conway Posted February 13, 2017 Posted February 13, 2017 (edited) As an addition to all current field axioms. "For every A in S there exists a z1 and a z2 constituting A. Such that any A in operation of a binary expression of mulitpilcation or divison is only representing z1 or z2. Such that z1 and z2 for all A's other than zero equal A. Such that z1 for zero equals zero. Such that z2 for zero equals 1. " Edited February 13, 2017 by conway
wtf Posted February 13, 2017 Posted February 13, 2017 (edited) Can you give examples for familiar fields such as the rationals, the reals, and the integers mod 5? Edited February 13, 2017 by wtf
studiot Posted February 13, 2017 Posted February 13, 2017 (edited) As an addition to all current field axioms. "For every A in S there exists a z1 and a z2 constituting A. Such that any A in operation of a binary expression of mulitpilcation or divison is only representing z1 or z2. Such that z1 and z2 for all A's other than zero equal A. Such that z1 for zero equals zero. Such that z2 for zero equals 1. " How would this work for the F4 field, comprising four non numeric elements {A,B,C,D}, defined below by the two binary operations, # and * , which satisfy all the standard Field Axioms? Edited February 13, 2017 by studiot
conway Posted February 13, 2017 Author Posted February 13, 2017 wtf a rational example .5 contains (.5 z1, and .5 z2 ) a real example 2 contains (2 z1, and 2 z2) a integer example -1 contains (-1 z1, and -1 z2 ) studiot If the "non numeric elements" are not zero... no new affects occurs. If the non numeric elements are zero, then the multiplication of reals by zero become relative to which non numeric element is z1 and which is z2 in the given binary expression. For example A = any numeric element other than zero 0(z1) x A(z2) = 0 0(z2) x A(z1) = A
studiot Posted February 13, 2017 Posted February 13, 2017 studiot If the "non numeric elements" are not zero... no new affects occurs. If the non numeric elements are zero, then the multiplication of reals by zero become relative to which non numeric element is z1 and which is z2 in the given binary expression. For example A = any numeric element other than zero 0(z1) x A(z2) = 0 0(z2) x A(z1) = A You completely failed to understand my question. I deliberately chose a field, that contains neither zero nor one. In fact I specified a non numeric field and emboldened that fact to help you. So please read my question again reconsider your answer.
conway Posted February 13, 2017 Author Posted February 13, 2017 studiot I apologize. If then the four elements given by you are A,B,C,D........and if then you chose to not define any element as a zero, then the answer to your question is.... There is no new affects by this axiom... Are you suggesting that I must adapt it? The axiom is specially designed for the use of zero in binary expressions of multiplication and division. If then the variables in an equation or expression are not zero... no adaption is possible...or necessary. I hope you will continue to help me....at least for a while. The axiom only yields new results regarding binary expressions of zero. It does so... in such a way as to leave all other expressions and equations the same. I hope I have done a better job of understanding and answering your question. I appreciate your time and patience studiot.
wtf Posted February 13, 2017 Posted February 13, 2017 (edited) I deliberately chose a field, that contains neither zero nor one. Every field contains 0 and 1 and they are distinct. A = 0 and B = 1 in your example, which can be read directly off your plus and times tables. As a check you see that B + B = A, or 1 + 1 = 0. That's because the field of order 4 has characteristic 2. Surely you understand that changing the names of the elements of an algebraic structure makes no difference at all. Edited February 13, 2017 by wtf 1
conway Posted February 13, 2017 Author Posted February 13, 2017 wtf I feel that Studiot and myself merely had a slight misunderstanding. Thank you for the support. I would also point out that the proposed axiom has no relevance to addition or subtraction. Were the examples I gave you sufficient? I can try to be more specific. Do you have any more curiosity in this regard? If not I understand and thank you for your time. studiot If I may say again. z1 and z2 for any "non numeric element" is the element itself. That is to say z1 for A is A z2 for A is A so then if no zero is defined..... there is no relevance to the given proposed axiom. I can not use the field as you gave it to either prove or disprove this proposed axiom. I hope I am not still misunderstanding you. I can show how it still works...but this then is a pointless exercise, as I am sure you can see......it only becomes relevant regarding zero.
