conway Posted July 11, 2015 Author Share Posted July 11, 2015 (edited) John I am under the impression (possible falsely)....that I have followed all forum rules. "Doing nothing, isn't doing something" In fact it is John. You are standing still, You are existing, but not moving(relative to earth)(still moving, which is something). You are breathing. Your heart is beating. You are still doing "nothing".... but all the other statements are also truth. Therefore "Doing nothing, is doing something" When a lawyer "does nothing" he is still holding a pie (if he started with one). That is my point. I understand your point John. You do not feel that I make any sense. OK. I understand that from you. I think the entire forum understands that you think I make no logical, intellectual, or verbal sense whatsoever. Edited July 11, 2015 by conway Link to comment Share on other sites More sharing options...
Greg H. Posted July 11, 2015 Share Posted July 11, 2015 0veloctiy * 50meters = 0 0value * 50space = 0 = (x,x,x,x,x,...50 times, then put 0 in all and add) Made for a good example! So by not moving, I can cross the room instantly? Wow. My morning commute is about to get a lot faster. Link to comment Share on other sites More sharing options...
conway Posted July 11, 2015 Author Share Posted July 11, 2015 Greg I am confused. I did not think I implied at all that you crossed the room instantly. Or that you crossed the room at all. 0v * 50s = 0........therefore you did not cross the room. Link to comment Share on other sites More sharing options...
John Cuthber Posted July 11, 2015 Share Posted July 11, 2015 I understand your point John. You do not feel that I make any sense. OK. I understand that from you. I think the entire forum understands that you think I make no logical, intellectual, or verbal sense whatsoever. If nobody thinks you make sense then you don't make sense. Link to comment Share on other sites More sharing options...
Strange Posted July 11, 2015 Share Posted July 11, 2015 Strange I have studied basic math since I was a child. I need not study it any further. I came here to ask individuals with Advanced math education, to see what happens when this idea is applied to these particular fields. Again "my opinion" is that this idea only affects basic mathematics....which I know all I need to know of. And that is the problem right there. You ask people with an advanced math education, they tell you it isn't as simple as you think (and that the idea won't work) and you dismiss this based on your "opinion". I can only assume this is because you don't have a deeper knowledge of mathematics. You claimed I have Dunning Kruger's effect, a mental illness. What!! That is not a mental illness. Whatever gave you that idea !? It is just one of the many psychological biases that we are all prey to in one form or another. People trained in science and engineering are taught (or learn) to be aware of and, as far as possible, ignore these personal biases in favour of the data. The idea is either sound or not. So attack that. The trouble is, multiple people have tried to explain why the idea will not work in so many different ways, ranging from the theoretical to the practical but you simply reject or ignore them all. It seems that you have made your mind up and nothing will change it. But that is a good question: what would persuade you that you are wrong about this? Out of curiosity, why did you refuse to continue "helping" me (thanks by the way), when the process of "q into r" came up. Because you were just writing meaningless "equations". I tried to formalise you descriptions in a fairly standard notation but you changed things around so they made no sense. I didn't think there was much point trying to work out what you were trying to express or trying to educate you in algebraic notation. I hoped that if you/we could express what you were trying to say in formal notation then: 1. It would be clear that your claims are inconsistent. In other words, what you say about it in one moment differs from what you say a few minutes later. I think this is because you haven't really clarified what the idea is in your won head. This is the advantage for formalising it: it forces you address any gaps or inconsistencies in the idea. 2. It would then be possible to show that your scheme leads to inconsistent or contradictory results. (I don't know this, of course, because the idea isn't formalised. But from what I have seen so far, I am fairly certain.) As I have already stated, nothing changes in regards to division and multiplication outside of zero. And I am almost certain this cannot possibly be true. Link to comment Share on other sites More sharing options...
conway Posted July 11, 2015 Author Share Posted July 11, 2015 (edited) Strange While my algebraic notations have changed this has been do to yours and others request. The idea however has NOT changed. I have given extensive amounts of thought to this idea. It is because of this , that we are still chatting. As you yourself stated .... "I am almost certain this cannot be true" I am almost certain that it is. Now that you admit that there is the smallest of chances that it is.....then we should be able to continue. As well as understanding that I have not ignored ANY idea that has been offered for contradiction. It is that I have countered all of them. (albeit) You may think otherwise. This is fine. Further...at the risk of the ire of the mighty moderators.....the original paper entitled "Relative Mathematics", was anything but un-formalized. All numbers represent two things in one. Space ( a quantity of dimension), Value ( A quantity of existence). They are defined accordingly and then when any symbol is multiplied or divided, it is representing only one piece of value or space from each given number. Edited July 11, 2015 by conway Link to comment Share on other sites More sharing options...
