doesnt* immediately imply thatall elliptic curves are modular. It has to be combined with othertheorems in a way that isnt fully explained on the website (whichafter all is just a sketchy overview). So before you can justlycriticize the proof, you need to look up the other theorems and seehow they are used together with the L-series statement to deduce thatevery elliptic curve is modular.>Besides even if what you claim is correct, itd mean that modular>forms are Weil curves. Is that your assertion?No. You need to have the courage of your convictions! You have correctlynoted that there is no superset that contains both elliptic curves andmodular forms. Saying that an elliptic curve *is* a modular form istherefore, as you correctly point out, simply wrong, and even Wiles hasnever claimed that. Similarly, saying that modular forms are Weil curvesis a mistake of the same type, and I dont make that assertion; neitherdoes Wiles make that assertion. It may follow from the claim that youare erroneously *attributing* to Wiles, but thats irrelevant.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === what he said! > Ah, well, this quotation helps explain where you got your impression of> what Wiles was claiming. You are perfectly correct in analyzing this> sentence by pointing out that on the one hand we have elliptic curves,> and on the other hand we have modular forms, and there is no superset> of objects of which elliptic curves and modular forms are both members.> So just because you get this L-series thingy from an elliptic curve,> and you can get the same L-series thingies from modular forms, how can> this possibly imply anything like all elliptic curves are modular?> The answer is that you are right, it *doesnt* immediately imply that> all elliptic curves are modular. It has to be combined with other> theorems in a way that isnt fully explained on the website (which> after all is just a sketchy overview). So before you can justly> criticize the proof, you need to look up the other theorems and see> how they are used together with the L-series statement to deduce that> every elliptic curve is modular.--A church-school McCrusade (Blairs ideals?):Harry-the-Mad-Potter wants US to kill Iraqis?...For a 1000-year anglo-american hegemony?HEY, JIMMY; LETS US and SU FIGHT -then-PM of England & Zbiggyhttp://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/les/curriculum/Cosmo.PCX== ==sorry, for breaking my recent vow, again, but ...this reminds me of that British math-guys absurd denitionof leftwing and rightwing proofs. of course, althoughI believe Ribet, that Wiles did prove *le prochaine theoremde Fermat* (a-hem), I also am (currently!) incompetentto comprehend Ribets founding work (oranyones elses, beyond the nding of Sophie G.)upon which it rests (his being the last in the chainof results that Wiles needed to use). actually,Wiles proof might really be considered to be rightwing,in the sense of Tory, in the sense thathe may have been *encouraged* to undertake his closet-proving,for the greater glory of Cool Britannia (althoughpossibly not with his conscious notice .-)personally, I think that it was proven in 76,ten years before I got wind of the problem,using comparitively elementary methods(trigonometric series plus modular arithmetic .-)now, other recent proofs may be more-opento overturning. > Wiles produced a proof (solid), not a proof (suspect). That proof> has been vetted to an extent comparable to, and probably greater than,> notions such as electrons, AIDS, relativity, etc. Of course, you> have given no reason to disbelieve in anything Wiles did.--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...La Troi Phases dExploitation de la Protocolsdes Grises de Kyoto:(FOSSILISATION [McCainanites?] (TM/sic))/BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm.Http://www.tarpley.net/ bushb.htm (content partiale, below): 17 -- LATTEMPTER de COUP DETAT, 3/30/81 23 -- Le FIN dHISTOIRE 24 -- LORDEUR du MONDE NOUVEAU 25 -- THYROID STORK !?! === [...]| Fair enough. In fact, perhaps I should have said Weil curve rather| than modular curve since thats the more common term, and sometimes| people use modular curve for something else.When I last regularly attended number theory seminars in the Bostonarea, in the 1990s, Weil curve was less common than modular curve.For one thing, the reason the name Weil was associated with it isthat some people called the Taniyama-Shimura conjecture the Weilconjecture. As Serge Lang has amply pointed out, it doesnt makesense to attribute the conjecture solely or mainly to Weil. (I thinkhe might say, at all, although Ive read that Weil made some usefulrenements on it relating to the level of the modular form and theconductor of the curve, or something like that.)I had the impression in fact that the term modular curve had largelysupplanted Weil curve, and that the Taniyama-Shimura(-Weil) conjecturewas typically being called the modularity conjecture, now of course atheorem.Keith Ramsay === >Wiles produced a proof (solid), not a proof (suspect). That proof>has been vetted to an extent comparable to, and probably greater than,notions such as electrons, AIDS, relativity, etc.>Of course, you> >>have given no reason to disbelieve in anything Wiles did.> >>To the contrary, Ive made a specic assertion based on logical>principles.>> Your assertion was no reason to disbelieve in anything Wiles did, as it > (your> assertion, that is) was illogical and mistaken.Lets see if you prove that assertion in this post. > In terms of your analogies, Wiles proved a statement of the form> for every man there is a suit that ts him. Nowhere does he claim> that the set of Men is equal (or equivalent, isomorphic, etc) to the> set of Suits. It would be preposterous to demand that for his proof to> be valid, he nd some superset(as opposed to discussing tailors,> clothing stores and so on). No. Something closer would be that every man must wear a suit thatts him.Now if you keep looking and notice that every man you see wears a suitthat ts him, then decide that every man must wear a suit that tshim, youre making the mistake of Wiles. > All he needs to prove, logically speaking, is that for every item of the > rst> kind (a man) there is at least one item of the second kind (a suit) > bearing a> certain relationship (tting) to the rst item. Replace the words > [man, suit, ts]> by [semistable elliptic curve, modular curve, modular parametrization] > and you get> exactly the statement proved by Wiles.The claim is that every semistable ellipic curve is modular.What mathematicians noticed is that they could associate variouselliptic curves with modular forms. Taniyam and Shimura conjecturedthat every elliptic curve had a corresponding modular form. Ribetpuportedly proved that if FLT were false thered be an elliptic curvethat did NOT have a corresponding modular form. Wiles did acomparison check where he tried to count out an innite number ofmodular forms against an innite number of elliptic curves.However, as Ive been pointing out, you could associate an innitenumber of elliptic curves with an innite number of modular forms andstill have an innite number of elliptic curves that were notmodular, if you dont have an inclusion set.Going back to your analogy, if you nd that you have an innitenumber of men with suits that t, where theyre *wearing* the suitsthat t, you cant assume that if you nd an innite number wearingtting suits that there cant be some oddball somewhere, like Kramerfrom Seinfeld, wearing an ill-tting suit.The logical fallacy is simple: Association is not a proof of acondition.I think the actual name for it is: Cum hoc ergo propter hoc.(Source http://users.tru.eastlink.ca/~brsears/reafault.htm )> Wiles purportedly proves >> Wiles actually, not purportedly, proved Fermats Last Theorem.>> If you have any reason to disbelieve that which is not also a reason>> to disbelieve in photons, DNA, and the Jurassic era, nothing prevents>> it from being posted.>Yet Wiless work is whats called pure math and I think it telling>that given that mathematicians have *bragged* about pure math not>being relevant to the real world that your attempt now is to link>Wiless work to the real world.>> of the> real-world relevance or irrelevance of Wiles work or any other mathematics.Your assertion appeared to me to be an attempt at claiming that thetruth of real world phenomena is dependent on the diligence of peoplechecking. As Wiless work rests on the acceptance by a relativelysmall group of people who *supposedly* have checked it thoroughly sothat there cant be an error.Remember mathematics requires perfection.> What I actually stated is that there is nothing to differentiate your > complaints> about Wiles from complaints about photons, DNA, the Jurassic era, > evolution,> relativity, and the rest.Which either shows youre extraordinarily naive, or you think thereaders of sci.logic are. Wiless supposed accomplishment rests*solely* on the assertion of a relatively small group of people thathis work is correct.Yet to take one of your examples--photons--and consider that theexistence of photons has been theorized for some time, but was provenby experiment.Since that time from lasers to spectral analysis the theory has twith reality.What I want readers to appreciate is that mathematicians and peoplewho support them can be as illogical as any true believers.My assertion that a restriction shared between members of disparatesets requires an inclusion set is not refutable.Therefore, Wiless work is logically awed at its outset, and itslength and complexity are actually irrelevant.James Harris === > >http://mathworld.wolfram.com/Taniyama-ShimuraConjecture.html >In effect, the conjecture says that every rational elliptic curve is>a modular form in disguise.> The words in disguise should tip you off to the fact that this sentence is> not intended to be a strictly accurate mathematical statement, but is an> informal statement in which the writer has taken some poetic license.And that statement should tip off readers, but the smoking gun comesfrom that site and I have it below in the relevant spot. >Equivalently, for every elliptic curve, there is a modular form with>the same Dirichlet L-series.> Ah, well, this quotation helps explain where you got your impression of> what Wiles was claiming. You are perfectly correct in analyzing this> sentence by pointing out that on the one hand we have elliptic curves,> and on the other hand we have modular forms, and there is no superset> of objects of which elliptic curves and modular forms are both members.> So just because you get this L-series thingy from an elliptic curve,> and you can get the same L-series thingies from modular forms, how can> this possibly imply anything like all elliptic curves are modular?Good question.> The answer is that you are right, it *doesnt* immediately imply that> all elliptic curves are modular. It has to be combined with other> theorems in a way that isnt fully explained on the website (which> after all is just a sketchy overview). So before you can justly> criticize the proof, you need to look up the other theorems and see> how they are used together with the L-series statement to deduce that> every elliptic curve is modular.Now thats fascinating given what else is on that webpage, which Imcalling the smoking gun.>Besides even if what you claim is correct, itd mean that modular> >forms are Weil curves. Is that your assertion?> No. You need to have the courage of your convictions! You have correctly> noted that there is no superset that contains both elliptic curves and> modular forms. Saying that an elliptic curve *is* a modular form is> therefore, as you correctly point out, simply wrong, and even Wiles has> never claimed that. Similarly, saying that modular forms are Weil curves> is a mistake of the same type, and I dont make that assertion; neither> does Wiles make that assertion. It may follow from the claim that you> are erroneously *attributing* to Wiles, but thats irrelevant.Hey, I go by what mathematicians say in talking about Wiless work,and heres a pertinent quote from the webpage I gave before:As of the early 1990s, most mathematicians believed that theTaniyama-Shimura conjecture was not accessible to proof. However, A.Wiles was not one of these. He attempted to establish thecorrespondence between the set of elliptic curves and the set ofmodular elliptic curves by showing that the number of each was thesame. Wiles accomplished this by counting Galois representations andcomparing them with the number of modular forms.(Source http://mathworld.wolfram.com/Taniyama-ShimuraConjecture.html )My assessment is that Wiles commits the logical fallacy of Cum hocergo propter hoc.(Source http://users.tru.eastlink.ca/~brsears/reafault.htm )My suggestion for readers is that they look back over comments bymathematicians about Wiless work, and pay attention to things likeWiless deception about what he was doing for all those years.Why would Wiles deceive his colleagues? Why havent more peoplethought that relevant? Why be surprised that a man obsessed andisolated for several years living a deception might delude himselnto believing a logically awed approach might work?James Harris === > ... Why be surprised that a man obsessed and> isolated for several years living a deception might delude himself> into believing a logically awed approach might work?>> James HarrisYoure certainly more than qualied to answer that question yourself, James Harris. Do yourself afavor and stop aunting your ignorance all over the internet.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === >>As of the early 1990s, most mathematicians believed that the>Taniyama-Shimura conjecture was not accessible to proof. However, A.>Wiles was not one of these. He attempted to establish the>correspondence between the set of elliptic curves and the set of>modular elliptic curves by showing that the number of each was the>same. Wiles accomplished this by counting Galois representations and>comparing them with the number of modular forms.>[...]>My assessment is that Wiles commits the logical fallacy of Cum hoc>ergo propter hoc.Popular, secondhand sources inevitably oversimplify technical statements.Here they even cue you to the fact by putting counting in scare quotes.What youre doing is to take an informal statement in a secondary sourceliterally, noticing that it is not perfectly accurate mathematically, andthen concluding that the formal mathematics in the primary sources mustbe logically awed.Its illegitimate to fault Wiless argument on the basis of secondarysources. If you think there is something wrong with Wiless argument,tell us specically which claims in his paper, or in his joint paperwith Richard Taylor, are wrong. I assume you *have*, of course, readand understood both papers? That you are not simply relying on secondarysources because the primary sources are too advanced for you?-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === >>As of the early 1990s, most mathematicians believed that the> >Taniyama-Shimura conjecture was not accessible to proof. However, A.>Wiles was not one of these. He attempted to establish the>correspondence between the set of elliptic curves and the set of>modular elliptic curves by showing that the number of each was the>same. Wiles accomplished this by counting Galois representations and>comparing them with the number of modular forms.>> [...]>My assessment is that Wiles commits the logical fallacy of Cum hoc>ergo propter hoc.> Popular, secondhand sources inevitably oversimplify technical statements.> Here they even cue you to the fact by putting counting in scare quotes.> What youre doing is to take an informal statement in a secondary source> literally, noticing that it is not perfectly accurate mathematically, and> then concluding that the formal mathematics in the primary sources must> be logically awed.Well in another thread someone posted a link to Wiless paper, so Ivestarted looking over its introduction.Heres an intriguing quote which Im focusing on, though it may dragme into greater details.The key development in the proof is a new and surprising link betweentwo strong but distinct traditions in number theory,the relationshipbetween Galois representations and modular forms on the one hand andthe interpretation of special values of L -functions on the other.p.2Source: http://modular.fas.harvard.edu/21n/papers/Wiles,Modular_ Elliptic_Curves_and_Fermats_Last_Theorem.pdfThe special valus of L-functions are those 4 descriptors popping upagain. My understanding is that mathematicians have reworked thatapproach having found something shorter, but Ill focus on theoriginal. > Its illegitimate to fault Wiless argument on the basis of secondary> sources. If you think there is something wrong with Wiless argument,> tell us specically which claims in his paper, or in his joint paper> with Richard Taylor, are wrong. I assume you *have*, of course, read> and understood both papers? That you are not simply relying on secondary> sources because the primary sources are too advanced for you?Oh the primary source is *way* too advanced for me in detail.However, its intriguing to see if the logical error pops out aseasily as I expect it should, or if its buried behind a lot oftechnical jargon.Im considering that question now with the source.James Harris === James Harris, it appears that your current strategy for saving face (read: posting crap) to the internet, havingcompletely failed to provide a valid proof of FLT, is to attack the proof provided by Wiles. This is standardfare for cranks -- if you cant create, try to destroy. But please spare us the disassociated hand-wavingarguments. Stick to a specic point instead of continually leaping triumphantly to false conclusions. You havebeen so thoroughly discredited that the only hope of redeeming yourself is to actually do something right.