conway Posted July 15, 2015 Author Posted July 15, 2015 (edited) Ajb Time and space is dictated by our perspective of the "body of reference". The sum of multiplication is dictated by our perspective of "value and space" In any case I will certainly not argue with you here. As it is slightly off topic. And as my only education in this filed comes from reading "relativity" written by Einstein. And as I am sure you know that book is meant for the "philosopher" not the "mathematician". As he even suggest himself, paraphrased of course. Also to say that if nothing else I should "think" I have chosen the right name in any case. As multiplication is indeed relative. Relative to what number is space, and what number is value. Albeit, "pointing" this out makes no difference except where as 0 is involved. Strange 1. (2dv)= (2 defined values) = (1v,1v )........no space present (2ds) = (2 defined spaces) = (x,x).......no value present (2dv,X,X) = 2 defined values and spaces "sort of" put together (1,1) is the 2 defined values placed into the 2 defined spaces. When a value is placed into a space it becomes a number so that. (1+1) = 2 So that (2dv,X,X) = 2 2. 2 defined values do not "necessarily" refer to 2 defined spaces, or (x,x). If however I put the values into the spaces then I have a number. The number 2. Again placing value into space while creating a number is slightly different then placing value into space when multiplying or dividing, that is the "quantities" of space and value are different when multiplying and dividing as opposed to "generating" a number. 3. Yes this is exactly what I am saying 4. An undefined value has no "further" defining. That is I can not say if the value is finite, or infinite. Fractional or Whole. Value is still present, but it is undefined. Where as a defined value may be "further" defined as finite, infinite, fractional, whole. 5. Like answer as above 6. and 7. Yes (0v,X) = (1undefindvalue,X) = 0 The significance being only that the first representation is the simplest. But the explanation for the symbol 0 existing here instead of 1 is that the value is undefined as opposed to 1's defined value. 8. and 9. "The value is represented by the "position" on the number line. A position is a measure of space. One is within the other. So that the actual "length" or space of all numbers on a number line is exactly the same. It is only the labeling of the values that change. \----------\----------\----------\ It is either that the first "vertical" line is -1, the second line 0, the third line 1 and so on, or it is that from the first "vertical" line to the second line is -1 from the second line to the third line is 0 and from the third line to fourth line is 1 but in all cases the "physically measurable space" of the line, and or "between" the lines is equal for all numbers including 0. 10. (F) stands for finite. (I) stands for Infinite (S) stands for small, or "fractional" (L) stands for large, or "whole" This extra "further defining" is only necessary when showing that some values and spaces are infinite as opposed to finite. And that some values and spaces are fractional as opposed to whole. I used this only because it was required to represent pi in AJB's equations. 11. I have to label all numbers accordingly to "get" across what I am trying to communicate. It is the "nature" of multiplication that the "commutative" property exists. Only with 0 do we need to point out the difference between value and space. Because if we don't , then we can't multiply and therefore divided by zero. I hope I have replied effectively and efficiently. You previously post was highly engaging and most appreciative. Edited July 15, 2015 by conway
John Cuthber Posted July 16, 2015 Posted July 16, 2015 So, as far as I understand it after 13 pages, you have a scenario that can be summarised as follows. I have invented a thing that isn't a number and a thing that isn't division (it's doing nothing) and a thing that isn't zero. Using these you can "divide" a "thing" by "zero". Which may well be tautologically true- but doesn't actually mean that you can divide by zero as shown by your repeated failure to say how to do so. 2
