> ==== > 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 field F it appears, F(u,v) can't be reduced to F(w) for some w. ==== > > I'm trying to find integer coefficients 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 algebraic integers 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 integers a 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 shows you how to relate the matrix to the minimal polynomial. - Randy ==== > Thus for field F it appears, F(u,v) can't be reduced to F(w) for some w. There is a theorem that if F is a field 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)) isn't isomorphic to Q(sqrt(3)). -- Jim Heckman ==== >> Thus for field F it appears, F(u,v) can't be reduced to F(w) >There is a theorem that if F is a field 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? Let's consider u = sqr 2, v = sqr 3, n = degree of w, F = Reals Who's 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)) isn't isomorphic to Q(sqrt(3)). > Nifty. ==== > > >> Thus for field F it appears, F(u,v) can't be reduced to F(w) > >There is a theorem that if F is a field 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? Let's consider > u = sqr 2, v = sqr 3, n = degree of w, F = Reals > Who's the w with R(w) = R(u,v) ? > What's 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 just any old element WILL do. Was that your point? ==== > >> Thus for field F it appears, F(u,v) can't be reduced to F(w) > >There is a theorem that if F is a field 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? Let's consider > u = sqr 2, v = sqr 3, n = degree of w, F = Reals > Who's the w with R(w) = R(u,v) ? What's 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? How complicated is the proof? ==== > >> Thus for field F it appears, F(u,v) can't be reduced to F(w) >There is a theorem that if F is a field 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? Let's consider > u = sqr 2, v = sqr 3, n = degree of w, F = Reals >Who's the w with R(w) = R(u,v) ? > >What's 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? > > It's in the chapter on Galois theory. If the book is conversational, it's easy. If it's formal, it could be pretty hard to understand. (YMMV, some people prefer formal to conversational.) It's not complicated at all, it's straightforward and usually done right where it makes sense. It's just that some books are hard to follow because of the way they're written. (Or the way I process what I read.) I like Van der Waerden, _Algebra_ or _Modern Algebra_, depending on your version. I'm 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 field F it appears, F(u,v) can't be reduced to F(w) > >There is a theorem that if F is a field 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? Let's consider > u = sqr 2, v = sqr 3, n = degree of w, F = Reals > Who's the w with R(w) = R(u,v) ? (Let's 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, +31205924131 home: 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 of a primitive element for any finite separable extension in those fields. ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== [in re: the Primitive Element Theorem,] > ...any algebra book that covers Galois theory will do. I think the exception here is Artin's 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)) isn't 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) it's (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)) isn't 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) it's > (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 Artin's 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 you're talking about Michael Artin's _Algebra_, he does indeed prove the Primitive Element Theorem. In fact, he then goes on to use it in his proof that For any finite extension K/F, the order |G(K/F)| of the Galois group divides the degree [K:F] of the extension. One thing I find somewhat disconcerting about Artin's exposition of Galois Theory is that he proves several important theorems out of order, i.e., he states them early on, proving them only in later sections. To my mind this makes following the logical development 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 Artin's 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 you're talking about Michael Artin's _Algebra_, No, he's talking about Emil Artin's 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 Artin's 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 field is generated by a single element in his book Galois Theory. It appears on page 64 (of 82 pages) in section M called Simple Extensions. In this section, Artin gives two theorems. The first theorem gives a necessary and sufficient condition for the existence of primitive elements: Theorem 26. A finite extension E of F is primitive over F if and only if there are only a finite number of intermediate fields. 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 polynomials for sqrt(2) and sqrt(3) do not have repeated roots. I have not yet grasped the impact that separable elements has in Galois theory. I just usually assume that I am working over extensions of Q, where separable follows because the characteristic of Q is 0. To find a couterexample to having a primitive element, I think that you need to have a finite extension E of F, where F is an infinite field of characteristic p. But, I haven't 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 Artin's 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 field is generated by a single element in his book Galois Theory. It appears on page 64 (of 82 pages) in section M called Simple Extensions. In this section, Artin gives two theorems. The first theorem gives a necessary and sufficient condition for the existence of primitive elements: Theorem 26. A finite extension E of F is primitive over F if and only if there are only a finite number of intermediate fields. 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 polynomials for sqrt(2) and sqrt(3) do not have repeated roots. I have not yet grasped the impact that separable elements has in Galois theory. I just usually assume that I am working over extensions of Q, where separable follows because the characteristic of Q is 0. To find a couterexample to having a primitive element, I think that you need to have a non-finite extension E of Z_p, the finite field of order p. But, I haven't pursued this. -- Bill Hale ==== 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 reflect 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 floor 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 ==== > 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 reflect 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 floor 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 G'day Doug, Perhaps too late at night, let's see if I can still get this right (too early in the day? ) x^2 = h^2 - y^2 doesn't 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 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 you're learning rules instead of math. y proportional to x when there's 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. ==== >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 you're describing _rates_of_change_, but not when you're describing relations of the original variables. In this case the constant h is (by definition 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 find yourself amusing, Blackadder. I try not to fly in the face of public opinion. ==== Julian Mattay typed: > Perhaps too late at night, let's see if I can still get this right (too > early in the day? ) > > x^2 = h^2 - y^2 doesn't 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 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 defines clearly the proportional relation of _x_ and _y_. -- Ayaz Ahmed Khan Yours 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 defines 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 ==== exercise. I'm 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 which y (the height) is decreasing when h is at say 2 o'clock can be found using the following method. ---------------------- In a right angled triangle YHX where the opposite sides are labelled yhx, then at 2 o'clock angle Y=30 at the centre, H=90 near 3 on the clock and X=60 near 2 on the clock. y/sin y = h/sin h y = (hsin y)/sin h y = 1 x = sqrt(h^2 - y^2) = sqrt(3) Y is changing at the rate of (pi/30)/second so dY/dt = pi/30 We know that y = (hsin y)/sin H and since h and sin H are constant at 2 and 1 respectively y = 2sin Y and dy/dY = 2cos Y To find 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 o'clock dy/dt = 2cos(pi/6)(pi/30) = 0.1814 y is decreasing at a rate of 0.1814 cm/s ---------------------- To find the rate at which x is increasing dx/dt = (dx/dy)(dy/dt) We know x = sqrt(h^2 - y^2) dx/dy = -y/[sqrt(h^2 - y^2)] So the rate at which x is increasing is dx/dt = -[y/(sqrt(h^2 - y^2))](0.1814) = -0.1047 x is increasing at a rate of 0.1047 cm/s. ---------------------- x is not proportional to y because the fraction x/y should remain unchanged 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 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 > you're describing _rates_of_change_, but not when you're 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 we're trying to discern. For example: 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. In this case the constant h is (by definition of a right triangle) > larger than either x or y. x and y are never proportional for > constant h. I must say that I don't 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 doesn't change as the other variables change. -- > Stan Brown, Oak Road Systems, Cortland County, New York, USA > http://OakRoadSystems.com/ > You find yourself amusing, Blackadder. > I try not to fly in the face of public opinion. Doug ==== 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 significant amount, the constant added on would start to become negligable, and thus allow for what I'm 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 you're learning rules instead of math. > y proportional to x when there's 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 the G'Day), I'll assume you're 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. I'm guessing that you're 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=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 find the derivative of d/dy...this time you'll get...(and I'll 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 reflect 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 floor 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 G'day Doug, Perhaps too late at night, let's see if I can still get this right (too > early in the day? ) x^2 = h^2 - y^2 doesn't 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 don't think this holds either...go over the example I just put up to Julian Mattay's comment and tell me if I'm still lost in the woods on this? Thanx doug > 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 defines 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)) = constant and now that I think about it, what Rich said makes complete and utter sense with respect to proportionality equations having constants when ones change is compared to another. And I know that taking the derivative of d/dy or d/dx will not yield a constant for y^2 + x^2 = h^2, but why? Simply because we have a constant that is being added to both? Doug > So seeing how you were waking up when me was hitting the sack, (and by the > G'Day), I'll assume > you're 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. > I'm guessing that you're 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=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 find the derivative of d/dy...this time you'll get...(and I'll > 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 reflect 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 floor 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 > > G'day Doug, > > Perhaps too late at night, let's see if I can still get this right (too > early in the day? ) > > x^2 = h^2 - y^2 doesn't 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 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 sufficient* 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 not working out to be proportional, what's bothering me is that I can't see why the wouldn't be proportional. The answer seems to be simply because there's a constant that is added on to one side (the h), and this seems to through everything off balance. With respect to the simple example you gave below (y = mx +b) you see, y and x ARE proportional if b = 0. Meaning, if we transform the cartesian plane by making sure that our segment goes through the origin, we'll be fine, and able to calculate the proportionality of x w.r.t y. And this further clerifies (and perhaps finally makes me see) why in the pythagoras eqtn x and y are not proportional....it seems, you can't simply eliminate constants that aren't related via a multiplication to the variable 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 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 > sufficient* 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 >> you're describing _rates_of_change_, but not when you're 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 we're 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 it's not proportional to r. >> In this case the constant h is (by definition of a right triangle) >> larger than either x or y. x and y are never proportional for >> constant h. I must say that I don't 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 doesn't 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 find yourself amusing, Blackadder. I try not to fly 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 isn't proportional to x, it's 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 significant amount, the > constant added on would start > to become negligable, and thus allow for what I'm 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 you're learning rules instead of math. > y proportional to x when there's 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 finding with the help of the given equation the values of _y_, I find that _x_ and _y_ and indeed proportional, but only inversely. -- Ayaz Ahmed Khan Yours Forever in, Cyberspace. ==== >Upon feeding arbitrary values of _x_ in the equation, and finding with >the help of the given equation the values of _y_, I find 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 there's 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 find yourself amusing, Blackadder. I try not to fly in the face of public opinion. ==== I'm trying to solve an algebra problem and I came up with four equations (and some side conditions) that I think can be condensed. Here goes: Let G be an abelian group, finite, 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. I've shown that blahblahblahblahblah 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. Here's what I've 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 can't 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 proficient with Mathematica want to give this one a whirl and post any simplifications you discover? (I don't have access to it or any other symbolic algebra package from where I currently work). ==== Let G be an abelian group, finite, 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. I've shown that > I'll presume G is additive as it's Abelian group. H is subgroup. What's 'c'? Some integer? What's 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 > What's that? Bush's 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. Here's what I've 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 can't 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 > proficient with Mathematica want to give this one a whirl and post any > simplifications you discover? (I don't 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 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 + b' can be written since there's no slope as the slope would be undefined. My teacher won't show the steps to this as he says he's not giving it for the test but he told me that m is undefined and should be treated as 0 in this type of problem. That's ok that he did not want to work this out on the board but I'm 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 undefined. This is why he did not want to work this out on the board. I understand how to do these problems when there's 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 undefined? For (a) parallel m = m2 (m2 is undefined) 1. y - y1 = m(x - x1) where m is undefined... 2. y - y1 / m = x - x1 ==> (y - 5) / m = x - 2 where m is undefined -or- infinity. 3. Since anything over infinity = 0 (according to my instructor) , then (y - 5)/{infinity} = 0 4. So rewrite equation as 0 = x - 2 5. x = 2 For (b) parallel m = -1 / m2 (m2 is undefined) 1. y - y1 = m(x - x1) where m is undefined therefore m(x - x1) (according to my instructor) would be rewritten as 0 in this case so go to step 2 2. y - y1 = 0 3. y - 5 = 0 4. y = 5 thanks 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 + b' can > be written since there's no slope as the slope would be undefined. Correct. > My > teacher won't show the steps to this as he says he's not giving it for the > test but he told me that m is undefined and should be treated as 0 in > this type of problem. _Absolutely_ not. A slope of 0 and an undefined slope are two VERY different things. The two lines x=2 and y=5 are perpendicular. The first has undefined slope while the second has 0 slope. Looks like this teacher is the one who should be sitting in on this class. > That's ok that he did not want to work this out on > the board but I'm 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 undefined. 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 there's 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 undefined? Slope intercept form y=mx+b applies only when you *have* a slope. Vertical lines don't have a slope, 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? It's just not there... > For (a) parallel m = m2 (m2 is undefined) 1. y - y1 = m(x - x1) where m is undefined... The right side is therefore undefined as a result of m being undefined. ...and if the left side is equal to the right, then the lhs is also undefined. 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 defined slope. Here's 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 defined 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 undefined expression, so this line doesn't even have a slope, nor does any vertical line. You will get a similar undefined expression 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 since you will get a delta(y) of 0 and a delta(x) something nonzero. Next, he could have taken that same definition 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 defined 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 *defined*. With m being undefined, as in the case of the line x=5, you simply don't *get* to the form y=mx+b. Either your teacher is being extremely nonhelpful on purpose (perhaps he doesn't have time to go over such basics in a precalc class) or he flat out doesn't know what he's doing. Normaly, I would give the benefit 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. It's 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 flat out doesn't know what he is doing and shouldn't be allowed to teach mathematics, or at best he likes to lie alot. It's 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 undefined slope to be a 0 slope. > 2. y - y1 / m = x - x1 ==> (y - 5) / m = x - 2 where m is > undefined -or- infinity. > 3. Since anything over infinity = 0 (according to my instructor) , then > (y - 5)/{infinity} = 0 You say this is a precalc course? In such a context, infinity doesn't even _exist_, much less has a recipricol of 0. Anything over infinity is not a defined expression, at least not in this context (which is real numbers). It's only when you study alternative number systems (as opposed to the standard real and imaginary numbers) that you get into things like 1/infinity. > 4. So rewrite equation as 0 = x - 2 > 5. x = 2 For (b) parallel m = -1 / m2 (m2 is undefined) 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 above explanation I offered is more appropriate for your context, surely. Doesn't 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 else's class instead, assuming the act of audit is for your benefit. If the audit truly is an audit, as in let's see if anything's wrong here then report your findings. -- Darrell ==== > <... > For (b) parallel m = -1 / m2 (m2 is undefined) 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 of parallel with perpendicular. -- Darrell ==== [cc'd 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 there's no slope as the slope would be undefined. My >teacher won't show the steps to this as he says he's not giving it for the >test but he told me that m is undefined 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 undefined slope and that can _not_ be treated as 0. Here's how they were worked out. The line x = 4 has no slope (meaning that its slope is undefined, 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 find 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 undefined. So for the parallel line you need another line with undefined slope. The parallel line must go through given point (2, 5). Other points with a run of 0 from that first 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 undefined slope and the other has slope = 0. So any line perpendicular to x = 4 must have slope of 0 (since x=4 has undefined 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 = 5 But really, I tell my students it's 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 y's 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 x's 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 justified. 