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === clearly, for the greater glory of his church-school sponsors, soas not to tip-off the many, competent to beat himto the punch (such as Ribet). of course,it could also be awed; or,simply inelegant.how would you characterize the sum-totalof your now 8-year mission, monsieur Harris,minus all of the vituperative garbarge? of course,such an approach may be feasible for teaching math, althoughits hard to imagine the student-body thatd tolerate that sortof harangue. (of course,in real life, one probably would be forced (or happy)to modify ones approach, if the students were at-all hominid .-) > Ah, well, this quotation helps explain where you got your impression of> what Wiles was claiming. You are perfectly correct in analyzing this> sentence by pointing out that on the one hand we have elliptic curves,> and on the other hand we have modular forms, and there is no superset> of objects of which elliptic curves and modular forms are both members.> So just because you get this L-series thingy from an elliptic curve,> and you can get the same L-series thingies from modular forms, how can> this possibly imply anything like all elliptic curves are modular?> Good question. > As of the early 1990s, most mathematicians believed that the> Taniyama-Shimura conjecture was not accessible to proof. However, A.> Wiles was not one of these. He attempted to establish the> correspondence between the set of elliptic curves and the set of> modular elliptic curves by showing that the number of each was the> same. Wiles accomplished this by counting Galois representations and> comparing them with the number of modular forms. > My assessment is that Wiles commits the logical fallacy of Cum hoc> ergo propter hoc.> (Source http://users.tru.eastlink.ca/~brsears/reafault.htm ) > Why would Wiles deceive his colleagues? Why havent more people> thought that relevant? Why be surprised that a man obsessed and> isolated for several years living a deception might delude himself> into believing a logically awed approach might work?--A church-school McCrusade (Blairs ideals?):Harry-the-Mad-Potter wants US to kill Iraqis?...For a 1000-year anglo-american hegemony?HEY, JIMMY; LETS US and SU FIGHT -then-PM of England & Zbiggy http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/les/curriculum/Cosmo.PCX== ==> James Harris, it appears that your current strategy for saving face (read: posting crap) to the internet, having> completely failed to provide a valid proof of FLT, is to attack the proof provided by Wiles. This is standard> fare for cranks -- if you cant create, try to destroy. But please spare us the disassociated hand-waving> arguments. Stick to a specic point instead of continually leaping triumphantly to false conclusions. You have> been so thoroughly discredited that the only hope of redeeming yourself is to actually do something right.> Human beings have a aw: they can be thoroughly convinced of falsethings.The reason the primary newsgroup for this thread is sci.logic is thatlogicians are tasked with being careful in a way that evenmathematicians arent.Mathematicians can get away with leaps and assertions because of therelevance of mathematics to the real world, where physics has beenpowered by mathematical models.But logic is cold, hard, and less amenable to social pressure.Ive outlined a specic logical aw in Wiless approach, which isthat it depends on the logical fallacy of trying to nd conditionsthrough association.That is the logical fallacy is, Cum hoc ergo propter hoc.So does Wiles get around the lack of a logical basis by using a nitesubset to nd a restriction on an innite set?The technique is called lifting and is something like innitedescent made somewhat famous by Fermat himself.It is an intriguing question, so Ill back off somewhat while Iconsider whether or not he somehow gets around the logical fallacy tond a way to break it.Now mathematicians apparently are quite certain that Wiles succeededand I applaud their energy. However, I must also rely on myunderstanding that human beings have a certain aw: an ability tobelieve almost anything.Logic, however, is perfect.James Harris === > Human beings have a aw: they can be thoroughly convinced of false> things.AS is James Steven Harris mistakenly convinced of his own genius. === > Human beings have a aw: they can be thoroughly convinced of false> things.I just love the way you state the most commonplace trivialities as ifthey were some profound new insight.> So does Wiles get around the lack of a logical basis by using a nite> subset to nd a restriction on an innite set?> The technique is called lifting and is something like innite> descent made somewhat famous by Fermat himself.> It is an intriguing question, so Ill back off somewhat while I> consider whether or not he somehow gets around the logical fallacy to> nd a way to break it.You are not qualied to determine whether or not Wiles was successful.Only a limited number of people on the face of the earth are qualied(and willing to take the necessary time and effort) to do so. Until youare able to read (and understand!) his proof for yourself, youll justhave to take their word for it.Or not. Your opinion has no importance to anyone but yourself, sobelieve whatever you want.-- Wayne Brown | When your tails in a crack, you improvisefwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === [snip]>Wiless supposed accomplishment rests>*solely* on the assertion of a relatively small group of people that>his work is correct.James Harriss supposed accomplishments rest *solely* on the assertionof one person that his work is correct.-- Yup, you guessed it. If worse comes to worse, I *will* turn to the Army to help me with mathematicians. -- James Harris <3c65f87.0304191552.511ad5b4@posting.google.com> === > [snip]>Wiless supposed accomplishment rests>*solely* on the assertion of a relatively small group of people that> >his work is correct.> James Harriss supposed accomplishments rest *solely* on the assertion> of one person that his work is correct.What can you do? Time after time Ive faced false assertions, andtime after time people have been unreasonable in the face ofrationality. Sure I set out a few years back to nd some answers tosome math problems, and made a LOT of mistakes along the way. Yup,Ive made a lot of mistakes.But right now Id like some cogent answers to what follows:Ive presented a problem with the logical foundation of Wiless workas it relies on association to prove a condition. More usefuldiscussion has worked things down to the assertion that Wiles used anite set, and lifting to prove the equality of two innite sets,where the equality supposedly forces the condition.There is, however, no reason for the condition given, and no claim ofa reason, where the condition, from my understanding, is that everyelliptic curve is a modular elliptic curve.The question raised before Wiless work being whether or not anelliptic curve could not be modular, where mathematicians had relatedvarious elliptic curves that they tested to something called modularforms, and found that every one they tested was modular. Then themathematicians Taniyama and Shimura conjectured that every ellipticcurve was modular, which my understanding means, they are associatedwith a modular form, where that association can be described by 4descriptors.My understanding is that Wiless work depends on association.The logical fallacy Ive put forward as a challenge against his workis,Cum hoc ergo propter hoc.James Harris === [snip]> My understanding is that Wiless work depends on association.>> The logical fallacy Ive put forward as a challenge against his work> is,>> Cum hoc ergo propter hoc.>> James HarrisYour understanding is a misunderstanding. What you have put forward is,Argumentum ad ignorantum.You have no standing to challenge anyone about anything. Get over it, James Harris. Youhave been thoroughly debunked. Remove your faulty and error-ridden attempts at proving FLTfrom public view. You are polluting the internet with your posts.--The only thing more pitiful than beating a dead horse is trying to ride one.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === > there is nothing to differentiate your complaints about Wiles> from complaints about photons, DNA, the Jurassic era,> evolution, relativity, and the rest.>> [...] Wiless supposed accomplishment rests *solely*> on the assertion of a relatively small group of people that> his work is correct.There is no difference in that respect between Wiles and photons,DNA, etc. Where differences exist, they tend to favor Wiles proofover the other situations. Experimental evidence, for example, can bechecked more easily, unambiguously and objectively in Wiles situationthan the others.> Yet to take one of your examples--photons--and consider that the> existence of photons has been theorized for some time, but was proven> by experiment.It was not *proven* by experiment: another respect in which Wileswork is qualitatively more reliable than photons, DNA, and the rest.The physics experiments were consistent with certain theoretical models,but of course, you have not even come close to verifying the immensechain of experimental and theoretical reasoning leading to the current modelswith photons. Instead, you rely on textbooks, fourth-hand (if that)accounts, and the assertions of the Science Establishment.And the 64 dollar question is why you happily parrot the party lineon matters of photons, DNA, relativity, evolution, the existence ofthe Iraq War and Sikkim and Napoleon --- but intone high skepticismconcerning Wiles.> Since that time from lasers to spectral analysis the theory has t> with reality.Lasers only hurt your cause, as to check that a laser (resp. spectrometry)experiment actually corroborates photons you would have to check mattersof chemistry, crystallography (geology!), engineering, manufacture, and soon all the way down. The only way out of this is to accept various assertionson faith from the Evil Scientic Establishment, and the question arises why youare such a sheep and conformist when it comes to non-Wiles but raise hightenedstandards concerning Wiles (who of courses passes all the scientic standardsfor photons, etc, and then some). === > there is nothing to differentiate your complaints about Wiles> from complaints about photons, DNA, the Jurassic era,> evolution, relativity, and the rest.>> [...] Wiless supposed accomplishment rests *solely*> on the assertion of a relatively small group of people that> his work is correct.> There is no difference in that respect between Wiles and photons,> DNA, etc. Where differences exist, they tend to favor Wiles proof> over the other situations. Experimental evidence, for example, can be> checked more easily, unambiguously and objectively in Wiles situation> than the others.> Yet to take one of your examples--photons--and consider that the> existence of photons has been theorized for some time, but was proven> by experiment.> It was not *proven* by experiment: another respect in which Wiles> work is qualitatively more reliable than photons, DNA, and the rest.> The physics experiments were consistent with certain theoretical models,> but of course, you have not even come close to verifying the immense> chain of experimental and theoretical reasoning leading to the current models> with photons. Instead, you rely on textbooks, fourth-hand (if that)> accounts, and the assertions of the Science Establishment.My degree is in physics. I did physics experiments in school. > And the 64 dollar question is why you happily parrot the party line> on matters of photons, DNA, relativity, evolution, the existence of> the Iraq War and Sikkim and Napoleon --- but intone high skepticism> concerning Wiles.Wiless work would mean a workaround to the logical fallacy called,Cum hoc ergo propter hoc.Ultimately, if Wiless work is correct then it does not have anylogical aws, but checking it potentially involves going through eachstep in his work, which is a formidable task. If he did nd a proof,then I think it interesting on logical grounds that there is aworkaround i.e. that Cum hoc ergo propter hoc is not actually alogically fallacious approach.Now as for physics results, like many people trained in physics, Ikeep a skeptical eye on theory, and depend on things Ive personallychecked, or that are very unlikely to be wrong that have been checkedby others. Physicists can be hard-liners to the extent that theydont believe physics they havent personally checked. Im not. Likehow I believe that nuclear weapons work. But still realize that theabsolute truth may be something other than what Ive learned.In mathematics, absolute truth *can* be determined, just like alogical argument can be checked against certain rules for internalconsistency. > Since that time from lasers to spectral analysis the theory has t> with reality.> Lasers only hurt your cause, as to check that a laser (resp. spectrometry)> experiment actually corroborates photons you would have to check matters> of chemistry, crystallography (geology!), engineering, manufacture, and so> on all the way down. The only way out of this is to accept various assertions> on faith from the Evil Scientic Establishment, and the question arises why you> are such a sheep and conformist when it comes to non-Wiles but raise hightened> standards concerning Wiles (who of courses passes all the scientic standards> for photons, etc, and then some).As a person with a science degree, I guess youd consider me a part ofthe Evil Scientic Establishment.Its actually more fun attacking them than just sitting aroundbelieving in them. Because you learn a lot in the attack, and yourguarantee from math and logic is that the proof doesnt care.To a math proof, you do not exist as a relevant entity.James Harris === > Wiless work would mean a workaround to the logical fallacy called,> Cum hoc ergo propter hoc.No, it wouldnt.[snip]> ... I feel like I can> speak condently on the subject.You speak condently whether you know what youre talking about or not. Condence is not yourproblem, honesty and credibility are.> To a math proof, you do not exist as a relevant entity.You do not exist as a relevant entity.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === > there is nothing to differentiate your complaints about Wiles> from complaints about photons, DNA, the Jurassic era,> evolution, relativity, and the rest.>> [...] Wiless supposed accomplishment rests *solely*> on the assertion of a relatively small group of people that> his work is correct.> There is no difference in that respect between Wiles and photons,> DNA, etc. Where differences exist, they tend to favor Wiles proof> over the other situations. Experimental evidence, for example, can be> checked more easily, unambiguously and objectively in Wiles situation> than the others.> Yet to take one of your examples--photons--and consider that the> existence of photons has been theorized for some time, but was proven> > by experiment.> It was not *proven* by experiment: another respect in which Wiles> work is qualitatively more reliable than photons, DNA, and the rest.> The physics experiments were consistent with certain theoretical models,> but of course, you have not even come close to verifying the immense> chain of experimental and theoretical reasoning leading to the current models> with photons. Instead, you rely on textbooks, fourth-hand (if that)> accounts, and the assertions of the Science Establishment.> My degree is in physics. I did physics experiments in school.> And the 64 dollar question is why you happily parrot the party line> on matters of photons, DNA, relativity, evolution, the existence of> the Iraq War and Sikkim and Napoleon --- but intone high skepticism> concerning Wiles.> Wiless work would mean a workaround to the logical fallacy called,> Cum hoc ergo propter hoc.> Ultimately, if Wiless work is correct then it does not have any> logical aws, but checking it potentially involves going through each> step in his work, which is a formidable task. If he did nd a proof,> then I think it interesting on logical grounds that there is a> workaround i.e. that Cum hoc ergo propter hoc is not actually a> logically fallacious approach.> Now as for physics results, like many people trained in physics, I> keep a skeptical eye on theory, and depend on things Ive personally> checked, or that are very unlikely to be wrong that have been checked> by others. Physicists can be hard-liners to the extent that they> dont believe physics they havent personally checked. Im not. Like> how I believe that nuclear weapons work. But still realize that the> absolute truth may be something other than what Ive learned.> In mathematics, absolute truth *can* be determined, just like a> logical argument can be checked against certain rules for internal> consistency.