Strange Posted July 16, 2015 Posted July 16, 2015 (edited) Strange 1. Why is "2defindedvalues" written with no spaces? (2dv)= (2 defined values) = (1v,1v )........no space present (2ds) = (2 defined spaces) = (x,x).......no value present (2dv,X,X) = 2 defined values and spaces "sort of" put together (1,1) is the 2 defined values placed into the 2 defined spaces. When a value is placed into a space it becomes a number so that. (1+1) = 2 So that (2dv,X,X) = 2 I am still struggling to get to grips with this. (As an aside, it doesn't answer my question, which was about the significance of "2defindedvalues" being written as one word.) If I understand correctly, neither "2dv" nor "2ds" represent numbers. Your "spaces" (x,x) seem to be like boxes into which you put values. Only when values are placed into boxes do they represent numbers. So when you say "(1,1) is the 2 defined values placed into the 2 defined spaces. When a value is placed into a space it becomes a number so that (1+1) = 2" you are putting the value "1" into two boxes. This represents the number 2; the sum of the values in the the boxes. Is that correct? If so, can we also represent the number 2 by putting the value 2 into one space/box, which I assume you would represent as "(2)" ? Can we also represent the number 2 by putting 0.5 into four spaces/boxes: "(0.5,0.5,0.5,0.5)" ? 2. Does "2 defined values" refer to the following x,x ? 2 defined values do not "necessarily" refer to 2 defined spaces, or (x,x). If however I put the values into the spaces then I have a number. The number 2. Again placing value into space while creating a number is slightly different then placing value into space when multiplying or dividing, that is the "quantities" of space and value are different when multiplying and dividing as opposed to "generating" a number. (Again, you haven't really answered the question which was, again, specifically about your notation. But never mind. Let's move on.) 3. Or are you saying that 2 is represented by the combination of two things: the "defined" value 2 combined with "x,x" ? Yes this is exactly what I am saying So (somehow) you are combining the "value" 2 with two spaces. How do you do that? But above you said: "(1,1) is the 2 defined values placed into the 2 defined spaces. When a value is placed into a space it becomes a number so that (1+1) = 2" Which seems to be saying that 2 is represented by the combination of two things: the "defined" value 1 combined with two spaces. So which of these is correct: a) 2 is represented by the combination of two things: the "defined" value 2 combined with "x,x" (two spaces/boxes) b) 2 is represented by the combination of two things: the "defined" value 1 placed in two spaces/boxes. I think I will stop there, before I get more confused! Edited July 16, 2015 by Strange 1
conway Posted July 16, 2015 Author Posted July 16, 2015 (edited) Strange Thanks! I hope to do better this time around. Your first question is correct. Your second question..... A) is correct.....and exactly so. It is possible to represent fractions, but I will wait. It is possible to put "different" quantities of values and spaces together. Such as in multiplication and division. 2dv * 3ds = 2 * 3 = (2dv or (1v,1v) placed into 3ds or (x,x,x)....then all "numbers" are added.) The values are now numbers because they have been placed into spaces. The above has different "quantities" of value and space. Where as when we are talking of making a "number" instead of multiplication and division, the values and spaces are the same. 1 = (1dv,1ds) = (1v,X,) = 1 2 = (2dv,2ds) = (1v,1v,X,X) = 2 so on..... Notice in 2 * 3 the "values" are placed "together" into all spaces (2dv,2dv,2dv) in the space of 3....then add...after removing the "dv" because they are now actually numbers.... lol...... Edited July 16, 2015 by conway
Strange Posted July 20, 2015 Posted July 20, 2015 (edited) A) is correct.....and exactly so. Let's just look at (a) again: a) 2 is represented by the combination of two things: the "defined" value 2 combined with "x,x" (two spaces/boxes) My interpretation of this was something like (2, [][]) - where [] represents your spaces/boxes But then you say: 2 = (2dv,2ds) = (1v,1v,X,X) = 2 And "(1v,1v,X,X)" doesn't look like option (a); it looks like option (b): b) 2 is represented by the combination of two things: the "defined" value 1 placed in two spaces/boxes. My interpretation of this was something like ([1][1]) But I am still very confused. (I am also busy and tired - I will try and come back to this later; see if I can make sense of it...) Just as an aside, why add all this complexity about "defined values" and "spaces"? Why not just say, "when there is a division by zero, I am going to treat it as division by 1". It seems a lot simpler and achieves the same result. One place I can see this might cause problems is in calculus: many functions which are currently differentiable and therefore continuous, might not be continuous and therefore not differentiable. Edited July 20, 2015 by Strange 1