John Cuthber Posted July 11, 2015 Share Posted July 11, 2015 "As well as understanding that I have not ignored ANY idea that has been offered for contradiction. It is that I have countered all of them. (albeit) You may think otherwise. This is fine. " No it isn't.. "All numbers represent two things in one. Space ( a quantity of dimension), Value ( A quantity of existence)." Not in any ordinary use of the words. For example, they may represent a probability, a duration, or a concentration. " They are defined accordingly " No they are not. Why do you keep posting things like that which are plainly not true and which a few minutes' thought will show not to be true? Link to comment Share on other sites More sharing options...
Greg H. Posted July 11, 2015 Share Posted July 11, 2015 Greg I am confused. I did not think I implied at all that you crossed the room instantly. Or that you crossed the room at all. 0v * 50s = 0........therefore you did not cross the room. We're not multiplying, we're dividing. Now, according to your axioms, which, if I am quoting this right, yield the following results 0/0=0 a/0=a 0/a=0 So if I have a 50 meter room to cross, at 0 meters/second, the time required can be determined by the formula. [math]t = \frac {d}{v}[/math] Inserting the numbers from my simple example, we get [math]t = \frac {50m}{0ms^{-1}}[/math] Cancelling out all the unnecessary measures, we're left with [math]t = \frac {50}{0}[/math] seconds. According to your statements, a/0 = a, so then 50/0 = 50 seconds. Which means it takes me 50 seconds to cross a room when I'm not moving at all. Awesome. But wait - 50 meters is roughly 164 feet. And if I'm moving 0 meters per second, then I am also moving 0 feet per second. So it takes me 164 seconds to cross the room if we measure it in feet. Even though it's the same room. So is it 50 seconds(meters), or 164 seconds (feet), or 54 seconds (yards), or 5000 seconds (centimeters). This is just one example of why division by zero doesn't work. It gives nonsense answers, like I can cross a room by not moving, but the time it will take depends on the unit you use to measure the room. Well, at least you were right about not crossing the room instantly. This is the point at which I usually refer people to my signature and say If the predictions of your theory do not match reality, it is not reality that is wrong. In this case, I will simply say - you cannot cross a room by not moving (relative to the room), no matter what your "math" tells you. Link to comment Share on other sites More sharing options...
John Cuthber Posted July 11, 2015 Share Posted July 11, 2015 In fact it is John. You are standing still, You are existing, but not moving(relative to earth)(still moving, which is something). You are breathing. Your heart is beating. You are still doing "nothing".... but all the other statements are also truth. Therefore "Doing nothing, is doing something" When a lawyer "does nothing" he is still holding a pie (if he started with one). So, when he has finished doing nothing, he is in exactly the same position he was before- he still has to divide up the inheritance. How does he do that? Clearly, doing nothing won't work because after doing nothing, he hasn't finished. Repeating it doesn't help. So, what does he do (The answer can't be "nothing" because, as I have shown, that doesn't work.) Link to comment Share on other sites More sharing options...