wtf Posted February 13, 2017 Posted February 13, 2017 (edited) I would also point out that the proposed axiom has no relevance to addition or subtraction. Well if z1 and z2 are elements in the field, then you are required to say what is their sum and product with each element of the field including themselves and each other. That's because a field is closed under addition and multiplication. In other words if x and y are any two elements of a field, not necessarily distinct, then you have to define x + y and xy. If you have elements that can't be added or multiplied then you might have some algebraic structure, but it wouldn't be a field. https://en.wikipedia.org/wiki/Field_(mathematics) I don't know if this would be helpful to you but for example suppose we define two formal symbols x and y, and we want to consider the collection of all finite sums of elements ax + by where a and b are real numbers. Then the structure we get would be the vector space of dimension two, in other words the usual Cartesian plane. Perhaps this is the kind of structure you have in mind. You have your formal gadgets z1 and z2, whatever they are, and you want to be able to multiply them by scalars. So you want to look at vector spaces or their generalization modules over a ring. https://en.wikipedia.org/wiki/Module_(mathematics) But if you don't tell me what z1 times z2 is, then you haven't got a field, by definition. Edited February 13, 2017 by wtf 1
conway Posted February 13, 2017 Author Posted February 13, 2017 wtf Excellent response! Thank you! z1 and z2 are NOT elements in a field. It is that they are elements of an element in a field. So then if 1 is an element in a field, then 1 is composed of the two elements z1 and z2. So then if A is an element in a field, then A is composed of the two elements z1 and z2. In all cases of A (not being zero) then z1 and z2 are both A. It is the case in addition and subtraction that the elements of an element are both inherent in the number, and are NOT separated. That is to say addition and subtraction stay the same. It is only in a binary expression of multiplication and division, that z1 and z2 of an element are separated. I hope this is clearer. I can continue to discuss the multiplication and division of z1 and z2 supposing I have clarified these earlier topics...... Thanks
wtf Posted February 13, 2017 Posted February 13, 2017 z1 and z2 are NOT elements in a field. It is that they are elements of an element in a field. Ok I can live with that. What is the purpose of characterizing 1 and 0 as being sets that contain z1 and z2? What happens next?
conway Posted February 13, 2017 Author Posted February 13, 2017 wtf We then apply to actual binary expressions. We should start with non zeros. If then the expression is ( 2 x 3 ) We declare "according to the proposed axiom" that neither symbol given is a number....rather that one symbol is z1 and one symbol is z2 As the commutative property exists, it does not matter which is labeled which. So then...... ( 2(is actually z1) x 3(is actually z2) ) where z1 for 2 is 2, z2 for 3 is 3.....we then return to an expression of 2 x 3 so again regarding non zero's this seems pointless....albeit functional........ I will hold here before addressing 1 and zero......to ensure we still remain on the same page.... It may be helpful to consider z1 and z2 by there philosophical names as opposed to there mathematical z1 = abstract value without space z2 = abstract space without value space = a labeled quantity of dimension value = a labeled quantity of existence other than a dimension. where as when a value(z1) is "placed" into a space(z2) a "NUMBER" is generated. again in addition and subtraction z1 and z2 are NOT separated.
studiot Posted February 14, 2017 Posted February 14, 2017 (edited) studiot If I may say again. z1 and z2 for any "non numeric element" is the element itself. That is to say z1 for A is A z2 for A is A so then if no zero is defined..... there is no relevance to the given proposed axiom. I can not use the field as you gave it to either prove or disprove this proposed axiom. I hope I am not still misunderstanding you. I can show how it still works...but this then is a pointless exercise, as I am sure you can see......it only becomes relevant regarding zero. What you seem to be misunderstanding is the meaning of an axiom. 1) You cannot either prove or disprove and axiom. Axioms are taken as true without proof. 