conway Posted July 20, 2015 Author Posted July 20, 2015 (edited) Strange I agree that the interpretation that you gave is 100 percent correct. ([1][1]) = 2 The only difference in the original options (a) and (b), was how we chose to represent the value. Really it was that both your statements were correct. It is only that (a) was (in my opinion) more accurate. While the value is 2v, and it is equal to 2 defined values, if I take a single defined value or 1v....and put it "additionally" into two boxes, then I get the number 2. The original number we were talking about "or using for value" was 2. So that 2v is "more accurate" as opposed to "two" 1v's, put together. I am not capable of adding complexity. To be honest there are two particular consequences to Relative Mathematics, that I for the most part have no idea how to make use of. LOL. I could say what you suggested in regards to division by zero, but not with multiplication at all. I can not say that "whenever" zero is used it functions as 1. As this is not correct. Sometimes it does however. 0 * 1 = 0 0 * 1 = 1 Both statements are correct. The only difference was how "zero" was perceived, used, and or represented in the equation. 0v * 1s = 0 0s * 1v = 1 Av/0s = A......(zero in this operation of division is always space, therefore it always acts as 1). However in division value is always first, space is always second in the equation. How we perceive zero , that is as "space" or as "value" dictates the sum of multiplication and division by zero. Your patience is greatly appreciated. Thank you. Edited July 20, 2015 by conway
Strange Posted July 21, 2015 Posted July 21, 2015 0 * 1 = 0 0 * 1 = 1 Both statements are correct. The only difference was how "zero" was perceived, used, and or represented in the equation. How do you decide which to use?
conway Posted July 21, 2015 Author Posted July 21, 2015 (edited) Strange The intention of the equation given. Truly it is inherent in the question that is being asked. Take John and I's debate for an example. We constantly talked of pies. Well if I have a pie, and divided it by zero, then clearly in this statement, the pie is the value, while zero is the space. Of if you take the lawyer joke, 120K and he had no one to divide it amongst, then 120k is the value and 0 is the space. This is the same in multiplication. Intention, or reasoning for asking the question in the first place, dictates which is which. Surely this does nothing for pure abstraction, that is NON word math problems. But then the "dealer" decides. Simply placing a V or and S, in front of each number to show us how they arrived at their answer. Edited July 21, 2015 by conway
Strange Posted July 21, 2015 Posted July 21, 2015 So for pure maths, you are arbitrarily allowed to treat 0 as either 0 or 1 for the purposes of multiplication? And you think that will have no wider implications? The canonical form of a polynomial equation is: [math]a_n x^n + a_{n-1}x^{n-1} + ... a_1x + a_0 = 0[/math] But, according to you, that is the same as: [math]a_n x^n + a_{n-1}x^{n-1} + ... a_1x + a_0 = 1[/math] I don't think so.
John Cuthber Posted July 21, 2015 Posted July 21, 2015 Strange The intention of the equation given. Truly it is inherent in the question that is being asked. Take John and I's debate for an example. We constantly talked of pies. Well if I have a pie, and divided it by zero, then clearly in this statement, the pie is the value, while zero is the space. Of if you take the lawyer joke, 120K and he had no one to divide it amongst, then 120k is the value and 0 is the space. This is the same in multiplication. Intention, or reasoning for asking the question in the first place, dictates which is which. Surely this does nothing for pure abstraction, that is NON word math problems. But then the "dealer" decides. Simply placing a V or and S, in front of each number to show us how they arrived at their answer. The joke isn't the question about the lawyer, but your answer. You still haven't actually said what the lawyer does. You only suggestion is that he does nothing, but that's plainly not helping. It's clear that what you are talking about is either total nonsense or "dividing " something that isn't a number by something that isn't zero.