conway Posted July 11, 2015 Author Share Posted July 11, 2015 (edited) Greg This is a matter of units and semantics, for example..... 50/0=undefined What we do know is that "undefined" is not how long it takes you to cross the room What we do know is that "you do NOT" cross the room, which is a defined "sum". 50/0=50 What the case here is, is that units are involved. Some units are space units, some units are value units, some units are both at the same time. I take it you can figure out which. So that in the equation what I actually have at all times is the following t = d/v 50(meters)/0(velocity) = 50(meters with 0 velocity) or (time), (but t can not operate with out a velocity). This then shows that no meters was crossed at all. As opposed to say 50(meters)/1(velocity) = 50(meters with 1 velocity) or (time), (t in this case can operate), therefore indicating a meter was crossed, and the new final unit being time. So in the first case I do have 50 meters, but 0 velocity which is my unit (time). Indicating no meters are crossed So in the second case I do have 50 meters, but 1 velocity which is my unit (time). Indicating meters were crossed It is not that I have 50(time) out of the operation d/v, therefore I crossed the room with out any velocity, It is that I have 50 (meters with no velocity) therefore I did not cross. Simplest form 50m/0v=50seconds,....of no movement 50m/1v=50seconds....of movement I found your last post engaging, and directly on a proper assault of the idea, thank you. I will listen carefully to what you have to say in regards to this reply. Edited July 11, 2015 by conway Link to comment Share on other sites More sharing options...
John Cuthber Posted July 11, 2015 Share Posted July 11, 2015 You keep saying things like this "50/0=undefined What we do know is that "undefined" is not how long it takes you to cross the room" Do you understand that the thing that is undefined isn't the answer to "what is 50 divided by zero?" The thing that is not defined is "how do you divide something by zero?". And pretending that doing nothing is doing something because you fart while you do nothing does not help. Link to comment Share on other sites More sharing options...
conway Posted July 11, 2015 Author Share Posted July 11, 2015 (edited) John The answer to any equation is "not what did we do" ergo multiply, divide ,subtract, add, the question is what does the operation equal. I understand the point you have been making john. It is that the "answer" to division by zero is undefined because the "operation" is undefined. It is also that I do not believe that the operation or the sum is undefined. Name one other equation John where the sum, is answered by defining or (lack there of) of the operation. Not one. The operation exists to create a sum. The sum is the answer..... not the definition of the operation. Edited July 11, 2015 by conway Link to comment Share on other sites More sharing options...
John Cuthber Posted July 11, 2015 Share Posted July 11, 2015 You miss the point. Until you can define how to perform the operation, you can not know what the outcome is. And, since you have repeatedly failed to explain how to perform the operation, you have got nowhere (though you have wasted a lot of time). So your question is the wrong way round. You asked "Name one other equation John where the sum, is answered by defining or (lack there of) of the operation. " Well, I'm turning that round Name one equation, Conway, where the sum is answered by not defining the operation. Link to comment Share on other sites More sharing options...
conway Posted July 12, 2015 Author Share Posted July 12, 2015 (edited) John If you believe that I have not offered a definition for division then you haven not been really paying attention to me. Division is the same action in all cases John, as it should be. Zero involved or other wise. All equations require a definition to their operations. I have defined all opearations. And generated sums. Edited July 12, 2015 by conway Link to comment Share on other sites More sharing options...
John Cuthber Posted July 12, 2015 Share Posted July 12, 2015 John If you believe that I have not offered a definition for division then you haven not been really paying attention to me. Division is the same action in all cases John, as it should be. Zero involved or other wise. All equations require a definition to their operations. I have defined all opearations. And generated sums. Good. Then you can show me what my lawyer friend should do and you should show how it avoids an inconsistency where a/0 =a/1 hence 1=0. Link to comment Share on other sites More sharing options...
conway Posted July 12, 2015 Author Share Posted July 12, 2015 (edited) John Ok no problem. Any number in operation of multiplication or division is only representing a value or a space in any given equation. So that in multiplication the first symbol(number) is value only, the second symbol(number) is space only. It is then that the value is "placed" equally into all given spaces then all values added in all spaces. 2v*3s = (2+2+2) = 6.....(x+x+x) = space of 3 As we have already talked about, in multiplication either number given can be value or space. That is 3v*2s = (3+3) = 6......(x+x) = space of 2 So that in division the first symbol is Always value, the second symbol is Always space. It is then that the value is "placed" equally and subtraction-ally into all given spaces then all values are subtracted but one. 6v/3s = (2-2-2)...subtract all BUT one = 2 So that... 6v/0s = (6).....V for 0 = 0, S for 0 = 1. That is (x) = the space of zero or the space of 1, if a value is placed into the space then it becomes a number. If the value is defined it becomes the number 1. If the value is undefined it becomes the number 0. Edited July 12, 2015 by conway Link to comment Share on other sites More sharing options...
