The Indeterminate Nature of 0/0Date: 12/21/2000 at 16:02:44 From: Rob Subject: Ludicrous new 0/0 theory? I have been fiddling around with dividing 0 by 0, and have come up with an interesting theory. I have not yet seen any proof that this theory cannot work, so as of now I think it is valid. It goes as follows: My theory is that: 0/0 = any number To start, use the definition of division equation where: a/b = c because c*b = a In the situation of 0/0 = c, then c*0=0 for any real value of c (I am not familiar with how to work with infinity or non-real numbers so I have left them out.) Some have argued that this would simply be an indeterminate form, but I counter with this; the equation of the line y = 1 in standard form is: y = 0x+1 If you were to solve for x, then x = (y-1)/0 By my theory only 0/0 can satisfy every value of x, thus y must always be equal to 1 (otherwise it is undefined). In this situation, where y always equals one, x can be any number of values. If you say that 0/0 is indeterminate, then the equation y = 1 would also be indeterminate because there are an infinite number of x values for that single y value. That is, unless 0/0 can be equal to any one of the infinite numbers at any given time. For example at point x = 5 in the x = (y-1)/0 equation, the value 0/0 = 5. This theory would radically change certain views of math. Consider the equation y = 0/[(x-2)(x+3)] using my theory. The points x = 2 and x = -3 in this equation would stretch forever straight up and down, since y would equal 0/0. Under traditional math laws, these points would be asymptotes. Also consider the equation y = (0/0)x. Using traditional math this could simplify to 0y = 0x, making the graph cover every point. This corresponds exactly to my original theory. I hope my writing has made sense. If you can think of a way to tear down this theory, I am anxious to hear it. Please keep in mind that I am not arguing 1/0, because that is still undefined, I am only arguing for 0/0. Thank you, Rob Date: 12/21/2000 at 17:35:39 From: Doctor Ian Subject: Re: Ludicrous new 0/0 theory? Hi Rob, This is an interesting theory, but there is a fly in the ointment. Let's say for a moment that we accept that 0/0 = any number. Then we can do this: 0 = 0 Trivially true 0/0 = 0/0 Dividing both sides by the same thing preserves equality 1 = 2 Substitute for 0/0 Oops! It appears that your definition leads to a contradiction. This is a problem, because once you can 'prove' a contradiction, you can use that contradiction to prove anything, which means that the concept of proof becomes meaningless. Perhaps the most common type of proof used in mathematics is "proof by contradiction." To prove something by contradiction, you assume that the thing is false; then you show that the assumption leads to a contradiction, which means that the thing can't be false; which means that it must be true. The classic example is the following proof that the square root of 2 is irrational: 1. Assume that it is rational, of the form a/b, where a/b is reduced to lowest terms. 2. Then (a/b)*(a/b) = 2, or a^2 = 2b^2, so a^2 must be even. 3. Since a^2 is even, a must be even, say a = 2c. 4. Then (2c)^2 = 2b^2, or 4c^2 = 2b^2 or 2c^2 = b^2, so b must also be even. 5. So in a/b, both a and b are even, but we assumed we'd reduced the fraction to lowest terms. We have a contradiction, so sqrt(2) must be irrational. With the addition of your new definition, proof by contradiction becomes trivial, regardless of what we want to prove. For example, let's prove that the square root of 2 is rational. 1. Assume that it is irrational. 2. Let 0/0 = 0/0, where 0/0 is any number. 3. Then 1 = 2, which is a contradiction. Therefore, sqrt(2) must be rational. Do you see the problem? If 0/0 = any number, then the square root of 2 is both rational and irrational; pi is the ratio of circumference to diameter, and it's also not that ratio; there is no largest integer, and there is a largest integer; every number is prime, and no number is prime... It's great that you're thinking about this stuff! What you've actually stumbled onto is one of the deepest issues in mathematics, which is the irresolvable tension between completeness and consistency, a subject explored by the mathematician Kurt Godel, who showed that you have to choose one or the other, because you can't have both in the same system. With that in mind, I strongly recommend that you take a look at a book called _Godel, Escher, Bach_, by Doug Hofstadter. You should be able to find it at any bookstore or library. Ignore the chapters the first time through, and just read the dialogs in between the chapters. Then go back and read the chapters. It's a fabulous book, which will give you a lot of new material to think about. I hope this helps. Let me know if you'd like to talk about this some more, or if you have any other questions. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 12/28/2000 at 01:27:03 From: Rob Subject: Re: Re: Ludicrous new 0/0 theory? Thanks for writing back, Doctor Ian. I hope I don't sound (too) stubborn, but I have a couple of follow-up questions about your answer. First a summary of my points: -> 0/0 = any number -> a/b = c because c*b = a, likewise 0/0 = any number because (any number)*0 = 0 -> for the equation y = 1: table of values for x table of values for y x y (only y = 1 will work) -2 1 y x -1 1 1 ? 