2) Every Field must obey every Field axiom completely or it is not a Field. By way of explanation a simpler algebraic structure is a group which has one binary operation by axiom. Some groups have additional structure such as a second operation or commutatively. We do not add the additional structure as axioms, just as definitions of the special restricted type of groups. It would be the same for Fields if you wished to introduce a special type of Field with some additional structure. Your special fields would then 'inherit' all the standard axioms to which you would add your definitions. But these definitions would not qualify for the elevated status of axioms. In particular none of the standard field axioms deal with numbers since some fields deal with objects other than numbers. Nor could you add an axiom the required numbers for the same reason. I have stated my version of the standard field axioms in my response to wtf, below. Every field contains 0 and 1 and they are distinct. A = 0 and B = 1 in your example, which can be read directly off your plus and times tables. As a check you see that B + B = A, or 1 + 1 = 0. That's because the field of order 4 has characteristic 2. Surely you understand that changing the names of the elements of an algebraic structure makes no difference at all. Since my use of letters was not clear enough I have redrawn the tables as below. The field contains four geometric shapes and two combining operations which take in any two shapes and output one of the four shapes. I hope you will agree that there is nothing numeric about turning a square and a triangle into an axehead (or the inverse operation for those inverses that are specified in the tables). So, given a set F and two combining operations, # and * the field axioms are 1) # and * are closed operations with unique outputs. 2) # and * both obey the associative law 3) There is a (different) identity element for both # and * in F. Combination of the identity element for either # or * on any element leaves that element unchanged. In my geometric example combination of the square with any element leaves it unchanged under # combination of the triangle with any element leaves it unchanged under * 4) There exists an inverse operation for every element in one of the combining operations and for every element except perhaps one in the second. In my geometric example For the # operation, having combined any pair of shapes to reach a third it is possible this third one with another shape to restore the original element. So if I combine a circle with a triangle under #I get an axehead. To get back to a circle I need to combine the axehead with a triangle and so on. For the * operation inverses work the same way, except for combinations with the square, for which there are no inverses. There is a final optional axiom which states that 5) Both # and * obey the commutative law. Any field also obeying axiom (5) is called a commutative field. Edited February 14, 2017 by studiot
wtf Posted February 14, 2017 Posted February 14, 2017 (edited) Studiot you must be making a point I don't understand to double down on this line of argument. I'm mystified to see you laboriously enumerate the field properties in the pursuit of demonstrating a manifestly false understanding of what a field is. Surely you don't think there's any difference between your ABCD field, your squares and triangles, and any other presentation of the unique field with four elements. If someone asks you how many fields there are of order four, and you give any answer other than one, you are missing the entire point of how algebraic structures work. We do not care about multiple representations of the same isomorphism class of objects. If a set of widgets satisfies the field axioms, then it has a 0 and a 1, no matter what you call them. That's why I don't object to Conway telling me that 0 and 1 are objects that contain z1 and z2. It makes no difference. 0 and 1 are defined by their behavior, not their representation. Perhaps this point is not brought out clearly enough in elementary presentations. Isomorphism isn't a map between two different things. It's a statement that we have two different representations for the same thing. It's really no different than the distinction between 1/2 and 2/4. Two different expressions for the exact same thing. Perhaps you'll give this some thought. To pick a simpler example, because to be fair the field of order 4 is a bit of a counterintuitive object, how many vector spaces are their of dimension two? I hope you'll agree that there is only one, even though it can appear in different guises. In your latest picture, the square is 0 and the triangle is 1. Every field has a 0 and a 1. It's part of the definition. I hope you will agree that there is nothing numeric about turning a square and a triangle into an axehead (or the inverse operation for those inverses that are specified in the tables). That's an interesting remark. I see nothing about being "numeric," whatever that means, in the definition of a field. If you insist I could do the same trick with the real numbers (replacing each real number by one of uncountably many symbols) and then I could say that the real numbers aren't numeric either. That's just wordplay. The real numbers are the unique complete ordered field. Any representation will do. Edited February 14, 2017 by wtf