>How? How can absolute truth be determined? About a month ago youessentially:1) proof of an absolute kind, presumably stated in the symbolism offormal logic, and2) proof that merely convinces other mathematicians, presumably statedin some meta-language (like English).Furthermore, your position is that proof of the second kind is whatmost mathematicians produce, and is not good enough. You go on to saythat you produce proofs of the 1st kind.Question: How does one determine that a proof or mathematicalargument is absolutely and irrefutably correct?The validity must be checked by 1) God, 2) a machine, or 3) a humanbeing, as a proof cannot check itself.I think (1) is out, for the time being anyway.What about (2)? Well, we could encode some axioms and rules onference, but it occurs to me that a few problems could arise. First,the algorithm may take an unreasonable amount of time to reach adecision. Second, hardware failure, electrical surges, sunspotactivity, running the program under Microsoft Windows, etc. couldcause erroneous results. Third, and perhaps most importantly, a humanbeing (or beings) must write the software. Therefore, any errorscaused by people could conceivably appear here. That leaves us withoption (3). As we all know, people make mistakes. They make mistakeswriting proofs. The publisher/editors of a journal may make a mistakemistake by erroneously believing the proof.Who has the nal and ultimate authority to say that a given argumentis valid or not? Surely, not one person. There is so much mathematics,no one person can know it all.So, a proof then must be judged by the readers. If there is adisagreement, then the sides may argue their cases until one sideprevails and convinces the other, at least within a given mathematicalsystem.Therefore, in this sense, all proofs are of the second type. We muststrive to convince other mathematicians. That is all there is --simply because there is no other means of asserting the validity of amathematical argument. It really is an appeal to the gallery.We must also consider that mathematics may be inconsistent. Accordingto Kurt Godel, this is a possibility (at least for mathematicalsystems strong enough to support integer arithmetic.)So much for proofs being irrefutable, absolute, perfect, eternal, etc.ad nauseum.> Since that time from lasers to spectral analysis the theory has t> with reality.> Lasers only hurt your cause, as to check that a laser (resp. spectrometry)> experiment actually corroborates photons you would have to check matters> of chemistry, crystallography (geology!), engineering, manufacture, and so> on all the way down. The only way out of this is to accept various assertions> on faith from the Evil Scientic Establishment, and the question arises why you> are such a sheep and conformist when it comes to non-Wiles but raise hightened> standards concerning Wiles (who of courses passes all the scientic standards> for photons, etc, and then some).> As a person with a science degree, I guess youd consider me a part of> the Evil Scientic Establishment.> experiences in the military, where I actually had the honor of giving> a lecture on the physics of lasers to the medical personnel at Madigan> Army Medical Center, including the surgeons, other doctors and nurses,> for their medical continuing education credits, I feel like I can> speak condently on the subject.> My position on Wiles is about logic. Emotional response is not> necessary as I assure you that if Wiles found a proof then there is no> need for concern. If he did not, why ght for a false belief?> Math proofs are indestructible, incorruptible, and irrefutable.> Its actually more fun attacking them than just sitting around> believing in them. Because you learn a lot in the attack, and your> guarantee from math and logic is that the proof doesnt care.> To a math proof, you do not exist as a relevant entity.> James Harris === > there is nothing to differentiate your complaints about Wiles> from complaints about photons, DNA, the Jurassic era,> evolution, relativity, and the rest.> Yet to take one of your examples--photons--and consider that the> existence of photons has been theorized for some time, but was> proven by experiment.> >> It was not *proven* by experiment: another respect in which Wiles> work is qualitatively more reliable than photons, DNA, and the rest.> The physics experiments were consistent with certain theoretical models,> but of course, you have not even come close to verifying the immense> chain of experimental and theoretical reasoning leading to the current models> with photons. Instead, you rely on textbooks, fourth-hand (if that)> accounts, and the assertions of the Science Establishment.>> My degree is in physics. I did physics experiments in school.Ask the school for a refund.First, the existence of photons is not a precisely formulated statementas in the case of Wiles proof, let alone one provable by experiment. Thereare of course theoretical models (not necessarily well-dened or knownto be logically consistent, by the way) within which one can single outcertain objects as photons.Second, your student experiments in optics could not possibly replicatethe mountain of theoretical and experimental steps involved in buildingup any of the theoretical model(s) involving photons. Instead, youaccepted on trust assertions by textbook authors, professors and similarpurveyors of the Social Truth that you like to castigate, amounting toa certication-by-authority that the apparatus you were doing the experimentswith actually corresponded to the theory in the manner claimed.You did not produce the relevant gases, crystals, apparatus, electricity, ..involved in laser experiments, nor did you do the experimentationneeded to corroborate the values of relevant physical and chemicalparameters listed in the CRC handbook, and so on all the way down.What actually happened is that a long and social processof knowledge-accumulation occurred and you took the results on trust.In particular, if your experiments gave a wrong result, the conclusionwould be that you made a mistake, not that photons existence is in doubt;a pure assertion of authority by the Scientic Establishment concerning itsSocial Truth, which you accept without any objection in all the non-FLTsituations.Note that your repeatedly discredited objections in this thread aboutWiles logic are irrelevant, as you also object to Ribets proof withoutgiving any particular reason to doubt it. The matter is simply one of anobvious double standard produced for the occasion, where social certicationby a small network of experts counts as OK for photons, DNA,evolution, relativity, the Jurassic era (or the existence of Napoleon andGeorge W Bush), etc --- but somehow the information that expertshave certied Ribets and Wiles work is cast as suspicious. === > My degree is in physics. I did physics experiments in school.Yet you seem never to have encountered the SR thoughtexperiment called the superluminal scissors or had anyidea what I was talking about in sci.physics when Iexplained how a 5 m/sec water jet can be used to createan illusion of arbitrarily fast, even superluminalmotion. - Randy === > there is nothing to differentiate your complaints about Wiles> from complaints about photons, DNA, the Jurassic era,> evolution, relativity, and the rest.> Yet to take one of your examples--photons--and consider that the> existence of photons has been theorized for some time, but was> proven by experiment.>> It was not *proven* by experiment: another respect in which Wiles> work is qualitatively more reliable than photons, DNA, and the rest.> The physics experiments were consistent with certain theoretical models,> but of course, you have not even come close to verifying the immense> chain of experimental and theoretical reasoning leading to the current models> with photons. Instead, you rely on textbooks, fourth-hand (if that)> accounts, and the assertions of the Science Establishment.>> My degree is in physics. I did physics experiments in school.> Ask the school for a refund.I had a full tuition scholarship. > First, the existence of photons is not a precisely formulated statement> as in the case of Wiles proof, let alone one provable by experiment. There> are of course theoretical models (not necessarily well-dened or known> to be logically consistent, by the way) within which one can single out> certain objects as photons.> Second, your student experiments in optics could not possibly replicate> the mountain of theoretical and experimental steps involved in building> up any of the theoretical model(s) involving photons. Instead, you> accepted on trust assertions by textbook authors, professors and similar> purveyors of the Social Truth that you like to castigate, Oh please, youve been beaten. Thats whats annoying about Usenet assome loser will state a case, get their ass kicked, but STILL keepcoming back as if nothing happened.My *degree* is in physics. I went to school on a full-tuitionscholarship, and you stepped into my eld with your assertions, gotyour ass kicked but refuse to back down.Now *emotion* is not necessary when it comes to Wiless work. If hefound a proof I can assure you that it is irrefutable. Thats whyitd be a proof. All this emotion just annoys me, as part of the funof science and mathematics is attacking things that are supposedlyproven.Its GREAT fun not just accepting what people tell you. But thatswhat really annoys me about mathematicians as time after time I getyahoos replying back in defense of mathematics, using tactics.But you see, not a single REAL mathematician in the world gets excitedabout an attack on a proof. No mathematician worth the title wouldget even a little concerned, nor would they lose sleep, or ndthemselves emotional about some person--any person--any time--anyplace--who decides to go after a math proof.Thats because a math proof is indestructible, incorruptible, andirrefutable.It just doesnt care if you attack it, and no real mathematician wouldcare either.Now Ive discovered math proofs, which is why Im not concerned aboutpeople refuting them because they are proofs. And in fact people whocall themselves mathematicians cant touch them, so they come up withextraneous stuff, or make claims of nding their own proofs to refutemy proofs, but you see, proofs dont duel.And you know what? I think that feature of mathematics terries somepeople who call themselves mathematicians. Mathematics does NOT carewhat you call yourself. It DOES NOT care that you have a mortgage. It DOES NOT CARE that you really, really, really want people to likeyou and think youre a great mathematician.Now Ive made a specic claim against Wiless work. If he found aproof the claim can be answered, but even if it is answerable then itmust be true that he has found a way around what is considered to be alogically fallacious approach. Logicians should thank him in thatcase for correcting them.My challenge is a logical one. Wiless work fails and is not a proofas it is an argument by Cum hoc ergo propter hoc.James Harris === the truth-value of any logical statement is *algebraically*of no importance. true,a proof with an error of inference may be false, ifthe error was also not nistakenly canceled (andthe value ipped, F for T, say, once more), buttaht doesnt mean that its not syntactically properin some sense. not taht yours necessarily are, two often!I dont have to know any Latin, whatsoever,to follow a pattern of inference. (in particular,see Lou Kauffmans slight rephrasing of G. Spencer-Brown,on his site; I htought of the same diagrammatic convention, butusing circles (to represent spheres .-)) and, yous still have to prove thatWiles faked his proof; have you? > But you see, not a single REAL mathematician in the world gets excited> about an attack on a proof. No mathematician worth the title would> get even a little concerned, nor would they lose sleep, or nd> themselves emotional about some person--any person--any time--any> place--who decides to go after a math proof.> Thats because a math proof is indestructible, incorruptible, and> irrefutable. > My challenge is a logical one. Wiless work fails and is not a proof> as it is an argument by Cum hoc ergo propter hoc.--Dec.2000 WAND Chairman Paul ONeill, reelectedto Board. Newsish?http://www.rand.org/publications/randreview/issues/rr .12.00/http://members.tripod.com/~american_almanac === I agree, that a REAL mathematician wouldnt waste your timeby objecting to the *statement* of a proof, althoughhe certainly might wish to attack the very proceedingsof the suposed proof, to get at the real truth of it (which maybe false, due to an error in the chain of inference; or,the truth-value could be correct, by a ip of the coin (say),in spite of any breaks in the chain). your problem is that you insist upon hacking-awayat the algebra, without any noticable hypothesisas to any simpler proof than Wiles (or onethat is akin to Fermats .-)... at this rate,there will not be enough room in the marginsof sci.math, alt.math etc.,within the loud ticking of your 10-year mission! proofs cannot come from just doing the math,any more than they can come from computerswithout substantial hypotheses being encoded,insofar as possible. after all, becausethe predicate logic or boolean algebra can be usedto remove as many redundant inferences as you wish, butcant change the truth-value of the statement. not even fuzzy logic can escape the necessary polesof the application of combinatorics to predicate logic,which is not any different than the syllogismsof the Older Greeks.well, the more that I think about it, the more it appears thatPierre et son l faisons un connundrum mirabilepour lediacion de generations futeur, etlexposition eventuel de La Methode du Fermat --together with that one of his few, proven results,for the exponent of 4 *par une absurdite*.does tha not seem clear to you, monsieur Harris? > I dont have to know any Latin, whatsoever,> to follow a pattern of inference. (in particular,> see Lou Kauffmans slight rephrasing of G. Spencer-Brown,> on his site; I htought of the same diagrammatic convention, but> using circles (to represent spheres .-))> My challenge is a logical one. Wiless work fails and is not a proof> as it is an argument by Cum hoc ergo propter hoc.--A church-school McCrusade (Blairs ideals?):Harry-the-Mad-Potter wants US to kill Iraqis?... http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/les/curriculum/Cosmo.PCX== ==L O N G12 15 14 7 = 48 Angie Tysseland was playing piano on the NW corner of 21st Streetand 4th Ave., I stopped to listen and to collect her stats. Angie wasteamed up with Teresa Long, Teresa was providing vocals and doing avery good job of it, I was pretty well astounded by Terri Longssinging ability. This was an incredibly good free concert at theSaskatoon Jazz Festival.213 Terri 4 5 59 124/241 806Teresa 68 Lorrelle 97 Long 48Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 <-7th-> 12 -- -- 58 50 Teresa was born in 59, its the 17th prime while 17 in turn is the7th prime, while the primes up to 7 add to 17 (59 is the 7th prime inprime position). She was born on day 124 (Numbers 7). She was born 117days closer to the beginning of the year than to the end of the year.Her rst name adds to 68 (4x17 and is the 7x7th non-prime). Hermiddle name adds to 97 (Leviticus 7). Her rst and last names average58 (the 7 primes up to 17). Her 68 (4x17 and is the 7x7th non-prime)valued rst name is 58 (the 7 primes to 17) short of her 124 (Numbers7) day of birth. Her given names differ in value by 29, its the 7thprime (17) plus the 7th non-prime (12), or simply 7p+7np. She has 7different letters in her given names. The vowels in her given namesadd to 36 (7+7p+7np). She is missing 17 letters from her full name.Her vowels add to 51 (17+17+17). Her odd valued letters add to 77 andare in positions adding to 84 (7 times the 7th non-prime), it is adifference of 7. Her even valued letters add to 136 (8x17) and are inpositions adding to 87, its a difference of 49 (7x7 and is the17+17th non-prime). Her even valued letters add to 136 (8x17) andexceed her odd valued letters by 59 (the 7th prime in prime positionand is her year of birth). Her even valued letters add to 136, itsthe 104th non-prime while 104 in turn is the 77th non-prime (136 isthe 77th non-prime in non-prime position). Her Fibonacci valuedletters add to the 21 (7+7+7) chapters of Bible Book 7 and are inpositions adding to 36 (7+7p+7np). Her Lucas valued letters add withtheir positions for the 108 verses of Bible Book 59 (the 17th prime,her year of birth). Her unrepeated letters add with their positionsfor the 108 verses of Bible Book 59 (the 17th prime, her year ofbirth). Her names add to the 49th non-prime, 25th prime and to the33rd non-prime, together for 107 (the perfect 28th prime while thenumbers 1 through 7 add to 28). Her 213 valued name is 171.77% of her124th (Numbers 7) day of birth. Her repeating letters are in positionsadding to 124 (Numbers 7, her day of birth). Her last 7 letters add to77. Her last two names differ in value by 49 (7x7 and is the 7x7thnon-prime). She abbreviates her rst name to Terri (70). She iscommonly known as Terri (70) Long (48), these names average 59 (heryear of birth).Non-Primes 1 27 50 72 4 28 51 74 6 30 52 75 8 32 54 76 9 33 55 7710 34 56 7812 35 57 8014 36 58 8115 38 60 8216 39 62 8418 40 63 8520 42 64 8621 44 65 8722 45 66 8824 46 68 9025 48 69 9126 49 70 92 Teresa adds to 68, her day, month and year of birth adds to 68. Shehas four Es and four Ls, these most frequently repeated letters addtogether for 68. Her 17 missing letters add to 240 (68.37% of thepossible total). Her 213 valued name adds with her 124th day of birthfor 337 (the 68th prime). She was born on day 124 (Numbers 7) whileher rst name adds to 68 (the 7x7th non-prime). Her rst name addsto 68 (49th non-prime) while her last two names differ in value by 49. Terri was born on day 124, corresponding to Numbers 7, the chaptercontains 89 verses. Her 213 valued name exceeds her 124th day of birthby 89. The Gospels are Bible Books 40, 41, 42 and 43, together for 166.Terris name adds to 213, its the 166th non-prime while 166 is the128th non-prime, while 128 is 2 to the 7th. Chapter 124 is Numbers 7, it is the 7x7th prime (227) short of thenumbers up to the 17th non-prime (351). Genesis 7, Exodus 7, Leviticus7 and Numbers 7 are chapters 7, 57, 97 and 124, together for 285 (the7x7th chapter of The Samuels). Numbers 7 and Deuteronomy 7 arechapters 124 and 160, together for 284 (Second Samuel 17, the 7thprime). First Samuel 7 and Second Samuel 7 are chapters 243 and 274,together for 517 (17 is the 7th prime, the primes up to 7 add to 17).First Kings 7 and Second Kings 7 are chapters 298 and 320, togetherfor 618 (the number of verses in Bible Book 7). First Chronicles 7 andSecond Chronicles 7 are chapters 345 and 374, together for the 719verses of Bible Book 12 (7th non-prime), this 719 is the 128th (2 tothe 7th) prime. First Samuel 7 and Second Samuel 7 are chapters 243and 274, together for 517, while First Chronicles 7 and SecondChronicles 7 are chapters 345 and 374, together for 719, its anaverage of the 618 verses of Bible Book 7.213 Terri 4 5 59 124/241 806Teresa 68 Lorrelle 97 Long 48189 Angie 27 2 61 58/307 1471Angie 36 Grace 34 Tysseland 119Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 Terri Long and Angie Tysseland are teamed up and producing a CD(7), they were born in months adding to 7. Terri was born on day 124(Numbers 7), Angie on day 58 (the 7 primes up to 17). Their 6 nameshave an average value of 67 (Exodus 17). Angies names add to 36(7+7p+7np), 34 (17+17) and 119 (7x17), together for 189 (the rst 17primes minus the rst 17 non-primes). Their rst names add togetherfor 104 (77th non-prime). Their last names add together for the 167verses of Bible Book 17 (the 7 primes in prime positions up to the17th prime add together for the 167 verses of Book 17). Esther becomesQueen in Book 17 and Q is the 17th letter of the alphabet. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 2310 2911 31 <- 3112 3713 41 <- 4114 4315 4716 5317 59 <- 59 --- 167 Esther Book 17 Terri and Angie (6x6) are separated by 665 days (Ecclesiastes 6).Teresa (6 letters) was born 66 days further into the year than Angie(6x6). They were together born 366 days closer to the beginning oftheir years than to the end of their years. Terris full name adds to213 (166th non-prime). Their given names add together for 235, it isthe 184th non-prime, pretty as the 184th prime (1097) and the 184thnon-prime (235) averages 666. Their initials have an average value of36 (6x6, and 1 through 36 adds to 666). Terri was born on the 4th(Numbers with 36 chapters). I showed them both gems and neither hadthanx for showing them evidence that their very names are a gift fromGod. My math was used repeatedly as an excuse to arrest and chemicallylobotomize me (torture me) in psychiatric facilities, and whenever Imeet with people and show them patterns that please them, they neverhave the decency to offer to buy me a measly cookie for my work norsend me a cheap letter expressing thanx. Man oh man, you people arecompassionless turds, you are the of the earth, soon God willspread you out over the surface of the earth like the dung that youare, and in this I rejoice.1-50 - Genesis51-90 - Exodus91-117 - Leviticus118-153 - Numbers154-187 - Deuteronomy188-211 - Joshua930-957 - Matthew958-973 - Mark974-997 - Luke998-1018 - John1019-1046 - Acts1047-1062 - Romans 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 661062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 61070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6)1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime)1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters)Daryl Shawn KabatoffBox 7134Saskatoon SaskatchewanCanadaS7K 4J1Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === I think youre ill mannered, boring and part of the reason why manypeople are not interested in religion.Dave. === > I think youre ill mannered, boring and part of the reason why many> people are not interested in religion.>> Dave.Its numerology, a religionPlonk-a-dork!!!RJP === Just testing - I havent been getting any new topics from this newsgroup lately (the last one I have is point estimate VS probability distribution from 6/15) === alt.math.undergrad, William Springer Just testing - I havent been getting any new topics from this newsgroup >lately (the last one I have is point estimate VS probability >distribution from 6/15)Why not just check in Google -- thats what its there for.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/You nd yourself amusing, Blackadder.I try not to y in the face of public opinion. === THANK you both Stan/Darrell for your posts... They were both an extremelybig help!!!thanks again!!!! === The Pyramid Paper Challenging & Ranking Game is the new type puzzlefor all ages.Puzzle sheet page is http://www5.ocn.ne.jp/~pachalle/paperchalleran213.htmlRule page is http://www5.ocn.ne.jp/~pcjapan/paperchalleran270.htmPlease try this game!!Please tell your family, friends, and students of your school thisinteresting contest puzzle as many as possible and enjoy to solve iteach other.Best puzzling,Ryosuke Ito === I have a mathproblem that I have been trying to solve for some time now.Since Im from sweden, and my english isnt the best, mathterms I use might bewrong.The problem is:Find every real number that satises:Squareroot( |x-1| ) = xThis is what I thought would be the right way to solve it:If I take the square of both sides Ill get two equations.-|x-1|=x^2 and |x-1|=x^2Ant then each of those will give me two equations:-|x-1|=x^2 <-> -x+1=x^2 and x+1=x^2|x-1|=x^2 <-> -x-1=x^2 and x-1=x^2But since they are both second degree polynoms I will get two answeres toeach. So there will be a total of 8 x`s (real and komplex).Im a doing it all wrong? How should it be done?There is another problem Ive tried to solve:Find a relation wich denes a triangle wich corners are: (2,1), (2,-3) and(-4,-2).I worked on this a while and came to the conclusion.The equations for the three lines that outline the triangle is:y=x/2x=2y=(-x-16)/6Therefore the triangle can be dened as:-(x+16)/6 < y < x/2and-4 < x < 2But is there anyway to combine theese two so that I get only onerelation/equation for the triangle?Said Aspen === > I have a mathproblem that I have been trying to solve for some time now.> Since Im from sweden, and my english isnt the best, mathterms I use might be> wrong.> The problem is:> Find every real number that satises:> Squareroot( |x-1| ) = x> This is what I thought would be the right way to solve it:> If I take the square of both sides Ill get two equations.> -|x-1|=x^2 and |x-1|=x^2Your rst equation cannot happen. A negative will never equal a positive.> Ant then each of those will give me two equations:> -|x-1|=x^2 <-> -x+1=x^2 and x+1=x^2This line goes away. Note: it should have been -x+1=x^2, x-1=x^2, just like below.> |x-1|=x^2 <-> -x-1=x^2 and x-1=x^2x-1 = x^2 or -x+1 = x^2. Be careful with the signs.> But since they are both second degree polynoms I will get two answeres to> each. So there will be a total of 8 x`s (real and komplex).> Im a doing it all wrong? How should it be done?Problems involving squareroots can generate extraneous solutions. You will have to check all (possibly) 4 solutions in the original equation to see which ones work. I believe you will get 2 solutions.> There is another problem Ive tried to solve:> Find a relation wich denes a triangle wich corners are: (2,1), (2,-3) and> (-4,-2).> I worked on this a while and came to the conclusion.> The equations for the three lines that outline the triangle is:> y=x/2> x=2> y=(-x-16)/6These are correct but could use boundary conditions on x or y.> Therefore the triangle can be dened as:> -(x+16)/6 < y < x/2> and> -4 < x < 2Are you trying to dene the interior of the triangle or the edges?> But is there anyway to combine theese two so that I get only one> relation/equation for the triangle?For the interior, you will have at least 2 equations. In this case you have found them.-- Will Twentyman === >I have a mathproblem that I have been trying to solve for some time now.>Since Im from sweden, and my english isnt the best, mathterms I use might be>wrong.>>The problem is:>Find every real number that satises:>Squareroot( |x-1| ) = x>>This is what I thought would be the right way to solve it:>>If I take the square of both sides Ill get two equations.>-|x-1|=x^2 and |x-1|=x^2Why do you think you get two equations here? (sqrt(a))^2 = a is alwaystrue; perhaps you are confusing it with sqrt(a^2)=|a|?If you square both sides, you will get that |x-1| = x^2, from whichyou get that either x^2 = x-1 or else x^2 = 1-x.>Ant then each of those will give me two equations:>-|x-1|=x^2 <-> -x+1=x^2 and x+1=x^2>|x-1|=x^2 <-> -x-1=x^2 and x-1=x^2This is completely wrong. |a|= b if and only if a=b OR -a=b; notand. And in any case you did the operations wrong. Presumably youmeant -x+1 and x-1 on the rst line, and x-1 and 1-x on the second.>But since they are both second degree polynoms I will get two answeres to>each. So there will be a total of 8 x`s (real and komplex).>Im a doing it all wrong? How should it be done?Yes, you are doing it wrong; in addition, you are forgetting thatafter squaring, you may be introducing extraneous solutions; you needto go back to the original equation to decide which solutions to yournew problem are solutions to the original one.You know that x^2 = x-1 or else that x^2 = 1-x. In addition, since x =sqrt(|x-1|), that means that x must be both real and nonnegative (notethat |x-1| is always nonnegative, so the square root is the realoperator and always a nonnegative real. what I accept as reality. --- Calvin (Calvin and Hobbes) === = === ==Arturo Magidinmagidin@math.berkeley.edu === > The problem is:>> Find every real number that satises:>> Squareroot( |x-1| ) = x> This is what I thought would be the right way to solve it:> If I take the square of both sides Ill get two equations.>> -|x-1|=x^2 and |x-1|=x^2> Thats not what you get when you take the square of both>sides. The square of the square root of |a| is just a^2.>Do it that way. -- Mike HardyEhr, no; the square of the square root of |a| is |a|. In general, forany a, (sqrt(a))^2 = a (because sqrt(a) implies that a isnonnegative), but sqrt(a^2))=|a|.So (sqrt(|a|))^2 = |a|, not a^2. The square of the square of thesquare root of |a| is a^2, however... === === Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === = === ==Arturo Magidinmagidin@math.berkeley.edu===> I have a mathproblem that I have been trying to solve for some time now.> Since Im from sweden, and my english isnt the best, mathterms I use mightbe> wrong.>> The problem is:> Find every real number that satises:> Squareroot( |x-1| ) = x>> This is what I thought would be the right way to solve it:>> If I take the square of both sides Ill get two equations.> -|x-1|=x^2 and |x-1|=x^2>> Ant then each of those will give me two equations:> -|x-1|=x^2 <-> -x+1=x^2 and x+1=x^2> |x-1|=x^2 <-> -x-1=x^2 and x-1=x^2>> But since they are both second degree polynoms I will get two answeres to> each. So there will be a total of 8 x`s (real and komplex).> Im a doing it all wrong? How should it be done?> There is another problem Ive tried to solve:>> Find a relation wich denes a triangle wich corners are: (2,1), (2,-3)and> (-4,-2).>> I worked on this a while and came to the conclusion.> The equations for the three lines that outline the triangle is:> y=x/2> x=2> y=(-x-16)/6>> Therefore the triangle can be dened as:> -(x+16)/6 < y < x/2> and> -4 < x < 2>> But is there anyway to combine theese two so that I get only one> relation/equation for the triangle?>> Said Aspen>>Thanx a bunch, all of you!a lot once again!Said Aspen === > I have a mathproblem that I have been trying to solve for some time now.> Since Im from sweden, and my english isnt the best, mathterms I use might be> wrong.> The problem is:> Find every real number that satises:> Squareroot( |x-1| ) = xStart with some basics (which will be needed toweed out extraneous roots later on):* |x-1| >=0 by the denition of |z|* Therefore, sqrt(|x-1|) >= 0* Therefore x>= 0Now square both sides to get: |x-1| = x^2We need to get x-1 out of the absolute value. Usethe face that |z|=z where z>=0 and |z|=-z where z<0.So for where 0 <= x < 1 (remember, the problem constructionguarantees that x must be greater than or equal to zero),we have -(x-1) = x^2or x^2 + x - 1 = 0which gives x = -1/2 + sqrt(5)/2 or x = -1/2 - sqrt(5)/2We can immediately throw away the second solutionbecause it is negative.But we can keep the rst solution because(-1 + sqrt(5))/2 is positive and is lessthan 1.Going back to |x-1| = x^2, we now takethe branch where x>=1. This gives us x-1 = x^2 x^2 - x + 1 = 0which has no real solutions.Thus the only potential solution is x=(-1 + sqrt(5))/2.Finally, substitute that back into sqrt(|x-1|)=x and see it is indeed valid.-- Rich Carreiro rlcarr@animato.arlington.ma.us === >> If I take the square of both sides Ill get two equations.>> -|x-1|=x^2 and |x-1|=x^2> Thats not what you get when you take the square of both>sides. The square of the square root of |a| is just a^2.>Do it that way. -- Mike Hardy> Ehr, no; the square of the square root of |a| is |a|. In general, for> any a, (sqrt(a))^2 = a (because sqrt(a) implies that a is> nonnegative), but sqrt(a^2))=|a|. === Im having trouble guring these out:1) Is the set of all functions from the boolean set {0,1} to thenatural numbers countable or uncountable?2) Let P(A) denote the power set for some set A.Given a set B, a subset A of P(B) is called an antichain if no elementof A is a subset of any other element of A. Does P(N) contain anuncountable antichain? === on sci.math:> Im having trouble guring these out:> 1) Is the set of all functions from the boolean set {0,1} to the> natural numbers countable or uncountable?Hmm, Im not an expert in this by any means, more like a beginner,but I think that the set youre after can be expressed as theCartesian product between the set of all functions from {0} to N,and the set of all functions from {1} to N. As the sets {0} and {1}are singletons, those sets of functions have the same cardinality asN itself. Therefore the Cartesian product has... what cardinality?> 2) Let P(A) denote the power set for some set A.> Given a set B, a subset A of P(B) is called an antichain if no element> of A is a subset of any other element of A. Does P(N) contain an> uncountable antichain?Im too lazy to try to gure out this one.-- /-- Joona Palaste (palaste@cc.helsinki.) ---------------------------| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki./~palaste W++ B OP+ |----------------------------------------- Finland rules! ------------/We sorcerers dont like to eat our words, so to say. - Sparrowhawk === > Im having trouble guring these out:> 1) Is the set of all functions from the boolean set {0,1} to the> natural numbers countable or uncountable?> 2) Let P(A) denote the power set for some set A.> Given a set B, a subset A of P(B) is called an antichain if no element> of A is a subset of any other element of A. Does P(N) contain an> uncountable antichain?In this question it makes no difference if we substitute N byany countably innite set. So lets replace N by Q, the setof all rationals. Can you see an uncountable antichain in P(Q)?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === >on sci.math:>> Im having trouble guring these out:>> 1) Is the set of all functions from the boolean set {0,1} to the>> natural numbers countable or uncountable?>Hmm, Im not an expert in this by any means, more like a beginner,>but I think that the set youre after can be expressed as the>Cartesian product between the set of all functions from {0} to N,>and the set of all functions from {1} to N. Not quite. A function from {0,1} to N is a set of the form{(0,n), (1,m)} for some n and m in N, not necessarily distinct,so the set of all such functions is {{(0,n), (1,m)} : n,m in N},a set of sets of ordered pairs.The sets of functions from {0} and {1} to N {{(0,n)} : n in N}and {{(1,n)} : n in N}, respectively, and their Cartesian productis {({(0,n)}, {(1,m)}) : n,m in N}, a set of ordered pairs ofsingletons of ordered pairs. Theyre denitely not the sameset. They are, however, the same size, and as you suggested inthe bit that I snipped below, theyre the same size as N x N. Inparticular, it should be easy to write down a bijection between{{(0,n), (1,m)} : n,m in N} and N x N, thereby showing that theset in question is countable.[...]>> 2) Let P(A) denote the power set for some set A.>> Given a set B, a subset A of P(B) is called an antichain if no element>> of A is a subset of any other element of A. Does P(N) contain an>> uncountable antichain?Yes. I could simply tell you how to construct one, but youlllearn more if you work at it a bit yourself. Heres a hinttowards one possible solution: nd an uncountable antichain inP(Q x Q), where Q is the set of rational numbers. Since Q x Q iscountably innite, this automatically gives you an uncountableantichain in P(N). You may also want to consider that for anyreal numbers c and d with 0 <= c < d, the set of points in Q x Qwith polar coordinates (r, t) satisfying the conditions r > 0 andc < t < d is non-empty.Brian === > Im having trouble guring these out:>> 1) Is the set of all functions from the boolean set {0,1} to the> natural numbers countable or uncountable?>> 2) Let P(A) denote the power set for some set A.> Given a set B, a subset A of P(B) is called an antichain if no element> of A is a subset of any other element of A. Does P(N) contain an> uncountable antichain?>Havent sci.mathers done about enough homework for this guy? Now he demandsproof, not just hints. === > Im having trouble guring these out:>> 1) Is the set of all functions from the boolean set {0,1} to the> natural numbers countable or uncountable?>> 2) Let P(A) denote the power set for some set A.> Given a set B, a subset A of P(B) is called an antichain if no element> of A is a subset of any other element of A. Does P(N) contain an> uncountable antichain?> Havent sci.mathers done about enough homework for this guy? Now he demands> proof, not just hints.