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 = 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 there's no slope as the slope would be undefined. Correct. > My > teacher won't show the steps to this as he says he's not giving it for > the test but he told me that m is undefined and should be treated as > 0 in this type of problem. _Absolutely_ not. A slope of 0 and an undefined slope are two VERY > different things. The two lines x=2 and y=5 are perpendicular. The > first has undefined slope while the second has 0 slope. Looks like this > teacher is the one who should be sitting in on this class. You'd be right, Darrell, _if_ that's what the teacher actually said. 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 undefined, treat its _reciprocal_ as being 0. > That's ok that he did not want to work this out on > the board but I'm 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 undefined. 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 there's 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 > undefined? Slope intercept form y=mx+b applies only when you *have* a slope. Specifically, a _finite_ slope. > Vertical lines don't have a slope, At least, not a finite 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? It's just > not there... Right. > For (a) parallel m = m2 (m2 is undefined) > > 1. y - y1 = m(x - x1) where m is undefined... The right side is therefore undefined as a result of m being undefined. > ...and if the left side is equal to the right, then the lhs is also > undefined. 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 defined slope. Rather, one could say the form applies only when m is finite. > Here's 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 defined 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 undefined expression, so this line doesn't even have > a slope, nor does any vertical line. You will get a similar undefined > expression regardless of which two points you select, for any vertical > line. Of course, that's undefined if you refuse to deal with an infinite value. But it seems that the teacher is willing to deal with an infinite value, in which case it is defined. BTW, IMO, the best answer to music's 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, it's much faster just to realize that a vertical line's equation can always be written as x = constant. But if you don't want to have to deal with a special case, then I think the form * is the best choice.] > 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 definition 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 defined 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 > *defined*. With m being undefined, as in the case of the line x=5, you > simply don't *get* to the form y=mx+b. Either your teacher is being extremely nonhelpful on purpose (perhaps he > doesn't have time to go over such basics in a precalc class) or he flat > out doesn't know what he's doing. Normaly, I would give the benefit 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. It's 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 flat out doesn't know what he is doing and > shouldn't be allowed to teach mathematics, or at best he likes to lie > alot. It's 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 > undefined slope to be a 0 slope. > 2. y - y1 / m = x - x1 ==> (y - 5) / m = x - 2 where m > is undefined -or- infinity. ??? Undefined and infinity are, in general, very different! However, in this context of slopes, what you've written is understandable. If we only allow real numbers, then a vertical line cannot have a defined slope. OTOH, if we allow an unsigned infinity in our system, then we do have a defined slope for vertical lines. > 3. Since anything over infinity = 0 (according to my instructor) , > then (y - 5)/{infinity} = 0 You say this is a precalc course? In such a context, infinity doesn't > even _exist_, If the instructor chooses to consider it, it does exist. [But of course, that does not address the question of whether the instructor's choice would then be pedagogically sound for students at that level.] > much less has a recipricol of 0. Anything over infinity > is not a defined expression, at least not in this context (which is real > numbers). If you _have_ something over infinity, then infinity is obviously involved, which means that the context _cannot_ be just the reals. And, yes, any nonzero quantity divided by infinity yields 0. > It's only when you study alternative number systems (as opposed > to the standard real and imaginary numbers) that you get into things like > 1/infinity. > 4. So rewrite equation as 0 = x - 2 > 5. x = 2 > > For (b) parallel m = -1 / m2 (m2 is undefined) 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 infinity and 1/(unsigned infinity) = 0, then we can state neatly that reciprocals. And, BTW, we can also state neatly that David Cantrell > The above explanation I offered is more appropriate for your context, > surely. Doesn't 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 > else's class instead, assuming the act of audit is for your benefit. > If the audit truly is an audit, as in let's see if anything's wrong > here then report your findings. ==== Your derogation of the instructor may not be justified. See below. Well we may disagree which is fine, but if he is telling them to treat a vertical line with slope 0 then there's a serious rpblem there. And it ain't 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 the problem > 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 there's no slope as the slope would be undefined. > > Correct. > > My > teacher won't show the steps to this as he says he's not giving it for > the test but he told me that m is undefined and should be treated as > 0 in this type of problem. > > _Absolutely_ not. A slope of 0 and an undefined slope are two VERY > different things. The two lines x=2 and y=5 are perpendicular. The > first has undefined slope while the second has 0 slope. Looks like this > teacher is the one who should be sitting in on this class. You'd be right, Darrell, _if_ that's what the teacher actually said. Well, it goes without saying, David. Of course, it that's not what was said then all bets are off. I stand by my remarks as written, since they obviously apply only when that's what the teacher actually *said.* if he actually said what was written, that m being undefined should be treated as the slope is 0, then something ain't 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 undefined, treat > its _reciprocal_ as being 0. Well, perhaps, being that he talks about 0 being the reciprocol of infinity. But rather than play guessing games as to what was actually said, I am simply responding to what was written. 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/infinity is 0 is at all appropriate. > That's ok that he did not want to work this out on > the board but I'm 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 undefined. > > 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 to have time to actually address this person's question directly in some standard fashion, which is a VERY legitimate concern BTW, but somehow he found the time to enter into discussions of infinity and 1/infinity=0 and such. and we wonder why so many people, when they take a calculus course, actually *believe* 1/infinity=0. > This is why he did not want to work this out on the board. > I understand how to do these problems when there's 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 > undefined? > > Slope intercept form y=mx+b applies only when you *have* a slope. Specifically, a _finite_ slope. > Vertical lines don't have a slope, At least, not a finite one. Note to OP-- these distinctions are valid, but not that important to your issue. A defined slope would fall into David's category of finite slope. There's just more than one way to say it, that's 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? It's just > not there... Right. > For (a) parallel m = m2 (m2 is undefined) > > 1. y - y1 = m(x - x1) where m is undefined... > > The right side is therefore undefined as a result of m being undefined. > ...and if the left side is equal to the right, then the lhs is also > undefined. 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 defined slope. Rather, one could say the form applies only when m is finite. Of course. > Here's 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 defined 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 undefined expression, so this line doesn't even have > a slope, nor does any vertical line. You will get a similar undefined > expression regardless of which two points you select, for any vertical > line. Of course, that's undefined if you refuse to deal with an infinite value. > But it seems that the teacher is willing to deal with an infinite value, in > which case it is defined. My point exactly. You may disagree, but I really don't think he should. This is a precalculus class, where they are reviewing basic algebra concerning lines and slope. BTW, IMO, the best answer to music's 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, it's much faster just to realize that a vertical line's > equation can always be written as x = constant. But if you don't 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 definition 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 defined 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 > *defined*. With m being undefined, as in the case of the line x=5, you > simply don't *get* to the form y=mx+b. > > Either your teacher is being extremely nonhelpful on purpose (perhaps he > doesn't have time to go over such basics in a precalc class) or he flat > out doesn't know what he's doing. Normaly, I would give the benefit 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. It's 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 flat out doesn't know what he is doing and > shouldn't be allowed to teach mathematics, or at best he likes to lie > alot. It's 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 > undefined slope to be a 0 slope. Again, just to clarify my position on your earler remarks, note the second word in this paragraph is if. > > 2. y - y1 / m = x - x1 ==> (y - 5) / m = x - 2 where m > is undefined -or- infinity. ??? Undefined and infinity are, in general, very different! However, in > this context of slopes, what you've written is understandable. If we only > allow real numbers, ...which of course, they do > then a vertical line cannot have a defined slope. OTOH, > if we allow an unsigned infinity in our system, then we do have a defined > slope for vertical lines. ...which is a nonstandard system, at least n the sense that they do not use only the standard real numbers. Are you sure what you infer may be the actual context of this particular person's class he is taking? I have serious doubts... > 3. Since anything over infinity = 0 (according to my instructor) , > then (y - 5)/{infinity} = 0 > > You say this is a precalc course? In such a context, infinity doesn't > even _exist_, If the instructor chooses to consider it, it does exist. I see. There's no way in hell he could *possibly* be wrong. >[But of course, > that does not address the question of whether the instructor's choice would > then be pedagogically sound for students at that level.] Translate: He was wrong. > much less has a recipricol of 0. Anything over infinity > is not a defined expression, at least not in this context (which is real > numbers). If you _have_ something over infinity, then infinity is obviously > involved, which means that the context _cannot_ be just the reals. And, > yes, any nonzero quantity divided by infinity yields 0. Well, no, any nonzero quantity divided by infinity 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. The simple and obvious conclusion is, the teacher simply failed to adequately (meaning appropriate for the scope of the class) address theis person's question. Instead, he chose to tell him things that were simply invalid _in such a contect_. > It's only when you study alternative number systems (as opposed > to the standard real and imaginary numbers) that you get into things like > 1/infinity. > > 4. So rewrite equation as 0 = x - 2 > 5. x = 2 > > For (b) parallel m = -1 / m2 (m2 is undefined) 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 infinity and 1/(unsigned infinity) = 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. -- 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/infinity is 0 is at all > appropriate. You're certainly entitled to your opinion as to what is pedagogically appropriate -- and so is the instructor. [snip] > and we wonder why so many people, when they take a calculus > course, actually *believe* 1/infinity=0. Well, it's _very good_ they believe it; it's true. I don't think I get your point. [snip] > Vertical lines don't have a slope, > > At least, not a finite one. Note to OP-- these distinctions are valid, but not that important to your > issue. A defined slope would fall into David's category of finite > slope. There's just more than one way to say it, that's all. Not quite. With the slope of a vertical line being unsigned infinity, the line has a defined slope. A defined slope need not be finite. [snip] > Are you sure what you infer may be > the actual context of this particular person's class he is taking? I > have serious doubts... I'd be almost certain that the text doesn't mention any such thing as an infinite slope, but the instructor apparently has mentioned such, thereby broadening the context. > 3. Since anything over infinity = 0 (according to my > instructor), then (y - 5)/{infinity} = 0 > > You say this is a precalc course? In such a context, infinity > doesn't even _exist_, > > If the instructor chooses to consider it, it does exist. I see. There's no way in hell he could *possibly* be wrong. Of course he could be wrong _pedagogically_, but there's certainly nothing wrong mathematically with, for example, saying 1/infinity = 0. > [But of course, that does not address the question of whether the > instructor's 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 don't think the instructor was wrong. > If you _have_ something over infinity, then infinity is obviously > involved, which means that the context _cannot_ be just the reals. And, > yes, any nonzero quantity divided by infinity yields 0. Well, no, any nonzero quantity divided by infinity 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 don't _need_ to, but IMO (and probably in that of the instructor) it's very nice to. Isn't it nice for _all_ lines to have slopes? But of course you're 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 person's 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 you're objecting to is actually valid. David > It's only when you study alternative number systems (as opposed > to the standard real and imaginary numbers) that you get into things > like 1/infinity. > > 4. So rewrite equation as 0 = x - 2 > 5. x = 2 > > For (b) parallel m = -1 / m2 (m2 is undefined) > > 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 infinity and 1/(unsigned > infinity) = 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 to your > issue. A defined slope would fall into David's category of finite > slope. There's just more than one way to say it, that's all. Not quite. With the slope of a vertical line being unsigned infinity, the > line has a defined slope. A defined slope need not be finite. David, I don't 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 undefined slope as a defined slope when he is reviewing basic algebra. It's either defied, or it's not. Because it may be defined as unsigned infinity in certain contexts (usually much later than a precalculus course) does not make it defined 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, 1/infinity 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 undefined 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 person's class he is taking? I > have serious doubts... I'd be almost certain that the text doesn't mention any such thing as an > infinite slope, but the instructor apparently has mentioned such, thereby > broadening the context. everything I've read, not the least of which is the OP's 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. that's 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 he's dealing with here, else he ain't an effective instructor. it's that simple. <...> -- Darrell ==== We need to end our dialog. I'll make just two brief comments below. > Note to OP-- these distinctions are valid, but not that important to > your issue. A defined slope would fall into David's category of > finite slope. There's just more than one way to say it, that's all. > > Not quite. With the slope of a vertical line being unsigned infinity, > the line has a defined slope. A defined slope need not be finite. David, I don't 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 > undefined slope as a > defined slope when he is reviewing basic algebra. It's either defied, or > it's not. Because it may be defined as unsigned infinity in certain > contexts (usually much later than a precalculus course) does not make it > defined 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! > 1/infinity 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 undefined 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 person's class he is taking? I > have serious doubts... > > I'd be almost certain that the text doesn't mention any such thing as > an infinite slope, but the instructor apparently has mentioned such, > thereby broadening the context. We know, David, but it doesn't make it *appropriate* necessarily. And of course I never said it was appropriate, from a pedagogical standpoint, for students at that level. David > everything I've read, not the least of which is the OP's 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. that's 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 > he's dealing with here, else he ain't an effective instructor. it's that > simple. ==== We need to end our dialog. I'll make just two brief comments below. ? Either we need to end it or we don't. If you feel we need to end it, then you should feel no urge to include more comments. If you include more comments, 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 the conversation. At least you're consistent, since an undefined slope is, well, defined. > > Note to OP-- these distinctions are valid, but not that important to > your issue. A defined slope would fall into David's category of > finite slope. There's just more than one way to say it, that's all. > > Not quite. With the slope of a vertical line being unsigned infinity, > the line has a defined slope. A defined slope need not be finite. > > David, I don't 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 > undefined slope as a > defined slope when he is reviewing basic algebra. It's either defied, or > it's not. Because it may be defined as unsigned infinity in certain > contexts (usually much later than a precalculus course) does not make it > defined 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! Huh? So you were _not_ the one that corrected my explanation that a vertical line has an undefined slope? So you were _not_ the one that kept on saying that if we accept unsigned infinity as the slope of a vertical line? David, this isn't James Harris you're talking to. I am not searching for nor do I enjoy a fruitless argument. You seem to completely understand that within the context of my remarks, my remarks are correct. Likewise, I believe you understand (you should, at least) that I feel your remarks are correct in the context in which you present them. Some way, can we get back to the context of the OP's question? Remember, just because his instructor delved into 1/infinity and the like, does not imply that he answered the OP's question in the context the OP presented it. Very clearly, the OP felt that a vertical line does not have a defined slope, therefore the equation of such line does not posses a slope-intercept form, despite the instructions in his book. The proper response to his question CONSIDERING THE CONTEXT IN WHICH IT WAS PRESENTED is not to somehow dance around the issue and present a context in which a vertical line HAS slope, but rather to simply explain to the OP what it is he is specifically asking for. That was done, and