Bignose Posted July 13, 2015 Posted July 13, 2015 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 ) conway, please don't take this as a personal attack or insulting in any way. But if you honestly do not know enough basic algebra to understand the above steps, then you owe it to yourself to educate yourself on modern mathematics. You need to know this stuff before you can convince anyone that the modern mathematical development needs to be modified in any way. You have to know what the current method says before you try to change it. There are free classes online. You can do this if you want to. Or you can continue to lecture us about things you are obviously ignorant on. If have any true desire to improve your communication, I cannot recommend this path enough. As before, this advice is worth exactly what your paid for it. You may chose to ignore it. It is just my opinion on what you can do to help your own knowledge and communication. 4
conway Posted July 13, 2015 Author Posted July 13, 2015 (edited) hypervalent_iodine, Bignose I believe you guys are forgetting a step. (Pemdas) demands that multiplication occur before any division so that.... 2(a-b) = 1(a-b) (2a-2b)/1 = (a-b)/1 0/1 = 0/1 0 = 0 You must distribute the 2 and the 1 before you can divide...... Edited July 13, 2015 by conway -2
ajb Posted July 13, 2015 Posted July 13, 2015 Conway, the 2=1 thing is a 'trick' it comes from forgetting that a=b, so forgetting that a-b =0 and then doing the undefined operation of dividing by zero. This is a standard fallacy to show school children the dangers of dividing by zero. Indeed you see that all that has really happened is one is showing 0=0 and then suggesting that 2 0/0 = 0/0 together with 0/0 =1; which is really where the problems start. Anyway, there is no reason why you need to 'distribute the 2' in your calculation, but you are to do so. You end up with (2a - 2b) = (a-b) and then divide both sides by a-b -> (2a - 2b)/(a-b) = 1 = 2 (a-b)/(a-b) etc.. As pointed out, it is the division by (a-b) that is problematic. Moreover, you idea that a*b is not the same as b*a is interesting, but seems to be trivially false. Well at least if we understand '=' in the standard way. 2
Strange Posted July 13, 2015 Posted July 13, 2015 (edited) 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. This makes zero sense. There is, however, a correct way of dealing with units (by "correct" I mean one that works): https://en.wikipedia.org/wiki/Dimensional_analysis [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] I just knew that not making one step explicit would cause problems! [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] divide through by (a-b): [math]\frac{2(a - b)}{(a-b)} = \frac{a - b}{a-b} [/math] cancel: [math]2 = 1 [/math] Edited July 13, 2015 by Strange 1
Greg H. Posted July 13, 2015 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. hypervalent_iodine, Bignose I believe you guys are forgetting a step. (Pemdas) demands that multiplication occur before any division so that.... 2(a-b) = 1(a-b) (2a-2b)/1 = (a-b)/1 0/1 = 0/1 0 = 0 You must distribute the 2 and the 1 before you can divide...... These two posts, if you will forgive me for saying so, show that you have a lot to learn about how math currently works, and they also explain why we're repeatedly covering the same ground with you. The steps involved in my example are basic first year algebra involving reduction and cancellation - I would expect a first year college or university student to understand these steps and how to apply them before they step foot on a campus. For the record, I am well aware that 2 = 1 is a false statement. The example given shows the dangers inherent in trying to make division by zero a defined operation. You end up with nonsensical answers. Edited July 13, 2015 by Greg H.