conway Posted February 14, 2017 Author Posted February 14, 2017 (edited) Studiot I would like to repeat back to you what it is I "think" you are trying to tell me, in order I hope to better communicate with you. You are saying that all current field axioms can be shown to exists with "non numberic values". That is (A,B,C), and or (squiggly lines, and triangles). If this is so then my reply is as follows. The proposed field axioms states that z1 and z2 for A's or "any non numberic value", OTHER THAN zero, equal A or (squiggly line, and triangle). So I can not show any operations differently than how they already exist, unless you distinguish a given A or (squiggly line, and triangle) as zero. If then you do the proposed field axiom becomes relevant and I can show you an operation containing an A, or (squiggly line, and triangle.) wtf So then regarding zero and 1. z2 for 1 = 1 z2 for 0 = 1 z1 for 1 = 1 z1 for 0 = 0 So then philosophically or semantically I am saying The space of zero and 1 are equivalent The value of zero and 1 are not equivalent Depending on whether zero is used as value, or as space, in any binary expression, determines the sum of the expression. Or 0(z1) x 1(z2) = 0 1(z2) x 0(z1) = 0 1(z1) x 0(z2) = 1 0(z2) x 1(z1) = 1 A(z1) x 0(z2) = A 0(z2) x A(z1) = A Edited February 14, 2017 by conway
Bignose Posted February 14, 2017 Posted February 14, 2017 (edited) Depending on whether zero is used as value, or as space, in any binary expression, determines the sum of the expression. Or 0(z1) x 1(z2) = 0 1(z2) x 0(z1) = 0 1(z1) x 0(z2) = 1 0(z2) x 1(z1) = 1 A(z1) x 0(z2) = A 0(z2) x A(z1) = A So we're back with this 'space' and 'value' stuff, eh? Been a few years. Any chance you can define them better this time around? Last time -- see http://www.scienceforums.net/topic/89692-understanding-by-zero/?p=874173 -- you never ever could tell me how to know which one was which. Except you always seemed to know which was which because you could get the 'right' answers. But somehow could never explain it to me. Here is an example: A ball moving at 10 m/s is observed for 0 s, how far has it moved during this observation? A second ball moving at 0 m/s is observed for 10 seconds, how for does it move during this observation? I have no idea which number is the z1 or which is the z2 here, but the z1 HAS to be one of the velocity or the time, and the z2 has to be the other. So, let's see: if the velocity is the z1, then first ball moves 10 m/s * observed for 0 seconds (literally no time) = 10 meters (per your 2nd to last equation). WTF?!?! If the velocity if the z2, then first ball moves 10 m/s * observed for 0 seconds = 0 meters (per 6th eqn from bottom). Ok. Good. So velocity must be the z2. But now, we look at second ball. ball moving a 0 m/s (at rest!!!!) * observed for 10 seconds = 10 meters (per last equation). Again, WTF?!?! Your math leads to obviously wrong conclusions. How can you continue to pursue this? Edited February 14, 2017 by Bignose 1
conway Posted February 14, 2017 Author Posted February 14, 2017 (edited) Bignose I gratefully appreciate your reply and time. It is possible to not consider space or value at all. I have only done so in hopes that it may help me communicate with Wtf. It is possible to always and only say z1 and z2. Perhaps as you point out it is best to continue to do so. If not I will always refer back to post #12 for the defentions for space and value. That has not changed from other uses of them by me. It is my opnion that z1 and z2 is inherent in the eqution. For example.... If neither z1(velocity/ball) or z2(velocity/ball) equal zero, than it does not matter which is z1 or z2 the sum is always the same. If either z1(velocity/ball) or z2(velocity/ball) equal zero, it then is inherent in the equation. If I have 1 ball traveling a velocity of 0. Then inherently the ball is the z1, the velocity is z2 1(z1) x 0(z2) = 1 (ball) If I have 0 ball traveling a velocity of 1, Then inherently the ball is z1, velocity is z2. 0(z1) x 1(z2) = 0 (ball) "If I have" is the semantical argument dictating "value, z1" or "space, z2". If again as I agree we should stay away from semantics then z1 is and z2 are simply chosen based upon the needs of the mathematician. All four equations are true, relative to what is defined z1 or z2, at the will of the observer. A x 0 = A 0 x A = A A x 0 = 0 0 x A = 0 Edited February 14, 2017 by conway
wtf Posted February 14, 2017 Posted February 14, 2017 A x 0 = A 0 x A = A 0 times anything is 0. That's provable from the field axioms. So the above can't be true in any field.
conway Posted February 14, 2017 Author Posted February 14, 2017 wtf I agree 0 times anything is 0. I also showed that eqution as exisiting as true. I am also suggesting that 0 times anyting is anyting.
wtf Posted February 14, 2017 Posted February 14, 2017 I am also suggesting that 0 times anyting is anyting. You could invent a system like that but it wouldn't be a field, since 0 times anything is 0 in a field. The reason is that it's provable from the field axioms.
