Buych778 1 Posted April 20, 2015 This article was started as more of a debate than an answer to the question. According to Chaos Theory, any number, n, is defined as being higher than anything with lower value, or higher than any value from 0 to n, and lower than anything with a higher value, or lower than any value from n to infinity. Now, this may seem quite obvious, but also according to Chaos Theory, infinity cannot exist as a number. Both of these are stated in Chaos Theory, and are noted as a flaw in the number system, as how can a definite number be described as being below infinity, yet infinity is not a number in itself? Chaos Theory also states that any number has no definite point of "accuracy", meaning that no number can be exact. Lets say you divide one by three, you get .333 repeated infinitely, but if you plan on actually using this number, you cannot use it as exactly .333 repeated, as no computational device can have them as infinite. The only way that the computer can use it infinitely, is to make it 1/3, which therefore, is not exactly .333 repeated. I am not referring to the value of 1/3, but I am referring to the exact form of .333 repeated. This means that no computer can actually calculate anything of value. Another thing to discuss, what defines the value of a number? What makes the value of 10, 10? Why doesn't the value of 10 equal 4? What defines the space between each number? What defines a number as large, or small? How can our mathematical functions have any distinction on what actually happens? If every number is not exact, how can one predict anything to actually either exist or occur? 0 Share this post Link to post Share on other sites

Strange 4267 Posted April 20, 2015 Both of these are stated in Chaos Theory, and are noted as a flaw in the number system Chaos Theory also states that any number has no definite point of "accuracy", meaning that no number can be exact. Citation needed. That doesn't appear to have anything to do with chaos theory. (And that last one is plainly wrong.) Lets say you divide one by three, you get .333 repeated infinitely, but if you plan on actually using this number, you cannot use it as exactly .333 repeated, as no computational device can have them as infinite. If you write that in base 3, then it is 0.1 so there is no problem. Don't confuse the representation with the number (or even the practicality of using the number). Another thing to discuss, what defines the value of a number? The nature of numbers (the natural numbers, anyway) can be defined in set theory: http://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory 1 Share this post Link to post Share on other sites

studiot 2039 Posted April 20, 2015 (edited) What you are describing is known as the well ordering property. Integers, rational numbers and real numbers can be put into a continuous list in order of size a<b<c........ and so on., Where a, b, c are integers, rational or real numbers. But this is not true of the complex numbers that do not possess this property. Edited April 20, 2015 by studiot 1 Share this post Link to post Share on other sites

MonDie 142 Posted April 20, 2015 According to Chaos Theory, any number, n, is defined as being higher than anything with lower value, or higher than any value from 0 to n, and lower than anything with a higher value, or lower than any value from n to infinity. Now, this may seem quite obvious, but also according to Chaos Theory, infinity cannot exist as a number. Both of these are stated in Chaos Theory, and are noted as a flaw in the number system, as how can a definite number be described as being below infinity, yet infinity is not a number in itself? I prefer to think of infinity as the absence of a limit, rather than a quantity. Another thing to discuss, what defines the value of a number? What makes the value of 10, 10? Why doesn't the value of 10 equal 4? What defines the space between each number? What defines a number as large, or small? How can our mathematical functions have any distinction on what actually happens? If every number is not exact, how can one predict anything to actually either exist or occur? This depends on the level of measurement. All quantities on an interval scale are arbitrary. There's still heat at zero Celsius, and 1C only corresponds to an arbitrarily defined difference in heat. On a ratio scale, zero indicates absence. It's even less arbitrary when the thing is countable. 0 Share this post Link to post Share on other sites

swansont 7338 Posted April 20, 2015 The only way that the computer can use it infinitely, is to make it 1/3, which therefore, is not exactly .333 repeated. I am not referring to the value of 1/3, but I am referring to the exact form of .333 repeated. This means that no computer can actually calculate anything of value. And yet we have calculators that do this very thing. Proof by contradiction — it means one of your assumptions is wrong. The key here is precision. You can't compute some numbers to arbitrary precision. But that's a far cry from being unable to calculate anything at all. 0 Share this post Link to post Share on other sites

Endy0816 444 Posted April 20, 2015 1/3 and 0.333... are one and the same. Another way to look at it is that 1/3 contains all the necessary information to fully describe 0.333... Anything that can be encapsulated like this our devices(and we ourselves) can work with, at 100% precision. You can multiply 6 by 1/3, for instance. You probably just skipped the addition of the infinite series of (0.3*6)+(0.03*6)+(0.003*6) + ... and instead leapt to the result of 2. 0 Share this post Link to post Share on other sites

MWresearch 40 Posted April 21, 2015 (edited) What you are describing is known as the well ordering property. Integers, rational numbers and real numbers can be put into a continuous list in order of size a<b<c........ and so on., Where a, b, c are integers, rational or real numbers. But this is not true of the complex numbers that do not possess this property. Why is that not true of complex numbers when their absolute value yields a real distance along the complex plane? Can I not say that 2*i is a greater distance from the origin than 1/2? 1/3 and 0.333... are one and the same. Another way to look at it is that 1/3 contains all the necessary information to fully describe 0.333... The way you define that is with a limit, like the sum of 3/10^n with n going to infinity, which, approaches 1/3. Edited April 21, 2015 by MWresearch 0 Share this post Link to post Share on other sites

studiot 2039 Posted April 21, 2015 MWresearch Why is that not true of complex numbers when their absolute value yields a real distance along the complex plane? Can I not say that 2*i is a greater distance from the origin than 1/2? Because you have just described a circle with an infinity of complex numbers. Since there are an infinity of such circles, your would have an infinity of complex numbers filling every place in the list. 2 Share this post Link to post Share on other sites