conway Posted July 21, 2015 Author Posted July 21, 2015 (edited) Strange No for pure mathematics you can NOT treat it arbitrarily. I stated that you MUST label space and value in pure mathematics. I have also said that to the best of "my" ability it affects nothing but multiplication, and division by zero. I am of course looking for other affects. You will note that both equations you gave contained zero. Therefore it is affected by this idea. If however you label your variables "space" and "value" you can solve for both sums given. So long as 0 in one equation is value, while in the other equation it is space. John If the question is "what is the lawyer doing"....then "doing nothing" is an applicable answer. I don't understand why you think it is not. Have you not answered friends and family the same when they have asked you what you were doing? Why do you still post? I understand you consider me to have mental issues. So why debate a crazy man? Sit back..... laugh at me.....wait for the thread to die. Lol thought it almost was the other day.....no worry's John I am sure Ajb and Strange are on the verge of moving on..........CHEERS! Edited July 21, 2015 by conway
studiot Posted July 21, 2015 Posted July 21, 2015 (edited) Strange No for pure mathematics you can NOT treat it arbitrarily. I stated that you MUST label space and value in pure mathematics. I have also said that to the best of "my" ability it affects nothing but multiplication, and division by zero. I am of course looking for other affects. You will note that both equations you gave contained zero. Therefore it is affected by this idea. If however you label your variables "space" and "value" you can solve for both sums given. So long as 0 in one equation is value, while in the other equation it is space. The obvious question your answer to Strange raises is Can you convert a 'space' into a 'value' and vice versa? If your answer is yes, how do you do this? Edited July 21, 2015 by studiot 1
conway Posted July 21, 2015 Author Posted July 21, 2015 (edited) Studiot Awesome question. I have not considered this. Of the top I would say yes. It is that a number is composed of both. One is within the other. The value is in the space. When a single value occupies a single space they are one thing. So .....as how to convert one to the other I can't think of anything other to say then..... change the variable assigned 1s changed into a value is 1v. 1v changed into a space is 1s. Consequently "placing values into spaces" are the axioms of multiplication and division. Number creation aside. Upon further thought This can not be done with zero. That is zero has a value that is "undefined", therefore I can not switch it with a "defined space". This is the reasoning for the symbol 0. 0 = (undefined value, defined space ) 1 = (defined value, defined space ) All numbers other than zero being (dv + ds), so then possible to switch value into spaces so long as all things being "switched" are defined. Edited July 21, 2015 by conway
studiot Posted July 21, 2015 Posted July 21, 2015 (edited) OK so going back the the post#259 by Strange that prompted this He has and Where we can see the left hand sides of the equations are identical. We can also see that the last term, a0 do not include an x. So let us put a number to a0 say 7 for instance. Now we have (an expression in a and x) + a0 = 1 or (an expression in a and x) + 7 = 1 Subtract 1 from both sides of the equation (an expression in a and x) + 6 = 0 Which contradicts the original equation How do you explain this? and have I converted a value into a space? Edit you could avoid zero altogether, by rewriting Strange's pair of equations as equal to some other pairs of numbers than 1 and 0 Edited July 21, 2015 by studiot
conway Posted July 21, 2015 Author Posted July 21, 2015 (edited) Why are you subtracting 1 from both sides of the equation? Well I will have to agree with Strange the rules according to exponents and logarithms by zero with also have to be altered. So I was wrong here, something else is affected. But in my defense logs and exponents are multiplication and division so..... I see now my confusion in regards to your post Studiot. In light of the information Strange has pointed out, I extend the idea. That is... Zero used as and exponent is zero as space. Zero used as a log is zero as value. x0 = x x0 = 0 Strange Cleary I glanced through your post #259, I apologize. I have been quit busy lately. Thanks for helping address this oversight. With any luck there will be more! Studiot Could you extend on what you mean? Edited July 22, 2015 by conway
John Cuthber Posted July 22, 2015 Posted July 22, 2015 If the question is "what is the lawyer doing"....then "doing nothing" is an applicable answer. I don't understand why you think it is not. Have you not answered friends and family the same when they have asked you what you were doing? Why do you still post? I understand you consider me to have mental issues. So why debate a crazy man? Sit back..... laugh at me.....wait for the thread to die. Lol thought it almost was the other day.....no worry's John I am sure Ajb and Strange are on the verge of moving on..........CHEERS! The point about doing nothing is not that it can't be done. The problem is that it doesn't achieve anything. I'm still posting to make sure that people don't lose sight of the fact that, after 14 pages, you still haven't divided a number by zero. You (like the imaginary lawyer) have achieved nothing, but I fear that you believe that you have. Do you understand that all you have done is put a lot of characters on a screen? 2
Strange Posted July 22, 2015 Posted July 22, 2015 I'm still posting to make sure that people don't lose sight of the fact that, after 14 pages, you still haven't divided a number by zero. You (like the imaginary lawyer) have achieved nothing, but I fear that you believe that you have. Do you understand that all you have done is put a lot of characters on a screen? And despite all that, the idea still hasn't been clearly explained. Any attempt to pin it down causes it to slip away. This conforms, to me, that conway has this vague idea in his head which he thinks works ("well, I tried it out with a one example") but even he can't define it in a way that anyone can work with.
