ajb Posted June 5, 2015 Posted June 5, 2015 Spell out what you would mean by a*(b+c) Do we have a*b + a*c ? I understand your v and s are labels that mean something. But I am not sure exactly what. Does a without v or s make sense in your algebraic structure?
conway Posted June 5, 2015 Author Posted June 5, 2015 (edited) Thank you, and ok. a*(b+c) If then b and c are added, per rules of order of operations, then we can assume we have a new number Z. a*z If then (a) is the value, then (z) is the space. It is then that (a) is placed additionally into all spaces of (z). Then all values are added in all spaces. If you prefer the distributive method. a*(b + c). If then b and c are actual numbers (that is the given symbols represents both value and space), then the symbol (a), is representing space only. so then assuming a=3 ((b+c)+(b+c)+(b+c)).........(the space of 3) If (a) were 2... ((b+c)+(b+c))......(the space of 2) So then in each of the last two examples it is that I have two numbers (b+c) that are placed into the space of (a) Then added. Edited June 5, 2015 by conway
ajb Posted June 6, 2015 Posted June 6, 2015 (edited) Have a look at http://en.wikipedia.org/wiki/Field_(mathematics) See what axioms you can keep and what is inconsistent with your /0. Really you should do this. You have declared we have an well-defined algebraic structure, but we don't know what it is. What we know for sure is that /0 is not really 'division' by zero. You cannot recover x from x0 =0 using this operation as the equation holds for all x. We do not have an inverse to multiplication by zero. You can include /0 as some extra 'thing'. One thing that people have tried is to define /0 as some 'formal number', so something formal that we can manipulate. However, if you do this you see that /0 cannot obey the standard rules of numbers. So you have some extension of the number line. It is generally considered not a useful thing to do. Edited June 6, 2015 by ajb
conway Posted June 6, 2015 Author Posted June 6, 2015 I will do so and post a list of the results. This will take me time. Until then. If (x) is representing only a space then there never was any "thing" to recover from x0=0. If (x) is a value then it is possible to recover it from x0. According to Relative Mathematics we do have an inverse to multiplication by zero. /0 is the inverse. Funny thing is........ is that this is so for anything other than zero. I will give you that under current axioms /0 cannot obey the "standard rules of numbers". Change the axioms, change how it obeys. (a(b+c) = (ab + ac)=( a(as value only)* b(as space only) + a(as value only) * c(as space only) ) = ( (b+c)+(b+c)+(b+c)....in as many quantities as in the space on a) I have read both articles on "Fields" and "Rings." It is stated in both cases that division by zero is not allowed. I can keep all given axioms except for this one, in both field and rings. There is a link to another article in "Fields" but not in "Rings" addressing division by zero. It is stated.... " As there is no number which, multiplied by zero gives a, " Apparently a man named George Berkeley has something more to say as to "why". I am going to see if it's anything new to me. I have offered that there is a number (representing only space, or only value), when multiplied by zero equals "a". While still assuming (a) does not equal 0. Or that is ( 1 does not equal 0). Which we talked about and I was refereeing that it did, but this was really only from a philosophical view point to help get my view point across. That is to say 1 and 0 are more alike than they are different(both have 1 single defined space, both have 1 single value, only one value is defined and one is undefined)...but still not equivalent. I am digressing. We can then ask why the axiom "multiplicative identity property of zero" exists. But that is axiomatic. This also is an axiom I would have to change. So then Only two axioms change the rest are kept.
ajb Posted June 7, 2015 Posted June 7, 2015 I have read both articles on "Fields" and "Rings." It is stated in both cases that division by zero is not allowed. I can keep all given axioms except for this one, in both field and rings. There is a link to another article in "Fields" but not in "Rings" addressing division by zero. Okay, so prove this carefully. If you drop impossibility of division by zero can you really keep all the other axioms?
conway Posted June 7, 2015 Author Posted June 7, 2015 Through extension I must also drop multiplicative identity property of zero. But I know of no other axioms affecting this. If this is not the case please let me know. How do you wish that I prove it carefully. Will listing a copy of all given axioms in the articles be sufficient? Associative of addition and multiplication. Commutative of addition and multiplication. Additive identity of zero but NOT multiplicative identity of zero. Additive and Multiplicative inverses Distributive. It may be because the article is a wiki article, but I should think it helps my argument. See how, again, it only says division by zero is not allowed. Then offers a link as to why. The link stating that it is not "done" simply because of the multiplicative identity property of zero. Which again is an axiom. It exist as it is with out proof of why it is. So if there are no other reason's (other that Multiplicative Identity Property of Zero), then I have ground to continue the debate.
uncool Posted June 7, 2015 Posted June 7, 2015 Zero is not a multiplicative identity; the multiplicative identity is 1. The fact that 0 is the additive identity, as well as additive inverses, distributivity, and associativity, does demonstrate that 0*a = 0 for any a. Proof: 0*a = 0*a + 0 = 0*a + (0*a + -(0*a)) = (0*a + 0*a) + -(0*a) = ((0 + 0)*a) + -(0*a) = 0*a + -(0*a) = 0. First step: additive identity of 0. Second step: Additive inverse for 0*a Third step: Associativity Fourth step: Distributivity Fifth step: Additive identity of 0 Sixth step: Additive inverse for 0*a Division really is about "undoing" multiplication. For any x not equal to 0, x*a = x*b if and only if a = b, so we define c/x as the unique element such that c/x * x = c. But for 0, that doesn't work - 1*0 = 2*0, so if we can freely divide by 0, we should get that 1 = 2.