may I say in a matter consistent with most precalculus texts. You yourself even acknowledged that in all probability his book does not approach the issue the way you and his instructor has. Apparently you are of the opinion that whatever an instructor says, is appropriate BY DEFINITION, regardless of the cpontext in which the question was asked of him. he can know all the math in the world but unless he can get it across, he is worth nothing as a teacher. Undefined slope means zero slope? C'mon. Undefined slope means defined slope? C'mon. If his approach is so effective, then why is the OP asking a newsgroup the same question he asked his instructor? did the OP not later acknowledge that the responses he received from myself and another were helpful? Of course. I say that not looking for any glory, but simply to point out which was the more effective (and APPROPRIATE) answer to his question. > 1/infinity 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 undefined 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 person's class he is taking? I > have serious doubts... > > I'd be almost certain that the text doesn't mention any such thing as > an infinite slope, but the instructor apparently has mentioned such, > thereby broadening the context. > > We know, David, but it doesn't 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 not necessarily appropriate. I go so far as to say it was actually unappropriate and I have clearly justified 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 justified the reasons why. Wow, did I really say unappropriate? You're right, it must be time to end the conversation ;-). ==== We need to end our dialog. I'll make just two brief comments below. >>Note to OP-- these distinctions are valid, but not that important to >>your issue. A defined slope would fall into David's category of >>finite slope. There's just more than one way to say it, that's all. >>>Not quite. With the slope of a vertical line being unsigned infinity, >the line has a defined slope. A defined slope need not be finite. >David, I don't 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 >> undefined slope as a >>defined slope when he is reviewing basic algebra. It's either defied, or >>it's not. Because it may be defined as unsigned infinity in certain >>contexts (usually much later than a precalculus course) does not make it >>defined 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/infinity 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 undefined 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 person's class he is taking? I >>have serious doubts... >>>I'd be almost certain that the text doesn't mention any such thing as >an infinite slope, but the instructor apparently has mentioned such, >thereby broadening the context. >We know, David, but it doesn't 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. There's 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 isn't any help at all. But when they're 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 infinity. It seems to be a magnet to the human imagination. There's no reason for us to shy away from it, including the fact that we can't actually *grasp* infinity, so we have to make some assumptions about it. But as long as we make those assumptions explicit, then there's 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 office, we can't take up class time for this. Jon Miller ==== > The line x = 2 has a slope of 0; >the line y = 5 has undefined 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 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 undefined; the horizontal change of zero gives a useless(?) divisor, so the slope is undefined. 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), the vertical change, change in y, is zero. Here, slope is (0/anyRealNumber), which is equal to zero. In the simple, crudest language, no slope means horizontally flat, no up and down as you move accross. As I say, I did not read the rest of your message; I just comment on the stuff quoted above. G C ==== > The line x = 2 has a slope of 0; >the line y = 5 has undefined 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 undefined; the horizontal change of zero gives a useless(?) divisor, > so the slope is undefined. Yes indeed, assuming that we're 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 nonzero real 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 flat, no up > and down as you move accross. That is so crude that it is simply incorrect. I assume that you're dealing with only real numbers for slopes. If so, then no slope means vertical, rather than horizontal. By contrast, slope of 0 means horizontal. There's a huge difference between having no slope and having a slope which happens 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 undefined 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 undefined (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, I'll add my pittance. First, in regard to OP's 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 like a perfectly valid slope-(x-)intercept form. Likewise, for the line through (2, 5) parallel to the x-axis, y = 5 seems fine for the slope-(y-)intercept form. Now, for the oo (= infinity sysmbol). Topologically, there is no problem with R / { oo }. It is the one point compactification 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? If so, then oo = 1 if you still want mutiplicative inverses. 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 can't think of _any_ algebra that works as one would like it to 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 If a*b = a*c then b = c if a =/= 0 but with oo the qualifier would have to be used virtually _all_ the time. 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 defined, oo is not considered 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 L'Hopital's rule) should, IMHO, be viewed as macros; the oo by itself is meaningless. -- Paul Sperry Columbia, SC (USA) ==== > With some hesitation, I'll add my pittance. I'll add a few comments -- without any hesitation. (Fools rush in where angels fear to tread.) > First, in regard to OP's 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 one's perspective. I would expect that some people would balk, claiming that A line without a slope couldn't possibly have a valid slope-anything form. I, OTOH, would suggest that, taking the slope of vertical lines to be unsigned infinity, we have y = m*x + b, slope-(y-intercept) form, valid for all nonvertical lines and symmetrically x = 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 fine for the slope-(y-)intercept form. Now, for the oo (= infinity sysmbol). Topologically, there is no problem with R / { oo }. It is the one > point compactification 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 similar big time problems. Merely replacing oo with 0 in what you said above gives: For example: 0 * 0 = 0? If so, then 0 = 1 if you still want mutiplicative inverses. > oo - oo = ? Right. And similarly 0/0 = ? [Actually, in the computer algebra system Derive, that's precisely what you get: 0/0 simplifies 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 compactification of R. Then certainly x/oo = 0 for all finite x, and x/0 = oo for all nonzero x. > But then (1/0)*0 = ? Right. That's normally taken to be undefined. > In fact, I can't think of _any_ algebra that works as one would like it > to if oo is included. And I can't think of _any_ algebra that works as one would like it to if 0 is included either. (But then we've grow accustomed to dealing with zero's 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 qualifier would have to be used virtually _all_ the time. Hmm. virtually _all_ the time? I don't 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 defined, oo is not > considered as any kind of number: it screws things up. Some mathematicians (and computer scientists) do consider it to be a number. Zero screws things up too, and for a long time was not regarded as a number either. But whether something is called a number or not is not really important, at least IMO. > Terminology like lim(f(x), x -> oo) or lim(f(x), x -> a) = oo or oo/oo > (for L'Hopital's rule) should, IMHO, be viewed as macros; the oo by > itself is meaningless. ??? It _can_ be given precise meaning. If you're 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. 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 L'Hopital's 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 *isn't* given precise meaning, OK? I wish for one moment you would pretend--just oo was given precise meaning. Please put things into proper context. Clearly, in the proper context, Paul's above statement should be as deserving of a ??? response about the same as a claim of a vertical line having no defined slope should be deserving of the same response. ...but yet, you seem to insist without hesitation that we should just go ahead and use infinity as though it was a well defined mathematical object, saying things like 1/oo=0 and the like. 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 defined slope: a. The slope of a vertical line is infinite (which *is* defined), 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 undefined, the slope of a horizontal line is 0, therefore the rule that perpendicular lines have negative recipricol slopes doesn't apply for vertical/horizontal lines. My money's on b. -- Darrell ==== If anyone else wants the last word, they are welcome to it. [...] > First, in regard to OP's 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 one's > perspective. I would expect that some people would balk, claiming that A > line without a slope couldn't possibly have a valid slope-anything form. Who says x = 2 doesn't have a slope? Maybe I choose to draw the y-axis horizontally 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 compactification of > R. Then certainly Certainly? Sez who? Not the topologists AFIK. > x/oo = 0 for all finite x, and > x/0 = oo for all nonzero x. > > But then (1/0)*0 = ? > > Right. That's normally taken to be undefined. > In fact, I can't think of _any_ algebra that works as one would like it > to if oo is included. > > And I can't 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 field. > (But then we've grow accustomed to dealing with > zero's 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 qualifier would have to be used virtually _all_ the time. > > Hmm. virtually _all_ the time? I don't think I agree. I guess we could (but won't) 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 horizontal line 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 is the only point on the line?) Adding fractions is really easy: 1/2 + 1/3 = (1*oo)/(2*oo) + (1*oo)/(3*oo) = undefined? Do you recommend that algebra refer to the Real Circle rather than the Real Line? [...] > Some mathematicians (and computer scientists) do consider it [ oo ] to be a > number. Sure, topologists and there's no accounting for computer scientists :-) [...] > Terminology like lim(f(x), x -> oo) or lim(f(x), x -> a) = oo or oo/oo > (for L'Hopital's rule) should, IMHO, be viewed as macros; the oo by > itself is meaningless. > > ??? It _can_ be given precise meaning. If you're 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 Sperry Columbia, 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 compactification > of R. Then certainly Certainly? Sez who? Not the topologists AFIK. Surely anyone who's ever considered arithmetic on R*. The two statements below are also valid, of course, in C*, and you should be able to find them in 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 finite x, and > x/0 = oo for all nonzero x. [...] > In fact, I can't think of _any_ algebra that works as one would like > it to if oo is included. > > And I can't 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 field. True, of course. But then it depends on what your priorities are. IMO, wheels are pretty nice too. [Of course, I'm referring to the algebraic structure 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, there's just one solution of course. But in R* (or C*), there are two. > 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 undefined if x = oo, much as x/x would normally be considered undefined if x = 0 or oo. [But in a wheel, these would all be defined. But that's 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 I'd covered that matter earlier: y = m*x + b, slope-(y-intercept) form, valid for all nonvertical lines x = 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) = undefined? Adding fractions is easy, of course. But the example is just as wrong as saying 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. I've also seen it called the Riemann circle. > Terminology like lim(f(x), x -> oo) or lim(f(x), x -> a) = oo or > oo/oo (for L'Hopital's rule) should, IMHO, be viewed as macros; the > oo by itself is meaningless. > > ??? It _can_ be given precise meaning. If you're 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 L'Hopital's 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 > *isn't* given precise meaning, OK? I certainly agree with that. Indeed, even once students have finished the basic calculus sequence, they don't know how to give precise meaning to something as simple as 1/2 + 1/3. [Related to this, you might enjoy the Preface to the Student in Landau's classic text _Foundations of Analysis_. At one point he says I speak only of such numbers as you have already dealt with in high school. ... Please forget what you have learned in school; you haven't learned it.] > I wish for one moment you would > paper, where oo was given precise meaning. OK, fine. I just did as you asked. That moment has now passed. (And I must now wonder what the purpose of that little exercise was.) > Please put things into proper context. Clearly, in the proper context, > Paul's above statement should be as deserving of a ??? response about > the same as a claim of a vertical line having no defined slope should be > deserving of the same response. ...but yet, you seem to insist without > hesitation that we should just go ahead and use infinity as though it was > a well defined mathematical object, saying things like 1/oo=0 and the > like. I have never _insisted_ that one _should_ do such. Your misrepresentations (inadvertent though I'm sure they are) of what I have said are very annoying. That's why I had tried to stop our dialog earlier in this thread. I hope that readers will read what I've said, rather than what you've said that I've said! Please feel free to continue to respond, Darrell. But if you do, as far as 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 defined slope: a. The slope of a vertical line is infinite (which *is* defined), 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 undefined, the slope of a horizontal > line is 0, therefore the rule that perpendicular lines have negative > recipricol slopes doesn't apply for vertical/horizontal lines. My money's on b. Of course. I'd wager that more than 90 of them would choose b. I wouldn't be 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 confidently on the subject. Radiation Protection Office, Madigan Army Medical Center who posted (which is archived at Vanderbilt). Now the question is: Are you the same James Harris who later posted to RADSAFE as James.Harris@rfets.gov, K-H Manager, 771 Radiological Safety in February 2001? If so, I'd *love* to hear all about the safety violations in Building 771 (including workers being fined $385,000 a few months after you posted that message. -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're 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 definition 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 flip a coin several times, each time you flip it is a trial, and each trial is independent of the other trials. If you were to flip two coins, the probability of getting two heads would be equal to the probability of getting a head on the first 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. - Stephen http://pages.prodigy.net/stephenmorais ==== Okay, I will make the assumption the Mr. Harris truly wants to learn about this field. I understand the history, here, but will try to make no assumptions this once. Wile's 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. 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, unramified outside 2p, and flat at p. These are four very unique properties. 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 unramified at all sufficiently large primes. And this can be generalized, so that we may call Galois representations modular if they are unramified 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 field that is the base of the representation. 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. 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. 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 fields and, in particular, the theory required to understand the proofs (elliptic curves, Galois representations, and modular 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. > > > I'm 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 that's 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? Here's a way to turn Wiles' proof into a proof in PA. Take a sufficiently strong, but crippled system (say an ad hoc subset of ZFC chosen so that only the things used in the proof are included) that suffices 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 Fermat's last theorem should be true. Of course, if Fermat's 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 codification of arithmetic and that all undecidable propositions are really higher order statements suitably coded. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) 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... I'm sure that regardless of how James Harris responds, there will be others who appreciate your effort. It might be worth pointing out that there is a somewhat more elementary way to state the modularity conjecture, that doesn't require understanding all that jazz about Galois representations. For every positive integer N, let Gamma_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 sending the complex number z to (az+b)/(cz+d). If we now take the quotient of the upper half plane by this group action and then compactify suitably (glossing over some details here ...not difficult, 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 out that there is a natural way to give it the structure of an algebraic curve over Q, the rationals (glossing over more details here...this time a little more substantial). Then the modularity conjecture (now theorem) says that given any elliptic curve E over Q, there exists an integer N such that there exists a nonconstant morphism (of algebraic curves, defined over Q) from X_0(N) to E. Modular forms live naturally on things like X_0(N). They have a lot of structure that lets you compute the coefficients of their Fourier expansions. Roughly speaking, the modularity conjecture carries this structure over to E and among other things tells you that these Fourier coefficients count solutions to certain polynomial equations that may not be easy to analyze otherwise. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences ==== let a billion flowers bloom! (I don't think that it should become a requirement for 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 coefficients of their Fourier > expansions. Roughly speaking, the modularity conjecture carries this > structure over to E and among other things tells you that these Fourier > coefficients count solutions to certain polynomial equations that may > not be easy to analyze otherwise. --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... For a 1000-year anglo-american hegemony? HEY, JIMMY; LET'S 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/files/curriculum/Cosmo.PCX ==== > Okay, I will make the assumption the Mr. Harris truly wants to learn > about this field. I understand the history, here, but will try to > make no assumptions this once. I've read expositions about Wiles's work before, and even glanced at the actual papers. What readers should notice is the verification below of my saying that 4 descriptors are used that are the link between elliptic curves and modular forms. 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 Wiles's mistake is so clear that there's simply no correcting for it. 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 defines how they are in some sense both the same thing. That's the identity relationship. 