conway Posted February 14, 2017 Author Posted February 14, 2017 I agree its provable from the field axioms that 0 times anything is 0. I suggest that the opposite is also true....with the given "proposed" axiom. So then both statements are true in any given field. Is there anything proving or suggesting that BOTH statements can NOT be true. (obviously not at the same time)
Bignose Posted February 14, 2017 Posted February 14, 2017 (edited) If neither z1(velocity/ball) or z2(velocity/ball) equal zero, than it does not matter which is z1 or z2 the sum is always the same. If either z1(velocity/ball) or z2(velocity/ball) equal zero, it then is inherent in the equation. If I have 1 ball traveling a velocity of 0. Then inherently the ball is the z1, the velocity is z2 1(z1) x 0(z2) = 1 (ball) If I have 0 ball traveling a velocity of 1, Then inherently the ball is z1, velocity is z2. 0(z1) x 1(z2) = 0 (ball) "If I have" is the semantical argument dictating "value, z1" or "space, z2". If again as I agree we should stay away from semantics then z1 is and z2 are simply chosen based upon the needs of the mathematician. All four equations are true, relative to what is defined z1 or z2, at the will of the observer. A x 0 = A 0 x A = A A x 0 = 0 0 x A = 0 What is velocity/ball? I didn't introduce that at all. I'm not trying to find out how many balls I have. All I am trying to do is basic introduction to physics stuff: measuring displacements given a velocity and observation time. And pointing out that following your rules, either an object a rest has moved when you observe it for some non-zero amount of time or an object moves even when observed for zero time. Either result is unacceptable as it gives an obviously wrong answer. Furthermore, you still, still, still didn't tell me how to tell which is z1 and which is z2. Why can't you just provide some rules on how to know which is which? Or if they are all 'relative', then how can someone use them and NOT get a ridiculously wrong result? Edited February 14, 2017 by Bignose
uncool Posted February 14, 2017 Posted February 14, 2017 (edited) I agree its provable from the field axioms that 0 times anything is 0. I suggest that the opposite is also true....with the given "proposed" axiom. So then both statements are true in any given field. Is there anything proving or suggesting that BOTH statements can NOT be true. (obviously not at the same time) How can they both be true, but "not at the same time"? The only way this can work is if you are trying to define a new operation, call it "z-multiplication", above and beyond the standard 2 structures, that does not satisfy some of the usual field axioms for multiplication. Only then can you have that 0 *z a is not 0. And more importantly: If you're proposing an axiom, you should show some examples of things that satisfy that axiom, and (more importantly) how the extra axiom makes them more useful. Edited February 14, 2017 by uncool
conway Posted February 14, 2017 Author Posted February 14, 2017 (edited) Bignose I placed the words ball and velocity next to each other in order to suggest that it does not matter which is labled which. Perhaps not the most efficient means of communicate as it has lead to misunderstadings here. I do not understand where you think I have implied an object moves or does not move over zero time. I have never labeled time as zero. Is there some reason why you feel that human consciousness can not chose which to label z1 and z2. Is there some rule stating human consciousness is NOT a factor in mathematics. It certainty appears so to me. Again there is no need to argue over which is z1 or which is z2......either labeling is correct....we chose relative to our needs. In any case I will consider what you have suggested to me and contain to ponder.... thank you for your time uncool ok.......rewrite the axiom to be a "secondary" form of multiplication, in addition to the original. I see that this changes nothing. As to it's usefulness.....and examples as to it.........that should be obvious. The equations already posted up to this point show how to satisfy the axiom. If I tangibly held apples in my hand and multiplied them by zero I still hold apples in my hand, and so on..... Edited February 14, 2017 by conway
Bignose Posted February 14, 2017 Posted February 14, 2017 (edited) I placed the words ball and velocity next to each other in order to suggest that it does not matter which is labled which. But it does matter. Your own rules state that. 0(z1) x 1(z2) = 0 1(z2) x 0(z1) = 0 A(z1) x 0(z2) = A 0(z2) x A(z1) = A Per your own rules here, it is CRITICALLY important to know which is the z1 and which is the z2. I do not understand where you think I have implied an object moves or does not move over zero time. I have never labeled time as zero. I implied it. I gave you an example that specifically brings that up. I made that example to specifically point out that trying to follow your rules here leads to ridiculously wrong conclusions. Is there some reason why you feel that human consciousness can not chose which to label z1 and z2. Is there some rule stating human consciousness is NOT a factor in mathematics. Simply put, when I program my computer to do some math, I don't want to also have to program some consciousness into it. And even more simply put, the rules of multiplication that we've used for thousands of years now without 'consciousness' seem to have been pretty successful. You have introduced something here that makes it demonstrably far, far less useful. To what end? I have no idea. Your previous threads, you were trying to fix what you perceive as the problem of dividing by zero. Not sure if that is where you're going here. But to date, you haven't shown how this idea improves any situation, and I and others have demonstrated how it makes understanding far, far worse. So again I ask, to what end? Edited February 14, 2017 by Bignose
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