John Cuthber Posted July 12, 2015 Share Posted July 12, 2015 (edited) You need to translate this "the first symbol(number) is value only" into plain English Also, even if you are redefining the whole of arithmetic (which is possible, but brave) you need to explain what you mean by this "Any number in operation of multiplication or division is only representing a value or a space in any given equation. ". If you are not seeking to invent a new arithmetic, then you are wrong in making the assertion and also youhave not addressed the real world problem of division by zero in conventional arithmetic. It is possible that you will come up with some system where there is a definition of zero and where you can divide by it. It's even possible that you will find a way round the contradictions that flow from any conventional definition of what you call zero. it still won't help the lawyer divide up the £120,000. So it still won't serve much purpose. Edited July 12, 2015 by John Cuthber Link to comment Share on other sites More sharing options...
Greg H. Posted July 12, 2015 Share Posted July 12, 2015 Simplest form 50m/0v=50seconds,....of no movement And this is exactly the issue I have - you have to quantify your answer with extraneous information that isn't required under the present mathematical system. Under the present system, I don't have to quantify the answer is any way. I can look at the question and immediately know what the answer means. It also doesn't help that the answer changes meaning when a zero is involved - this also increases the complexity of the system, without providing any additional benefit, since the end result is the same. I know I'm not moving, the velocity value tells me this - I shouldn't need to clarify the answer to indicate it. Let me give you another rather famous example of what happens when you try and divide by zero. [math]a = b [/math] [math]a + a = a + b [/math] [math]2a = (a + b) [/math] [math]2a -2b = a + b - 2b [/math] [math]2(a - b) = a + b - 2b [/math] [math]2(a - b) = a - b [/math] [math]2 = 1 [/math] 1 Link to comment Share on other sites More sharing options...
conway Posted July 12, 2015 Author Share Posted July 12, 2015 (edited) Greg I do not have to qualify "technically". It is still entirely possible to do it all in one's head. It decreases the systems complexity, while adding the benefit of DEFNINING all equations. That makes it worth it , even if it is more complicated. Greg if the current answer to the question you posed (about the room) is undefined, then you do have an "extraneous information". If then I proved an answer and that answer is 50(units of time). Then you still have "time" it does not disappear, nor is that time undefined. What is the case is that time passes with no movement or it passes with movement. That MUST be defined. Ant that is NOT erroneous. 2(a-b) = a - b.....does not equal to 2=1 2a-2b = a - b if a and b are 1 then 2-2 = 1-1 0=0 if a and b are 0 then 0=0 with out solving for the variables 2(a-b) = a - b 2a-2b = a - b....a true statement where did you come up with 2=1 ? You can't just drop your variables into thin air. John I am not sure how else to say it. If you look at an equation with two numbers in it. And it involves multiplication or division. Then one of the numbers is ONLY representing value. While the other number is ONLY representing space. Again 2 * 3 = 6 2(as a value only) * 3(as a space only) = 6 (x, x, x ) = the space of 3 I then put my value, or 2 into all given spaces then add. 2 + 2 + 2 = 6 I know that you insist that it does not help your lawyer with his 120k. I agree, he still has no one to give it to. But at lest we generate a sum other that undefined when talking of what he does (or doesn't) do with the 120k. Edited July 12, 2015 by conway Link to comment Share on other sites More sharing options...
imatfaal Posted July 12, 2015 Share Posted July 12, 2015 1. Start with this proposition 2. Add a to both sides to give this 3. Simplify to this 4. Substract 2b from both sides to give this 5. Simplify LHS of equation to this 6. Simplify RHS of the equation to this 7. Divide both sides of the equation by (a-b) to give this Everything is fine except step 7 - because it is division by zero and that is undefined. 1 Link to comment Share on other sites More sharing options...
