I was listening to the radio a while ago and they were talking about a set theory. The theory basically said that everything belongs in a set, because if an element does not belong in a set, it belongs in the set of elements that do not belong in a set (does that make sense?). What is this theorem called? Does anyone have any resources on the web for it? Taras PS: I hope what I have written above is correct/makes sense :) ==== Yes this is makes sense I was listening to the radio a while ago and they were talking about a > set theory. The theory basically said that everything belongs in a > set, because if an element does not belong in a set, it belongs in the > set of elements that do not belong in a set (does that make sense?). > What is this theorem called? Does anyone have any resources on the > web for it? > Taras PS: I hope what I have written above is correct/makes sense :) ==== > I was listening to the radio a while ago and they were talking about a > set theory. The theory basically said that everything belongs in a > set, because if an element does not belong in a set, it belongs in the > set of elements that do not belong in a set (does that make sense?). > What is this theorem called? Does anyone have any resources on the > web for it? > There's no such theorem. Whether that's so depends upon the brand of set theory. It's not so in ZFC, but it is so in NF An easier way to express the same notion is everything is a member of the set that contains only itself. Even so, in ZFC, there are some boogie men that refuse membership in any and all sets. In NF those boogie men are exterminated, ie denied a visa of existence. > PS: I hope what I have written above is correct/makes sense :) > Some. I suggest you find a better source for popularize math and enjoy the new adventure. ==== Yes, this all makes sense. Thers is something called the universal set. The universal set is the set of all object that we are concerned about at the moment. For example if car is not in the universal set then for this problem there is no such thing as a car. Nothing that is not in the universal set exists. Now suppose the universal set, U, contains 1, 2, 3, 4, 5, 6, 7, 8, egg, and sponge. That is U = {1, 2, 3, 4, 5, 6, 7, 8, egg, sponge}. Now suppose set A = {1, 3,5,8, egg}--note that everything that is in is also in U. Now we can talk about A', the set that contains everything exact what is in A. (A' is called A complement or A prime)In our case A' = {2, 4, 6, 7, sponge).Notice that everything ( in U ) is either in A or not in A, that is A', by definition of A'. In the statement everything belongs in a set, because if an element does not belong in a set, it belongs in the set of elements that do not belong in a set when it says it belongs in the set of elements that do not belong in a set it is referring to the complement of the set. I was listening to the radio a while ago and they were talking about a > set theory. The theory basically said that everything belongs in a > set, because if an element does not belong in a set, it belongs in the > set of elements that do not belong in a set (does that make sense?). > What is this theorem called? Does anyone have any resources on the > web for it? > Taras PS: I hope what I have written above is correct/makes sense :) ==== > Yes, this all makes sense. Thers is something called the universal set. The > universal set is the set of all object that we are concerned about at the > moment. Makes sense yes. If it's true, depends upon the brand of set theory. The universal set U = { x | x = x } Now in ZFC set theory, currently the most accepted brand, it's not a set, it's a class. If were a set, then U would be in U, but that would contradict the axiom of regularity, so U can't be a set, nor can U belong to U I was listening to the radio a while ago and they were talking about a > set theory. The theory basically said that everything belongs in a > set, because if an element does not belong in a set, it belongs in the > set of elements that do not belong in a set (does that make sense?). > What is this theorem called? Does anyone have any resources on the > web for it? > Taras PS: I hope what I have written above is correct/makes sense :) > ==== > And as I pointed out before there have been people who have been > attacking algebra itself in posts. Now it still bothers me that I was > the one person standing up to defend algebra itself from what I > noticed. The image of JSH as a stand up guy defending any part of mathemetics from the attacks of others is hysterical. ==== > My paper does rest fully on the following rather basic lemma: > > Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c > is a factor of the contant term P(0), while r=g-c, so it varies if g > varies. That simple lemma not only anchors my paper, and my proof of Fermat's > Last Theorem, it can be used in lots of places. Now I'll use it with > the roots of > > y^3 + 3y - 2 I find its roots fascinating. > and the three roots are > > y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} > > y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - > > (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} > > y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - > > (-1+sqrt(-3))/2(1+sqrt(2))^{1/3} As simple as they are you can't just take the roots and *look* at them to tell whether or not any one of them is coprime to 2. It is just impossible. Yet I work that out and still I keep coming back, as the simplicity and elegance of those roots keeps calling me. Sure I know that the cuberoot is ambiguous, so depending on how you take the cuberoot, you shift between the solutions, which is why you can't prove whether or not one of them is coprime to 2, but still it seems like there should be some way to see by looking. But there isn't. Given the impossibility of proving that none of them are coprime to 2, you'd think that people objecting would sit back, but nope, they claim Galois Theory eliminates a possibility. But that can't be. Galois Theory is based on that shifting. Actually, they can neither prove nor disprove the assertion with the tools they're using. > where the polynomial I'm interested in is > > P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}. > > Letting > > g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1, > > I have c=1, so r=g-1. > > (Some may worry about my choice for c, I'll come back to that later.) > > Now using the substitution g=r+1, with x=sqrt(2), with my roots, I > have > > y_1 = r+1 - 1/(r+1) > > y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1) > > y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1) > > so I have > > y_1 = (r^2 + 2r)/(r+1) > > y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1) > > y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1) So once again I found myself drawn to that simplicity, looking for some way, some how to make the math definite to the eye--to make it eliminate a possibility on sight. But it doesn't. You can't look at those roots and tell. There's just no way to limit the math in that way, and intellectually I know why: The math has the flexibility to handle several possibilities. For some polynomials, none of roots would be coprime to integer factors of the constant term, while for others, like y^3 + 3y - 2, several are coprime. But you have to use advanced techniques to prove it. Still something keeps drawing me back to stare at those roots, looking for what cannot be found--a limitation that I can see. It's like in physics where you have quarks that can't be pulled out one by one. You can determine they're there experimentally in pairs or threesomes. You can bounce things off them, but you can't *see* them. And there's such a need to see. James Harris ==== > My paper does rest fully on the following rather basic lemma: Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c > is a factor of the contant term P(0), while r=g-c, so it varies if g > varies. That simple lemma not only anchors my paper, and my proof of Fermat's > Last Theorem, it can be used in lots of places. Now I'll use it with > the roots of y^3 + 3y - 2 I find its roots fascinating. You are again answering your own posts with narcissistic monologues instead of answering your critics. If you had any integrity, you would deal with the challenges not carry on conversations with yourself. Talk to a mirror if you want, but why use a newgroup to talk to yourself. > and the three roots are y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(1+sqrt(2))^{1/3} As simple as they are you can't just take the roots and *look* at them > to tell whether or not any one of them is coprime to 2. It is just > impossible. You don't have to just look. The observation that y_1 y_2 y_3 = 2 is sufficient to prove that each root is a divisor of 2 in the ring of algebraic integers and, since none of these is a unit, none are coprime to 2. > Yet I work that out and still I keep coming back, as the simplicity > and elegance of those roots keeps calling me. Didn't someone 'call you home' once? If so, how come you came back? > Sure I know that the cuberoot is ambiguous, so depending on how you > take the cuberoot, you shift between the solutions, What the hell does that mean? > which is why you > can't prove whether or not one of them is coprime to 2, Oh yes, you can. The product of the roots, y_1 y_2 y_3 equals 2. Hence, each of the roots divides 2 in the ring of algebraic integers. This is because the quotient equals the product of the other two roots and the product of two algebraic integers is an algebraic integer. Hence none of the roots are coprime to 2. > but still it > seems like there should be some way to see by looking. But there isn't. So what? If mathematician's could prove everything by 'looking' they wouldn't need theorems and lemmas. > Given the impossibility of proving that none of them are coprime to 2, Hold it. This is proven by noting that the product of the roots, y_1 y_2 y_3, equals 2 and that each of the roots divides 2 in the ring of algebraic integers -- meaning the quotient is an algebraic integer. Hence none of the roots are coprime to 2. QED > you'd think that people objecting would sit back, but nope, they claim > Galois Theory eliminates a possibility. I don't. > But that can't be. Galois Theory is based on that shifting. > Actually, they can neither prove nor disprove the assertion with the > tools they're using. Galois Theory isn't necessary in this case. 'Basic algebra' does the trick. You wouldn't want to attack 'basic algebra', would you? > So once again I found myself drawn to that simplicity, looking for > some way, some how to make the math definite to the eye--to make it > eliminate a possibility on sight. But it doesn't. You can't look at those roots and tell. As has been shown above, it isn't necessary to 'look at those roots and tell'. None of them is coprime to 2. You have been refuted, defeated, vanquished, humiliated, crushed, conquered, exposed and undone. Take down your false proof and acknowledge your errors. > There's just no way to limit the math in that way, and intellectually > I know why: The math has the flexibility to handle several possibilities. For > some polynomials, none of roots would be coprime to integer factors of > the constant term, while for others, like y^3 + 3y - 2, several are > coprime. None of them are coprime. The product of all the roots, y_1 y_2 y_3, is equal to 2. Each root divides 2 producing an algebraic integer quotient. Hence, in the ring of algebraic integers, none of the roots, count 'em again!, none is coprime to 2. Your claim is false. > But you have to use advanced techniques to prove it. You can't prove your claim at all, because it is false. > Still something keeps drawing me back to stare at those roots, looking > for what cannot be found--a limitation that I can see. The limitation is between your ears, not in your eyes. > It's like in physics where you have quarks that can't be pulled out > one by one. You can determine they're there experimentally in pairs > or threesomes. You can bounce things off them, but you can't *see* > them. No, it isn't like quarks. It is like an LSD trip. > And there's such a need to see. James Harris Open your eyes. You have been shown the way. To paraphrase Samuel Johnson: I can give you a solution, but I can't give you an understanding. -- 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 ==== [snip] >The math has the flexibility to handle several possibilities. For >some polynomials, none of roots would be coprime to integer factors of >the constant term, while for others, like y^3 + 3y - 2, several are >coprime. (I assume your talking about _algebraic integers_.) The constant term of y^3 + 3y - 2 is -2. If Y is _any_ root of y^3 + 3y - 2, then (-2)/Y is root of x^3 + 3x^2 + 4. That's a monic polynomial, so (-2)/Y is an algebraic integer, and therefore Y divides -2 in the ring of algebraic integers. So Y in _not_ coprime to -2. >But you have to use