conway Posted July 13, 2015 Author Posted July 13, 2015 (edited) Ajb, Strange, Greg Again I do not see how there is a problem here at all. If at all times we follow the rules of (PEMDAS), then one NEVER arrives at 2 = 1. Ajb why do you not "have" to distribute. Clearly you break a rule by NOT distributing. Clearly multiplication comes before division. What am I missing here? Division by (a-b) is not a problem at all when the numerator is zero. The numerator is Zero. NOT 2. 0/A = 0. This "cancel" like Strange has done, is not correct. We may be able to cancel like divisors in expressions if we only had + and - in the numerator. Again PEMDAS. Ajb While a*b and b*a are not the same but yield the same result is not trivial. I understand why one would think so. It is in fact trivial until a or b or both contain 0. Depending on if 0 is used as space or value is dependent on the sum. 0 or A. Edited July 13, 2015 by conway
Strange Posted July 13, 2015 Posted July 13, 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.... I do not admit that. All I mean is that I am not able to prove it, because it is not possible to formulate a consistent definition of your idea. The problem with your attempts at algebraic notation is that, because you don't understand them you end up writing meaningless strings of symbols. Greg I do not have to qualify "technically". It is still entirely possible to do it all in one's head. And I think that is part of the problem: because this just exists as a vague half-defined notion in your head, you have convinced yourself it works. If you were to attempt to use it in reality, you would find it was very different. Clearly multiplication comes before division. What am I missing here? I guess not knowing the difference between arithmetic and mathematics? Edited July 13, 2015 by Strange
ajb Posted July 13, 2015 Posted July 13, 2015 Ajb why do you not "have" to distribute. Clearly you break a rule by NOT distributing. Clearly multiplication comes before division. What am I missing here? There is no need to 'distribute' where you did so. You would simply 'undistribute' to cancel the terms. Division by (a-b) is not a problem at all when the numerator is zero. The numerator is Zero. It is still problematic. NOT 2. 0/A = 0. This "cancel" like Strange has done, is not correct. We may be able to cancel like divisors in expressions if we only had + and - in the numerator. Again PEMDAS. The cancellation is wrong as the step before is wrong. While a*b and b*a are not the same but yield the same result is not trivial. I understand why one would think so. It is in fact trivial until a or b or both contain 0. Depending on if 0 is used as space or value is dependent on the sum. 0 or A. I don't quite follow your 'space' and 'value', anyway if a*b = c and b*a = c then a*b = b*a. So I am confused as why they are not the same while being the same?
John Cuthber Posted July 13, 2015 Posted July 13, 2015 Ajb, Strange, Greg Again I do not see how there is a problem here at all. Yet everyone else does. Do you think that's because everyone else is wrong? At the risk of sounding like a stuck record. https://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
Greg H. Posted July 13, 2015 Posted July 13, 2015 Ajb, Strange, Greg Again I do not see how there is a problem here at all. If at all times we follow the rules of (PEMDAS), then one NEVER arrives at 2 = 1. Ajb why do you not "have" to distribute. Clearly you break a rule by NOT distributing. Clearly multiplication comes before division. What am I missing here? Division by (a-b) is not a problem at all when the numerator is zero. The numerator is Zero. NOT 2. 0/A = 0. This "cancel" like Strange has done, is not correct. We may be able to cancel like divisors in expressions if we only had + and - in the numerator. Again PEMDAS. PEDMAS doesn't apply until I start trying to solve the equation, IOW actually trying to work the problem. I'm simplifying the equation, not solving it at this point Factoring and cancellation can occur outside of a standard PEDMAS approach, in an effort to make the problem simpler prior to solving it.
Strange Posted July 13, 2015 Posted July 13, 2015 Do you think that's because everyone else is wrong? Someone in Speculations thinking that they are right and everyone else in the world is wrong? Surely not.