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. Wiles's work requires a superset, and as given would need to apply to objects with the 4 descriptors. Um, that's an infinite number of objects other than elliptic curves and modular forms, and there is no effort made from what I've seen to show a more defined set. For instance, many of you have 4 fingers on your hand. Write certain numbers on each finger, and your fingers are included in the set that Wiles's work covers. > Wile's 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 I've 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. I'd 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, > unramified outside 2p, and flat 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 Fermat's 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 unramified at all > sufficiently large primes. And this can be generalized, so that we > may call Galois representations modular if they are unramified > 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 field > that is the base of the representation. What makes that fascinating is that it's possible that internally Wiles's 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 flying past you. Some of you labor under the misapprehension that knowledge proves truth. But say I know what the length of the Nile river is? So what? Mathematicians can rattle off all kinds of knowledge which in context is as useless as the length of the Nile river generally is. I mean, it's neat to know the answer, like for the TV show Jeopardy, but a lot of the time, it doesn't 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 doesn't 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 isn't logically correct if you wish to prove that a limitation 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 fields and, in particular, the theory required > to understand the proofs (elliptic curves, Galois representations, and > modular forms). The other fascinating thing I find 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. 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. Don't worry, unlike the work of Wiles, we know a lot of things work because they connect to the real world, like thermodynamic theory and cars. Now a lot of pure mathematics may be total crap, but for most of you it probably won't 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 the truth as simply as I can and face people who get away with stating things that are false using really complicated stuff. The mathematical truth doesn't change day-to-day people. Wiles's argument was wrong when he presented it, and it's wrong today. It'll be wrong tomorrow, as well. And mathematicians can write reams of technical jargon which won't change the truth. And yes, I do feel a certain sense of satisfaction at being someone who actually *believes* in logic. James Harris ==== >Wiles's 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 a curve. I recognize that I have not referred to the 4 descriptors or used the exact term modular form. That's because neither of these is needed in order to address the specific question you are asking. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences ==== >Wiles's 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. That's the result he *wanted* if I'm to accept that your claim is correct. But it's meaningless unless a modular form is a modular curve. The reader should note that this poster did not define modular curve. I'm assuming that he means the result that every elliptic curve is modular. Unfortunately I've noticed a tendency within the math community to react to correct challenges by replying with unsupported statements. My assessment is that when faced with the truth, as they can't refute the 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 elliptic curve would exist that's 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. Like I said in the post to which you're replying, if you have 4 fingers, you can put your fingers in the SAME set by writing certain numbers on them. And like I said in my post to which you're replying I think some readers who may think the mathematicians *must* have something would get sick if they actually saw some of those numbers. Why doesn't some mathematician post an example giving the 4 descriptors? > I recognize that I have not referred to the 4 descriptors or used the > exact term modular form. That's because neither of these is needed > in order to address the specific question you are asking. 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. Debates won't change logic. Note: Part of the reason I included the newsgroup sci.logic is to emphasize that point as I've noticed posters on math newsgroups to so easily switch to non-logic statements and get general *agreement* from math newsgroups like sci.math that it made *me* sick. James Harris ==== >> Every elliptic curve is a modular curve. But it's meaningless unless a modular form is a modular curve. No. If you've gotten the impression that Wiles claimed that elliptic curves and modular forms are the same type of object, then either you or the account you've read is mistaken. The simplest definition 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 doesn't make sense to say that they are the same thing. But the fault lies not with Wiles or other establishment mathematicians, since they haven't claimed that elliptic curves and modular forms are the same thing. >The reader should note that this poster did not define modular curve. >I'm assuming that he means the result that every elliptic curve is >modular. That's right. >Unfortunately I've 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 can provide them if necessary. I don't think it's relevant to the argument. >Remember Ribet purportedly proved that if FLT were false an elliptic >curve would exist that's 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. That's 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. Notice that even in the way you've 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. >Why doesn't 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 don't follow you here. If I want to prove that every integer is an algebraic integer, how do I do this without mentioning algebraic integers? -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences ==== nevermind. let's start another item! > But the fault lies not with Wiles or other establishment mathematicians, > since they haven't claimed that elliptic curves and modular forms are the > same thing. > That's 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 don't 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 d'Enron! http://quincy4board.homestead.com/ Funny.html (schoolboard stuffin') ==== > Okay, I will make the assumption the Mr. Harris truly wants to learn > about this field. I understand the history, here, but will try to > make no assumptions this once. I've read expositions about Wiles's work before, and even glanced at > the actual papers. What readers should notice is the verification > 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 the proof, does not mean you understand what they mean or even what they apply to (the Galois representations). And this is more historical result to Wiles' 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 Wiles's mistake is so clear that there's > simply no correcting for it. Wiles' proved (with the help of the previous results) that A has property x A does not have property x Therefore not A I 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 are attempting to prove and certain other well known mistakes mentioned on other threads) > 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 defines how they are in some sense both the > same thing. That's the identity relationship. Do you mean you cannot say that a particular algebraic number over a field cannot be mapped to an ideal of the ring of polynomials over that field? 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. Wiles's work requires a superset, and as given would need to apply to > objects with the 4 descriptors. Um, that's an infinite number of > objects other than elliptic curves and modular forms, and there is no > effort made from what I've seen to show a more defined set. There is a functor from the category of elliptic curves to that of Galois representations constructed thus: One constructs the Tate module of the groups of torsion points on the elliptic curve at a given prime (in the limit over the powers). The p-adic Galois representation is given by the mapping from the Galois group over Q to the general linear group over the Tate module on E. > For instance, many of you have 4 fingers on your hand. Write certain > numbers on each finger, and your fingers are included in the set that > Wiles's work covers. ??? Wiles work covers semistable elliptic curves. No matter how many fingers I hold up or write numbers on, I do not see semistable elliptic curves (maybe if I squint harder?). > Wile's 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 I've 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. I'd 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 you should 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, > unramified outside 2p, and flat 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 Fermat's Last Theorem. I could explain absolutely irreducible, odd, unramified at an ideal, and flat, but I would suggest spending some time with it yourself and coming back when you have difficulties. If you are questioning big words now, I have no idea how far I would have to go back to stop the ad hominem and 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 doesn't 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 isn't 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 prove consequences of those properties, then? 10? 100? Or is it impossible to prove properties of an object? The linking between object categories is done by a constructed functor, so we are talking only about properties on one 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 fields and, in particular, the theory required > to understand the proofs (elliptic curves, Galois representations, and > modular forms). The other fascinating thing I find 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 the benefit. Instead, you explored nothing. You are indeed either a troll (a bad one at that, since you have to keep posting to keep a thread alive) or seriously psychologically avoidant. Which is a shame, since looking at your website I see that you do appear to enjoy manipulating equations, and if you could just focus yourself on seeing where you were making mistakes (which others have pointed out in other threads), I am sure you could derive results on your own that are truly unique, as many do every day. I hope that you are a troll, and that you do real math somewhere behind your screen, and that this is just your way of letting out steam. The alternative is rather sad. Unfortunately, I won't be responding to any more of your threads. ==== >> Every elliptic curve is a modular curve. > >But it's meaningless unless a modular form is a modular curve. > > No. If you've gotten the impression that Wiles claimed that elliptic > curves and modular forms are the same type of object, then either you > or the account you've read is mistaken. The simplest definition 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 doesn't make sense to say that they > are the same thing. I didn't say that as I noted that your statement was meaningless unless 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 set where you include both of the other sets. For instance, Mustang convertibles are cars, and Volkswagen Bugs are cars. Cars are the inclusion set, or the identity set, or the superset, or the superclass, as interestingly enough there are any number of ways to describe. Cars as a set is the inclusion set because it includes both Mustang convertibles and Volkswagen Bugs. Cars as a set is the identity set because Mustang convertibles are cars, and Volkswagen Bugs are cars. It can be considered the superset because it's a higher order set, and similarly from object oriented thinking it's a superclass because it's a higher order class. My point is that Wiles needs a superset. > But the fault lies not with Wiles or other establishment mathematicians, > since they haven't claimed that elliptic curves and modular forms are the > same thing. I didn't say they had. But what mathematicians *have* to have is a superset. So far they have modular forms and elliptic curves, and having read further down I can now add Galois Representations. However, I still don't see mention of a superset. >The reader should note that this poster did not define modular curve. >I'm assuming that he means the result that every elliptic curve is >modular. > > That's right. I was making certain that modular curve isn't a phrase that might mean something other than curve that is modular, as mathematics can get quirky. The verification is that the obvious meaning is correct. I conclude then that modular curve does not necessarily mean modular elliptic curve. >Unfortunately I've 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 that comment: objects with the 4 descriptors. The conjecture that Wiles (partially) proved can be restated as follows. Every elliptic curve is a modular curve. My post was challenging that Wiles had a proof because he doesn't identify a superset, in reply you simply restated the claim, and added in the modular curve which had to be further defined, though I could have *assumed* what you meant. Let's say that I have a proof that the sun orbits the earth. Also let's say that world opinion is that it does, so I have that on my side, 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 the earth at a loci in its orbit. You reply that my statement is unsupported. The context is that given your disproof of the original statement, I can't simply restate the same claim in a different way, as that is an unsupported 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 don't think it's relevant to the argument. My point is, what superclass includes elliptic curves *and* modular forms. Just like there's a superclass that includes Mustang convertibles *and* Volkswagen bugs. The problem is that if you're not dealing with the superclass, if you notice something about one it doesn't necessarily apply to the other. For instance, let's say Volkswagen Bugs have single-side mirrors, and I notice Mustang convertibles with single-side mirrors. Now I imagine a Mustang convertible with dual-side mirrors, look at all the Volkswagen Bugs and find they are all single-side, and proclaim I've proven that this particular Mustang convertible doesn't exist!!! That's what Wiles's and Ribet's 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 that's 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. > > That's 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 4 descriptors, and you have the same problem. The logic is rather direct and absolutely, as always, rigid. > Notice that even in the way you've 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 THAT'S THE problem!!! >Why doesn't some mathematician post an example giving the 4 descriptors? > > Example of what? Of a Galois representation? Just give the 4 numbers, isn't 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 don't 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. It's all quite logical. Wiles needs to show a superset. Without it, he can't have a proof. How about this? Ask yourself, where's the identity relationship in Wiles's work? James Harris ==== >I was making certain that modular curve isn't 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 rather than modular curve since that's the more common term, and sometimes people use modular curve for something else. Let me use Weil curve from now on. So the (allegedly unsupported) claim is, Every elliptic curve 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 doesn't some mathematician post an example giving the 4 descriptors? >> Example of what? Of a Galois representation? >Just give the 4 numbers, isn't 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 an algebraic integer. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences ==== >That brings me to Wiles's proof > 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. >Wiles purportedly proves > Wiles actually, not purportedly, proved Fermat's 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. ==== >Wiles never finds 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, that's 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. >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? >Wiles set out to map infinity against infinity and claimed proof that >a limitation on modular forms was a limitation on elliptic curves. Nope, that's not what he set out to do. You were misinformed. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the 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, I've made a specific 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. In terms of your analogies, Wiles proved a statement of the form for every man there is a suit that fits 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 find 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 first kind (a man) there is at least one item of the second kind (a suit) bearing a certain relationship (fitting) to the first item. Replace the words [man, suit, fits] by [semistable elliptic curve, modular curve, modular parametrization] and you get exactly the statement proved by Wiles. > Wiles purportedly proves > Wiles actually, not purportedly, proved Fermat's 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 Wiles's work is what's 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 >Wiles's work to the real world. > of the real-world relevance or irrelevance of Wiles work or any other mathematics. 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. ==== >Wiles never finds 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, that's 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 is http://mathworld.wolfram.com/Taniyama-ShimuraConjecture.html and from that page I quote: In effect, the conjecture says that every rational elliptic curve is a modular form in disguise. A little further down it's stated: Equivalently, for every elliptic curve, there is a modular form with the same Dirichlet L-series. Note for readers: The Dirichlet L-series goes back to the 4 descriptors I've repeatedly mentioned. Besides even if what you claim is correct, it'd mean that modular forms are Weil curves. Is that your assertion? A good example for those who feel confused about the logical flaw I've Knowles with the context of having children or capable of having children. The superset in that context is the set of women, as women are the sex capable of having children. Notice that with the superset identified I don't have to talk talk about women and having children. For Wiles to have a proof, he needs a superset, and if he had a superset, you wouldn't have to talk about elliptic curves or modular forms, as you could talk about the superset. You could define the limitation on the superset, and it'd automatically 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 descriptors could be matched between objects that were members of these sets. Technically (or maybe grammarians can correct me) the commas don't belong for a subordinate clause beginning with that. >Wiles set out to map infinity against infinity and claimed proof that >a limitation on modular forms was a limitation on elliptic curves. > > Nope, that's not what he set out to do. You were misinformed. Well that goes back to your earlier claim, where you mention Weil curve. Continuing on, some readers may think that there's no way that mathematicians would go on for years making a claim that I can so easily show to be false as it is based on a logically flawed approach, but in my experience mathematicians live in a society that is very conforming as well as being strictly hierarchical. The following is speculation, but I think it outlines what may have happened. Wiles decided to prove Fermat's Last Theorem. Isolating himself he set out to find a way to show that the Taniyama-Shimura Conjecture was while letting his colleagues believe he was working on something else. That's a deception that's important in context. Assume that he really wanted to prove Fermat's Last Theorem, and consider that he was spending a lot of time by himself. Let's suppose that he had an approach and decided that he'd take it, and convinced himself that it'd work. My understanding is that Wiles was a well established and respected mathematician: well within the ranks of the mathematical elite. When his fellow colleagues within his circle heard what he was working on, and that he felt that he'd succeeded, they probably felt elation. Now let's suppose that slowly it sinks in that his path is logically flawed. To tell him would mean forcing him to realize that he'd thrown away all those years, and would probably humiliate him. He'd be like some crank, like so many others who'd thought they'd proven the great problem. So instead they look carefully and find a gap, which possibly would have let him out with some dignity. But Wiles gets a former student, and they work for a year to find a workaround. It's easier after so much time for mathematicians to give in, as their society is so against confrontation on such matters, as their social order would not handle one of their elite being so humbled. Now I have personal experience to add, as I contacted Ken Ribet a few years back when I had an unwise bet about one of my earlier flawed attempts at proving FLT (yeah, I lost the bet). He not only replied but 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 the bet. I suggest to you that there may have been more to it. Advanced Polynomial Factorization and he replied back with encouragement and some pointed questions. I replied back of course including the final version, but haven't heard from him since. What I've presented is speculation and circumstantial evidence based on limited contact that I've had with people most of you probably only read about, if you know about FLT, where you can read about Ken Ribet's paper, or Barry Mazur's role in the FLT saga. But the people you just read about, I've contacted, and in some cases, as I've mentioned, they've answered back. Who knows, maybe they *want* the truth to be known, but are trapped somewhat by circumstance and training. What might have began in a time of heavy emotion has continued now to a point where they may feel incapable of stopping it. I suggest to you the possibility that they were trapped in the wiles of Wiles. But that of course is speculation, and you may consider it wild speculation, which I understand. I'm merely trying to find some rational reason for a situation which is rather bizarre. For what there is no doubt about is that without a superset, Wiles does not have a proof of the Taniyam-Shimura Conjecture, and therefore, has not proven Fermat's 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 is not intended to be a strictly accurate mathematical statement, but is an informal 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 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 *doesn't* immediately imply that all elliptic curves are modular. It has to be combined with other theorems in a way that isn't 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. >Besides even if what you claim is correct, it'd 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 don't make that assertion; neither does Wiles make that assertion. It may follow from the claim that you are erroneously *attributing* to Wiles, but that's irrelevant. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the 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 *doesn't* immediately imply that > all elliptic curves are modular. It has to be combined with other > theorems in a way that isn't 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 (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... For a 1000-year anglo-american hegemony? HEY, JIMMY; LET'S 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/files/curriculum/Cosmo.PCX ==== sorry, for breaking my recent vow, again, but ... this reminds me of that British math-guy's absurd definition of leftwing and rightwing proofs. of course, although I believe Ribet, that Wiles did prove *le prochaine theorem de Fermat* (a-hem), I also am (currently!) incompetent to comprehend Ribet's founding work (or anyones else's, beyond the finding of Sophie G.) upon which it rests (his being the last in the chain of results that Wiles needed to use). actually, Wiles proof might really be considered to be rightwing, in the sense of Tory, in the sense that he may have been *encouraged* to undertake his closet-proving, for the greater glory of Cool Britannia (although possibly 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-open to 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 d'Exploitation de la Protocols des 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 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81 23 -- Le FIN d'HISTOIRE 24 -- L'ORDEUR du MONDE NOUVEAU 25 -- THYROID STORK !?! ==== [...] | Fair enough. In fact, perhaps I should have said Weil curve rather | than modular curve since that's the more common term, and sometimes | people use modular curve for something else. When I last regularly attended number theory seminars in the Boston area, in the 1990s, Weil curve was less common than modular curve. For one thing, the reason the name Weil was associated with it is that some people called the Taniyama-Shimura conjecture the Weil conjecture. As Serge Lang has amply pointed out, it doesn't make sense to attribute the conjecture solely or mainly to Weil. (I think he might say, at all, although I've read that Weil made some useful refinements on it relating to the level of the modular form and the conductor of the curve, or something like that.) I had the impression in fact that the term modular curve had largely supplanted Weil curve, and that the Taniyama-Shimura(-Weil) conjecture was typically being called the modularity conjecture, now of course a theorem. 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, I've made a specific 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. Let's 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 fits 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 find some superset(as opposed to discussing tailors, > clothing stores and so on). No. Something closer would be that every man must wear a suit that fits him. Now if you keep looking and notice that every man you see wears a suit that fits him, then decide that every man must wear a suit that fits him, you're making the mistake of Wiles. > All he needs to prove, logically speaking, is that for every item of the > first > kind (a man) there is at least one item of the second kind (a suit) > bearing a > certain relationship (fitting) to the first item. Replace the words > [man, suit, fits] > 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 various elliptic curves with modular forms. Taniyam and Shimura conjectured that every elliptic curve had a corresponding modular form. Ribet puportedly proved that if FLT were false there'd be an elliptic curve that did NOT have a corresponding modular form. Wiles did a comparison check where he tried to count out an infinite number of modular forms against an infinite number of elliptic curves. However, as I've been pointing out, you could associate an infinite number of elliptic curves with an infinite number of modular forms and still have an infinite number of elliptic curves that were not modular, if you don't have an inclusion set. Going back to your analogy, if you find that you have an infinite number of men with suits that fit, where they're *wearing* the suits that fit, you can't assume that if you find an infinite number wearing fitting suits that there can't be some oddball somewhere, like Kramer from Seinfeld, wearing an ill-fitting suit. The logical fallacy is simple: Association is not a proof of a condition. 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 Fermat's 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 Wiles's work is what's 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 >Wiles's 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 the truth of real world phenomena is dependent on the diligence of people checking. As Wiles's work rests on the acceptance by a relatively small group of people who *supposedly* have checked it thoroughly so that there can't 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 you're extraordinarily naive, or you think the readers of sci.logic are. Wiles's supposed accomplishment rests *solely* on the assertion of a relatively small group of people that his work is correct. 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. Since that time from lasers to spectral analysis the theory has fit with reality. What I want readers to appreciate is that mathematicians and people who support them can be as illogical as any true believers. My assertion that a restriction shared between members of disparate sets requires an inclusion set is not refutable. Therefore, Wiles's work is logically flawed at its outset, and its length 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 comes from 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 *doesn't* immediately imply that > all elliptic curves are modular. It has to be combined with other > theorems in a way that isn't 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 that's fascinating given what else is on that webpage, which I'm calling the smoking gun. >Besides even if what you claim is correct, it'd 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 don't make that assertion; neither > does Wiles make that assertion. It may follow from the claim that you > are erroneously *attributing* to Wiles, but that's irrelevant. Hey, I go by what mathematicians say in talking about Wiles's work, and here's a pertinent quote from the webpage I gave before: 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. (Source http://mathworld.wolfram.com/Taniyama-ShimuraConjecture.html ) My assessment is that Wiles commits the logical fallacy of Cum hoc ergo propter hoc. (Source http://users.tru.eastlink.ca/~brsears/reafault.htm ) My suggestion for readers is that they look back over comments by mathematicians about Wiles's work, and pay attention to things like Wiles's deception about what he was doing for all those years. Why would Wiles deceive his colleagues? Why haven't 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 flawed 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 flawed approach might work? James Harris You're certainly more than qualified to answer that question yourself, James Harris. Do yourself a favor and stop flaunting 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 ==== >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 you're 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 flawed. It's illegitimate to fault Wiles's argument on the basis of secondary sources. If you think there is something wrong with Wiles's argument, tell us specifically 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? -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the 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 you're 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 flawed. Well in another thread someone posted a link to Wiles's paper, so I've started looking over its introduction. Here's an intriguing quote which I'm focusing on, though it may drag me into greater details. The key development in the proof is a new and surprising link between two strong but distinct traditions in number theory,the relationship between Galois representations and modular forms on the one hand and the interpretation of special values of L -functions on the other. p.2 Source: http://modular.fas.harvard.edu/21n/papers/Wiles,Modular_Elliptic_Curves_and_F ermats_Last_Theorem.pdf The special valus of L-functions are those 4 descriptors popping up again. My understanding is that mathematicians have reworked that approach having found something shorter, but I'll focus on the original. > It's illegitimate to fault Wiles's argument on the basis of secondary > sources. If you think there is something wrong with Wiles's argument, > tell us specifically 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, it's intriguing to see if the logical error pops out as easily as I expect it should, or if it's buried behind a lot of technical jargon. I'm 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, 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 can't create, try to destroy. But please spare us the disassociated hand-waving arguments. Stick to a specific 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. -- 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, so as not to tip-off the many, competent to beat him to the punch (such as Ribet). of course, it could also be flawed; or, simply inelegant. how would you characterize the sum-total of your now 8-year mission, monsieur Harris, minus all of the vituperative garbarge? of course, such an approach may be feasible for teaching math, although it's hard to imagine the student-body that'd tolerate that sort of harangue. (of course, in real life, one probably would be forced (or happy) to modify one's 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 haven't 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 flawed approach might work? --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... For a 1000-year anglo-american hegemony? HEY, JIMMY; LET'S 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/files/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 can't create, try to destroy. But please spare us the disassociated hand-waving > arguments. Stick to a specific 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 flaw: they can be thoroughly convinced of false things. The reason the primary newsgroup for this thread is sci.logic is that logicians are tasked with being careful in a way that even mathematicians aren't. Mathematicians can get away with leaps and assertions because of the relevance of mathematics to the real world, where physics has been powered by mathematical models. But logic is cold, hard, and less amenable to social pressure. I've outlined a specific logical flaw in Wiles's approach, which is that it depends on the logical fallacy of trying to find conditions through 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 finite subset to find a restriction on an infinite set? The technique is called lifting and is something like infinite descent made somewhat famous by Fermat himself. It is an intriguing question, so I'll back off somewhat while I consider whether or not he somehow gets around the logical fallacy to find a way to break it. Now mathematicians apparently are quite certain that Wiles succeeded and I applaud their energy. However, I must also rely on my understanding that human beings have a certain flaw: an ability to believe almost anything. Logic, however, is perfect. James Harris ==== > Human beings have a flaw: they can be thoroughly convinced of false > things. AS is James Steven Harris mistakenly convinced of his own genius. ==== > Human beings have a flaw: they can be thoroughly convinced of false > things. I just love the way you state the most commonplace trivialities as if they were some profound new insight. > So does Wiles get around the lack of a logical basis by using a finite > subset to find a restriction on an infinite set? > The technique is called lifting and is something like infinite > descent made somewhat famous by Fermat himself. > It is an intriguing question, so I'll back off somewhat while I > consider whether or not he somehow gets around the logical fallacy to > find a way to break it. You are not qualified to determine whether or not Wiles was successful. Only a limited number of people on the face of the earth are qualified (and willing to take the necessary time and effort) to do so. Until you are able to read (and understand!) his proof for yourself, you'll just have to take their word for it. Or not. Your opinion has no importance to anyone but yourself, so believe whatever you want. -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock ==== [snip] >Wiles's supposed accomplishment rests >*solely* on the assertion of a relatively small group of people that >his work is correct. James Harris's supposed accomplishments rest *solely* on the assertion of 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] > >Wiles's supposed accomplishment rests >*solely* on the assertion of a relatively small group of people that >his work is correct. > > James Harris's supposed accomplishments rest *solely* on the assertion > of one person that his work is correct. What can you do? Time after time I've faced false assertions, and time after time people have been unreasonable in the face of rationality. Sure I set out a few years back to find some answers to some math problems, and made a LOT of mistakes along the way. Yup, I've made a lot of mistakes. But right now I'd like some cogent answers to what follows: I've presented a problem with the logical foundation of Wiles's work as it relies on association to prove a condition. More useful discussion has worked things down to the assertion that Wiles used a finite set, and lifting to prove the equality of two infinite sets, where the equality supposedly forces the condition. There is, however, no reason for the condition given, and no claim of a reason, where the condition, from my understanding, is that every elliptic curve is a modular elliptic curve. The question raised before Wiles's work being whether or not an elliptic curve could not be modular, where mathematicians had related various elliptic curves that they tested to something called modular forms, and found that every one they tested was modular. Then the mathematicians Taniyama and Shimura conjectured that every elliptic curve was modular, which my understanding means, they are associated with a modular form, where that association can be described by 4 descriptors. My understanding is that Wiles's work depends on association. The logical fallacy I've put forward as a challenge against his work is, Cum hoc ergo propter hoc. James Harris ==== [snip] > My understanding is that Wiles's work depends on association. The logical fallacy I've put forward as a challenge against his work > is, Cum hoc ergo propter hoc. James Harris Your '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. You have been thoroughly debunked. Remove your faulty and error-ridden attempts at proving FLT from 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. [...] Wiles's 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. 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. > Since that time from lasers to spectral analysis the theory has fit > 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 Scientific 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 scientific standards for 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. > > [...] Wiles's 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. Wiles's work would mean a workaround to the logical fallacy called, Cum hoc ergo propter hoc. Ultimately, if Wiles's work is correct then it does not have any logical flaws, but checking it potentially involves going through each step in his work, which is a formidable task. If he did find 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 I've 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 don't believe physics they haven't personally checked. I'm not. Like how I believe that nuclear weapons work. But still realize that the absolute truth may be something other than what I've learned. In mathematics, absolute truth *can* be determined, just like a logical argument can be checked against certain rules for internal consistency. > Since that time from lasers to spectral analysis the theory has fit > 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 Scientific 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 scientific standards > for photons, etc, and then some). As a person with a science degree, I guess you'd consider me a part of the Evil Scientific Establishment. It's 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 doesn't care. To a math proof, you do not exist as a relevant entity. James Harris ==== > Wiles's work would mean a workaround to the logical fallacy called, > Cum hoc ergo propter hoc. No, it wouldn't. [snip] > ... I feel like I can > speak confidently on the subject. You speak confidently whether you know what you're talking about or not. Confidence is not your problem, 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. > > [...] Wiles's 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. > > Wiles's work would mean a workaround to the logical fallacy called, > Cum hoc ergo propter hoc. > > Ultimately, if Wiles's work is correct then it does not have any > logical flaws, but checking it potentially involves going through each > step in his work, which is a formidable task. If he did find 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 I've 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 > don't believe physics they haven't personally checked. I'm not. Like > how I believe that nuclear weapons work. But still realize that the > absolute truth may be something other than what I've 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 you essentially: 1) proof of an absolute kind, presumably stated in the symbolism of formal logic, and 2) proof that merely convinces other mathematicians, presumably stated in some meta-language (like English). Furthermore, your position is that proof of the second kind is what most mathematicians produce, and is not good enough. You go on to say that you produce proofs of the 1st kind. Question: How does one determine that a proof or mathematical argument is absolutely and irrefutably correct? The validity must be checked by 1) God, 2) a machine, or 3) a human being, 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 of inference, but it occurs to me that a few problems could arise. First, the algorithm may take an unreasonable amount of time to reach a decision. Second, hardware failure, electrical surges, sunspot activity, running the program under Microsoft Windows, etc. could cause erroneous results. Third, and perhaps most importantly, a human being (or beings) must write the software. Therefore, any errors caused by people could conceivably appear here. That leaves us with option (3). As we all know, people make mistakes. They make mistakes writing proofs. The publisher/editors of a journal may make a mistake mistake by erroneously believing the proof. Who has the final