>Tho 1 is easy, I dont get 2, even with the hint to use the bijectionbetween Q and N. Also I wonder if OP meant N^{0,1}, and not {0,1}^N.Whence |N^{0,1}| = (Aleph_0)^2 = Aleph_0,while |{0,1}^N| = 2^Aleph_0 = Aleph_1. === >> Im having trouble guring these out:>> 1) Is the set of all functions from the boolean set {0,1} to the>> natural numbers countable or uncountable?>> 2) Let P(A) denote the power set for some set A.>> Given a set B, a subset A of P(B) is called an antichain if no element>> of A is a subset of any other element of A. Does P(N) contain an>> uncountable antichain?>> Havent sci.mathers done about enough homework for this guy? Now he demands>> proof, not just hints.> Tho 1 is easy, I dont get 2, even with the hint to use the bijection> between Q and N. Also I wonder if OP meant N^{0,1}, and not {0,1}^N.> Whence |N^{0,1}| = (Aleph_0)^2 = Aleph_0,> while |{0,1}^N| = 2^Aleph_0 = Aleph_1.Think a bit about Dedekind cuts. They dont quite ll the bill as is,but a slight modication of the idea will work.By the way, 2^Aleph_0 is not equal to Aleph_1, unless you are assumingCH.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === >> Im having trouble guring these out:>> 1) Is the set of all functions from the boolean set {0,1} to the>> natural numbers countable or uncountable?>> > 2) Let P(A) denote the power set for some set A.>> Given a set B, a subset A of P(B) is called an antichain if no element>> of A is a subset of any other element of A. Does P(N) contain an>> uncountable antichain?[...]>Tho 1 is easy, I dont get 2, even with the hint to use the bijection>between Q and N. How about my hint to look at Q x Q and sectors from the origin?Its also easy to do by transnite recursion, but I didntexpect Stuck to be familiar with that. Use N x N instead of N;your antichain is going to be a family F of functions from N to Nsuch that if f and g are distinct members of F, there is ann(f,g) in N such that either f(n) > g(n) for all n >= n(f,g) -- fdominates g -- or g(n) > f(n) for all n >= n(f,g) -- g dominatesf. Start with F_0 = the family of constant functions. Given acountable family A of functions from N to N, its easy toconstruct a new function from N to N that dominates every memberof A.>Also I wonder if OP meant N^{0,1}, and not {0,1}^N.>Whence |N^{0,1}| = (Aleph_0)^2 = Aleph_0,>while |{0,1}^N| = 2^Aleph_0 = Aleph_1.That last step is invalid: the statement that 2^Aleph_0 = Aleph_1is the Continuum Hypothesis, which is independent of ZFC.Brian === > 2) Let P(A) denote the power set for some set A.>> Given a set B, a subset A of P(B) is called an antichain if no element>> of A is a subset of any other element of A. Does P(N) contain an>> uncountable antichain?>> Tho 1 is easy, I dont get 2, even with the hint to use the bijection> between Q and N. Also I wonder if OP meant N^{0,1}, and not {0,1}^N.>> Think a bit about Dedekind cuts. They dont quite ll the bill as is,> but a slight modication of the idea will work.>{ (r,r+1) / Q | r in RQ }/ intersect> By the way, 2^Aleph_0 is not equal to Aleph_1, unless you are assuming> CH.>c, senior. <3f18c949.11042797@enews.newsguy.com> === >> > 2) Let P(A) denote the power set for some set A.>> Given a set B, a subset A of P(B) is called an antichain if no element>> of A is a subset of any other element of A. Does P(N) contain an>> uncountable antichain?>>Tho 1 is easy, I dont get 2, even with the hint to use the bijection> >between Q and N.>> How about my hint to look at Q x Q and sectors from the origin?Too complicated. How about { (r,r+1) / Q | r in RQ } ?/ intersect> Its also easy to do by transnite recursion, but I didnt> expect Stuck to be familiar with that. Use N x N instead of N;> your antichain is going to be a family F of functions from N to N> such that if f and g are distinct members of F, there is an> n(f,g) in N such that either f(n) > g(n) for all n >= n(f,g) -- f> dominates g -- or g(n) > f(n) for all n >= n(f,g) -- g dominates> f. Start with F_0 = the family of constant functions. Given a> countable family A of functions from N to N, its easy to> construct a new function from N to N that dominates every member> of A.> <3f18c949.11042797@enews.newsguy.com> === ::::>> Im having trouble guring these out:::>> 1) Is the set of all functions from the boolean set {0,1} to the:>> natural numbers countable or uncountable?::>> 2) Let P(A) denote the power set for some set A.:>> Given a set B, a subset A of P(B) is called an antichain if no element:>> of A is a subset of any other element of A. Does P(N) contain an:>> uncountable antichain?:::[...]::>Tho 1 is easy, I dont get 2, even with the hint to use the bijection:>between Q and N.::How about my hint to look at Q x Q and sectors from the origin?:Its also easy to do by transnite recursion, but I didnt:expect Stuck to be familiar with that. Use N x N instead of N;:your antichain is going to be a family F of functions from N to N:such that if f and g are distinct members of F, there is an:n(f,g) in N such that either f(n) > g(n) for all n >= n(f,g) -- f:dominates g -- or g(n) > f(n) for all n >= n(f,g) -- g dominates:f. Start with F_0 = the family of constant functions. Given a:countable family A of functions from N to N, its easy to:construct a new function from N to N that dominates every member:of A.::>Also I wonder if OP meant N^{0,1}, and not {0,1}^N.:>Whence |N^{0,1}| = (Aleph_0)^2 = Aleph_0,:>while |{0,1}^N| = 2^Aleph_0 = Aleph_1.::That last step is invalid: the statement that 2^Aleph_0 = Aleph_1:is the Continuum Hypothesis, which is independent of ZFC.Maybe the problem isnt being solved in ZFC though. Perhaps ZFCH was beingused or ZFG.::Brian: === >>:>:>:>:>> Im having trouble guring these out:>:>:>> 1) Is the set of all functions from the boolean set {0,1} to the>:>> natural numbers countable or uncountable?>:>:>> 2) Let P(A) denote the power set for some set A.>:>> Given a set B, a subset A of P(B) is called an antichain if no element>:>> of A is a subset of any other element of A. Does P(N) contain an>:>> uncountable antichain?>:>:>:[...]>:>:>Tho 1 is easy, I dont get 2, even with the hint to use the bijection>:>between Q and N.>:>:How about my hint to look at Q x Q and sectors from the origin?>:Its also easy to do by transnite recursion, but I didnt>:expect Stuck to be familiar with that. Use N x N instead of N;>:your antichain is going to be a family F of functions from N to N>:such that if f and g are distinct members of F, there is an>:n(f,g) in N such that either f(n) > g(n) for all n >= n(f,g) -- f>:dominates g -- or g(n) > f(n) for all n >= n(f,g) -- g dominates>:f. Start with F_0 = the family of constant functions. Given a>:countable family A of functions from N to N, its easy to>:construct a new function from N to N that dominates every member>:of A.>:>:>Also I wonder if OP meant N^{0,1}, and not {0,1}^N.>:>Whence |N^{0,1}| = (Aleph_0)^2 = Aleph_0,>:>while |{0,1}^N| = 2^Aleph_0 = Aleph_1.>:>:That last step is invalid: the statement that 2^Aleph_0 = Aleph_1>:is the Continuum Hypothesis, which is independent of ZFC.>>Maybe the problem isnt being solved in ZFC though. Perhaps ZFCH was being>used or ZFG.Irrelevant to the problem -- only uncountability was required --and unlikely in any case.Brian === > 2) Let P(A) denote the power set for some set A.> Given a set B, a subset A of P(B) is called an antichain if no element> of A is a subset of any other element of A. Does P(N) contain an> uncountable antichain?>>Tho 1 is easy, I dont get 2, even with the hint to use the bijection>>between Q and N.>> How about my hint to look at Q x Q and sectors from the origin?>Too complicated. Depends on what you happen to see rst. Not intrinsically morecomplicated than the one below, in my opinion. In any case, Ithought that since my hint was a little broader than Robins, itmight point you in the right direction.>How about> { (r,r+1) / Q | r in RQ } ?>/ intersectBrian === Been doing some soul searching lately. Im 37 and have a BS in CIS (got itback in 1988) and have been in the computer eld as a software developerfor the last 15 years. Unfortunately, due to the lousy economy and theoverabundence of software developers, IT is not what it used to be anymore.For the longest time, I wanted to go and get an equivalent degree in CS tocompliment my CIS degree. Basically, what I lacked there is the Math,Chemistry and Physics. I took care of the Math portion by taking Calculusthru Differential equations and found that I really enjoyed it (got an A inall of those classes). Took 10 years off from college and worked my buttoff in the eld as a developer on some jobs. Recently working for acompany developing chemistry software, I nished an engineering CHM105course and got an A in it. However, I did not really enjoy the chemistryas much as Math. I basically need to take a Physics course and a LinearAlgebra course to get an Associates in Math. Im taking a refresher coursein PreCalc on audit (its coming back to me) since its been 13-14 yearssince I took Trig (you dont use it, you do lose it) and will take Calc 1 onaudit to ready myself for Linear Algebra next Jan.An advisor at the community college where Im taking this PreCalc recognizesthat I like Math and has suggested maybe that I consider a BS in Math(rather than CS) if I nish my Associates. The argument (and its a goodone) is that Math is more generalized than the specic Computer Sciencedegree and that it may open up more doors of opportunity. One things forsure, in the computer eld, they do not view a Math degree as negative.In-fact, they view it as a positive. I was just curious, *if* the computereld tends to continue on the same path its in now (ie, massive layoffs,salary cuts and an even more dwindling chance of nding rewarding jobs),what other areas could one (with a BS in Math) get into? I guess with anSchool. I dont know if I have the time to get a Masters as that wouldmean another 4 years. The BS in Math would be about 2 years away.Im not 100% convinced the computer eld is as great as they say it is (Imseeing alot of people losing their jobs and working in non-computer relatedwork with their CS degrees as CS degrees have denitely lost some worthover the last 2 years or so) and am wondering if the Math route (since I doenjoy it) could be a good one to take. At best case, I can still get theMath degree and remain a software developer. At worst case, Im forced tochange careers but can hope that an undergrad in Math would help.A Math degree is denitely not a useless degree. It can be applied to farmore different areas than a CS degree could. Im just wondering though whatan undergrad math degree could be useful for (other than in the computereld). thanks!!! === [...]> Im just wondering though what> an undergrad math degree could be useful for (other than in the computer> eld). thanks!!!> You might look at Scroll down a little to Careers...-- Paul SperryColumbia, SC (USA) === > tAn undergrad degree in pure math is useful in computer programming and actuary work. If you are interested in math then go for it, but if you are viewing it as a means to make money, I dont see that as a good idea. Why not a degree in business? === >> An undergrad degree in pure math is useful in computer programming and> actuary work. If you are interested in math then go for it, but if you> are viewing it as a means to make money, I dont see that as a good> idea. Why not a degree in business?>Because theres an oversupply of business majors as the economy is down,and whether CEOs like it or not, there will never be too few CEOs. === > An undergrad degree in pure math is useful in computer programming and> actuary work. If you are interested in math then go for it, but if you> are viewing it as a means to make money, I dont see that as a goodGood advice!!!> Why not a degree in business?Argg!! :-) With my CIS degree, I got a minor in business(which includedmicro¯o econ)/ and 2 years of accounting.In this day and age with corporate downsizing, business scandals, etc.., Imnot so sure a business degree gets you as far as it once did.I know 2 MBAs that are still looking for work. Since graduating with myCIS, I cant recall ever needing to use anything from mybusiness courses on the job. Accounting is a totally different storythough as alot of computer jobs out there require applications developmentof AR/AP/inventory, etc...As per the Math, I have never had to use the Calculus/diff-eq I learned onany jobs (which is why I forgot my math and am having to take it again)*but* I did have to use some of the trig functions (at a basic level) to dosome graphics programming. The one thing that I got out of math which waspriceless was the ability to solve problems systematically. The samewith Chemistry. As a programmer, its easy to fall into the trap of beinglazy and slamming out code off the top of your head (which usually ends upbeing sloppy) without setting things up on paper and doing it a better way.With the math and chemistry, it gets you into the habit of laying things outin front of you and coming up with the solution. To me, thats what makesMath a VERY useful subject to know. Thats something I did not get when Iwas taking my business courses. Accounting though was similar to the mathas youre forced to lay everything out in front of you so accounting is agood thing to know/do. Unless you aspire to being a CEO or middle manager(which I dont and never have), business courses tend to be a waste. Theydont necessarily even help you if you want to start your own business. Ifyou want to start your own business, you may be better off attendingseminars/lectures rather than wasting a few semesters of time takingbusiness 101 or what have you. If youre going to be a CEO or middle/uppermanagement, youll need the MBA and the business courses are pre-reqs so youhave no choice but to take them. For a software developer, you can learnall you really need to know about businesses just by working and observingthe techniques employed by middle/upper management. And even then, manymanagers dont have a business background.The one thing that fascinates me about Math is that its been around forcenturies and will still be around in the same form centuries from now.Calculus will still be calc. I cant say the same for computer andbusiness courses :-)Math denitely tends to sharpen your analytical and problem solving skills.The business courses I have taken in the past did not. But sadly, youreright in that theres not alot of money to be made in Math. But I dontthink theres any denying that its denitely a useful (perhaps one of themost useful) foundation skill to possess. === >Been doing some soul searching lately. Im 37 and have a BS in CIS (got it>back in 1988) and have been in the computer eld as a software developer>for the last 15 years. Unfortunately, due to the lousy economy and the>overabundence of software developers, IT is not what it used to be anymoreTrue, but I think job opportunities with a bare bachelors in math are even worse. I dont think having that degree would make CIS employers look on you any more favorably, and I cant think of any specic occupations where a bachelors in math plus a bachelors in CIS is any more likely to get you in the door than a bachelors in CIS. (Granted, I may be missing something here.)-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/You nd yourself amusing, Blackadder.I try not to y in the face of public opinion. === >Been doing some soul searching lately. Im 37 and have a BS in CIS (got it>>back in 1988) and have been in the computer eld as a software developer>>for the last 15 years. Unfortunately, due to the lousy economy and the>>overabundence of software developers, IT is not what it used to be anymore>True, but I think job opportunities with a bare bachelors in math >are even worse.>I disagree. CIS degrees are seen to be like engineering degrees, specialized.Math degrees, like science and liberal arts undergraduate degrees in general, are seen as providing evidence that the holder can either solve problems logically (science/math) or communicate (liberal arts, especially English and history). These are valuable skills that employers want. They will give you the chance to prove that you can do the other, and assume that you can learn to do business. So you can get in many doors with a math degree that you cant get into with a CIS degree.> I dont think having that degree would make CIS employers look on you any more favorably,>No -- either they want you for CIS or for business, but if you want to do both thats a long haul (except in the computer biz).> and I cant think of any specic occupations where a bachelors in math plus a bachelors in CIS is any more likely to get you in the door than a bachelors in >CIS. (Granted, I may be missing something here.)