conway Posted July 13, 2015 Author Posted July 13, 2015 (edited) Greg In this case it did NOT make the problem simpler did it. In fact it yielded an incorrect sum didn't it. But I will bite. Why then is "cancelation" allowed" outside of pemdas? If cancelation exists in an effort to help prepare to solve, than it is PART of solving it. Therefore the standard PEMDAS, which as you pointed out, exists to be used when solving. So why then can I cancel? In fact I just so happen to know an equation than when you do "cancel" it gives a wrong answer. Further if I solve with out "canceling" and I arrive at one sum, but if I cancel then arrive at another....isn't something wrong? Ajb If 0/x is always 0, then how is it still problematic? Is it really true that "cancelation" can occur before PEMDAS. If so why? Is the previously stated equation not an example of how doing this can lead to incorrect answers. Such as if I did solve the equation without "cancelation" I arrive at a completely different answer, one that also makes sense. a*b = c b*a = c a*b = b*a In all the above equations all variables but ( c ) represent only a space or a value. In each separate expression only 1 value and 1 space exists. So that..... a(is a value) * b(is a space)......or a(is a space) * b(is a value) Then the "value" is placed into the "space" given. Then all values are added. (a+a+a+a+a), where I have as many (a) as spaces in (b) 2 * 3 (3,3) this is the value 3 placed into the space of 2....then added. (3+3)=6 (2,2,2) this is the value 2 placed into the space 3....then added. (2+2+2)=6 This again is not my idea. This is current explanation.....different perspective albeit. Edited July 13, 2015 by conway
John Cuthber Posted July 14, 2015 Posted July 14, 2015 2 times 3 means ## ## ## 3 times two means ### ### They are the same
ajb Posted July 14, 2015 Posted July 14, 2015 If 0/x is always 0, then how is it still problematic? For the example given x = a-b =0. This was the 'illegal' step. After that whatever you do is going to end in nonsense. Is it really true that "cancelation" can occur before PEMDAS. If so why? Just as an example 2(a+b)/ (a+b) = 2 c/c = 2 where c = a+b is non-zero. You could think about (2a + 2b)/(a+b), but how would you now cancel the (a+b) term without using (2a +2b) = 2(a+b)? Is the previously stated equation not an example of how doing this can lead to incorrect answers. The division by zero was the problem. Such as if I did solve the equation without "cancelation" I arrive at a completely different answer, one that also makes sense. Indeed, as all we really have is 0=0. We know this trivially, but the 'trick' is to hind this in a-b and then do the 'illegal' operation of dividing by zero. We all know this is wrong and why you seemingly prove '2=1' or indeed you can modify this 'proof' to get any number equal to any other number. Again, this is all false and due to dividing by zero. If anything examples like this should warn you of the difficulties of /0. a*b = c b*a = c a*b = b*a In all the above equations all variables but ( c ) represent only a space or a value. In each separate expression only 1 value and 1 space exists. So that..... a(is a value) * b(is a space)......or a(is a space) * b(is a value) Then the "value" is placed into the "space" given. Then all values are added. (a+a+a+a+a), where I have as many (a) as spaces in (b) 2 * 3 (3,3) this is the value 3 placed into the space of 2....then added. (3+3)=6 (2,2,2) this is the value 2 placed into the space 3....then added. (2+2+2)=6 This again is not my idea. This is current explanation.....different perspective albeit. This seems quite standards as a definition of multiplication for integers. For all real numbers you need to be more clever. Anyway, you have simply shown that (for example) 2*3 = 3+3 = 2+2+2 = 3*2? Then I would conclude that the number 2*3 is the same as 3*2. I think I see what you are saying. It is common in algebra to associate with an element an endomorphism via left or right multiplication. So I can think of the number 3 as such an endomorphism as 3* or *3 (I then operate on any other number placed in the correct position). As the algebra of real numbers is commutative there is no difference here, so I will pick 3*. So more generally if a is a real number then a* : R -> R is given by a*x for all x. So 2* is not the same endomorphism as 3*. Okay, but still this does not help with /0. 2