and ultimate authority to say that a given argument is 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 a disagreement, then the sides may argue their cases until one side prevails and convinces the other, at least within a given mathematical system. Therefore, in this sense, all proofs are of the second type. We must strive to convince other mathematicians. That is all there is -- simply because there is no other means of asserting the validity of a mathematical argument. It really is an appeal to the gallery. We must also consider that mathematics may be inconsistent. According to Kurt Godel, this is a possibility (at least for mathematical systems 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 fit > 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 Scientific 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 scientific standards > for photons, etc, and then some). > > As a person with a science degree, I guess you'd consider me a part of > the Evil Scientific 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 confidently 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 fight for a false belief? > > Math proofs are indestructible, incorruptible, and irrefutable. > > It's 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 doesn't 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 statement as in the case of Wiles' proof, let alone one provable by experiment. There are of course theoretical models (not necessarily well-defined 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, amounting to a certification-by-authority that the apparatus you were doing the experiments with 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 experimentation needed to corroborate the values of relevant physical and chemical parameters listed in the CRC handbook, and so on all the way down. What actually happened is that a long and social process of knowledge-accumulation occurred and you took the results on trust. In particular, if your experiments gave a wrong result, the conclusion would be that you made a mistake, not that photons' existence is in doubt; a pure assertion of authority by the Scientific Establishment concerning its Social Truth, which you accept without any objection in all the non-FLT situations. Note that your repeatedly discredited objections in this thread about Wiles' logic are irrelevant, as you also object to Ribet's proof without giving any particular reason to doubt it. The matter is simply one of an obvious double standard produced for the occasion, where social certification by a small network of experts counts as OK for photons, DNA, evolution, relativity, the Jurassic era (or the existence of Napoleon and George W Bush), etc --- but somehow the information that experts have certified Ribet's 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 thought experiment called the superluminal scissors or had any idea what I was talking about in sci.physics when I explained how a 5 m/sec water jet can be used to create an illusion of arbitrarily fast, even superluminal motion. - 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-defined 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, you've been beaten. That's what's annoying about Usenet as some loser will state a case, get their ass kicked, but STILL keep coming back as if nothing happened. My *degree* is in physics. I went to school on a full-tuition scholarship, and you stepped into my field with your assertions, got your ass kicked but refuse to back down. Now *emotion* is not necessary when it comes to Wiles's work. If he found a proof I can assure you that it is irrefutable. That's why it'd be a proof. All this emotion just annoys me, as part of the fun of science and mathematics is attacking things that are supposedly proven. It's GREAT fun not just accepting what people tell you. But that's what really annoys me about mathematicians as time after time I get yahoo's replying back in defense of mathematics, using tactics. 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 find themselves emotional about some person--any person--any time--any place--who decides to go after a math proof. That's because a math proof is indestructible, incorruptible, and irrefutable. It just doesn't care if you attack it, and no real mathematician would care either. Now I've discovered math proofs, which is why I'm not concerned about people refuting them because they are proofs. And in fact people who call themselves mathematicians can't touch them, so they come up with extraneous stuff, or make claims of finding their own proofs to refute my proofs, but you see, proofs don't duel. And you know what? I think that feature of mathematics terrifies some people who call themselves mathematicians. Mathematics does NOT care what you call yourself. It DOES NOT care that you have a mortgage. It DOES NOT CARE that you really, really, really want people to like you and think you're a great mathematician. Now I've made a specific claim against Wiles's work. If he found a proof the claim can be answered, but even if it is answerable then it must be true that he has found a way around what is considered to be a logically fallacious approach. Logicians should thank him in that case for correcting them. My challenge is a logical one. Wiles's work fails and is not a proof as 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, if the error was also not nistakenly canceled (and the value flipped, F for T, say, once more), but taht doesn't mean that it's not syntactically proper in some sense. not taht yours necessarily are, two often! I don't have to know any Latin, whatsoever, to follow a pattern of inference. (in particular, see Lou Kauffman's slight rephrasing of G. Spencer-Brown, on his site; I htought of the same diagrammatic convention, but using circles (to represent spheres .-)) and, you's still have to prove that Wiles 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 find > themselves emotional about some person--any person--any time--any > place--who decides to go after a math proof. > > That's because a math proof is indestructible, incorruptible, and > irrefutable. > My challenge is a logical one. Wiles's work fails and is not a proof > as it is an argument by Cum hoc ergo propter hoc. --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac ==== I agree, that a REAL mathematician wouldn't waste your time by objecting to the *statement* of a proof, although he certainly might wish to attack the very proceedings of the suposed proof, to get at the real truth of it (which may be false, due to an error in the chain of inference; or, the truth-value could be correct, by a flip of the coin (say), in spite of any breaks in the chain). your problem is that you insist upon hacking-away at the algebra, without any noticable hypothesis as to any simpler proof than Wiles' (or one that is akin to Fermat's .-)... at this rate, there will not be enough room in the margins of 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 computers without substantial hypotheses being encoded, insofar as possible. after all, because the predicate logic or boolean algebra can be used to remove as many redundant inferences as you wish, but can't change the truth-value of the statement. not even fuzzy logic can escape the necessary poles of the application of combinatorics to predicate logic, which is not any different than the syllogisms of the Older Greeks. well, the more that I think about it, the more it appears that Pierre et son fil faisons un connundrum mirabile pour l'edifiacion de generations futeur, et l'exposition 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 don't have to know any Latin, whatsoever, > to follow a pattern of inference. (in particular, > see Lou Kauffman's 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. Wiles's work fails and is not a proof > as it is an argument by Cum hoc ergo propter hoc. --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's 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/files/curriculum/Cosmo.PCX ==== L O N G 12 15 14 7 = 48 Angie Tysseland was playing piano on the NW corner of 21st Street and 4th Ave., I stopped to listen and to collect her stats. Angie was teamed up with Teresa Long, Teresa was providing vocals and doing a very good job of it, I was pretty well astounded by Terri Long's singing ability. This was an incredibly good free concert at the Saskatoon Jazz Festival. 213 Terri 4 5 59 124/241 806 Teresa 68 Lorrelle 97 Long 48 Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 <-7th-> 12 -- -- 58 50 Teresa was born in 59, it's the 17th prime while 17 in turn is the 7th prime, while the primes up to 7 add to 17 (59 is the 7th prime in prime position). She was born on day 124 (Numbers 7). She was born 117 days closer to the beginning of the year than to the end of the year. Her first name adds to 68 (4x17 and is the 7x7th non-prime). Her middle name adds to 97 (Leviticus 7). Her first and last names average 58 (the 7 primes up to 17). Her 68 (4x17 and is the 7x7th non-prime) valued first name is 58 (the 7 primes to 17) short of her 124 (Numbers 7) day of birth. Her given names differ in value by 29, it's the 7th prime (17) plus the 7th non-prime (12), or simply 7p+7np. She has 7 different letters in her given names. The vowels in her given names add 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 and are in positions adding to 84 (7 times the 7th non-prime), it is a difference of 7. Her even valued letters add to 136 (8x17) and are in positions adding to 87, it's a difference of 49 (7x7 and is the 17+17th non-prime). Her even valued letters add to 136 (8x17) and exceed her odd valued letters by 59 (the 7th prime in prime position and is her year of birth). Her even valued letters add to 136, it's the 104th non-prime while 104 in turn is the 77th non-prime (136 is the 77th non-prime in non-prime position). Her Fibonacci valued letters add to the 21 (7+7+7) chapters of Bible Book 7 and are in positions adding to 36 (7+7p+7np). Her Lucas valued letters add with their positions for the 108 verses of Bible Book 59 (the 17th prime, her year of birth). Her unrepeated letters add with their positions for the 108 verses of Bible Book 59 (the 17th prime, her year of birth). Her names add to the 49th non-prime, 25th prime and to the 33rd non-prime, together for 107 (the perfect 28th prime while the numbers 1 through 7 add to 28). Her 213 valued name is 171.77% of her 124th (Numbers 7) day of birth. Her repeating letters are in positions adding to 124 (Numbers 7, her day of birth). Her last 7 letters add to 77. Her last two names differ in value by 49 (7x7 and is the 7x7th non-prime). She abbreviates her first name to Terri (70). She is commonly known as Terri (70) Long (48), these names average 59 (her year of birth). Non-Primes 1 27 50 72 4 28 51 74 6 30 52 75 8 32 54 76 9 33 55 77 10 34 56 78 12 35 57 80 14 36 58 81 15 38 60 82 16 39 62 84 18 40 63 85 20 42 64 86 21 44 65 87 22 45 66 88 24 46 68 90 25 48 69 91 26 49 70 92 Teresa adds to 68, her day, month and year of birth adds to 68. She has four E's and four L's, these most frequently repeated letters add together for 68. Her 17 missing letters add to 240 (68.37% of the possible total). Her 213 valued name adds with her 124th day of birth for 337 (the 68th prime). She was born on day 124 (Numbers 7) while her first name adds to 68 (the 7x7th non-prime). Her first name adds to 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 chapter contains 89 verses. Her 213 valued name exceeds her 124th day of birth by 89. The Gospels are Bible Books 40, 41, 42 and 43, together for 166. Terri's name adds to 213, it's the 166th non-prime while 166 is the 128th non-prime, while 128 is 2 to the 7th. Chapter 124 is Numbers 7, it is the 7x7th prime (227) short of the numbers up to the 17th non-prime (351). Genesis 7, Exodus 7, Leviticus 7 and Numbers 7 are chapters 7, 57, 97 and 124, together for 285 (the 7x7th chapter of The Samuels). Numbers 7 and Deuteronomy 7 are chapters 124 and 160, together for 284 (Second Samuel 17, the 7th prime). 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, together for 618 (the number of verses in Bible Book 7). First Chronicles 7 and Second Chronicles 7 are chapters 345 and 374, together for the 719 verses of Bible Book 12 (7th non-prime), this 719 is the 128th (2 to the 7th) prime. First Samuel 7 and Second Samuel 7 are chapters 243 and 274, together for 517, while First Chronicles 7 and Second Chronicles 7 are chapters 345 and 374, together for 719, it's an average of the 618 verses of Bible Book 7. 213 Terri 4 5 59 124/241 806 Teresa 68 Lorrelle 97 Long 48 189 Angie 27 2 61 58/307 1471 Angie 36 Grace 34 Tysseland 119 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 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 names have an average value of 67 (Exodus 17). Angie's names add to 36 (7+7p+7np), 34 (17+17) and 119 (7x17), together for 189 (the first 17 primes minus the first 17 non-primes). Their first names add together for 104 (77th non-prime). Their last names add together for the 167 verses of Bible Book 17 (the 7 primes in prime positions up to the 17th prime add together for the 167 verses of Book 17). Esther becomes Queen 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 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 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 of their years than to the end of their years. Terri's full name adds to 213 (166th non-prime). Their given names add together for 235, it is the 184th non-prime, pretty as the 184th prime (1097) and the 184th non-prime (235) averages 666. Their initials have an average value of 36 (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 had thanx for showing them evidence that their very names are a gift from God. My math was used repeatedly as an excuse to arrest and chemically lobotomize me (torture me) in psychiatric facilities, and whenever I meet with people and show them patterns that please them, they never have the decency to offer to buy me a measly cookie for my work nor send me a cheap letter expressing thanx. Man oh man, you people are compassionless turds, you are the shit of the earth, soon God will spread you out over the surface of the earth like the dung that you are, and in this I rejoice. 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - 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 66 1062 <-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 6 1070 <-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 Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! ==== I think you're ill mannered, boring and part of the reason why many people are not interested in religion. Dave. ==== > I think you're ill mannered, boring and part of the reason why many > people are not interested in religion. Dave. It's numerology, a religion Plonk-a-dork!!! RJP ==== Just testing - I haven't 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 haven't 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 -- that's what it's there for. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. I try not to fly in the face of public opinion. ==== THANK you both Stan/Darrell for your posts... They were both an extremely big help!!! thanks again!!!! ==== The Pyramid Paper Challenging & Ranking Game is the new type puzzle for all ages. Puzzle sheet page is http://www5.ocn.ne.jp/~pachalle/paperchalleran213.html Rule page is http://www5.ocn.ne.jp/~pcjapan/paperchalleran270.htm Please try this game!! Please tell your family, friends, and students of your school this interesting contest puzzle as many as possible and enjoy to solve it each 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 be wrong. The problem is: Find every real number that satisfies: 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 defines 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 defined 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 ==== > 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 satisfies: > 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 Your first 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^2 This 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^2 x-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 defines 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 These are correct but could use boundary conditions on x or y. > > Therefore the triangle can be defined as: > -(x+16)/6 < y < x/2 > and > -4 < x < 2 Are you trying to define 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 satisfies: >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 Why do you think you get two equations here? (sqrt(a))^2 = a is always true; perhaps you are confusing it with sqrt(a^2)=|a|? If you square both sides, you will get that |x-1| = x^2, from which you 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^2 This is completely wrong. |a|= b if and only if a=b OR -a=b; not and. And in any case you did the operations wrong. Presumably you meant -x+1 and x-1 on the first 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 that after squaring, you may be introducing extraneous solutions; you need to go back to the original equation to decide which solutions to your new 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 (note that |x-1| is always nonnegative, so the square root is the real operator and always a nonnegative real. what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > The problem is: >> Find every real number that satisfies: >> 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 > That's 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|. So (sqrt(|a|))^2 = |a|, not a^2. The square of the square of the square root of |a| is a^2, however... ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > 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 satisfies: > 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 defines 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 defined 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 satisfies: > Squareroot( |x-1| ) = x Start with some basics (which will be needed to weed out extraneous roots later on): * |x-1| >=0 by the definition of |z| * Therefore, sqrt(|x-1|) >= 0 * Therefore x>= 0 Now square both sides to get: |x-1| = x^2 We need to get x-1 out of the absolute value. Use the face that |z|=z where z>=0 and |z|=-z where z<0. So for where 0 <= x < 1 (remember, the problem construction guarantees that x must be greater than or equal to zero), we have -(x-1) = x^2 or x^2 + x - 1 = 0 which gives x = -1/2 + sqrt(5)/2 or x = -1/2 - sqrt(5)/2 We can immediately throw away the second solution because it is negative. But we can keep the first solution because (-1 + sqrt(5))/2 is positive and is less than 1. Going back to |x-1| = x^2, we now take the branch where x>=1. This gives us x-1 = x^2 x^2 - x + 1 = 0 which 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 if 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 > > That's 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|. ==== I'm having trouble figuring 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? ==== on sci.math: > I'm having trouble figuring these out: > 1) Is the set of all functions from the boolean set {0,1} to the > natural numbers countable or uncountable? Hmm, I'm not an expert in this by any means, more like a beginner, but I think that the set you're 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. As the sets {0} and {1} are singletons, those sets of functions have the same cardinality as N 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? I'm too lazy to try to figure out this one. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ We sorcerers don't like to eat our words, so to say. - Sparrowhawk ==== > I'm having trouble figuring 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 by any countably infinite set. So let's replace N by Q, the set of 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: >> I'm having trouble figuring these out: >> 1) Is the set of all functions from the boolean set {0,1} to the >> natural numbers countable or uncountable? >Hmm, I'm not an expert in this by any means, more like a beginner, >but I think that the set you're 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 product is {({(0,n)}, {(1,m)}) : n,m in N}, a set of ordered pairs of singletons of ordered pairs. They're definitely not the same set. They are, however, the same size, and as you suggested in the bit that I snipped below, they're the same size as N x N. In particular, 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 the set 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 you'll learn more if you work at it a bit yourself. Here's a hint towards one possible solution: find an uncountable antichain in P(Q x Q), where Q is the set of rational numbers. Since Q x Q is countably infinite, this automatically gives you an uncountable antichain in P(N). You may also want to consider that for any real numbers c and d with 0 <= c < d, the set of points in Q x Q with polar coordinates (r, t) satisfying the conditions r > 0 and c < t < d is non-empty. Brian ==== > I'm having trouble figuring 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? > Haven't sci.mathers done about enough homework for this guy? Now he demands proof, not just hints. ==== > I'm having trouble figuring 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? > > Haven't sci.mathers done about enough homework for this guy? Now he demands > proof, not just hints. > Tho 1 is easy, I don't 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. ==== >> I'm having trouble figuring 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? >> Haven't sci.mathers done about enough homework for this guy? Now he demands >> proof, not just hints. > Tho 1 is easy, I don't 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 don't quite fill the bill as is, but a slight modification of the idea will work. By the way, 2^Aleph_0 is not equal to Aleph_1, unless you are assuming CH. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== >> I'm having trouble figuring 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 don't 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? It's also easy to do by transfinite recursion, but I didn't 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, it's 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. 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 don't 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 don't quite fill the bill as is, > but a slight modification 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 don't 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 > It's also easy to do by transfinite recursion, but I didn't > 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, it's easy to > construct a new function from N to N that dominates every member > of A. > <3f18c949.11042797@enews.newsguy.com> ==== : : : :>> I'm having trouble figuring 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 don't 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? :It's also easy to do by transfinite recursion, but I didn't :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, it's 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 isn't being solved in ZFC though. Perhaps ZFCH was being used or ZFG. : :Brian : ==== : >: >: >:>> I'm having trouble figuring 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 don't 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? >:It's also easy to do by transfinite recursion, but I didn't >: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, it's 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 isn't 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 don't 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 first. Not intrinsically more complicated than the one below, in my opinion. In any case, I thought that since my hint was a little broader than Robin's, it might point you in the right direction. >How about > { (r,r+1) / Q | r in RQ } ? >/ intersect Brian ==== Been doing some soul searching lately. I'm 37 and have a BS in CIS (got it back in 1988) and have been in the computer field 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. For the longest time, I wanted to go and get an equivalent degree in CS to compliment my CIS degree. Basically, what I lacked there is the Math, Chemistry and Physics. I took care of the Math portion by taking Calculus thru Differential equations and found that I really enjoyed it (got an A in all of those classes). Took 10 years off from college and worked my butt off in the field as a developer on some jobs. Recently working for a company developing chemistry software, I finished an engineering CHM105 course and got an A in it. However, I did not really enjoy the chemistry as much as Math. I basically need to take a Physics course and a Linear Algebra course to get an Associates in Math. I'm taking a refresher course in PreCalc on audit (it's coming back to me) since it's been 13-14 years since I took Trig (you don't use it, you do lose it) and will take Calc 1 on audit to ready myself for Linear Algebra next Jan. An advisor at the community college where I'm taking this PreCalc recognizes that I like Math and has suggested maybe that I consider a BS in Math (rather than CS) if I finish my Associates. The argument (and it's a good one) is that Math is more generalized than the specific Computer Science degree and that it may open up more doors of opportunity. One things for sure, in the computer field, they do not view a Math degree as negative. In-fact, they view it as a positive. I was just curious, *if* the computer field tends to continue on the same path it's in now (ie, massive layoffs, salary cuts and an even more dwindling chance of finding rewarding jobs), what other areas could one (with a BS in Math) get into? I guess with an School. I don't know if I have the time to get a Master's as that would mean another 4 years. The BS in Math would be about 2 years away. I'm not 100% convinced the computer field is as great as they say it is (I'm seeing alot of people losing their jobs and working in non-computer related work with their CS degrees as CS degrees have definitely lost some worth over the last 2 years or so) and am wondering if the Math route (since I do enjoy it) could be a good one to take. At best case, I can still get the Math degree and remain a software developer. At worst case, I'm forced to change careers but can hope that an undergrad in Math would help. A Math degree is definitely not a useless degree. It can be applied to far more different areas than a CS degree could. I'm just wondering though what an undergrad math degree could be useful for (other than in the computer field). thanks!!! ==== [...] > I'm just wondering though what > an undergrad math degree could be useful for (other than in the computer > field). thanks!!! > > You might look at Scroll down a little to Careers... -- Paul Sperry Columbia, SC (USA) ==== > t 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 don't 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 don't see that as a good > idea. Why not a degree in business? > Because there's an oversupply of business majors as the economy is down, and whether CEO's like it or not, there will never be too few CEO's. ==== > 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 don't see that as a good Good advice!!! > Why not a degree in business? Argg!! :-) With my CIS degree, I got a minor in business(which included micro¯o econ)/ and 2 years of accounting. In this day and age with corporate downsizing, business scandals, etc.., I'm not so sure a business degree gets you as far as it once did. I know 2 MBA's that are still looking for work. Since graduating with my CIS, I can't recall ever needing to use anything from my business courses on the job. Accounting is a totally different story though as alot of computer jobs out there require applications development of AR/AP/inventory, etc... As per the Math, I have never had to use the Calculus/diff-eq I learned on any 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 do some graphics programming. The one thing that I got out of math which was priceless was the ability to solve problems systematically. The same with Chemistry. As a programmer, it's easy to fall into the trap of being lazy and slamming out code off the top of your head (which usually ends up being 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 out in front of you and coming up with the solution. To me, that's what makes Math a VERY useful subject to know. That's something I did not get when I was taking my business courses. Accounting though was similar to the math as you're forced to lay everything out in front of you so accounting is a good thing to know/do. Unless you aspire to being a CEO or middle manager (which I don't and never have), business courses tend to be a waste. They don't necessarily even help you if you want to start your own business. If you want to start your own business, you may be better off attending seminars/lectures rather than wasting a few semester's of time taking business 101 or what have you. If you're going to be a CEO or middle/upper management, you'll need the MBA and the business courses are pre-reqs so you have no choice but to take them. For a software developer, you can learn all you really need to know about businesses just by working and observing the techniques employed by middle/upper management. And even then, many managers don't have a business background. The one thing that fascinates me about Math is that it's been around for centuries and will still be around in the same form centuries from now. Calculus will still be calc. I can't say the same for computer and business courses :-) Math definitely tends to sharpen your analytical and problem solving skills. The business courses I have taken in the past did not. But sadly, you're right in that there's not alot of money to be made in Math. But I don't think there's any denying that it's definitely a useful (perhaps one of the most useful) foundation skill to possess. ==== >Been doing some soul searching lately. I'm 37 and have a BS in CIS (got it >back in 1988) and have been in the computer field 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 bachelor's in math are even worse. I don't think having that degree would make CIS employers look on you any more favorably, and I can't think of any specific occupations where a bachelor's in math plus a bachelor's in CIS is any more likely to get you in the door than a bachelor's in CIS. (Granted, I may be missing something here.) -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. I try not to fly in the face of public opinion. ==== > >Been doing some soul searching lately. I'm 37 and have a BS in CIS (got it >>back in 1988) and have been in the computer field 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 bachelor's 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 employer's 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 can't get into with a CIS degree. > I don't 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 that's a long haul (except in the computer biz). > and I can't think of any specific occupations where a bachelor's in math plus a bachelor's in CIS is any more likely to get you in the door than a bachelor's in >CIS. (Granted, I may be missing something here.) > > No. The reason for a bachelor's in math is that you've come to the realization that a computer career is a dead-end path, except in the field 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 (that's customers/bosses/vendors/whatever) in the vernacular that *they* are comfortable with. If that means wearing white tie and tails, then that's 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, that's what you do. Jon Miller ==== Very good points you raised there Jon!! Most of them fall into the reason why I am thinking about getting an undergrad in Math!!! thanks, Theron ==== > True, but I think job opportunities with a bare bachelor's in math > are even worse. I don't think having that degree would make CIS > employers look on you any more favorably, and I can't think of any > specific occupations where a bachelor's in math plus a bachelor's in > CIS is any more likely to get you in the door than a bachelor's in > CIS. (Granted, I may be missing something here.) I don't think you're missing anything! I feel that you're points are also valid. In the 15 years of working in IT, I have never seen a case where an employer 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 for coming up with some heavy duty dialing algorithms for a telephone switch. Poisson and Exponential Smoothing were the 2 tools he ended up using. A co-worker (friend) and myself simply stamped his algorithms into the source code and the system was off to the races :-) 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. 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. It's 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 I'm speaking for just that one course) and have not taken any physics (but plan to do so)). I really don't think Math is a useless degree. These days, CIS and CS degrees 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 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 can't 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. > It's 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 I'm speaking for just that one > course) and have not taken any physics (but plan to do so)). I really don't 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 qualifications are certainly less desireable than people with IT qualifications. 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 Microsoft's MCSE and related certifications--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 field on an entry level. Study a little, take a few tests, fork over some dough, and it's yours. This certification 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 certification. 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. -- Darrell ==== If you enjoy math you should major in it. A smart employer will not look down on a math degree (especially if you have good grades). It's absolutely ridiculous to choose a major based on whether someone perceives it as being nerdy or not. If you want to loose the stigma join a business club and make some contacts. Better yet, take a speech class or join a speech club. The great thing about math is that you can build on it. It is much easier to pick-up computer science, physics, and chemistry if you know math well. If you have the intellect for math or physics or even chemistry, do that instead of CIS. A math major is not trendy but it won't be as dependent on trends as one that is. I wouldn't do business either. You can get your MBA later and you'll be a much more attractive (and capable) employee. I think all those bosses who got lectured about not getting it when it came to IT are having a lot of fun now. Unless you live in India or China 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: I'm not saying this in a derogatory way. Auto mechanics is challenging but it's 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 can't 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. > It's 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 I'm speaking for just that one > course) and have not taken any physics (but plan to do so)). > > I really don't 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 qualifications are certainly less desireable than people with > IT qualifications. 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 Microsoft's MCSE and related certifications--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 field on an entry level. Study a little, take a few tests, > fork over some dough, and it's yours. This certification 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 > certification. 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 T 22 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 medical students at the U of S. 64+ Dad 15 3 /291 64+ Mom 4 8 /149 183 Victoria 29 2 76 60/306 6951 Victoria 97 Lee 22 Vogt 64 64+ Sis 8 3 78 67/298 7689 64+ Bro 29 6 83 180/185 9628 Victoria was born on the 60th day of the year, it is 11 plus the 11th prime (31) plus the 11th non-prime (18), or simply 11+11p+11np. Her first and last names both begin with the 22nd (11+11th) letter of the alphabet. Her first two letters add to 31 (11p). Her middle name adds to the 22 (11+11) chapters of Bible Book 11. Her given names add to 119. Her first and last names differ in value by 33 (3x11). Her names have an average value of 61 (Exodus 11), it is the 18th prime while 18 in turn is the 11th non-prime), it is the 11th prime in non-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 were born in months adding to 11, together for the 22 chapters of Bible Book 11. The kids were born on days of the month adding to 66 (6x11) and in years adding to 237, it's the opening chapter of The Samuels, pretty as The Samuels contain 55 (5x11) chapters. The kids were born on 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 of the year adding to 127 (the 31st prime while 31 in turn is the 11th prime), it is the 11th prime in prime position. The males were likely born on days of the year averaging 127 (the 11th prime in prime position). Generally the parents have their birthdays 150 days closer to 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 the beginning of their years than to the end of their years, pretty as there are 150 chapters in Bible Book 19 while the parents were born on days of the month adding to 19. Mom was born 67 days closer to the end of the year than to the beginning of the year (19th prime), keeping in mind 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 family was born on days of the year adding to 597 (Psalm 119), pretty that Victoria's first name adds to 97 while her given names would add together 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 first 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 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 18 61 19 67 <- 67 <-the 8th prime in prime position R U T H <-Book 8 18 21 20 8 = 67 The family was born on days of the month adding to the 85 verses of Bible Book 8, it is a combination of the first 8 primes (up to 19) plus 8 more. The little sister was born on the 8th day of the month and on the 67th (19th prime or the 8th prime in prime position) day of the year, Bible Book 8 is Ruth (67). Generally the parents have their birthdays on days of the year adding to 290. Dad was born with 291 days remaining in the year. The little sister was born with 298 days remaining in the year. The family was together born with 1229 days remaining in their years. The first and the last kids were both born on the 29th day of the month. There are 29 chapters in Bible Book 13 (6th prime) and 29 verses in chapter 666 (Ecclesiastes 7), pretty as 29 is a combination of 6 plus the 6th prime (13) plus the 6th non-prime (10). Dad generally has his birthday on the 74th day of the year (a factor of 666). Mom generally has her birthday on the 216th (6x6x6) day of the year. The kids were born on days of the month adding to 66. 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - 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 66 1062 <-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 6 1070 <-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 first two letters in Vogt add to 37, the first pair of kids were born on days of the month adding to 37 and also the last two kids were born on days of the month adding to 37. The Vo(37)gt kids were born 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 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 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 piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime Copper and Zinc are elements 29 and 30 (together for 59), and together they make Brass (59): B R A S S 2 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 really wanted to see her panties, and explained that her money was a gift from God. Anyway, she had 4 pennies, they were dated 89, 90, 96 and 02, together for 277 (the 59th prime or the 17th prime in prime position). The last two kids are separated by 1939 (7x277) days, or exactly 277 weeks. The kids are together separated by 2677 days. The kids were born on days 29, 8 and 29, these Bible Books contain an average of 77 verses. The kids were born on days and in months adding to 77 and in years adding to 237. The sisters were born on days of the month adding to 37. The kids were born on days of the year adding to 307. The kids were born on days of the month adding to the 66 Books of the Bible, it is 7x7+17 or 7 squared plus the 7th prime, or the 17th prime 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 was born with 149 days remaining in the year (the number of verses in Bible Book 48). The little sister was born with 298 (149+149) days remaining in the year. The last two kids are separated by 1939 (7x277) days, or exactly 277 weeks. Victoria's pennies were together 35 years old, 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 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 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 with 17+17 verses. There are 17 verses at chapters 1 and 3, and 59 (the 17 prime) verses at chapter 13, so the 17's and the 17th prime are at chapter numbers adding to 17 (1+3+13=17). The first 17 versed chapters in the Bible are at chapters 91 and 93, together for 184, or the 167 verses of Book 17 plus 17 more. Leviticus contains 859 verses, it ends in 59 (the 17th prime). The first 17's in the Bible surround chapter 92 (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 9 100 10 101 11 102 12 103 13 59 <-17th prime 104 14 105 15 106 16 107 17 108 18 109 19 110 20 111 21 112 22 113 23 114 24 115 25 116 26 117 27 34 <-17+17 The family was born on days of the month averaging 17. The kids were born in years adding to 237, it's a combination of the 17th, 17+17th, 17+17+17th and the 17+17+17+17th non-prime numbers (26, 49, 70 and 92). The last two kids are separated by exactly 277 weeks, Victoria's pennies had years adding to 277 (the 17th prime in prime position). It's another true story. Non-Primes 1 27 50 72 4 28 51 74 6 30 52 75 8 32 54 76 9 33 55 77 10 34 56 78 12 35 57 80 14 36 58 81 15 38 60 82 16 39 62 84 18 40 63 85 20 42 64 86 21 44 65 87 22 45 66 88 24 46 68 90 25 48 69 91 26 49 70 92 <-the 17th level Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! ==== Stand back, everyone! Dar is winding up again. The pitch should come in the 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.com A nice site - simple & informative. gtoomey ==== I am searching for the definition of the term k-algebra. I have multiplication that is a bilinear operator, however it does not elaborate on this. Any assistance would be greatly appreciated. A book or online Sincerely, LS Thomas ==== > I am searching for the definition 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 definitions & 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 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 you'd have to force it into unequal intervals (we are already doing unnecessary cumulative frequency curves to find 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 ==== > 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 you'd have to force it into unequal intervals (we are already > doing unnecessary cumulative frequency curves to find 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 don't 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 you'd have to force it into unequal intervals (we are >> already >> doing unnecessary cumulative frequency curves to find 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 don't 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 don't 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 you'd have to force it into unequal intervals (we are already > doing unnecessary cumulative frequency curves to find 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 I'll be shot down in flames as I generally take the simpleton's 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, fire 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 for discrete. -- 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. > > -- > Dave Not really. A histogram is a graph of a frequency (or probability) density function. The integral of (ie area bounded by, more or less) a frequency (probability) density function then gives the frequency (probability). For example, one probability density function (pdf) that most people have 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 area beneath the Normal pdf gives probabilities ... The graph of the Normal pdf is a histogram, because areas thereunder give probabilities. Fine, that's a continuous pdf. There *are* discrete pdf's, of course, such as the Binomial ... graph a Binomial pdf and you have a histogram for discrete data. One good (pedagogic) reason (there are other reasons, mostly not so good) for distinguishing between bar charts that aren't histograms (graphs of *frequency*, mostly) and bar charts that are histograms (graphs of *frequency density*, as I said) is that it makes the conceptual step to probability density functions easier to make. 