>No. The reason for a bachelors in math is that youve come to the realization that a computer career is a dead-end path, except in the eld of computers.So, the reason for getting a math degree (or English, history, chemistry, or Chinese literature) is because you enjoy the subject (at the undergraduate level), you want to polish your analytical and/or communication skills, and/or you just want to be more generally educated. This last is incredibly important: you want to be able to talk to your clients (thats customers/bosses/vendors/whatever) in the vernacular that *they* are comfortable with. If that means wearing white tie and tails, then thats what you do. (Not good for auto mechanics, not necessary for computer gurus, could be incredibly important for corporate or municipal bond underwriters.) And if it means wearing blue jeans to work instead of a suit, thats what you do.Jon Miller === Very good points you raised there Jon!! Most of them fall into the reasonwhy I am thinking about getting an undergrad in Math!!!thanks, Theron === > True, but I think job opportunities with a bare bachelors in math> are even worse. I dont think having that degree would make CIS> employers look on you any more favorably, and I cant think of any> specic occupations where a bachelors in math plus a bachelors in> CIS is any more likely to get you in the door than a bachelors in> CIS. (Granted, I may be missing something here.)I dont think youre missing anything! I feel that youre points are alsovalid. In the 15 years of working in IT, I have never seen a case where anemployer of mine treated someone with a math degree unfavorably. In fact,it was the opposite in one case. He was my boss and was responsible forcoming up with some heavy duty dialing algorithms for a telephone switch.Poisson and Exponential Smoothing were the 2 tools he ended up using. Aco-worker (friend) and myself simply stamped his algorithms into the sourcecode and the system was off to the races :-)Math is a good degree to have. At least employers cannot accuse you of notbeing analytical if you have had 2-3 years of Calculus and Diff-eq andperhaps beyond. And if you do have a Math degree, employers would beignorant if they thought you did not possess criticalanalytical/thinking/problem-solving skills.Math may not help me get promoted or what have you. But I do know this. Ienjoy it and the satisfaction of solving complex problems is very rewarding.Its easy to measure your progress in learning as you progress from mathcourse to math course. I think the same for perhaps Physics and Chemistry(although I only have 1 chem course(so Im speaking for just that onecourse) and have not taken any physics (but plan to do so)).I really dont think Math is a useless degree. These days, CIS and CSdegrees seem to be useless (when you talk to some employers out there). === <...>> Math is a good degree to have. At least employers cannot accuse you ofnot> being analytical if you have had 2-3 years of Calculus and Diff-eq and> perhaps beyond. And if you do have a Math degree, employers would be> ignorant if they thought you did not possess critical> analytical/thinking/problem-solving skills.Yes, but a considerable portion of employers (often off the record ofcourse) will actually look unfavorably upon this type of trait (analytical).Employers are more and more looking for people with good human interactionskills, and less and less looking for geeks. To many, a paradigm stillexists that says if you are a geek (eg a computer whiz of other analyticaltype) then you necessarily lack some of these important people skills. IOW,many associate analytical with I cant work with this person. Truth is,often times that is exactly right.>> Math may not help me get promoted or what have you. But I do know this.I> enjoy it and the satisfaction of solving complex problems is veryrewarding.> Its easy to measure your progress in learning as you progress from math> course to math course. I think the same for perhaps Physics and Chemistry> (although I only have 1 chem course(so Im speaking for just that one> course) and have not taken any physics (but plan to do so)).>> I really dont think Math is a useless degree. These days, CIS and CS> degrees seem to be useless (when you talk to some employers out there).They are useless in the sense that everyone seems to have one, not in thesense that it is a less desireable quality. In general, people withmathematics qualications are certainly less desireable than people withIT qualications. True, there have been many IT layoffs but there stillremains many, many IT positions (many more than math positions if I had tospeculate).Also true is that Microsofts MCSE and related certications--onceprestigious--seem to be possessed now by many people who never actuallyworked in IT (much less engineered any real networks) but rather are tryingto get into this eld on an entry level. Study a little, take a few tests,fork over some dough, and its yours. This certication does not carrynear as much weight as it used to, except to some of the more naive ITmanagers (yes, they exist in abundance) and executive types (CFO/CIO)charged with IT recruitment. On paper, it certainly looks good to have thatcertication. Everyone knows its not what you can do but what someone elsesays you can do, that often times lands a job. So everyone and theirbrother has an MCSE now, thus lessoning the value of having one.-- Darrell === If you enjoy math you should major in it. A smart employer will notlook down on a math degree (especially if you have good grades). Itsabsolutelyridiculous to choose a major based on whether someone perceives it asbeingnerdy or not. If you want to loose the stigma join a business club andmake some contacts. Better yet, take a speech class or join a speechclub. The great thing about math is that you can build on it. It ismuch easier to pick-up computer science, physics, and chemistry if youknow math well. If you have theintellect for math or physics or even chemistry, do that instead ofCIS. A math major is not trendy but it wont be as dependent on trendsas one that is. I wouldnt do business either. You can get your MBAlater and youll be a much more attractive (and capable) employee. Ithink all those bosses who got lectured about not getting it when itcame to IT are having a lot of fun now. Unless you live in India orChina and will work for $5000 a year, IT is only going to get worse.In my mind studying IT is equivalent to studying auto mechanics. Note:Im not saying this in a derogatory way. Auto mechanics is challengingbut its limiting.Nathan > <...>> Math is a good degree to have. At least employers cannot accuse you of> not> being analytical if you have had 2-3 years of Calculus and Diff-eq and> perhaps beyond. And if you do have a Math degree, employers would be> ignorant if they thought you did not possess critical> analytical/thinking/problem-solving skills.> Yes, but a considerable portion of employers (often off the record of> course) will actually look unfavorably upon this type of trait (analytical).> Employers are more and more looking for people with good human interaction> skills, and less and less looking for geeks. To many, a paradigm still> exists that says if you are a geek (eg a computer whiz of other analytical> type) then you necessarily lack some of these important people skills. IOW,> many associate analytical with I cant work with this person. Truth is,> often times that is exactly right.>> > Math may not help me get promoted or what have you. But I do know this.> I> enjoy it and the satisfaction of solving complex problems is very> rewarding.> Its easy to measure your progress in learning as you progress from math> course to math course. I think the same for perhaps Physics and Chemistry> (although I only have 1 chem course(so Im speaking for just that one> course) and have not taken any physics (but plan to do so)).>> I really dont think Math is a useless degree. These days, CIS and CS> degrees seem to be useless (when you talk to some employers out there).> They are useless in the sense that everyone seems to have one, not in the> sense that it is a less desireable quality. In general, people with> mathematics qualications are certainly less desireable than people with> IT qualications. True, there have been many IT layoffs but there still> remains many, many IT positions (many more than math positions if I had to> speculate).> Also true is that Microsofts MCSE and related certications--once> prestigious--seem to be possessed now by many people who never actually> worked in IT (much less engineered any real networks) but rather are trying> to get into this eld on an entry level. Study a little, take a few tests,> fork over some dough, and its yours. This certication does not carry> near as much weight as it used to, except to some of the more naive IT> managers (yes, they exist in abundance) and executive types (CFO/CIO)> charged with IT recruitment. On paper, it certainly looks good to have that> certication. Everyone knows its not what you can do but what someone else> says you can do, that often times lands a job. So everyone and their> brother has an MCSE now, thus lessoning the value of having one. === V O G T22 15 7 20 = 64 In the late afternoon I went to Starbucks and met Roger Bristow,soon he was joined by a friend of his, Victoria Vogt, both are medicalstudents at the U of S.64+ Dad 15 3 /29164+ Mom 4 8 /149183 Victoria 29 2 76 60/306 6951Victoria 97 Lee 22 Vogt 6464+ Sis 8 3 78 67/298 768964+ Bro 29 6 83 180/185 9628 Victoria was born on the 60th day of the year, it is 11 plus the11th prime (31) plus the 11th non-prime (18), or simply 11+11p+11np.Her rst and last names both begin with the 22nd (11+11th) letter ofthe alphabet. Her rst two letters add to 31 (11p). Her middle nameadds to the 22 (11+11) chapters of Bible Book 11. Her given names addto 119. Her rst and last names differ in value by 33 (3x11). Hernames have an average value of 61 (Exodus 11), it is the 18th primewhile 18 in turn is the 11th non-prime), it is the 11th prime innon-prime position.Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 <-11th-> 18 <-11th-> 55 <-11th-> 199 --- --- --- --- 160 113 143 518 The parents were born in months adding to 11 and also the kids wereborn in months adding to 11, together for the 22 chapters of BibleBook 11. The kids were born on days of the month adding to 66 (6x11)and in years adding to 237, its the opening chapter of The Samuels,pretty as The Samuels contain 55 (5x11) chapters. The kids were bornon days 29, 8 and 29, these Bible Books contain an average of 77(7x11) verses. The males were born on days of the month adding to 44(4x11). The males were born on days of the month averaging the 22(11+11) chapters of Bible Book 11. The sisters were born on days ofthe year adding to 127 (the 31st prime while 31 in turn is the 11thprime), it is the 11th prime in prime position. The males were likelyborn on days of the year averaging 127 (the 11th prime in primeposition). Generally the parents have their birthdays 150 days closerto the beginning of their years than to the end of their years,corresponding to Deuteronomy 33 (3x11). Generally the parents have their birthdays 150 days closer to thebeginning of their years than to the end of their years, pretty asthere are 150 chapters in Bible Book 19 while the parents were born ondays of the month adding to 19. Mom was born 67 days closer to the endof the year than to the beginning of the year (19th prime), keeping inmind that the 2460 verses of Book 19 is 7x19x19 minus the 19th prime(67). If the parents were both born in non-leap years, then the familywas born on days of the year adding to 597 (Psalm 119), pretty thatVictorias rst name adds to 97 while her given names would addtogether for 119. I meet Victoria on the 19th.Primes Non-Primes 2 1 3 4 5 6 7 <-4th-> 8 <-Bible Book 8 contains 11 4 chapters, pretty as 13 8 is the 4th non-prime 17 19 -- 77 <-the rst 8 primes plus 8 more adds to the 85 verses of Bible Book 8 Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 2310 2911 31 <- 3112 3713 41 <- 4114 4315 4716 5317 59 <- 5918 6119 67 <- 67 <-the 8th prime in prime positionR U T H <-Book 818 21 20 8 = 67 The family was born on days of the month adding to the 85 verses ofBible Book 8, it is a combination of the rst 8 primes (up to 19)plus 8 more. The little sister was born on the 8th day of the monthand on the 67th (19th prime or the 8th prime in prime position) day ofthe year, Bible Book 8 is Ruth (67). Generally the parents have their birthdays on days of the yearadding to 290. Dad was born with 291 days remaining in the year. Thelittle sister was born with 298 days remaining in the year. The familywas together born with 1229 days remaining in their years. The rstand the last kids were both born on the 29th day of the month. Thereare 29 chapters in Bible Book 13 (6th prime) and 29 verses in chapter666 (Ecclesiastes 7), pretty as 29 is a combination of 6 plus the 6thprime (13) plus the 6th non-prime (10). Dad generally has his birthdayon the 74th day of the year (a factor of 666). Mom generally has herbirthday on the 216th (6x6x6) day of the year. The kids were born ondays of the month adding to 66.1-50 - Genesis51-90 - Exodus91-117 - Leviticus118-153 - Numbers154-187 - Deuteronomy188-211 - Joshua930-957 - Matthew958-973 - Mark974-997 - Luke998-1018 - John1019-1046 - Acts1047-1062 - Romans 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 661062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 61070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6)1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime)1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) The rst two letters in Vogt add to 37, the rst pair of kidswere born on days of the month adding to 37 and also the last two kidswere born on days of the month adding to 37. The Vo(37)gt kids wereborn in years adding to 237 (the opening chapter of The Samuels).Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29J O E L <-Bible Book 2910 15 5 12 = 42 <-29th non-primeC O P P E R <-29th element3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent pieceC E N T <-made out of 29th element3 5 14 20 = 42 <-29th non-primeCopper and Zinc are elements 29 and 30 (togetherfor 59), and together they make Brass (59):B R A S S2 18 1 19 19 = 59 Because Victoria was born on the 29th I asked to see her pennies(copper, the 29th element), although I was attracted to her and reallywanted to see her panties, and explained that her money was a giftfrom God. Anyway, she had 4 pennies, they were dated 89, 90, 96 and02, together for 277 (the 59th prime or the 17th prime in primeposition). The last two kids are separated by 1939 (7x277) days, orexactly 277 weeks. The kids are together separated by 2677 days. Thekids were born on days 29, 8 and 29, these Bible Books contain anaverage of 77 verses. The kids were born on days and in months addingto 77 and in years adding to 237. The sisters were born on days of themonth adding to 37. The kids were born on days of the year adding to307. The kids were born on days of the month adding to the 66 Books ofthe Bible, it is 7x7+17 or 7 squared plus the 7th prime, or the 17thprime plus 7 more.Primes 2 73 179 3 79 181 5 83 191 7 89 193 11 97 197 13 101 199 17 103 211 19 107 223 23 109 227 29 113 229 31 127 233 37 131 239 41 137 241 43 139 251 47 149 257 53 151 263 59 157 269 61 163 271 67 167 277 <-59th 71 173 281 Mom was born on the 4th day of the 8th month, and see that she wasborn with 149 days remaining in the year (the number of verses inBible Book 48). The little sister was born with 298 (149+149) daysremaining in the year. The last two kids are separated by 1939 (7x277)days, or exactly 277 weeks. Victorias pennies were together 35 yearsold, pretty as 149 is the 35th prime. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 <-17 is the 7th prime 5 11 <- 11 while the primes up 6 13 to 7 add to 17 7 17 <- 17 8 19 9 2310 2911 31 <- 3112 3713 41 <- 4114 4315 4716 5317 59 <- 59 <-the 7th prime in --- prime position 167 Esther Book 17 <-the 7th prime Leviticus begins with 17 verses and terminates at chapter 117 with17+17 verses. There are 17 verses at chapters 1 and 3, and 59 (the 17prime) verses at chapter 13, so the 17s and the 17th prime are atchapter numbers adding to 17 (1+3+13=17). The rst 17 versed chaptersin the Bible are at chapters 91 and 93, together for 184, or the 167verses of Book 17 plus 17 more. Leviticus contains 859 verses, it endsin 59 (the 17th prime). The rst 17s in the Bible surround chapter92 (the 4x17th non-prime):Leviticus--------- 91 1 17 92 2 <-68th (4x17th) non-prime 93 3 17 94 4 95 5 96 6 97 7 98 8 99 9100 10101 11102 12103 13 59 <-17th prime104 14 105 15106 16107 17108 18109 19110 20111 21112 22113 23114 24115 25116 26117 27 34 <-17+17 The family was born on days of the month averaging 17. The kidswereborn in years adding to 237, its a combination of the 17th, 17+17th,17+17+17th and the 17+17+17+17th non-prime numbers (26, 49, 70 and92). The last two kids are separated by exactly 277 weeks, Victoriaspennies had years adding to 277 (the 17th prime in prime position).Its another true story.Non-Primes 1 27 50 72 4 28 51 74 6 30 52 75 8 32 54 76 9 33 55 7710 34 56 7812 35 57 8014 36 58 8115 38 60 8216 39 62 8418 40 63 8520 42 64 8621 44 65 8722 45 66 8824 46 68 9025 48 69 9126 49 70 92 <-the 17th levelDaryl Shawn KabatoffBox 7134Saskatoon SaskatchewanCanadaS7K 4J1Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === Stand back, everyone! Dar is winding up again. The pitch should come inthe next day or so.--It takes a village to raise an idiot.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === ...> http://www.crbond.comA nice site - simple & informative.gtoomey === I am searching for the denition of the term k-algebra. I havemultiplication that is a bilinear operator, however it does notelaborate on this. Any assistance would be greatly appreciated. A book or onlineSincerely,LS Thomas === > I am searching for the denition of the >term k-algebra. I have>multiplication that is a bilinear operator, >however it does not>elaborate on this.You might want to take a look at this page for basic denitions & properties:http://www.math.harvard.edu/~elkies/M250.01/ kalgebra.html === The answer to this may be obvious but...when would you need to use ahistogram? Is there a really obvious case that I could present as anexample.The background to this is that I am a teacher in a school in England andrecently there has been a development that requires pupils to do aStatistics based coursework. For the higher marks the students need to beable to draw histograms but as the students have control over the data theycollect afaics youd have to force it into unequal intervals (we are alreadydoing unnecessary cumulative frequency curves to nd a mean of a data setof 100 values). I can only see the need for a histogram if you are presentedwith data that is already unequally grouped.cheersdd === > The answer to this may be obvious but...when would you need to use a> histogram? Is there a really obvious case that I could present as an> example.