conway Posted July 14, 2015 Author Posted July 14, 2015 (edited) Ajb I will take it that I can move forward? I am assuming then that you agree with the following. 1. Any number is composed of value and space, and can represent only it's space or only it's value. If then this is correct, I reiterate the following in regards to zero. Zero, must also then contain a value and a space. As you put it (0*, or *0). But it is in this case that there is a fundamental difference. Were as with any non zero there is no real difference. So then zero as value is represented as ( 0v ). So then zero as space is represented as (X). So that 0 as a number is ( 0v, X ). 0 = ( 0v, X ) 1 = ( 1v, X ) 2 = ( 2v, X,X) 3 = ( 3v,X,X,X) so on...... So that in multiplication and division with 0, I am also referring to 0 as either a space or as a value. 2 * 0 (x) is the space of 0. I place the given value of 2 into the space given (2)......then add. This is opposed to 0 being used as value 0 * 2 (x,x) is the space of 2. I place the given value of 0 into the space given. (0,0)....then add (0+0). It is zero used as space or value, that is important. 0v * As = 0 As * 0v = 0 Av * 0s = A 0s * Av = A Where as in division it is that value is always labeled first, space is always labeled second. Just as in regular definition of division. That is NOT commutative. It is also that placing the value into the space is "inverse" of addition. That is subtraction....I am sure you get the trend. I am also prepared to represent any rational, irrational, numbers. Edited July 14, 2015 by conway
ajb Posted July 14, 2015 Posted July 14, 2015 So generally you cannot just write a*b etc as you need these extra 'value' and 'space'. You are really changing the structure of integer numbers. You should not think of division as being separate from multiplication; they are really the same thing. That is, provided the algebra is commutative and for when it makes sense a/b = ab^{-1} = b^{-1}a. Meaning that division is really all about finding the element b^{-1} such that bb^{-1} = b^{-1}b = Id. In this sense it is all commutative. Now, in your constructions you are doing more than just simple arithmetic. You have to show that inverses make sense when you include 'space' and 'value'. 1
Strange Posted July 14, 2015 Posted July 14, 2015 (edited) Ajb I will take it that I can move forward? I am assuming then that you agree with the following. 1. Any number is composed of value and space, and can represent only it's space or only it's value. If then this is correct, I reiterate the following in regards to zero. Note that it is not a matter of this being "correct"; this is your definition of New Numbers. So for your number system it is correct by definition (as long as you can produce consistent maths based on it). It is, of course, not "correct" for numbers as we currently define and use them. Zero, must also then contain a value and a space. As you put it (0*, or *0). But it is in this case that there is a fundamental difference. Were as with any non zero there is no real difference. So then zero as value is represented as ( 0v ). So then zero as space is represented as (X). So that 0 as a number is ( 0v, X ). What is X in this new representation? Does it mean any value? (As 'x' is usually used) Edited July 14, 2015 by Strange 1
Greg H. Posted July 14, 2015 Posted July 14, 2015 Greg In this case it did NOT make the problem simpler did it. In fact it yielded an incorrect sum didn't it. But I will bite. Why then is "cancelation" allowed" outside of pemdas? If cancelation exists in an effort to help prepare to solve, than it is PART of solving it. Therefore the standard PEMDAS, which as you pointed out, exists to be used when solving. So why then can I cancel? In fact I just so happen to know an equation than when you do "cancel" it gives a wrong answer. Further if I solve with out "canceling" and I arrive at one sum, but if I cancel then arrive at another....isn't something wrong? You can use cancellation to simplify even when there is nothing to solve - indeed when you have no equals mark at all. For example Simplify [math]\frac {2x-2y}{x-y}[/math] By factoring out the two, it's possible to show that this is really just two, dressed up with some extra bits. But I didn't solve anything. There was no equation here, so PEDMAS doesn't even come into play. All I did was cancel out the parts that don't affect the actual value. Cancellation can also be combined with factoring to reduce fractions to more understandable values. Par exemple [math]\frac {99}{144}[/math] By factoring, I can remove a 3 from both values, and by cancellation I can drop the 3s. [math]\frac {33 \times 3}{48 \times 3}[/math] [math]\frac {33}{48}[/math] But wait, I can do that again. [math]\frac {11 \times 3}{16 \times 3}[/math] [math]\frac {11}{16}[/math] Seriously, I'm flabbergasted that you're literally trying to redefine division, while (seemingly) unable to grasp some of the more simple concepts of math itself.