advanced techniques to prove it. No, you don't. Solving cubics is not an advanced technique. >Still something keeps drawing me back to stare at those roots, looking >for what cannot be found--a limitation that I can see. It's like in physics where you have quarks that can't be pulled out >one by one. You can determine they're there experimentally in pairs >or threesomes. You can bounce things off them, but you can't *see* >them. If that's not seeing (in physics), then what is? >And there's such a need to see. Yes. And (almost) everybody _does_. -- Thomas Wasell | I'd love to go out with you, but I'm having all my wasell@bahnhof.se | plants neutered. ==== |> Sure I know that the cuberoot is ambiguous, so depending on how you |> take the cuberoot, you shift between the solutions, | |What the hell does that mean? it means that he's much closer to actually understanding something about mathematics in this post than he has been in just about any of his other posts. he's still horribly mixed up, but there's an important truth that he almost seems close to understanding here. |> But that can't be. Galois Theory is based on that shifting. |> Actually, they can neither prove nor disprove the assertion with the |> tools they're using. | |Galois Theory isn't necessary in this case. galois theory is highly relevant and perhaps crucial here. |> It's like in physics where you have quarks that can't be pulled out |> one by one. You can determine they're there experimentally in pairs |> or threesomes. You can bounce things off them, but you can't *see* |> them. | |No, it isn't like quarks. It is like an LSD trip. no, it's in fact very much like quarks, and in a surprisingly deep way. but yes, it's also like an lsd trip. -- ==== > I've written a paper and it is currently at a journal. > And I'm waiting to hear further from them. Why wait? Everyone already knows what's going to happen. They're going to reject it, and you're going to make excuses. Why not start rationalizing your failure *now* and get a head start? -- 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 ==== |> Sure I know that the cuberoot is ambiguous, so depending on how you > |> take the cuberoot, you shift between the solutions, > | > |What the hell does that mean? it means that he's much closer to actually understanding something > about mathematics in this post than he has been in just about any of > his other posts. he's still horribly mixed up, but there's an > important truth that he almost seems close to understanding here. Why not enlighten the rest of us then? Evidently James isn't quite up to it, and you could perform a service by identifying the 'important truth he almost seems close to understanding'. > |> But that can't be. Galois Theory is based on that shifting. > |> Actually, they can neither prove nor disprove the assertion with the > |> tools they're using. > | > |Galois Theory isn't necessary in this case. galois theory is highly relevant and perhaps crucial here. It may be relevant, but it certainly isn't necessary or crucial in proving that the roots of a monic polynomial with integer coefficients are divisors of the constant term within the ring of algebraic integers. > |> It's like in physics where you have quarks that can't be pulled out > |> one by one. You can determine they're there experimentally in pairs > |> or threesomes. You can bounce things off them, but you can't *see* > |> them. > | > |No, it isn't like quarks. It is like an LSD trip. no, it's in fact very much like quarks, and in a surprisingly deep > way. but yes, it's also like an lsd trip. Please provide an in-depth explanation. Why leave the less informed hanging when you are able to clarify why it is that proving/disproving coprimality within the ring of algebraic integers is like quarks? > -- > Looking forward to your follow-up, as I assume your post was not a 'hit and run' contribution. -- 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 ==== >> My paper does rest fully on the following rather basic lemma: >> >> Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c >> is a factor of the contant term P(0), while r=g-c, so it varies if g >> varies. >> That simple lemma not only anchors my paper, and my proof of Fermat's >> Last Theorem, it can be used in lots of places. Now I'll use it with >> the roots of >> >> y^3 + 3y - 2 I find its roots fascinating. > >> and the three roots are >> >> y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} >> >> y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - >> >> (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} >> >> y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - >> >> (-1+sqrt(-3))/2(1+sqrt(2))^{1/3} As simple as they are you can't just take the roots and *look* at them >to tell whether or not any one of them is coprime to 2. It is just >impossible. No, it is not impossible. It is amazingly trivial, as Keith Ramsay pointed out right after you posted them: NONE OF THEM ARE COPRIME TO 2. THEY ARE ALL DIVISORS OF 2: y_1*(y_1^2+3) = 2 y_2*(y_2^2+3) = 2 y_3*(y_3^2+3) = 2. Surely you agree with all three of those expressions, since they are all roots of y^3 + 3y - 2. So they are all divisors of 2, and since none of them are units, they are none of them coprime to 2. Simple, nay, trivial. [.snip.] >Given the impossibility of proving that none of them are coprime to 2, it follows that red is blue, 0 is 1, and I am the Pope (who, therefore, is not catholic). >you'd think that people objecting would sit back, but nope, they claim >Galois Theory eliminates a possibility. Not even Galois theory is needed here. ->THIS<- one is trivial. [.snip.] ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== [.snip.] Nitpicking... The first time you made the point, you correctly stated: >You don't have to just look. The observation that y_1 y_2 y_3 = 2 is sufficient to prove that each root is >a divisor of 2 in the ring of algebraic integers and, since none of these is a unit, none are coprime to 2. [.snip.] Later, you said, in several different versions, things like: >Oh yes, you can. The product of the roots, y_1 y_2 y_3 equals 2. Hence, each of the roots divides 2 in the >ring of algebraic integers. This is because the quotient equals the product of the other two roots and the >product of two algebraic integers is an algebraic integer. Hence none of the roots are coprime to 2. The last statement only follows once you add since none of them is a unit [in the ring of algebraic integers]. ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== [.snip.] Nitpicking... The first time you made the point, you correctly stated: You don't have to just look. The observation that y_1 y_2 y_3 = 2 is sufficient to prove that each root is >a divisor of 2 in the ring of algebraic integers and, since none of these is a unit, none are coprime to 2. [.snip.] Later, you said, in several different versions, things like: Oh yes, you can. The product of the roots, y_1 y_2 y_3 equals 2. Hence, each of the roots divides 2 in the >ring of algebraic integers. This is because the quotient equals the product of the other two roots and the >product of two algebraic integers is an algebraic integer. Hence none of the roots are coprime to 2. The last statement only follows once you add since none of them is a > unit [in the ring of algebraic integers]. off one or more repititions of the same, or similar, argument. I don't take it as nitpicking, by the way. The argument is so short that there is no reason (or excuse) for it to be incomplete. -- 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 ==== How to compute this limit: lim |z + h|/h h -> 0 where h is a real number and z is a complex number thanks ==== > How to compute this limit: lim |z + h|/h > h -> 0 where h is a real number and z is a complex number > Let z = x + iy |z + h|/h = ((x+h)^2 + y^2)/h = (x^2 + y^2)/h + 2x + h lim = oo if z /= 0 = 1 if z = 0 ==== William, I am totally confused. I easily see why you say the limit is 1 if z = 0--you plug in 0 for z in |z+h | to get |h| etc... Now I know that you can't divide by 0 and so on but when I take the lim h-->0 of (x^2 + y^2)/h + 2x + h where x = y =0 then I get lim h-->0 (0/h + 2*0 + h) = lim h-->0 (0 + 0 + h) = limh-->0 (h) = 0 not 1. What is going on here?? How to compute this limit: lim |z + h|/h > h -> 0 where h is a real number and z is a complex number Let z = x + iy > |z + h|/h = ((x+h)^2 + y^2)/h > = (x^2 + y^2)/h + 2x + h lim = oo if z /= 0 > = 1 if z = 0 ==== No, I don't agree that lim h-->0 ( |h|/h) = 1. This limit dne! How to compute this limit: lim |z + h|/h > h -> 0 where h is a real number and z is a complex number Let z = x + iy > |z + h|/h = ((x+h)^2 + y^2)/h > = (x^2 + y^2)/h + 2x + h lim = oo if z /= 0 > = 1 if z = 0 ==== > William, I am totally confused. I easily see why you say the limit is 1 if z > = 0--you plug in 0 for z in |z+h | to get |h| etc... Now I know that you > can't divide by 0 and so on but when I take the lim h-->0 of (x^2 + y^2)/h + > 2x + h where x = y =0 then I get lim h-->0 (0/h + 2*0 + h) = lim h-->0 (0 + > 0 + h) = limh-->0 (h) = 0 not 1. What is going on here?? William missed the Sqrt() when calculating |z+h|... For z=0, the expression becomes |h|/h which is +/-1 depending on whether h>0 or h<0. So the limit as h->0 does not exist, although separate limits exist if we consider h->0 from above or from below. When z /= 0, we have a similar situation but this time the expression tends to +/- infinity if h->0 from above or below respectively. Mike. ps. I think top-posting is frowned upon here! (doesn't worry me too much though..) How to compute this limit: lim |z + h|/h > h -> 0 where h is a real number and z is a complex number Let z = x + iy > |z + h|/h = ((x+h)^2 + y^2)/h > = (x^2 + y^2)/h + 2x + h lim = oo if z /= 0 > = 1 if z = 0 ==== > No, I don't agree that lim h-->0 ( |h|/h) = 1. This limit dne! How to compute this limit: lim |z + h|/h > h -> 0 where h is a real number and z is a complex number Let z = x + iy > |z + h|/h = ((x+h)^2 + y^2)/h > = (x^2 + y^2)/h + 2x + h lim = oo if z /= 0 > = 1 if z = 0 > > Top posting is very confusing. Let's consider lim h->0 (|h|/h). Suppose h is really really small, like 1/zillion. Then |h| is 1/zillion and h is 1/zillion and |h|/h is 1. This works out the same no matter how small h is. When evaluating this limit, it makes no difference whether |h|/h is defined for h=0. All that matter is what happens when h is CLOSE TO 0. And for any value of h close to 0, |h|/h is 1. That's the definition of a limit. ==== I actually have a Math disability and to graduate college I have to take this class which is about Math logic. It's going really fast for me and the other math phoebes. I have been to a hundred web sites and none of them have helped. I need to know the following: 1.What the hell dose the Q and the P stand for? true and false? 2. How do you read a truth table? 3. How do I find the principle of inference? 4. Principle of a syllogism? And on and on. If I don't pass I don't graduate and that's even with a proven disability. ==== > I actually have a Math disability and to graduate college I have to take > this class which is about Math logic. It's going really fast for me and the > other math phoebes. I have been to a hundred web sites and none of them have > helped. I need to know the following: > > 1.What the hell dose the Q and the P stand for? true and false? Some context would have been nice but I'll take a guess. I suspect that P and Q or more likely, p and q stand for staements. Informally a statement is a sentence that can be either true or false. P := My cat is black is a statement. Q := Get lost. is not. > 2. How do you read a truth table? A truth table is a way of determining the truth values of a compound statement involving P and Q (and may even a third statement R) like not(P and Q). The top row will have P, Q ,P and Q and not(P and Q). In the columns headed by P and Q, you put all possible truth values of P and Q: P Q T T T F F T F F. The second row (actually the third but the top one doesn't count) says P is true and Q is false. So, to complete the second row, you will consult your and truth table and discover that when P is true (T) and Q is false (F), P and Q is false so you'll put F in the second row of the P and Q column. Now, you'll consult your not truth table and discover that when P and Q is false (F), not(P and Q) is true (T). So the whole second row of the table is: T F F T. That is, when P is true and Q is false, not(P and Q) is true. An aside: your not table will probably have a column labeled P. _That_ P is _your_ P and Q. If P:= 2 + 3 = 5 and Q := All cats have 10 lives then not(P and Q) is true. > 3. How do I find the principle of inference? I've no idea. I need some context, example, something. > 4. Principle of a syllogism? Ditto. One doesn't, as a rule, find principles so I don't know what the last two questions are. -- Paul Sperry Columbia, SC (USA) ==== >I actually have a Math disability and to graduate college I have to take >this class which is about Math logic. It's going really fast for me and the >other math phoebes. >I have been to a hundred web sites and none of them have >helped. Statements like this always trigger my b---s--- detector. >I need to know the following: 1.What the hell dose the Q and the P stand for? true and false? >2. How do you read a truth table? >3. How do I find the principle of inference? >4. Principle of a syllogism? These are all answered in your textbook, and at much greater length than we could write out in a newsgroup. If the textbook doesn't help you, how can we? I don't understand any of it may be your honest feeling but it gives us nothing to work with, no way to know how to help you. If you can ask specific questions, we can provide specific answers. If you show evidence that you have actually tried to understand the material, we'll be more eager to. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry ==== Well finally there's resolution, but it probably won't be satisfying to many of you though it's greatly satisfying to me, as the problem I've highlighted is rather easy to demonstrate, and is a problem with *actual* decomposition of algebraic integers. Many of you may have considered decomposition of algebraic integers in the abstract, but if you'd dug into it, you might have made math history. Consider the polynomial y^3 + 12y - 65 which of course has 3 roots and is irreducible over Q. Now imagine dividing off all factors of 5 from its roots and getting the polynomial for which the results are roots. Give that polynomial. It turns out that you can't as there's a problem with decomposing algebraic integers in actuality. What I can do is start you out on the polynomial as you have x^3 + sx^2 + tx - 13 and what neither you, nor any other mortal in all of the Universe can do is give s, and t, to *see* the polynomial. Not even I can even though I know that only two of the roots of y^3 + 12y - 65 have a factor that is sqrt(5), but you see, you can't pick them out! You can prove that only two have that factor but because of the cuberoot operator, which must be used to display them, you have a fundamental ambiguity, like how sqrt(1) = +/- 1, which prevents you from picking out which ones. So you can't calculate, and neither can you display s or t. They are fundamentally unknowable, which is a fascinating result. Don't believe me? Well then, go find s and t. James Harris ==== >Well finally there's resolution, but it probably won't be satisfying >to many of you though it's greatly satisfying to me, as the problem >I've highlighted is rather easy to demonstrate, and is a problem with >*actual* decomposition of algebraic integers. The sort of decompisition you have in mind _seems_ to be something like a unique prime factorization. It's actually well-known that there is no such thing in the algebraic integers - many of your mistakes arise from assuming that there is such a thing even though there isn't (see below). This is not a problem with the algebraic integers, it's a problem with your understanding of them. >Many of you may have considered decomposition of algebraic integers in >the abstract, but if you'd dug into it, you might have made math >history. Just like you've made history. Right. >Consider the polynomial y^3 + 12y - 65 which of course has 3 roots and is irreducible over Q. Now imagine dividing off all factors of 5 from its roots and getting >the polynomial for which the results are roots. There are at least two things you might mean by this: (i) take each root and divide by 5 as many times as possible (while remaining within the algebraic integers), (ii) take each root and divide out all the factors (in the algebraic integers) that it shares with 5. In other words, divide each root r by GCD(r, 5), where GCD means greatest common divisor in the algebraic integers. You might note that I think that (i) is the most natural actually mean is (ii). But let's not worry about that - whichever one you meant, there's nothing historic here: Case 1: Assuming you meant (i): Then the things you say below we cannot do are very easy, because it's easy to see that none of the roots are divisible by 5: If r is a root then r/5 is a root of 25y^3 + 12y - 13, hence r is not an algebraic integer. So after doing (i) the roots are unchanged, and the polynomial you say we can't find is just y^3 + 12y - 65. Case 2: Assuming you mean (ii): Asking us to divide out all the factors of 5 in the sense of (ii) simply doesn't make any sense. Someone correct me if I'm wrong: There simply _is_ no GCD in the algebraic integers! (If there _were_ a unique prime factorization in the algebraic integers this would make sense.) >Give that polynomial. It turns out that you can't as there's a problem with decomposing >algebraic integers in actuality. What I can do is start you out on the polynomial as you have x^3 + sx^2 + tx - 13 and what neither you, nor any other mortal in all of the Universe can >do is give s, and t, to *see* the polynomial. Not even I can No. Surely _you_ can? >even though I know that only two of the roots of y^3 + 12y - 65 have a factor that is sqrt(5), It's amazing how you repeat the same errors over and over. You've never explained _why_ any of the roots have factors of sqrt(5), and you ignore repeated very simple proofs that _none_ of them do! Well, since I haven't been following the details maybe this is a new one, and nobody's yet showed you a simple proof that you're wrong. It's incredibly simple, so simple even I can do it: Say r is a root of y^3 + 12y - 65. Then r/sqrt(5) is easily seen to be a root of 25x^6 + 120x^4 + 144x^2 - 845. Since thus polynomial is irreducible (or so Mathematica claims) it follows that r/sqrt(5) is not an algebraic integer. So there you are. _None_ of the roots have sqrt(5) as a factor (in the algebraic integers). And this is not hard to see, it's only a tiny bit more complicated than seeing that none of the roots have 5 as a factor. HINT: when you mistakes are so basic that it's trivial for a guy like me to demonstrate the error, even though I really don't know anything about these things, it follows that you're making really basic mistakes. Over and over. And over. > but you see, you can't pick them out! Yeah. Just like you can't pick out the unicorn hiding among all the other animals at the zoo. (_Some_ people _can_ see the unicorn. They're not ordinary mortals. Of course above you imply that you're not an ordinary mortal