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 ==== > >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 fixed) Bob, is there any reason why the students can't 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 be used 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 data they > collect afaics you'd have to force it into unequal intervals (we are already > doing unnecessary cumulative frequency curves to find 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 ==== >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. That's it, I was not sure! Also another thing to note is that a histogram has a frequency density for the y-axis, so that when multiplied by the class width, the area given will be the frequency. The other difference between bar charts and histograms is that bar charts have 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 fixed) Bob, is there any reason why > the students can't use a histogram with all the groups the same size? > Sometimes the results just don't 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 fixed) Bob, is there any reason why the > students can't 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 to draw a normal distribution curve. i.e. a smooth curve through the top middle of each bar. Now calculate the standard deviation of the sample, draw vertical lines one standard deviation either side of your central bar and measure the area enclosed (i.e. count squares) if it's around 66% of the total area, then we have normal distribution. Obviously works best for height, 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 data they > collect afaics you'd have to force it into unequal intervals (we are already > doing unnecessary cumulative frequency curves to find 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 ==== This is loosely related to another thread on histograms. I am looking for some statistics I could use to analyse reaction times. Let's 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 I've got this to work the non-writing hand is quickest. cheers dd I realise I'm posting to asmp is there a suitable stats NG? I couldn't find one. ==== >This is loosely related to another thread on histograms. I am looking for some statistics I could use to analyse reaction times. >Let's 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 I've got this to work the non-writing hand is quickest. cheers dd I realise I'm posting to asmp is there a suitable stats NG? I couldn't find >one. > sci.stat.edu It's my recollection that one of sci.stat.math and sci.math.stat is busy and the other not, but I don't recall which sci.stat.consult is willing (or was in the past) to help with this sort of question. They'd 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. They'd much rather deal with educated consumers than idiots. But that cuts the base of questioners so drastically! :-) (I'm fresh from dealing with one student who phones five minutes before the midterm to ask if she could go to her brother's graduation instead, and another who insists on asking me questions that are answered in the syllabus because I don't 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. They'd much rather deal with educated consumers than idiots. But that cuts the base of questioners so drastically! :-) > I didn't say they insist on it, just that they prefer it. I've been known to adjust my rates depending on how interesting the assignment looks. (I wish I could adjust them base on how the cash flow looks!) Jon Miller ==== S T E V E N S O N 10 20 5 22 5 14 19 15 14 = 133 In the afternoon I went to the food court in the Wildwood Mall and met Buddy, he is the 6th of 8 kids. 133+ Dad 4 10 14 277/88 +15477 133+ Mom 9 4 16 100/266 +14924 278 Buddy 10 6 57 161/204 113 William 79 Lyon 66 Stevenson 133 The parents were together born 23 days closer to the end of their years than to the beginning of their years. Buddy and his parents were born on days of the month adding to 23, Buddy was born exactly 23 23 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 an average of 614 verses. Buddy was born 43 days closer to the beginning of the year than to the end of the year (14th prime). He was born in to 79, it is the 22nd prime while 22 in turn is the 14th non-prime (79 parents are separated by exactly 79 weeks while his first name adds to name 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 adding to 14. My birthday was 14 days ago while Buddy and I had our birthdays an average of 140 days ago. Mom was born in 16 and married Stevenson, pretty as the name adds to 133 while Bible chapter 133 is Numbers 16. This 133 (Numbers 16) exceeds it's 101st non-prime position by 32 (16+16). Buddy has 16 letters in his first and last names, his last name adds to 133 born 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 57 both contain 25 verses (the 16th non-prime). The parents were born in years adding to 30 (20th non-prime). Buddy and his parents were born in months adding to 20. Buddy has 20 letters. Dad and Buddy were born in years adding to 71 (20th prime). Buddy and his parents were together born 20 days closer to the beginning of their years than to the end of their years. Buddy was born 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, it's the 59th prime while 59 in turn is the 17th prime (277 is the 17th prime in prime position). The parents were born on days 277 and 100, these are the 59th prime and the 75th non-prime, together for 134 (Numbers 17). Dad was born 189 days closer to the end of the year than to the beginning of the year (the first 17 primes minus the first 17 non-primes). Dad was born 177 days further into the year than mom (3 times the 17th prime). Buddy was born with 204 days remaining in the year (Joshua 17), it is 12x17, or the 7th non-prime times the 7th prime, and is twice 17 plus twice the 17th 161 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 (6th prime). Dad was born in 14, Bible Book 14 contains 36 chapters (6x6 while 1 through 36 adds to 666). Mom was born in 16, Bible Book 16 contains 13 chapters (6th prime). Buddy's 278 valued name exceeds his letters in his given names add to 91 (1 through 13), the repeating letters in his given names add to 54 (13 plus the 13th prime), it is a difference of 37 (37 chapters in the Bible contain the length of 13 verses). I am 113 days older than Buddy, there are 113 verses in Bible Book 54 while his initials add to 54 (13 plus the 13th prime). He is the 6th of the kids, his first 6 letters add to 66, his middle name prime and the 184th non-prime averages 666). Buddy was born on the 10th (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 Buddy's first name adds to 79 (Exodus 29), his given names add together for 145 (5x29). In his given names, his consonants plus their positions exceed his vowels plus their positions by 29. The primes and squares in his given names add together for 79 (Exodus 29). Buddy and his parents were born in years averaging 29. Mom and Buddy were born in years adding to the 73 verses of Bible Book 29 (and is the Lucas numbers up to 29). Dad and Buddy were born on days of the year adding to 438 (6 times the 73 verses of Book 29). Dad and Buddy were together born with 292 days remaining in their years (4 times the 73 verses of Book 29). Mom and Buddy were together born 209 days closer to the beginning 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). Mom and Buddy were born on days of the century adding to 26923 (13x19x109), it is both a multiple of 13 (29 chapters in Bible Book 13) and a multiple of 109 (the 29th prime). Dad was 42 (29th non-prime) years old when Buddy was born. We meet when I am 16815 days old, my age ends in 815 (the first 29 primes minus the first 29 non-primes). Bible Book 13 (the 6th prime) contains 29 chapters and Bible chapter 666 (Ecclesiastes 7) contains 29 verses, pretty as 29 is 6 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 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 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 piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime Copper and Zinc are elements 29 and 30 (together for 59), and together they make Brass (59): B R A S S 2 18 1 19 19 = 59 Buddy is coming out of a family of 10, his dad was born in the 10th month while mom was born on the 10x10th day of the year. Buddy was born 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 Buddy's first name adds to 79 (the 11+11th prime), his middle name initials add to 31 (11th prime). He has 11 letters in his given names and his given names add to 145, corresponding to Numbers 28 with 31 primes, squares and cubes together add with their positions for 211. exceed his vowels by 110. In his given names, his odd valued letters exceed his even valued letters by 45 (the 11th non-prime in prime position). In his given names, his unrepresented letters exceed his represented letters by 127 (the 11th prime in prime position). Buddy was born 62 days after mom's birthday (twice the 11th prime). He was born 249 days after dad's birthday (First Samuel 11). He was born 311 days after his parent's birthdays. Mom was born on the 9th, corresponding to First Samuel with 31 chapters (11th prime). Buddy and his parents were born on days of the year adding to 538, corresponding to 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 first 11 non-primes). 389 <-77th prime 104 <-77th non-prime 77 <-77 --- 570 The Four 57's Genesis 41 -> 41 Leviticus 14 -> 104 Judges 9 -> 220 <-I dreamt of 220 roofs blown John 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th prime Chapter 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 part about 666 while Isaiah contains 66 chapters): Daniel - 357 verses Revelation - 404 verses <-57 plus the 57th prime plus the 57th non-prime Isaiah - 1292 verses <-an average of 19.575757... verses per chapter parents were born on days and in months and years adding to 57. Dad and 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 (they are 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 him with his name, it would only result in him giving money to a church that has an Egyptian penis on it's roof. Perhaps Buddy would only attend a church in late December, and he would reluctantly do so in order to please his wife and her parents. While at that church he would see their decorated evergreen trees, comment on their beauty, and give the church money. The churches teach the sheep to turn evergreen trees into decorated idols, the evergreen tree was worshipped for centuries as a fertility symbol for it remains green throughout the year. The Old Testament documents and condemns how pagans surrounding and opposed to ancient Israel worshipped the evergreen trees. The Old Testament also documents and condemns the obelisks, some version refer to these symbolic penises as the Towers of Bethshemesh. Like the evergreen trees, the penis was worshipped as a fertility symbol. Or if I could convince Buddy that the God of the Bible provided him with his name, he would seek out a priest that has a fish head hat, and give that person money. The ancient priests of fish god Dagon dressed up in fish outfits, over the years their costumes evolved so that all that now remains is the fish head hat, again the fish is being worshipped as a fertility symbol due to their large number of eggs. Sticking a penis on the roof of your church (or a fish head hat on the head of your priest) is a violation of God's Second Commandment. Turning trees into idol, bowing to decorated trees and worshipping trees, penises and fish is a violation of God's First Commandment. By December 25th the sun is visibly returning from the south, calling this pagan sunwhoreshipping holiday Christmas is a violation of God's Third Commandment (it is a pagan mass and not facilities in an attempt to make me shut up about the false traditions in your churches is nothing less than a violation of God's Sixth Commandment. They tortured me for years, I begged for years for assistance to get out of the country to no avail, and all you people can do is give money to the churches that teach you to violate Commandments. You people collectively spent millions of dollars having me tortured, then annually you people collectively spend billions of dollars in turning trees into idols. My math was used as an excuse to repeatedly arrest and torture me, and then when I manage to meet with you people in restaurants and show you evidence that your name is a gift from God, you are so cheap and ignorant that you don't even have taking the time out of my life to show you such. You are an incompassionate turd, you are the shit of the earth, and I am on my knees begging God to honor Exodus 20:5 and Hosea 4:6 as promises, and terminate your life, the lives of your siblinks, and the lives of your children. Buddy, if I find you or your family members in the obituaries, I will cheer, it will be in accordance to Scripture (Psalm 137:9), and I will post these stats again. All you are good for is to have your stats posted on the usenet and be used as an example to others, and look, here you are!!! And it should be mentioned that Buddy Stevenson is a native Indian, Joe Munroe once informed me that the Indians on his reserve will flip flop between traditional aboriginal and Christian beliefs, depending upon whom they are trying to get into their beds. Your only compassion is for your filthy can try and let your filthy traditions save you. Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 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 of at t=25s. (NOTE: s denotes unit of time seconds and m/s denotes meters per second.) The first part was rather simple for me (finding the velocity at t=25), and the second part (finding 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 I'd 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 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 figure 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 wasn't 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). Here's 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 I'm on the right track, because the first component matches, but the other two, while close, don't actually match. I figure 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 five seconds on that 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 first part was rather simple for me (finding the velocity at > t=25), and the second part (finding 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 I'd 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 p'' = a v = p' = a*(t - t0) + v0 p = a*(t - t0)^2 / 2 + v0 * (t - t0) + p0 Now t0 = 5s a = (0, 10.3, -1.5) ms^-2 v0 = (-12, 0, 21.4) m/s p0 = (134, -25.1, 124) m p' = (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 figure 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 wasn't 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). can't quite follow this Here's 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 so you have p(t) = a*t^2/2 + v(0)(t-5) + r0 + C, where C is constant stuff which I can't immediately see how you derived. There's no choice of C which will give the correct answer because the coefficient of (t-5) should be v(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>m p = (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>m This answer has been rounded to nearest integer for some bizarre reason. > I can tell I'm on the right track, because the first component matches, > but the other two, while close, don't actually match. I figure 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 five seconds on that part as well? And if > so, how do I go about doing that exactly? The initial conditions should be obviously satisfied; 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 and D have to be determined. Don't put specific numbers into the equation before you have to; it's more difficult to spot mistakes if you do it early. -- P.A.C. Smith 'If 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 gives v = (t-5)a + v_5 d = (t^2 / 2)a - 5ta + t v_5 - (25/2)a + 25a - 5v_5 + d_5 d_25 = 300a - 100a + 20v_5 + d_5 = 200a + 20v_5 + d_5 Starting from the beginning v = ta + v_0 d = (t^2 / 2)a + t v_0 + d_0 from v_5 find v_0 and from d_5 find d_0, hence find d_25 > The first part was rather simple for me (finding the velocity at > t=25), and the second part (finding 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 I'd 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? It's easier just to do it than to juggle all those itchy numbers. How's 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 > figure 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 wasn't 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). Here's 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 I'm on the right track, because the first component > matches, but the other two, while close, don't actually match. I > figure 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 five 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 can't 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? Before you spend too much time on this, the problem is from Larson- Hostetler-Edwards /College Algebra with Trigonometry/. If it's 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 can't 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 its zeroes are algebraic (by defintion) they are not expressible in radicals. To see the group is S_5, note that the polynomial is irreducible modulo 5 and splits as an irreducible cubic multiplied by an irreducible quadratic modulo 3. Tus its Galois group contains elements of cycle structure 5 and 3 2. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > > A fellow teacher asked me this, and I can't 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 see http://mathworld.wolfram.com/QuinticEquation.html and http://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 it's 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. ==== It's a bit late, but I want to say THANKS to Robin and George for posting responses to my query. I've passed them on to the teacher who originally asked me. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. 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