> The background to this is that I am a teacher in a school in England and> recently there has been a development that requires pupils to do a> Statistics based coursework. For the higher marks the students need to be> able to draw histograms but as the students have control over the data they> collect afaics youd have to force it into unequal intervals (we are already> doing unnecessary cumulative frequency curves to nd a mean of a data set> of 100 values). I can only see the need for a histogram if you are presented> with data that is already unequally grouped.> > cheers> ddorganized way. Consider presenting the results of rolling 3 six-sided dice 1000 times. The mean/standard deviation dont give as much information as the histogram of the data will. The raw data is nearly useless for conveying anything. It also introduces them to the concept of a bar chart (as in Excel) and may help them better interpret results in the future.-- Will Twentyman === > The answer to this may be obvious but...when would you need to use a>> histogram? Is there a really obvious case that I could present as an>> example.>> The background to this is that I am a teacher in a school in England and>> recently there has been a development that requires pupils to do a>> Statistics based coursework. For the higher marks the students need >> to be>> able to draw histograms but as the students have control over the >> data they>> collect afaics youd have to force it into unequal intervals (we are >> already>> doing unnecessary cumulative frequency curves to nd a mean of a >> data set>> of 100 values). I can only see the need for a histogram if you are >> presented>> with data that is already unequally grouped.>> cheers>> dd> organized way. Consider presenting the results of rolling 3 six-sided > dice 1000 times. The mean/standard deviation dont give as much > information as the histogram of the data will. The raw data is nearly > useless for conveying anything. It also introduces them to the > concept of a bar chart (as in Excel) and may help them better > interpret results in the future.Now think of summarizing the experience of hundreds of thousands of insureds, over several product lines, over several years. A picture is worth a thousand words (at least), so you use a histogram.For setting grading standards (assuming you let performance on the exam tell you something about the exam).For anything where the data is traditionally presented using a histogram. Including tire failure (warranty) data. Battery warranty data. Maybe other warranty data, I dont know. Sometimes drug trials.The point of histograms is also the point of summarizing. Grouping the data into meaningful groups, and analyzing the groups.Jon Miller === > The answer to this may be obvious but...when would you need to use a> histogram? Is there a really obvious case that I could present as an> example.> The background to this is that I am a teacher in a school in England and> recently there has been a development that requires pupils to do a> Statistics based coursework. For the higher marks the students need to be> able to draw histograms but as the students have control over the data they> collect afaics youd have to force it into unequal intervals (we are already> doing unnecessary cumulative frequency curves to nd a mean of a data set> of 100 values). I can only see the need for a histogram if you are presented> with data that is already unequally grouped.> Ok, this is my understanding - and I guess Ill be shot down in ames as I generally take the simpletons approach.I always that even if the groups are the same size the chart is still a histogram - just a special case if you like.Now, as part of the student work he/she may make the choice to keep the group sizes equal. If the student explains the choice in the coursework and explains that this removed the need to divide the frequency by whatever then surely that shows (a) that the student has a grasp of what a histogram is and (b) they can make good choices on the collection of data and provide evidence of planning.Parachute ready, re away! === >I always that even if the groups are the same size the chart is still a >histogram - just a special case if you like.Absolutely. A histogram is used for continuous data a bar chart fordiscrete.-- Dave === >I always that even if the groups are the same size the chart is still a>histogram - just a special case if you like.> Absolutely. A histogram is used for continuous data a bar chart for> discrete.> --> DaveNot really. A histogram is a graph of a frequency (or probability)density function. The integral of (ie area bounded by, more or less) afrequency (probability) density function then gives the frequency(probability).For example, one probability density function (pdf) that most peoplehave seen is the Normal pdf - that familiar bell curve whose(standard) equation is f(x)=(1/Sqrt(2*pi))*e^(-0.5*x^2). The areabeneath the Normal pdf gives probabilities ... The graph of the Normalpdf is a histogram, because areas thereunder give probabilities. Fine,thats a continuous pdf. There *are* discrete pdfs, of course, such asthe Binomial ... graph a Binomial pdf and you have a histogram fordiscrete data.One good (pedagogic) reason (there are other reasons, mostly not sogood) for distinguishing between bar charts that arent histograms(graphs of *frequency*, mostly) and bar charts that are histograms(graphs of *frequency density*, as I said) is that it makes theconceptual step to probability density functions easier to make.So keep using histograms for preference (rather than simple frequencybar charts), and emphasising the difference between them ... yourstudents will thank you for it later.Bob === >I always [thought] that even if the groups are the same size the chart is still a>histogram - just a special case if you like.>>snip> So keep using histograms for preference (rather than simple frequency> bar charts), and emphasising the difference between them ... your> students will thank you for it later.> > BobSo going back to my comment (now xed) Bob, is there any reason why the students cant use a histogram with all the groups the same size? === I shall look it up in my textbook, but from memory, I think that they can beused in situations where the data is of the continuous frequency variety.-- MESSAGE ENDS.John Porcella> The answer to this may be obvious but...when would you need to use a> histogram? Is there a really obvious case that I could present as an> example.>> The background to this is that I am a teacher in a school in England and> recently there has been a development that requires pupils to do a> Statistics based coursework. For the higher marks the students need to be> able to draw histograms but as the students have control over the datathey> collect afaics youd have to force it into unequal intervals (we arealready> doing unnecessary cumulative frequency curves to nd a mean of a data set> of 100 values). I can only see the need for a histogram if you arepresented> with data that is already unequally grouped.>> cheers>> dd>> === >I always that even if the groups are the same size the chart is still a>histogram - just a special case if you like.>> Absolutely. A histogram is used for continuous data a bar chart for> discrete.Thats it, I was not sure!Also another thing to note is that a histogram has a frequency density forthe y-axis, so that when multiplied by the class width, the area given willbe the frequency.The other difference between bar charts and histograms is that bar chartshave columns which do not touch each other, whereas histograms do.-- MESSAGE ENDS.John Porcella === > I always [thought] that even if the groups are the same size the >> chart is still a>> histogram - just a special case if you like.>> snip> So keep using histograms for preference (rather than simple frequency>> bar charts), and emphasising the difference between them ... your>> students will thank you for it later.>> Bob> So going back to my comment (now xed) Bob, is there any reason why > the students cant use a histogram with all the groups the same size?>Sometimes the results just dont make sense.Your decision on what size groupings to make should be based on what you intend to say with the results, and not just some rule you always use.Jon Miller === >>I always [thought] that even if the groups are the same size the chart is still a> >histogram - just a special case if you like.> snip> > So keep using histograms for preference (rather than simple frequency> bar charts), and emphasising the difference between them ... your> students will thank you for it later.>> Bob> So going back to my comment (now xed) Bob, is there any reason why the> students cant use a histogram with all the groups the same size?No reason at all. Perfectly acceptable. Bob === To achieve some of the higher marks, you can use an equal bar histogram todraw a normal distribution curve. i.e. a smooth curve through the top middleof each bar. Now calculate the standard deviation of the sample, drawvertical lines one standard deviation either side of your central bar andmeasure the area enclosed (i.e. count squares) if its around 66% of thetotal area, then we have normal distribution. Obviously works best forheight, mass IQ etc.> The answer to this may be obvious but...when would you need to use a> histogram? Is there a really obvious case that I could present as an> example.>> The background to this is that I am a teacher in a school in England and> recently there has been a development that requires pupils to do a> Statistics based coursework. For the higher marks the students need to be> able to draw histograms but as the students have control over the datathey> collect afaics youd have to force it into unequal intervals (we arealready> doing unnecessary cumulative frequency curves to nd a mean of a data set> of 100 values). I can only see the need for a histogram if you arepresented> with data that is already unequally grouped.>> cheers>> dd>> === This is loosely related to another thread on histograms.I am looking for some statistics I could use to analyse reaction times.Lets say I have collected 30 reaction times for my left (non-writing) handand 30 for my right (writing) hand. The most obvious thing to do iscalculate the mean and see which was quickest. Beyond that you mighthypothesise that your non-writing hand would be more erratic and do standarddeviations to investigate. Can anyone here suggest anything else?Using this site http://www.jamsarts.com/reactiontime.htm (suggested in amuch earlier thread) you could hypothesise about reactions to differentcolours alongside writing/non-writing hand but what basic statistics wouldyou use? Any help gratefully appreciated.FWIW whenever Ive got this to work the non-writing hand is quickest.cheersddI realise Im posting to asmp is there a suitable stats NG? I couldnt ndone. === >This is loosely related to another thread on histograms.>>I am looking for some statistics I could use to analyse reaction times.>Lets say I have collected 30 reaction times for my left (non-writing) hand>and 30 for my right (writing) hand. The most obvious thing to do is>calculate the mean and see which was quickest. Beyond that you might>hypothesise that your non-writing hand would be more erratic and do standard>deviations to investigate. Can anyone here suggest anything else?>>Using this site http://www.jamsarts.com/reactiontime.htm (suggested in a>much earlier thread) you could hypothesise about reactions to different>colours alongside writing/non-writing hand but what basic statistics would>you use? Any help gratefully appreciated.>>FWIW whenever Ive got this to work the non-writing hand is quickest.>>cheers>>dd>>I realise Im posting to asmp is there a suitable stats NG? I couldnt nd>one.>sci.stat.eduIts my recollection that one of sci.stat.math and sci.math.stat is busy and the other not, but I dont recall whichsci.stat.consult is willing (or was in the past) to help with this sort of question. Theyd much rather deal with educated consumers than idiots.Jon Miller === >sci.stat.consult is willing (or was in the past) to help with this sort >of question. Theyd much rather deal with educated consumers than idiots.But that cuts the base of questioners so drastically! :-)(Im fresh from dealing with one student who phones ve minutes before the midterm to ask if she could go to her brothers graduation instead, and another who insists on asking me questions that are answered in the syllabus because I dont want to read a lot of words. This is college!)-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/Walrus meat as a diet is less repulsive than seal. === >sci.stat.consult is willing (or was in the past) to help with this sort >>of question. Theyd much rather deal with educated consumers than idiots.>>But that cuts the base of questioners so drastically! :-)>I didnt say they insist on it, just that they prefer it.Ive been known to adjust my rates depending on how interesting the assignment looks. (I wish I could adjust them base on how the cash ow looks!)Jon Miller === S T E V E N S O N10 20 5 22 5 14 19 15 14 = 133 In the afternoon I went to the food court in the Wildwood Mall andmet Buddy, he is the 6th of 8 kids.133+ Dad 4 10 14 277/88 +15477133+ Mom 9 4 16 100/266 +14924278 Buddy 10 6 57 161/204 113William 79 Lyon 66 Stevenson 133 The parents were together born 23 days closer to the end of theiryears than to the beginning of their years. Buddy and his parents wereborn on days of the month adding to 23, Buddy was born exactly 2323 with 66 chapters.Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 <-14th-> 22 --- --- 281 176 Dad was born in 14. The parents were born in months adding to 14.The parents were born in 14 and 16, these Bible Books contain anaverage of 614 verses. Buddy was born 43 days closer to the beginningof the year than to the end of the year (14th prime). He was born into 79, it is the 22nd prime while 22 in turn is the 14th non-prime (79parents are separated by exactly 79 weeks while his rst name adds toname exceeds his 161st day of birth by the 117 verses of Bible Book 22(14th non-prime). Dad and Buddy were born on days of the month addingto 14. My birthday was 14 days ago while Buddy and I had our birthdaysan average of 140 days ago. Mom was born in 16 and married Stevenson, pretty as the name addsto 133 while Bible chapter 133 is Numbers 16. This 133 (Numbers 16)exceeds its 101st non-prime position by 32 (16+16). Buddy has 16letters in his rst and last names, his last name adds to 133born on days 4, 9 and 10, together these Bible Books contain 96 (6x16)chapters. Buddy was born in 57, Bible chapter 57 and Bible Book 57both contain 25 verses (the 16th non-prime). The parents were born in years adding to 30 (20th non-prime). Buddyand his parents were born in months adding to 20. Buddy has 20letters. Dad and Buddy were born in years adding to 71 (20th prime).Buddy and his parents were together born 20 days closer to thebeginning of their years than to the end of their years. Buddy wasborn on day 161 (Deuteronomy 8 with 20 verses).Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 Dad was born on day 277, its the 59th prime while 59 in turn isthe 17th prime (277 is the 17th prime in prime position). The parentswere born on days 277 and 100, these are the 59th prime and the 75thnon-prime, together for 134 (Numbers 17). Dad was born 189 days closerto the end of the year than to the beginning of the year (the rst 17primes minus the rst 17 non-primes). Dad was born 177 days furtherinto the year than mom (3 times the 17th prime). Buddy was born with204 days remaining in the year (Joshua 17), it is 12x17, or the 7thnon-prime times the 7th prime, and is twice 17 plus twice the 17th161 day of birth by 117.Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 --- --- 238 154 <-Lamentations The parents were born on days of the month adding to 13 (6thprime). Dad was born in 14, Bible Book 14 contains 36 chapters (6x6while 1 through 36 adds to 666). Mom was born in 16, Bible Book 16contains 13 chapters (6th prime). Buddys 278 valued name exceeds hisletters in his given names add to 91 (1 through 13), the repeatingletters in his given names add to 54 (13 plus the 13th prime), it is adifference of 37 (37 chapters in the Bible contain the length of 13verses). I am 113 days older than Buddy, there are 113 verses in BibleBook 54 while his initials add to 54 (13 plus the 13th prime). He isthe 6th of the kids, his rst 6 letters add to 66, his middle nameprime and the 184th non-prime averages 666). Buddy was born on the10th (6th non-prime) day of the 6th month.Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 26 61 27 67 28 71 30 73 32 79 33 83 34 89 35 97 36 101 38 103 39 107 40 109 <-29th-> 42 --- ---1480 665 Buddys rst name adds to 79 (Exodus 29), his given names addtogether for 145 (5x29). In his given names, his consonants plus theirpositions exceed his vowels plus their positions by 29. The primes andsquares in his given names add together for 79 (Exodus 29). Buddy andhis parents were born in years averaging 29. Mom and Buddy were bornin years adding to the 73 verses of Bible Book 29 (and is the Lucasnumbers up to 29). Dad and Buddy were born on days of the year addingto 438 (6 times the 73 verses of Book 29). Dad and Buddy were togetherborn with 292 days remaining in their years (4 times the 73 verses ofBook 29). Mom and Buddy were together born 209 days closer to thebeginning of their years than to the end of their years (29.85 weeks).Buddy and I were born on days of the year adding to 209 (29.85 weeks).Buddy was born with 204 days remaining in the year (29.14 weeks). Momand Buddy were born on days of the century adding to 26923(13x19x109), it is both a multiple of 13 (29 chapters in Bible Book13) and a multiple of 109 (the 29th prime). Dad was 42 (29thnon-prime) years old when Buddy was born. We meet when I am 16815 daysold, my age ends in 815 (the rst 29 primes minus the rst 29non-primes). Bible Book 13 (the 6th prime) contains 29 chapters andBible chapter 666 (Ecclesiastes 7) contains 29 verses, pretty as 29 is6 plus the 6th prime (13) plus the 6th non-prime (10).Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29J O E L <-Bible Book 2910 15 5 12 = 42 <-29th non-primeC O P P E R <-29th element3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent pieceC E N T <-made out of 29th element3 5 14 20 = 42 <-29th non-primeCopper and Zinc are elements 29 and 30 (togetherfor 59), and together they make Brass (59):B R A S S2 18 1 19 19 = 59 Buddy is coming out of a family of 10, his dad was born in the 10thmonth while mom was born on the 10x10th day of the year. Buddy wasborn on the 10th day of the month.Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 <-11th-> 18 <-11th-> 55 <-11th-> 199 --- --- --- --- 160 113 143 518 Buddys rst name adds to 79 (the 11+11th prime), his middle nameinitials add to 31 (11th prime). He has 11 letters in his given namesand his given names add to 145, corresponding to Numbers 28 with 31primes, squares and cubes together add with their positions for 211.exceed his vowels by 110. In his given names, his odd valued lettersexceed his even valued letters by 45 (the 11th non-prime in primeposition). In his given names, his unrepresented letters exceed hisrepresented letters by 127 (the 11th prime in prime position). Buddywas born 62 days after moms birthday (twice the 11th prime). He wasborn 249 days after dads birthday (First Samuel 11). He was born 311days after his parents birthdays. Mom was born on the 9th,corresponding to First Samuel with 31 chapters (11th prime). Buddy andhis parents were born on days of the year adding to 538, correspondingto Psalm 60 (11+11p+11np). I meet Buddy on the 62nd day of the year(twice the 11th prime). I am 113 days older than him (the rst 11non-primes).389 <-77th prime104 <-77th non-prime 77 <-77---570 The Four 57sGenesis 41 -> 41Leviticus 14 -> 104Judges 9 -> 220 <-I dreamt of 220 roofs blownJohn 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th primeChapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime <-together for 57->Major Books of End-Times Prophecy (Daniel and Revelation are in partabout 666 while Isaiah contains 66 chapters):Daniel - 357 versesRevelation - 404 verses <-57 plus the 57th prime plus the 57th non-primeIsaiah - 1292 verses <-an average of 19.575757... verses per chapterparents were born on days and in months and years adding to 57. Dadand Buddy were born on days of the year adding to 438 (62.57 weeks).Daniel and Revelation are the major Books of end-times prophecy (theyare in part about 666), they contain 357 verses and 404 (57+57p+57np)verses. He and his parents were born on days of the month adding to 23(Isaiah with an average of 19.575757... verses per chapter). If I could convince Buddy that the God of the Bible provided himwith his name, it would only result in him giving money to a churchthat has an Egyptian penis on its roof. Perhaps Buddy would onlyattend a church in late December, and he would reluctantly do so inorder to please his wife and her parents. While at that church hewould see their decorated evergreen trees, comment on their beauty,and give the church money. The churches teach the sheep to turnevergreen trees into decorated idols, the evergreen tree wasworshipped for centuries as a fertility symbol for it remains greenthroughout the year. The Old Testament documents and condemns howpagans surrounding and opposed to ancient Israel worshipped theevergreen trees. The Old Testament also documents and condemns theobelisks, some version refer to these symbolic penises as the Towersof Bethshemesh. Like the evergreen trees, the penis was worshipped asa fertility symbol. Or if I could convince Buddy that the God of theBible provided him with his name, he would seek out a priest that hasa sh head hat, and give that person money. The ancient priests ofsh god Dagon dressed up in sh outts, over the years theircostumes evolved so that all that now remains is the sh head hat,again the sh is being worshipped as a fertility symbol due to theirlarge number of eggs. Sticking a penis on the roof of your church (ora sh head hat on the head of your priest) is a violation of GodsSecond Commandment. Turning trees into idol, bowing to decorated treesand worshipping trees, penises and sh is a violation of Gods FirstCommandment. By December 25th the sun is visibly returning from thesouth, calling this pagan sunwhoreshipping holiday Christmas is aviolation of Gods Third Commandment (it is a pagan mass and notfacilities in an attempt to make me shut up about the false traditionsin your churches is nothing less than a violation of Gods SixthCommandment. They tortured me for years, I begged for years forassistance to get out of the country to no avail, and all you peoplecan do is give money to the churches that teach you to violateCommandments. You people collectively spent millions of dollars havingme tortured, then annually you people collectively spend billions ofdollars in turning trees into idols. My math was used as an excuse torepeatedly arrest and torture me, and then when I manage to meet withyou people in restaurants and show you evidence that your name is agift from God, you are so cheap and ignorant that you dont even havetaking the time out of my life to show you such. You are anincompassionate turd, you are the of the earth, and I am on myknees begging God to honor Exodus 20:5 and Hosea 4:6 as promises, andterminate your life, the lives of your siblinks, and the lives of yourchildren. Buddy, if I nd you or your family members in theobituaries, I will cheer, it will be in accordance to Scripture (Psalm137:9), and I will post these stats again. All you are good for is tohave your stats posted on the usenet and be used as an example toothers, and look, here you are!!! And it should be mentioned thatBuddy Stevenson is a native Indian, Joe Munroe once informed me thatthe Indians on his reserve will ip op between traditionalaboriginal and Christian beliefs, depending upon whom they are tryingto get into their beds. Your only compassion is for your lthycan try and let your lthy traditions save you.Daryl Shawn KabatoffBox 7134Saskatoon SaskatchewanCanadaS7K 4J1Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === The book that I have gives me this problem:(-12.0, 0, 21.4)m/s at time t=5s and has a constant acceleration ofat t=25s.(NOTE: s denotes unit of time seconds and m/s denotes meters persecond.)The rst part was rather simple for me (nding the velocity att=25), and the second part (nding the position at t=25) should bejust as simple, but for some reason I keep coming up with the wronganswer. I know this because the answer is in the back of the book(thank god or Id never know I was wrong).My approach was this: starting at the acceleration (call it a(t))which is constant, I anti-differentiated adjusting for the 5 secondsto get the velocity function (call it v(t)), which gives me:v(t) = <-12.0, 10.3t - 51.5, -1.50t + 28.9>m/sI know this is right, because solving for v(25) gives me the answer<-12.0, 206, -8.6>m/s which is what is in the back of the book. Igure I repeat the process of anti-differentiating, using the answerfor v(0) and correcting for the 5 seconds again, since I have toadjust the position function for the 5 seconds as well (I wasnt sureif that was the case, but anti-differentiating just v(t) comes up withan answer that is not even close to what the answer in the book is, sothey obviously intended the 5 seconds to apply to both the initialposition and initial velocity).Heres what my work looks like:v(0) = <-12, -51.5, 28.9>m/sp(t) = <-12t+(134-(-12*5)), 5.15t^2-51.5t+(-25.1-(-51.5*5)),-0.75t^2+28.9t+(124-(28.9*5)) >msimplifying gives me:p(t) = <-12t+194, 5.15t^2-51.5t+232.4, -0.75t^2+28.9t-20.5>mUsing my answer above, I get:p(25) = <-106, 2163.65, 233.25>mHowever, the answer in the back of the book is:p(25) = <-106, 2035, 252>mI can tell Im on the right track, because the rst componentmatches, but the other two, while close, dont actually match. Igure I must be skipping a step that involves, for instance, the10.3t from the second component of the velocity function (the one thatsays 10.3t - 51.5). Am I supposed to adjust for the ve seconds onthat part as well? And if so, how do I go about doing that exactly?than himself. === > The book that I have gives me this problem:>> (-12.0, 0, 21.4)m/s at time t=5s and has a constant acceleration of> at t=25s.I think you have a mistyping; below you use a = (-12, 10.3, -1.5)>> (NOTE: s denotes unit of time seconds and m/s denotes meters per> second.)>> The rst part was rather simple for me (nding the velocity at> t=25), and the second part (nding the position at t=25) should be> just as simple, but for some reason I keep coming up with the wrong> answer. I know this because the answer is in the back of the book> (thank god or Id never know I was wrong).>> My approach was this: starting at the acceleration (call it a(t))> which is constant, I anti-differentiated adjusting for the 5 seconds> to get the velocity function (call it v(t)), which gives me:>> v(t) = <-12.0, 10.3t - 51.5, -1.50t + 28.9>m/sp = av = p = a*(t - t0) + v0p = a*(t - t0)^2 / 2 + v0 * (t - t0) + p0Nowt0 = 5sa = (0, 10.3, -1.5) ms^-2v0 = (-12, 0, 21.4) m/sp0 = (134, -25.1, 124) mp = (0, 10.3, -1.5)*(t - 5) + (-12, 0, 21.4) = (-12, 10.3*(t-5), -1.5*(t-5) + 21.4)so p(25) = (-12, 10.3*20, -1.5*20 + 21.4) = (-12, 206, -8.6) m/s>> I know this is right, because solving for v(25) gives me the answer> <-12.0, 206, -8.6>m/s which is what is in the back of the book.so far so good> I gure I repeat the process of anti-differentiating, using the answer> for v(0) and correcting for the 5 seconds again, since I have to adjust> the position function for the 5 seconds as well> (I wasnt sure if that> was the case, but anti-differentiating just v(t) comes up with an answer> that is not even close to what the answer in the book is, so they> obviously intended the 5 seconds to apply to both the initial position> and initial velocity).cant quite follow this>> Heres what my work looks like:>> v(0) = <-12, -51.5, 28.9>m/s> p(t) = <-12t+(134-(-12*5)), 5.15t^2-51.5t+(-25.1-(-51.5*5)),> -0.75t^2+28.9t+(124-(28.9*5))>mso you have p(t) = a*t^2/2 + v(0)(t-5) + r0 + C, where C is constant stuffwhich I cant immediately see how you derived. Theres no choice of Cwhich will give the correct answer because the coefcient of (t-5) should bev(5), and your t^2 term should be a*(t-5)^2 / 2>> simplifying gives me:> p(t) = <-12t+194, 5.15t^2-51.5t+232.4, -0.75t^2+28.9t-20.5>m>> Using my answer above, I get:> p(25) = <-106, 2163.65, 233.25>mp = (0, 10.3, -150)*(t - 5)^2 / 2 + (-12,0,21.4) * (t - 5) + (134, 25.1, 124) = (-12*(t-5)+134, 5.15*(t-5)^2 - 25.1, -0.75*(t-5)^2 + 21.4*(t-5) + 124)so p(25) = (-12*20 + 134, 5.15*400 - 25.1, -0.75*400 + 21.4*20 + 124) = (-106, 2034.9, 252) m>> However, the answer in the back of the book is:> p(25) = <-106, 2035, 252>mThis answer has been rounded to nearest integer for some bizarre reason.> I can tell Im on the right track, because the rst component matches,> but the other two, while close, dont actually match. I gure I must be> skipping a step that involves, for instance, the 10.3t from the second> component of the velocity function (the one that says 10.3t - 51.5). Am> I supposed to adjust for the ve seconds on that part as well? And if> so, how do I go about doing that exactly?The initial conditions should be obviously satised; thus, a*(t-5)^2 /2 +v0*(t-5) + p0 is a better form to use than a*t^2 / 2 + C*t + D where C andD have to be determined.Dont put specic numbers into the equation before you have to; itsmore difcult to spot mistakes if you do it early.-- P.A.C. SmithIf the Apocalypse comes, beep me. <*> http://www.srcf.ucam.org/~pas51 === > (-12.0, 0, 21.4)m/s at time t=5s and has a constant acceleration of> at t=25s.>Basic vector calculus givesv = (t-5)a + v_5d = (t^2 / 2)a - 5ta + t v_5 - (25/2)a + 25a - 5v_5 + d_5d_25 = 300a - 100a + 20v_5 + d_5 = 200a + 20v_5 + d_5Starting from the beginningv = ta + v_0d = (t^2 / 2)a + t v_0 + d_0from v_5 nd v_0 and from d_5 nd d_0, hence nd d_25> The rst part was rather simple for me (nding the velocity at> t=25), and the second part (nding the position at t=25) should be> just as simple, but for some reason I keep coming up with the wrong> answer. I know this because the answer is in the back of the book> (thank god or Id never know I was wrong).>> My approach was this: starting at the acceleration (call it a(t))> which is constant, I anti-differentiated adjusting for the 5 seconds> to get the velocity function (call it v(t)), which gives me:>> v(t) = <-12.0, 10.3t - 51.5, -1.50t + 28.9>m/s>Achoo. Did you forget mathematicans are allergic to numbers?Its easier just to do it than to juggle all those itchy numbers.Hows my method?> I know this is right, because solving for v(25) gives me the answer> <-12.0, 206, -8.6>m/s which is what is in the back of the book. I> gure I repeat the process of anti-differentiating, using the answer> for v(0) and correcting for the 5 seconds again, since I have to> adjust the position function for the 5 seconds as well (I wasnt sure> if that was the case, but anti-differentiating just v(t) comes up with> an answer that is not even close to what the answer in the book is, so> they obviously intended the 5 seconds to apply to both the initial> position and initial velocity).>> Heres what my work looks like:>> v(0) = <-12, -51.5, 28.9>m/s> p(t) = <-12t+(134-(-12*5)), 5.15t^2-51.5t+(-25.1-(-51.5*5)),> -0.75t^2+28.9t+(124-(28.9*5))>m>> simplifying gives me:> p(t) = <-12t+194, 5.15t^2-51.5t+232.4, -0.75t^2+28.9t-20.5>m>> Using my answer above, I get:> p(25) = <-106, 2163.65, 233.25>m>> However, the answer in the back of the book is:> p(25) = <-106, 2035, 252>m>> I can tell Im on the right track, because the rst component> matches, but the other two, while close, dont actually match. I> gure I must be skipping a step that involves, for instance, the> 10.3t from the second component of the velocity function (the one that> says 10.3t - 51.5). Am I supposed to adjust for the ve seconds on> that part as well? And if so, how do I go about doing that exactly?> than himself.> === A fellow teacher asked me this, and I cant see any algebraic approach. A numerical solution for the real root is easy enough, and a Mathematica front-end at http://mss.math.vanderbilt.edu/~pscrooke/MSS/ solvepoly.htmlgives me the four non-real roots. But is there any analytic solution method?Before you spend too much time on this, the problem is from Larson-Hostetler-Edwards /College Algebra with Trigonometry/. If its not just a misprint, any algebraic solution would have to be within the reach of students taking pre-calculus algebra.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/Walrus meat as a diet is less repulsive than seal. === > A fellow teacher asked me this, and I cant see any algebraic> approach. A numerical solution for the real root is easy enough, and> a Mathematica front-end at> http://mss.math.vanderbilt.edu/~pscrooke/MSS/solvepoly.html> gives me the four non-real roots. But is there any analytic solution> method?The Galois group of this equation is S_5. Thus although itszeroes are algebraic (by dention) they are not expressiblein radicals. To see the group is S_5, note that the polynomialis irreducible modulo 5 and splits as an irreducible cubicmultiplied by an irreducible quadratic modulo 3.Tus its Galois group contains elements of cycle structure 5and 3 2.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === > > A fellow teacher asked me this, and I cant see any algebraic> approach. A numerical solution for the real root is easy enough, and> a Mathematica front-end at> http://mss.math.vanderbilt.edu/~pscrooke/MSS/solvepoly.html> gives me the four non-real roots. But is there any analytic solution> method?For _analytic_ solutions seehttp://mathworld.wolfram.com/QuinticEquation.html andhttp://xxx.lanl.gov/abs/math.GM/0005026/.GC> Before you spend too much time on this, the problem is from Larson-> Hostetler-Edwards /College Algebra with Trigonometry/. If its not> just a misprint, any algebraic solution would have to be within the> reach of students taking pre-calculus algebra.Really?!> --> Stan Brown, Oak Road Systems, Cortland County, New York, USA> http://OakRoadSystems.com/> Walrus meat as a diet is less repulsive than seal. === Its a bit late, but I want to say THANKS to Robin and George for posting responses to my query. Ive passed them on to the teacher who originally asked me.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/You nd yourself amusing, Blackadder.I try not to y in the face of public opinion. === AO__.b9[CapitalAAcute]AFSAF.a6AHAO_[C urrency]kA.bd.bc[Currency]FH[Degree]dHAO_.a6][Hyphen ]D.be[Hyphen]P[Hyphen][Currency].beA.a6[Capi talThorn][Micro]oefH([DownExclamation]G.a6[CapitalThorn] ef[Copyright].a6U[Thorn]f)AO_.a6][Hyphen]D.a6.b3y. a6.a6U[Hyphen][PlusMinus].a6[LeftGuillemet]HHAQd.bdG [Currency]UH http://et.akkf.com