conway Posted June 7, 2015 Author Posted June 7, 2015 (edited) My apologies for the mistake on my part. It is then that the multiplicative identity property is kept....but it is the multiplicative property of zero, otherwise the zero product property that is not kept. I shall try to slow down. Also to say that the additive identity of 0 and the additive inverse of 0 * a are both kept. The entire list actually. If then a=1, then the additive identity , and the additive inverse axioms requires that (a) be defined as value and (a) be defined as space in a very specific fashion. ((0 + 0)(s) * A(v)) + -(0v * As )) = 0s * Av + -(0v * As) = A These are the "fifth" and "sixth" steps correct? Thank you Uncool, I must now add to the original idea that which symbol is labeled as space in which symbol is labeled as value, must be done so in accordance with all given axioms. I must add that I stated as an axiom that in division, value must be labeled first while space is labeled second. If then this would affect these above property's let me know. I have yet to see this is the case however. Actually I propose the following During the process of the "order of operations", Multiplication within parenthesis are solved in such fashion as which the order of the symbols value and space is adjusted according to all given axioms relative to the equation. If then axioms are not relevant to the equation, value is labeled first and space is labeled second. Edited June 7, 2015 by conway
conway Posted June 9, 2015 Author Posted June 9, 2015 I accept and appreciate the forums conclusions on this matter.
conway Posted November 4, 2015 Author Posted November 4, 2015 An update for any who may care. As an addition to all current field axioms. "For every zero in S there exists a Z1 and a Z2, such that any zero in operation of muliplication or division, (excluding exponets and logarithms), is only representing Z1 or Z2. Allowing that Z1 for zero equals 0, and that Z2 for zero equals 1. "
RuthlessOptimism Posted November 5, 2015 Posted November 5, 2015 This is a very interesting idea Conway, but I don't think it actually is all that different from what has already been created in Mathematics. I have two points that I would like to add to this discussion. The first is what I think is a very intuitive physical description as to how multiplication works in the mathematical system you have defined that I would like you to either agree or disagree with as to its accuracy. If you have a "saw tooth" ruler, a ruler with spikes sticking up out of it say every one centimeter. This ruler has a total number of spikes = say 10. You have another saw tooth ruler with ten total spikes each spaced 1 centimeter apart. Multiplication of ten by 2 is basically like taking both rulers and fitting them together so that each spike now fills the 1 centimeter gap between the other rulers respective spikes. Now along the single length of the combined rulers you have 20 spikes. This doesn't work out exactly like it should because the total length of the combined ruler pair will have increased by length a length of 2 centimeters, since one tooth will hang off each end, but the idea that one value is fitted between the values defined in one of the numbers space, without altering the space is what I am trying to physically demonstrate. Multiplication makes the values in a space more "dense", division makes them less dense. Does that make sense to you? The second point is this idea simply reminds me a lot about vectors and imaginary numbers. The idea that an imaginary number has a magnitude and a phase seems similar to your idea of value and space. The magnitude is basically its "value", and the phase its "space". The idea in relative mathematics that you can partially define the addition, subtraction, multiplication of numbers etc relative to either the value or space of a partially defined number reminds me a lot about that. Its basically like saying we can add two imaginary numbers, A: we know its magnitude and phase, B: we know its phase but not its magnitude, we then know that the addition A+B will have a phase equal to the addition of the phase of A and B and its magnitude will be some number relative to the magnitude of A. Or say if we multiply A and B we instead know that the resulting magnitude is some multiple of A with the phase of A+B. You might instead want to take a look at the axioms defined for imaginary numbers rather than real ones, I am not sure but they might be more relevant to what you are doing. 1
conway Posted November 6, 2015 Author Posted November 6, 2015 (edited) Ruthless If I may suggest ways in which it is different from current mathematics. 1.variable amounts of zero 2.solves for division by zero I could add others. As to your first part. I would agree. More or less. The point of this idea is that space must also be addressed when multiplying and dividing (not just value). It may or may not be the case that value or space ( only ) is what changes with a given operation. As you point out. That again is the purpose of relative mathematics. It is my thoughts that all things posses value and space, that would then include all things not imaginary. Including imaginary. Thanks for your time. If I could ask you a question, requiring only your agreeing or disagreeing..... As space and time is relative to the motion of the body of reference.....Multiplication and Division by zero are relative to the declaration of space or value. Edited November 6, 2015 by conway
hypervalent_iodine Posted November 9, 2015 Posted November 9, 2015 ! Moderator Note conway, We are not doing this again. I am going to be extremely lenient and give you one chance to back up your claims. If staff are not satisfied with your response, this is getting shut down.
conway Posted November 9, 2015 Author Posted November 9, 2015 (edited) I merely posted an update on a separate but related thread. This even MONTHS latter. One person asked me a question and I returned with and answer and a question. Lock it if you got a problem with it. I am happy with the information I posted in the update. Nothing further is necessary. Edited November 9, 2015 by conway
swansont Posted November 9, 2015 Posted November 9, 2015 Lock it if you got a problem with it. ! Moderator Note Right. Don't bring this topic up again.
Recommended Posts