conway Posted July 14, 2015 Author Posted July 14, 2015 (edited) Ajb, Strange Thank you guys, I understand nothing is "correct" yet. This is all relative, and tentative. And you guys are about to drop me like a hot potato....lol. Ajb, I do not have to use v and s "attached" to a number like variables. Except in the case where a variable is equal to 0. And for reasons all ready set forth. I can represent the idea of space and value in division quite easily. As you have said it is a matter of "undoing" multiplication.6/3. 6v/3s (x,x,x) space of 3 (2,,2,2) the value 6 evenly subtracted into all given spaces. (2) then all values subtracted (but one) 8/2 8v/2s (2,2), the space of 2 (4,4), the value * subtracted equally into all given spaces (4) then all values subtracted (but one) This can be done with any number combination, whole, natural, integers, rational, irrational. It is then that when using decimals, one converts to a fraction, then draws "space" accordingly. 1/2 (x/x,x) the space of 1/2 1/2 * 1/2 (x/x,x) (1/2,2) (1/2+2) all values added in all spaces (1/4 = .25) It is in this case Strange that (x) does not have value. It is a labeling of quantities of dimensions such as 1 has in it (x) = 1 single quantity of a spatial dimension 2 has in it (x,x) = 2 single quantity's of a spatial dimension 3 has in it (x,x,x) = 3 single quantity's of a spatial dimension 0 has in it (x) = 1 single quantity of a spatial dimension Edited July 14, 2015 by conway
ajb Posted July 14, 2015 Posted July 14, 2015 Ajb, I do not have to use v and s "attached" to a number like variables. I can see they are not variables, but they are still attached to every number (well integer so far). Except in the case where a variable is equal to 0. If you have a variable then you don't know if it is zero or not, it is variable. That said, it is usual to state for example that a variable cannot take certain values. I can represent the idea of space and value in division quite easily. As you have said it is a matter of "undoing" multiplication. 6/3. 6v/3s (x,x,x) space of 3 (2,,2,2) the value 6 evenly subtracted into all given spaces. (2) then all values subtracted (but one) Always 'v/s' and never 's/v '? This can be done with any number combination, whole, natural, integers, rational, irrational. It is then that when using decimals, one converts to a fraction, then draws "space" accordingly. So you have integers and rational, what about transcendental numbers? Say what is pi * 5.6763 and 5.6763 * pi ? (just as an example)
Strange Posted July 14, 2015 Posted July 14, 2015 Thank you guys, I understand nothing is "correct" yet. I think you missed my point: your definition of your new type of number is not really correct or incorrect: it is part of the definition. It is, of course, incorrect for numbers as currently defined. It is in this case Strange that (x) does not have value. It is a labeling of quantities of dimensions such as 1 has in it (x) = 1 single quantity of a spatial dimension 2 has in it (x,x) = 2 single quantity's of a spatial dimension 3 has in it (x,x,x) = 3 single quantity's of a spatial dimension 0 has in it (x) = 1 single quantity of a spatial dimension So how do you distinguish 0 from 1? What was wrong with the notation of representing each number by a pair of values (value,space)? Why suddenly change to lists of x's? This inconsistent and constantly changing description is what convinces me that you don't really have a clear concept. If did, you would be able to write it down clearly and consistently. If you have a variable then you don't know if it is zero or not, it is variable. That said, it is usual to state for example that a variable cannot take certain values. And this is exactly why I wanted to formalise the definition, so that we could see that it worked in general (with variables) and not just with you picking some special cases.
ajb Posted July 14, 2015 Posted July 14, 2015 And this is exactly why I wanted to formalise the definition, so that we could see that it worked in general (with variables) and not just with you picking some special cases. It does seem that a lot of this s and v construction is done with 'hindsight'; you know the answer so you make sure it works. A more formal definition is indeed required. I am very unsure how this would work outside of the integers. 1
conway Posted July 15, 2015 Author Posted July 15, 2015 (edited) Ajb, Strange I agree whole heartedly, that a more accurate formalization is still required. Thanks for helping me take these steps. I assure you that it only "appears" that my reasoning is hindsight. As you suggest ajb that I am "changing" the idea to yield the right answers. I have represented space and value in three different ways on this thread. Initially it was z1 and z2. This was done so as not to invoke the ire of the moderators. I switched to q and r, at your request Strange. I then went back to the original representation I have always used. I think that you can both find evidence of this in the original post "Relative Mathematics". I have struggled to find an appropriate way to represent this I agree. Yet assuming we can "momentarily" agree on space and value then I will restate all relative definitions, that can be found in "relative mathematics''. I am assuming this formalization is only a "launching" point for a better more accurate version. space = labeling of quantities of dimensions. value = labeling of quantities of existence, other that dimensions. Any given number is composed of space and value together (value inside space). x= single quantity of space. 