0 1 1 1 2 1 There is no single x value for this point it has many values. If having many values makes 0/0 indeterminate, then it should also make y = 1 indeterminate, which it isn't. If y = 1 is a legitimate graph, then 0/0 should be good also. Conversely, if 0/0 is not legitimate, then neither should is the equation y = 1. You might say that my conclusion is not based on my facts, but examine this: y = 1 --> y = 0x+1 --> y-1 = 0x --> x = (y-1)/0 According to my theory, 0/0 = any number, the only way to satisfy the equation for any value of x is for the other side of the equation to be 0/0, which can only happen if y = 1. This seems to be a logical extension of my theory and the manipulation of equations. Consider also this, from "Zero Laws and L'Hopital's Rule" in your archives at: http://mathforum.org/dr.math/problems/sheryl3.4.98.html "Say we have two functions, f(x) and g(x). If you take their quotient, f(x)/g(x), you get another function. Let's call this h(x). "You can also take limits of functions. For example, the limit of a function f(x) as x goes to 2 is the value that the function gets closest to as it takes on x's that approach 2. We would write this limit as: lim{x->2} f(x) "This is a pretty intuitive notion. Just draw a graph of your function, and move your pencil along it. As the x values approach 2, see where the function goes. "Now for our example. Let: f(x) = 2x - 2 g(x) = x - 1 and consider their quotient: h(x) = f(x)/g(x) "What if we want the lim{x->1} h(x)? There are theorems that say that lim{x->1} h(x) = lim{x->1} f(x)/g(x) = (lim{x->1} f(x))/(lim{x->1} g(x)) = 0 / 0 !!! "So we now have: lim{x->1} h(x) = 0/0 "But I can employ another theorem, called l'Hopital's rule, that tells me that this limit is also equal to 2. So in this case, 0/0 = 2 -Doctors Sonya and Celko, The Math Forum" You wrote: 0 = 0 Trivially true 0/0 = 0/0 Dividing both sides by the same thing preserves equality 1 = 2 Substitute for 0/0 "Oops! It appears that your definition leads to a contradiction." - Doctor Ian, The Math Forum To rebut your contradiction theory, I would say that it is not the 0/0 = any number that is wrong, but the substitution. If you may substitute any value in for 0/0 arbitrarily, then why wouldn't this be possible? 2x = 3y 2(5) = 3(6) 10 = 18 or even b 3x-1 = 2x+4 3(-7)-1 = 2(-7)+4 -22 = -10 Can you arbitrarily substitute in numbers that make no sense in the equation? It was also my theory that the value of 0/0 is dependent on the situation. 0/0 is equal to 1 and 2, but not in an equation at the same time necessarily. In the equation y = 1, can you not have the point P(4,1) and the point P(-3,1)? You just can't get them by substituting a single x coordinate, that is all. I hope I have raised some good points, I'll be grateful for whatever you can do for me. Thanks, Rob Date: 12/31/2000 at 18:18:16 From: Doctor Rick Subject: Re: Re: Ludicrous new 0/0 theory? Hi, Rob. I don't know exactly what your theory means, and neither did Dr. Ian. You say, "0/0 = any number." The natural way to interpret this phrase is, "You can choose any number x, arbitrarily, and the equation 0/0 = x is true." Dr. Ian read it this way, I believe, and from this he derived that 1 = 2. This causes big problems for math, as he said. Your response is that this isn't what you meant. You say that "the value of 0/0 is dependent on the situation." The word I would use for this is "indeterminate." Here is what we mean by "indeterminate." The value of 1/0 is called "undefined" because there is NO number x that satisfies the equation 1/0 = x, or equivalently, 0*x = 1. In contrast, EVERY number x satisfies the equation 0/0 = x, or equivalently, 0*x = 0. But this does not mean that you could substitute any number arbitrarily for 0/0. It depends on the context - on the details of the problem in which you encountered 0/0. When you run into a problem whose solution appears to be 0/0, there may be a solution, but you'll have to back up and try to find it by another approach. When you find it, you could say that 0/0 "equals" that value, but only for that particular problem - just as you say. The example you quote from our website, concerning L'Hopital's Rule, is a good example of how a problem in which 0/0 crops up can have a valid solution. L'Hopital's Rule arises in calculus, and if you haven't seen it yet, I couldn't explain it quickly, but the basic idea is that, if you encounter 0/0 in evaluating the limit of the quotient of two functions, there is another method by which you can find the value of the limit. You can't use the 0/0 to find the solution, but you can back up and take another route to find the solution. Thus, a good way to think of the "indeterminate" 0/0 is as a "detour sign". It says in effect, "The road ahead is washed out. The town beyond it is still there, but you'll have to find another way to reach it." To bring this around to the word that you don't like, it says, "The solution can NOT be DETERMINED by this method. There may be a solution, but you must determine it some other way." In your original submission, you cite the equation y = 0/[(x-2)(x+3)] as one in which your math gives a different answer than traditional math. You say that traditional math identifies x = 2 and x = -3 as asymptotes, whereas your math leads to a graph with vertical lines at x = 2 and x = -3. The fact is, your answer is the traditional answer to the equivalent equation y(x-2)(x+3) = 0 In order for the product to equal zero, at least one of the three expressions y, x-2, and x+3 must equal zero. This leads to the solution y = 0 or x = 2 or x = -3 which is graphed as three lines: a horizontal line at y = 0, and the two that you call "asymptotes." They are part of the solution set, NOT asymptotes. Finally, let's consider the equation y = 1, as you do. You say that, "if y = 1 is a legitimate graph, then 0/0 should be good also." I think you mean that, since the graph of y = 1 is a line consisting of all points with y = 1 and x = any number, then by analogy, the solution of the single-variable equation 0/0 = x should be the entire number line, x = any number. I'd say, yes, the solution of the equivalent equation 0x = 0 is the entire number line. In this particular problem, it's quite true that the solution is x = any number. But this does not mean that 0/0 = any number, in itself. The example about L'Hopital's Rule is a case in which the solution is NOT any number; it is the particular number 2, and no other. As I have said, you must consider each particular problem on its own. The fact that 0/0 = any number in one problem cannot be used to solve another problem. Have I made sense yet? I think the key concept you have been missing is this problem-specific nature of 0/0. The difference between "any number" and "indeterminate" is that "indeterminate" means "it MAY be any number, but you can't tell just by looking at 0/0 itself." - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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