either. Hmm.) >You can prove that only two have that factor but because of the >cuberoot operator, which must be used to display them, you have a >fundamental ambiguity, like how sqrt(1) = +/- 1, which prevents you >from picking out which ones. So you can't calculate, and neither can you display s or t. They are fundamentally unknowable, which is a fascinating result. Don't believe me? Well then, go find s and t. >James Harris ************************ David C. Ullrich ==== [cut] > because of the > cuberoot operator, which must be used to display them, You do not need the cube root operator to display such numbers. For example, if you use the ordered pairs of real numbers (r,s) to represent the complex numbers in the standard way, then the two square roots of -1 are (0, 1) and (0, -1). I don't need the square root operator. > you have a > fundamental ambiguity, like how sqrt(1) = +/- 1, which prevents you > from picking out which ones. There is no ambiguity for sqrt(1) = +/- 1 since x^2 - 1 is not irreducible. Even for the cube root operator, there is no ambiguity in displaying such numbers. The ambiguity that you are referring to is the following. You say you are thinking of one of the cube roots and you want me to tell you which one you are thinking of. Of course, I can't do that. But, that does not mean that I can't write down the three cube roots (in a given representation of the number system) unambiguously. Mine point is the following. You need to have already established the existence of the numbers that you are discussing. There can't be any ambiguity here. After having established the existence of these numbers, then you can start talking about some of them be roots of a particular irreducible cubic polynomial. Those three roots will be distinguishable in the pre-existing established number system. -- Bill Hale ==== [snip] > Yeah. Just like you can't pick out the unicorn > hiding among all the other animals at the zoo. -- 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 ==== > Well finally there's resolution, but it probably won't be satisfying to James as it shows that he is totally demented. ==== |Case 2: Assuming you mean (ii): Asking us to |divide out all the factors of 5 in the sense of |(ii) simply doesn't make any sense. Someone |correct me if I'm wrong: There simply _is_ |no GCD in the algebraic integers! | |(If there _were_ a unique prime factorization in the |algebraic integers this would make sense.) i wouldn't swear i'm not screwing up here, but the impression i have is that the algebraic integers have a good concept of gcd _despite_ lacking unique prime factorization. that is, arturo magidin mentioned in a post on april 13 that the algebraic integers form a bezout domain, which is a kind of ring in which (if i understand correctly) the sum of any two principal ideals is principal. the operation of sum of principal ideals corresponds to the operation of gcd of ring elements, so a bezout domain has a good concept of gcd. -- ==== > Well finally there's resolution, but it probably won't be satisfying > to many of you though it's greatly satisfying to me, as the problem > I've highlighted is rather easy to demonstrate, and is a problem with > *actual* decomposition of algebraic integers. > > Many of you may have considered decomposition of algebraic integers in > the abstract, but if you'd dug into it, you might have made math > history. > > Consider the polynomial > > y^3 + 12y - 65 > > which of course has 3 roots and is irreducible over Q. > > Now imagine dividing off all factors of 5 from its roots and getting > the polynomial for which the results are roots. > > Give that polynomial. > > It turns out that you can't as there's a problem with decomposing > algebraic integers in actuality. > > What I can do is start you out on the polynomial as you have > > x^3 + sx^2 + tx - 13 > > and what neither you, nor any other mortal in all of the Universe can > do is give s, and t, to *see* the polynomial. Damn. Um, I think you can give them. Oh well, back to the drawing board. James Harris ==== > Well finally there's resolution, but it probably won't be satisfying > to many of you though it's greatly satisfying to me, as the problem > I've highlighted is rather easy to demonstrate, and is a problem with > *actual* decomposition of algebraic integers. > > Many of you may have considered decomposition of algebraic integers in > the abstract, but if you'd dug into it, you might have made math > history. > > Consider the polynomial > > y^3 + 12y - 65 > > which of course has 3 roots and is irreducible over Q. > > Now imagine dividing off all factors of 5 from its roots and getting > the polynomial for which the results are roots. > > Give that polynomial. > > It turns out that you can't as there's a problem with decomposing > algebraic integers in actuality. > I concede it is difficult to do the computation (though not impossible). But the real question here is: so what? Does the fact that it is difficult to write down such a polynomial prove that you are right in your claims, or prove that the rest of us are wrong? Isn't it just an irrelevant side issue? > What I can do is start you out on the polynomial as you have > > x^3 + sx^2 + tx - 13 > > and what neither you, nor any other mortal in all of the Universe can > do is give s, and t, to *see* the polynomial. > > Not even I can even though I know that only two of the roots of > > y^3 + 12y - 65 > > have a factor that is sqrt(5), but you see, you can't pick them out! > Let y be any root of [1] y^3 + 12y - 65 = 0. Substitute y = x + 5. Then [1] is equivalent to x^3 + 15*x^2 + 75*x = -120, or [2] x*(x^2 + 15*x + 75) = -120. Now suppose x is coprime to 5. Then clearly x^2 + 15*x + 75 is also coprime to 5. Therefore in [2], both x and (x^2 + 15*x + 75) are coprime to 5. The right side, however, is 120, which is not coprime to 5. There is a theorem which says: If A and B are algebraic integers which are coprime to p, then both A and B are coprime to p. [see proof below {***}] Therefore a contradiction. Therefore x is NOT coprime to 5. But y = x + 5. Therefore y is also not coprime to 5. Therefore ALL roots of [1] are not coprime to 5. Therefore your statement that exactly two of the roots are divisible by sqrt(5) (and thus that the other one is necessarily coprime to 5) is wrong. {***} Proof of theorem: Assume A is coprime to p and B is coprime to p. Then there exist algebraic integers r and s, and r' and s' such that r*A + s*p = 1 and r'*B + s'*p = 1. Therefore r*A*B + s*p*B = B. Therefore r'*(r*A*B + s*p*B) + s'*p = 1 This can be rearranged as: r'*r*(A*B) + (r'*s*B + s')*p = 1, which implies that A*B and p are coprime. QED. Nora B. [more irrelevant stuff deleted] ==== There was a dumb misstatement in my previous message, corrected below at {###}. Nora B. > Well finally there's resolution, but it probably won't be satisfying > to many of you though it's greatly satisfying to me, as the problem > I've highlighted is rather easy to demonstrate, and is a problem with > *actual* decomposition of algebraic integers. > > Many of you may have considered decomposition of algebraic integers in > the abstract, but if you'd dug into it, you might have made math > history. > > Consider the polynomial > > y^3 + 12y - 65 > > which of course has 3 roots and is irreducible over Q. > > Now imagine dividing off all factors of 5 from its roots and getting > the polynomial for which the results are roots. > > Give that polynomial. > > It turns out that you can't as there's a problem with decomposing > algebraic integers in actuality. > I concede it is difficult to do the computation (though not impossible). But the real question here is: SO WHAT? Does the fact that it is difficult to write down such a polynomial prove that you are right in your claims, or prove that the rest of us are wrong? Isn't it just an irrelevant side issue? > What I can do is start you out on the polynomial as you have > > x^3 + sx^2 + tx - 13 > > and what neither you, nor any other mortal in all of the Universe can > do is give s, and t, to *see* the polynomial. > > Not even I can even though I know that only two of the roots of > > y^3 + 12y - 65 > > have a factor that is sqrt(5), but you see, you can't pick them out! > Let y be any root of [1] y^3 + 12y - 65 = 0. Substitute y = x + 5. Then [1] is equivalent to x^3 + 15*x^2 + 75*x = -120, or [2] x*(x^2 + 15*x + 75) = -120. Now suppose x is coprime to 5. Then clearly x^2 + 15*x + 75 is also coprime to 5. Therefore in [2], both x and (x^2 + 15*x + 75) are coprime to 5. The right side, however, is -120, which is not coprime to 5. There is a theorem which says: If A and B are algebraic integers which {###} are coprime to p, then A * B is coprime <--Misstatement was here> to p. [See proof below.] Therefore a contradiction. Therefore x is NOT coprime to 5. But y = x + 5. Therefore y is also not coprime to 5. Therefore ALL roots of [1] are not coprime to 5. Therefore your statement that exactly two of the roots are divisible by sqrt(5) (and thus that the other one is necessarily coprime to 5) is wrong. {***} Proof of theorem: Assume A is coprime to p and B is coprime to p. Then there exist algebraic integers r and s, and r' and s', such that r*A + s*p = 1 and r'*b + s'*p = 1. Therefore (multiplying the first equality by B), r*A*B + s*p*B = B. Therefore (using the second equality) r'*(r*A*B + s*p*B) + s'*p = 1. This can be rearranged as r'*r*(A*B) + (r'*s*B + s')*p = 1, which implies that A*B and p are coprime. QED. Nora B. [more irrelevant stuff deleted] ==== Sorry, my previous replies to this post were just plain wrong due to an algebraic error. Note however that David Ullrich has given a simple valid disproof of Harris's statement that two of the roots of y^3 + 12*y - 65 are divisible by sqrt(5). Nora B. > Well finally there's resolution, but it probably won't be satisfying > to many of you though it's greatly satisfying to me, as the problem > I've highlighted is rather easy to demonstrate, and is a problem with > *actual* decomposition of algebraic integers. > > Many of you may have considered decomposition of algebraic integers in > the abstract, but if you'd dug into it, you might have made math > history. > > Consider the polynomial > > y^3 + 12y - 65 > > which of course has 3 roots and is irreducible over Q. > > Now imagine dividing off all factors of 5 from its roots and getting > the polynomial for which the results are roots. > > Give that polynomial. > > It turns out that you can't as there's a problem with decomposing > algebraic integers in actuality. > > What I can do is start you out on the polynomial as you have > > x^3 + sx^2 + tx - 13 > > and what neither you, nor any other mortal in all of the Universe can > do is give s, and t, to *see* the polynomial. > > Not even I can even though I know that only two of the roots of > > y^3 + 12y - 65 > > have a factor that is sqrt(5), but you see, you can't pick them out! > > You can prove that only two have that factor but because of the > cuberoot operator, which must be used to display them, you have a > fundamental ambiguity, like how sqrt(1) = +/- 1, which prevents you > from picking out which ones. > > So you can't calculate, and neither can you display s or t. > > They are fundamentally unknowable, which is a fascinating result. > > Don't believe me? Well then, go find s and t. > > > James Harris ==== [.snip.] >(ii) take each root and divide out all the factors (in the >algebraic integers) that it shares with 5. In other words, >divide each root r by GCD(r, 5), where GCD means >greatest common divisor in the algebraic integers. [.snip.] >Case 2: Assuming you mean (ii): Asking us to >divide out all the factors of 5 in the sense of >(ii) simply doesn't make any sense. Someone >correct me if I'm wrong: There simply _is_ >no GCD in the algebraic integers! (If there _were_ a unique prime factorization in the >algebraic integers this would make sense.) There is a natural way of defining a gcd, but it is not uniquely determined; it is only determined up to units. Since the algebraic integers are a Bezout domain (so every finitely generated ideal is principal), then given any two algebraic integers a and b, the ideal (a,b) is principal. Let c be any generator of (a,b). Then c satisfies the following two properties: (1) c|a and c|b; (2) if d is any algebraic integer such that d|a and d|b, then d|c. So it makes sense to call c ->a<- greatest common divisor of a and b, and it is unique up to units. ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > I don't think this works...consider the group G/H = (Z_6 x > Z_4)/<(3,2)>. This group has order 12 and is generated by _(1,-1)_ = > _(1,0)_ - _(0,1)_ (I'm using underscoring to represent cosets of > <(3,2)> in Z_6 x Z_4. > > Then n = 12, x = 3, y = 2, so that would mean > > G/H = <_(1,0)_, _(0,1)_ | 4_(1,0)_ = _(0,0)_ and 2_(1,0)_ = > 3_(0,1)_>. > > But 4_(1,0)_ = _(0,0)_ is false....d'oh. > > Maybe I'm misunderstanding your original qustion. Suppose we have a cyclic group G of order n, generated by some element (g-h), and obeying the presentation (with g + h = h + g implicit): G = I thought you wanted values for a, b, c that would generate Z_n, and such that (g-h) is a generator. This does not imply that _every_ pair of elements g' and h' with (g'-h') = (g - h) will satisfy ag' = 0, cg' = bh'; it's a statement that, given that there exist elements g and h in G satisfying the relations, (g-h) will generate the group, which will be isomorphic to Z_n. In your example, let g = 3*_(1,-1)_ = _(3,-3)_; and let h = 2*_(1,-1)_ = _(2,-2)_. Clearly, 4*g = 12*_(1,-1)_ = _(0,0)_; and 2*g = 6*_(1,-1)_ = 3*h; and g-h = _(1,-1)_, which as you note generates the group. If you have _already selected_ g and h from G cyclic of order n, with the property that G = <(g-h)>, can we always get a good presentation? Let k = (g-h). Then since G = , for some x, g = xk; and for y = x-1, h = yk. Then the order of g is a = lcm(n,x)/x, and r = lcm(n,y)/y is the order of h. Then yg = yxk = xyk = xh. Let c = y mod a, and b = x mod r; then cg = bh. Assuming c and b are coprime (I think they always are, can't prove it off the top of my head, but I think it follows from x coprime to y), then g and h satisfy the presentation: G = and G is isomorphic to Z_n. In your example, k = _(1,-1)_. g = _(1,0)_ = {(4,2), (1,0)}; it has order a = 6, and g = 10*k = _(10,-10)_ = _(4,2)_ = _(1,0)_. h = _(0,-1)_ has order r = 4, and h = 9*k = _(9,-9)_ = _(3,-1)_ = _(3,3)_ = _(1,0)_. 9g = 10h, so 3g = 2h. So G = (3g - 2h) - h = 3g - 3h = 3(g-h) = -h (3g - 2h) - g = 2g - 2h = 2(g-h) = -g giving us (as we hoped and expected!) g = 10(g-h) h = 9(g-h) and so the group is cyclically generated by (g-h). Assuming that g+h = h+g is implicit... > > For any n, let the prime decomposition of n be n = p_1^r_1 p_2^r_2 ... > p_k^r_k; and let x and y be integers > 1 such that xy = prod(p_i) for > i = 1 to k and, WLOG, x > y. Then (x - y) will not be divisible by > p_i for any i; and thus (x-y) will be coprime to n, and generate Z_n. > > Let a = n/x, b = y, c = x. Then should give a > presentation (assuming g+h = h+g) of Z_n - I think!! > ==== > <(3,2)> = {(3,2),(0,0)} > <_(1,-1)_> = <_(1,3)_ = {_(1,3)_, _(2,2)_, _(3,1)_, _(4,0)_, _(5,3)_, _(0,2)_, > _(1,1)_, _(2,0)_, _(3,3)_, _(4,2)_, _(5,1)_, _(0,0)_} > > _(2,0)_ = 2_(1,0)_ = 3_(0,1)_ = _(0,3)_ ??? ^^^ ||| Why do you think this? _(3,3)_ = _(0,3)_ Perhaps you meant: _(3,0)_ = { (3,2) + (3,0), (0,0) + (3,0)} = ( (0,2), (3,0) } = ( (0,2) + (0,0), (0,2) + (3,2)} = _(0,2)_ > _(0,0)_ = _(2,0)_ - _(0,3)_ = _(2,-3)_ = _(2,1)_ ??? > _(0,0)_ = _(3,0)_ - _(0,2)_ = _(3,-2)_ = _(3,2)_ = _(0,0)_ ==== It's a reply to a line of yours below. > What does this line _(0,0)_ = _(2,0)_ - _(0,3)_ = _(2,-3)_ = _(2,1)_ ??? intend to show? > <(3,2)> = {(3,2),(0,0)} > <_(1,-1)_> = <_(1,3)_ = {_(1,3)_, _(2,2)_, _(3,1)_, _(4,0)_, _(5,3)_, _(0,2)_, > _(1,1)_, _(2,0)_, _(3,3)_, _(4,2)_, _(5,1)_, _(0,0)_} _(2,0)_ = 2_(1,0)_ = 3_(0,1)_ = _(0,3)_ ??? > _(0,0)_ = _(2,0)_ - _(0,3)_ = _(2,-3)_ = _(2,1)_ ??? As you top post, I presume you want me to top post also. I don't think this works...consider the group G/H = (Z_6 x > Z_4)/<(3,2)>. This group has order 12 and is generated by _(1,-1)_ = > _(1,0)_ - _(0,1)_ (I'm using underscoring to represent cosets of > <(3,2)> in Z_6 x Z_4. Then n = 12, x = 3, y = 2, so that would mean G/H = <_(1,0)_, _(0,1)_ | 4_(1,0)_ = _(0,0)_ and 2_(1,0)_ = > 3_(0,1)_>. But 4_(1,0)_ = _(0,0)_ is false....d'oh. ==== I don't, Heather did. > <(3,2)> = {(3,2),(0,0)} > <_(1,-1)_> = <_(1,3)_ = {_(1,3)_, _(2,2)_, _(3,1)_, _(4,0)_, _(5,3)_, _(0,2)_, > _(1,1)_, _(2,0)_, _(3,3)_, _(4,2)_, _(5,1)_, _(0,0)_} _(2,0)_ = 2_(1,0)_ = 3_(0,1)_ = _(0,3)_ ??? > ^^^ > ||| > Why do you think this? _(3,3)_ = _(0,3)_ Perhaps you meant: _(3,0)_ = { (3,2) + (3,0), (0,0) + (3,0)} > = ( (0,2), (3,0) } > = ( (0,2) + (0,0), (0,2) + (3,2)} > = _(0,2)_ _(0,0)_ = _(2,0)_ - _(0,3)_ = _(2,-3)_ = _(2,1)_ ??? > _(0,0)_ = _(3,0)_ - _(0,2)_ = _(3,-2)_ = _(3,2)_ = _(0,0)_ ==== > > <(3,2)> = {(3,2),(0,0)} > <_(1,-1)_> = <_(1,3)_ = {_(1,3)_, _(2,2)_, _(3,1)_, _(4,0)_, _(5,3)_, _(0,2)_, > _(1,1)_, _(2,0)_, _(3,3)_, _(4,2)_, _(5,1)_, _(0,0)_} _(2,0)_ = 2_(1,0)_ = 3_(0,1)_ = _(0,3)_ ??? > ^^^ > ||| > Why do you think this? _(3,3)_ = _(0,3)_ I don't, Heather did. > AH! I see... sorry! ==== > For any n, let the prime decomposition of n be n = p_1^r_1 p_2^r_2 ... > p_k^r_k; and let x and y be integers > 1 such that xy = prod(p_i) for > i = 1 to k and, WLOG, x > y. Then (x - y) will not be divisible by > p_i for any i; and thus (x-y) will be coprime to n, and generate Z_n. > > Let a = n/x, b = y, c = x. Then should give a > presentation (assuming g+h = h+g) of Z_n - I think!! > I think you want > a = n/y, b = y, c = x. In my example, this gives a = 12/2 = 6, and indeed 6*_(1,0)_ = _(0,0)_ ==== Actually you want G/H = . Can you get a symmetric condition in terms of the order of h? At first glance it would look like ord(h)=n/x, but in our example, this would mean ord(_(0,1)_)=12/3=4. We need 4*_(0,1)_ = _(0,0)_. ==> 4*[ (0,1) + <(3,2)> ] = <(3,2)> ==> 4<(3,2)>=<(3,2)> ==> (12,8)=(3,2) .... False. > For any n, let the prime decomposition of n be n = p_1^r_1 p_2^r_2 ... > p_k^r_k; and let x and y be integers > 1 such that xy = prod(p_i) for > i = 1 to k and, WLOG, x > y. Then (x - y) will not be divisible by > p_i for any i; and thus (x-y) will be coprime to n, and generate Z_n. > > Let a = n/x, b = y, c = x. Then should give a > presentation (assuming g+h = h+g) of Z_n - I think!! > > > I think you want > > a = n/y, b = y, c = x. > > In my example, this gives a = 12/2 = 6, and indeed 6*_(1,0)_ = _(0,0)_ ==== If we were to create a tree structure with predicate logic as the highest node on the tree what other disciplines would be under that tree? Also are their any disciplines that are of a higher priority then Predicate logic? Explain. If you have any web resources to illustrate the answer that would be great. -- David ==== > If we were to create a tree structure with predicate logic as the highest > node on the tree what other disciplines would be under that tree? Also are > their any disciplines that are of a higher priority then Predicate logic? > > Explain. _You_ should explain. The question seems incomprehensible. Is the tree structure supposed to indicate which courses are prerequisites to which in a graduate program in mathematical logic? Or is it about something else? If so, what? Mike Hardy