1 = (1defindvalue, x ) 2 = (2defindedvalues, x,x ) etc... 0 = (1undefinedvalue, x ) This is different from 1 in that 1's value is defined. Zero's value is not defined. Naturally the "symbol" that represents undefined value is 0. 0 = ( 0v, x ) It is not that the space of 0 is undefined. It is defined. The number line and "traversing" integers is proof of this. So then 0 has the same "space" as 1 but not the same value. (3.141dvIL, 3.141dsIL) = Pi.......use as many "actual" digits as you like. (5.6763dvFL, 5.6763dsFL) = 5.6763 5.6763dvFl * 3.141dsIL = pi * 5.6763 This then gets quite lengthy I will continue definition if anyone wishes. Otherwise I will hold here. For example I must first represent 5 as value, the represent the fraction with 5 as value, the represent 3 as space, then the infinite fraction as space. Then place all values into all spaces. Ajb Only with 0 do we need to declare space and value. I have taken many test in which at the beginning of the question it sates that the given variables are or are not 0. This may be a fallacy in my small level of education...that is in the higher grades they don't tell you a given variable is or is not zero. The point being is that I am under the assumption that if we want to solve any equation with pure variables we must then know if any given variables are or are not zero. Only that same information is required when dealing with variables in space and value form. In division it is always v/s. As division is not commutative. But that is because it is the "inverse" of multiplication which is commutative. Also....jokingly.... I would think a person attending the polish relativity conference would be excited at the prospect of "relative mathematics". If you would care to here.....It is my opinion that everything in this idea of "mine" is actually Einstein's idea. That is he came up with it (relativity). He proved it (relativity). I am only applying relativity to mathematics. Time and Space are two things in one. Time is a value. Numbers then also are composed of two things in one....etc ....etc........ Strange The only issue I took with your form of formalization is that it appeared to me "possibly falsely" that each equation then yielded two sums. While the sums are always the same in regards to non zero's, it will always yield two different sums in regards to zero. So how then do we distinguish? Edited July 15, 2015 by conway
ajb Posted July 15, 2015 Posted July 15, 2015 I would think a person attending the polish relativity conference would be excited at the prospect of "relative mathematics". If you would care to here.....It is my opinion that everything in this idea of "mine" is actually Einstein's idea. That is he came up with it (relativity). He proved it (relativity). I am only applying relativity to mathematics. Time and Space are two things in one. Time is a value. Numbers then also are composed of two things in one....etc ....etc........ In fact it was not Einstein who realised that we need some form of relativity, it was (at least) Galileo. Einstein worked out the details of what is needed in electromagnetic theory (others also worked on this) and he realised what form of relativity we need for gravity. I understand your comments are in jest, but I see no real link with your 'relative mathematics' and relativity in physics. 1
Strange Posted July 15, 2015 Posted July 15, 2015 Any given number is composed of space and value together (value inside space). x= single quantity of space. 1 = (1defindvalue, x ) 2 = (2defindedvalues, x,x ) It is still very unclear (to me) what this notation means. 1. Why is "2defindedvalues" written with no spaces? 2. Does "2 defined values" refer to the following x,x ? 3. Or are you saying that 2 is represented by the combination of two things: the "defined" value 2 combined with "x,x" ? 0 = (1undefinedvalue, x ) 4. What does "1undefinedvalue" mean and how does it differ from "1definedvalue" ? Zero's value is not defined. 5. What does "not defined" mean? (This appears to be a problem as all the formal definitions of arithmetic I have seen start with 0 as the only defined value.) 0 = ( 0v, x ) 6. How does this new notation relate to the previous "0 = (1undefinedvalue, x )" notation? 7. Does this mean that: 0v = 1undefinedvalue If so, what is the significance of that? It is not that the space of 0 is undefined. It is defined. The number line and "traversing" integers is proof of this. So then 0 has the same "space" as 1 but not the same value. On the number line, the "value" is represented by the position on the line: ... -3 ... -2 ... -1 ... 0 ... 1 ... 2 ... 3 ... So the value of zero seems quite clear and well defined in this representation. 8. Is your number line more like: ... -3 ... -2 ... -1 ????? 1 ... 2 ... 3 ... 9. How do you represent "space" on the number line in your system ? (3.141dvIL, 3.141dsIL) = Pi.......use as many "actual" digits as you like. (5.6763dvFL, 5.6763dsFL) = 5.6763 10. What do "dvIL", "dsIL", "dvFL" and "dsFL" mean? Only with 0 do we need to declare space and value. 11. Then why did you start with "2 = (2defindedvalues, x,x )" which appears to be defining both "value" and "space" for the number 2 ? 1
Recommended Posts