conway Posted July 22, 2015 Author Posted July 22, 2015 Strange and John Again......thanks for your time.
John Cuthber Posted July 22, 2015 Posted July 22, 2015 For what it's worth there are mathematical entities called matrices that have numbers in places, a bit like what Conway might be on about (it's hard to tell). However they don't seem to help much. You can't divide one by zero. Not only that, they have rules of arithmetic that are more restrictive then ordinary numbers. Conway's major objection is that you can divide a number by another number as long as you don't try to use zero for the second number. Well at least you always have the option of multiplying two numbers. no matter what numbers you choose for a and b you can form the product ab Not only that, but it's the same as the product ba. However- if you combine a place as well as a value (and thus get a matrix) you end up with a system where not only can you not always divide A by B, but you cant always even multiply them. In short, adding position as well as value to a number leads to a more restricted set of valid mathematical operations. So, if you can't divide by zero with ordinary numbers, it's even less likely that you can do so once you add the complexity of a "space" as well as a value. This whole thread seems to me most likely to have been barking up the wrong tree. That certainly explains why it got nowhere.
conway Posted July 22, 2015 Author Posted July 22, 2015 (edited) Strange At this point I only wish to understand what happened between you and I. I freely admitted that you pointed out an issue I had not addressed, that is log's and exponents. I then thanked you and proceed to offer definitions. That is... zero as an exponent is zero as space zero as a log is zero as value So then why did you feel that I was not making an effort to "pin this down"? John As far as I am concerned..... "Do you understand that all you have done is put a lot of characters on the screen?" This is how I feel towards your replies. Cleary you have pointed out this is how you feel about my replies. A thread does not reach 250 plus replies and 14 pages if it does not have SOME merit. You and others make sure of that. I think that this is why you truly keep coming back to this thread. You seem to have a harder time letting go than I do..... Edited July 22, 2015 by conway
John Cuthber Posted July 22, 2015 Posted July 22, 2015 "A thread does not reach 250 plus replies and 14 pages if it does not have SOME merit. " Yes, but the merit may only have existed in the posts patiently explaining why you were wrong. 2
Strange Posted July 22, 2015 Posted July 22, 2015 "A thread does not reach 250 plus replies and 14 pages if it does not have SOME merit. " Yes, but the merit may only have existed in the posts patiently explaining why you were wrong. Given the relentless ignorance and refusal to learn in this thread (and pretty vile and offensive comments in another thread) conway is now on ignore. In my experience, one of the main reasons for threads going on so long (apart from the rare genuinely interesting ones) is because people are committed to trying to educate and inform. Even when we give up on the OP as being incapable of learning, there are others who might need to have the facts explained. 2
Phi for All Posted July 22, 2015 Posted July 22, 2015 Kudos to those trying to teach someone it's sometimes better to float with the flow of the water than to carry your canoe upstream (especially when your destination is downstream anyway). I think conway might have been able to admit he was wrong back on page six, but he feels he has too much sunk cost now on page fourteen. Most everything since has been semantics. I'm going to close this now. 3
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