John Cuthber Posted July 12, 2015 Share Posted July 12, 2015 Greg I do not have to qualify "technically". It is still entirely possible to do it all in one's head. It decreases the systems complexity, while adding the benefit of DEFNINING all equations. That makes it worth it , even if it is more complicated. Greg if the current answer to the question you posed (about the room) is undefined, then you do have an "extraneous information". If then I proved an answer and that answer is 50(units of time). Then you still have "time" it does not disappear, nor is that time undefined. What is the case is that time passes with no movement or it passes with movement. That MUST be defined. Ant that is NOT erroneous. 2(a-b) = a - b.....does not equal to 2=1 2a-2b = a - b if a and b are 1 then 2-2 = 1-1 0=0 if a and b are 0 then 0=0 with out solving for the variables 2(a-b) = a - b 2a-2b = a - b....a true statement where did you come up with 2=1 ? You can't just drop your variables into thin air. John I am not sure how else to say it. If you look at an equation with two numbers in it. And it involves multiplication or division. Then one of the numbers is ONLY representing value. While the other number is ONLY representing space. Again 2 * 3 = 6 2(as a value only) * 3(as a space only) = 6 (x, x, x ) = the space of 3 I then put my value, or 2 into all given spaces then add. 2 + 2 + 2 = 6 I know that you insist that it does not help your lawyer with his 120k. I agree, he still has no one to give it to. But at lest we generate a sum other that undefined when talking of what he does (or doesn't) do with the 120k. The trouble is that 2 times 3 is the same as 3 times 2. Your suggestion is that, in some way they are different. but you don't say why you think that. Now there are mathematical entities such as matricies where the order of multiplication matters and where A times B is not always the same as B times A and you may have "invented" another sort of thing like that which is all fine and dandy, but it doesn't tell me anything about dividing ordinary numbers by zero- because these new rules don't apply to ordinary numbers. Link to comment Share on other sites More sharing options...
conway Posted July 12, 2015 Author Share Posted July 12, 2015 John in the equation 2 * 3 2 is the value, 3 is the space in the equation 3 * 2 3 is the value, 2 is the space Therefore the equations are different. The rules do apply to all numbers, ordinary or not. It is exactly the current definition of multiplication and division (except by zero). Only not described in terms of space or value. It is that we just "say" 2 * 3 is (2+2+2), or the other way around. Using terms of space and value explain WHY it is explained as such. Link to comment Share on other sites More sharing options...
Greg H. Posted July 13, 2015 Share Posted July 13, 2015 where did you come up with 2=1 ? You can't just drop your variables into thin air. The fact that you don't know how I did simple algebra really makes me question your ability to do math. I encourage you to read Imfataal's explanation of it to see where the variables went - they didn't just disappear (well, they did, but in a mathematically viable way). Therefore the equations are different. But they aren't. According to the rules of math, 2 * 3 is exactly the same 3 * 2, because if they aren't, multiplication breaks down. As has been pointed out to you multiple times. And while you keep saying they aren't the same, you've never given anyone a good reason why they aren't the same, except in your version of multiplication. Which is fine, except that your version isn't the commonly accepted version, and you haven't really provided anything approaching a thorough enough explanation to show us why it should overturn the version that's worked just fine for thousands of years just because you don't accept that you can't divide by zero. Link to comment Share on other sites More sharing options...
conway Posted July 13, 2015 Author Share Posted July 13, 2015 (edited) Greg Because it allows for division and multiplication by zero is the reason it "over turns" current mathematics. It is that 2 * 3 and 3 * 2 yield the same sum, but they are not exactly the same. For that matter, the only thing that really matters is which is value, and which is space. I may put space first. That is 2s * 3v. 2(a-b) = a-b does NOT equal 2 = 1 it equals 2a-2b = a-b so please......post this mathematically viable way of producing (2 = 1) out of ( 2(a-b) = a - b ) In the end it is not MY definition of multiplication and division. It is CURRENT definition of multiplication and division. It is only that I found a more accurate way of stating it. You have always been told that 2 * 3 = 2+2+2 = 3+3. But you where never told why. This is why Greg. One number only represents value, One number only represents space. The value is placed into the spaces and added. Edited July 13, 2015 by conway Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted July 13, 2015 Share Posted July 13, 2015 Conway, if I can understand Greg's reasoning, I can assure you that it's simple algebra. Take this equation. 2(a-b) = a-b You can rewrite that as: 2(a-b) = 1(a-b) Dividing both sides by (a-b) gives: 2 = 1 Which is obviously wrong. 1 Link to comment Share on other sites More sharing options...
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