MWresearch 40 Posted April 22, 2015 (edited) Even if your vague description was true which I don't see the reasoning for, I'm assuming you're indirectly referring to Euler's identity and a unit circle on the complex plane, how would an infinite set of numbers prevent those numbers from being greater than another number? There are infinitely greater numbers than 2, but that doesn't mean we can't measure that numbers of that infinite set, like 3 and pi and 4, are greater than two. Similarly, there may be an infinite arrangement of numbers which achieve the same coordinate as 2i+1, but that doesn't mean we cannot measure those specific complex numbers distances from the origin. You will have to explain the actual meaning of what you said. Edited April 22, 2015 by MWresearch 0 Share this post Link to post Share on other sites

John 165 Posted April 22, 2015 (edited) We need to be careful with our terminology here. The standard order we use with the real numbers is not a well-order, since a well-order requires that any non-empty subset has a least element, which is not true of the reals (consider, for example, the interval (0, 1)). It is, however, a total order.If we assume the axiom of choice, then we have the (logically equivalent) well-ordering theorem, which states that any set can be well-ordered. So while the standard order on the reals is not a well-order, if we assume the axiom of choice, then there does exist a well-order for the reals, as well as a well-order for the complex numbers. Whether the order is meaningful, much less useful, is another story.However, what we can't achieve with the complex numbers is an ordered field structure, since the square of any non-zero element of an ordered field must be positive, but we have i^{2} = -1.Edit: If we try to order the complex numbers simply according to their magnitude, then we find that infinitely many complex numbers share any given magnitude. Looking at the complex plane, what we see is each magnitude |z| represented by a circle of radius |z|. This is what studiot was saying. So for instance, looking strictly at magnitude, -1 = i = 1 = -i. What we can do in that case is define the equivalence relation ~ such that z_{1} ~ z_{2} if |z_{1}| = |z_{2}|. But all we accomplish with that is mapping each equivalence class to a real number, thus it's not exactly a mathematical breakthrough and doesn't really accomplish what we want. Edited April 22, 2015 by John 2 Share this post Link to post Share on other sites

studiot 2039 Posted April 22, 2015 Thank you, John for that full and frank exposition. +1 0 Share this post Link to post Share on other sites

MonDie 142 Posted April 22, 2015 (edited) I had to look up John's terms, and I'm still uncertain. Maybe he can clarify it for the both of us. In dumby terms, consider the set of all real numbers greater than 2. That set has no least element because 2.01 < 2.1 and 2.001 < 2.01 and 2.0001 < 2.001 and so on, so the real numbers are not wellordered. In an ordered field structure, a set of elements must maintain the same order after they're multiplied by another positive element. Imaginaries violate this rule. If x < y and z > 0, then xz < yz. https://en.wikipedia.org/wiki/Ordered_field#Properties_of_ordered_fields Edited April 22, 2015 by MonDie 0 Share this post Link to post Share on other sites

MWresearch 40 Posted April 22, 2015 (edited) We need to be careful with our terminology here. The standard order we use with the real numbers is not a well-order, since a well-order requires that any non-empty subset has a least element, which is not true of the reals (consider, for example, the interval (0, 1)). It is, however, a total order. If we assume the axiom of choice, then we have the (logically equivalent) well-ordering theorem, which states that any set can be well-ordered. So while the standard order on the reals is not a well-order, if we assume the axiom of choice, then there does exist a well-order for the reals, as well as a well-order for the complex numbers. Whether the order is meaningful, much less useful, is another story. However, what we can't achieve with the complex numbers is an ordered field structure, since the square of any non-zero element of an ordered field must be positive, but we have i^{2} = -1. Edit: If we try to order the complex numbers simply according to their magnitude, then we find that infinitely many complex numbers share any given magnitude. Looking at the complex plane, what we see is each magnitude |z| represented by a circle of radius |z|. This is what studiot was saying. So for instance, looking strictly at magnitude, -1 = i = 1 = -i. What we can do in that case is define the equivalence relation ~ such that z_{1} ~ z_{2} if |z_{1}| = |z_{2}|. But all we accomplish with that is mapping each equivalence class to a real number, thus it's not exactly a mathematical breakthrough and doesn't really accomplish what we want. I see more clearly where the infinite numbers are coming from, since imaginary numbers in a way act like their own dimension that also acts to treat fundamental operations differently, whereas just real numbers on a number line would not create a circle of possibilities. Is there any manner of expressing that one complex number is greater than or less than another at all? Edited April 22, 2015 by MWresearch 0 Share this post Link to post Share on other sites

studiot 2039 Posted April 22, 2015 MWresearch Is there any manner of expressing that one complex number is greater than or less than another at all? No because say one real number, a is equal to another real number is equivalent to saying that a>b and b>a means that a=b. Now for a complex number the definition that A_{complex} = B_{complex} is more complicated. 0 Share this post Link to post Share on other sites

MWresearch 40 Posted April 22, 2015 (edited) Then let's start with 0. What is i compared to 0i? i seems to possess some nonzero distance from the origin, doesn't it have to be greater than or less than 0i? Perhaps it is just that i isn't comparable to real numbers, and thus complex numbers containing both a real and imaginary component are inherently incomparable, but that doesn't answer why imaginary numbers aren't comparable in the first place. So you have the square root of a negative number that's going in the opposite direction of the positive direction on a number line, big deal, why should that create all this anti-symmetry and contradictions of logic? It's like i is its own universe somehow. If you graph sqrt(x) the imaginary range in the negative domain still makes some physical sense, it still functions as some base unit you need to multiply by itself to achieve that specific length, which, is symmetrical to the principal roots when you overlap the plots of the function in both the real and complex plane. For some reason, these lengths transfer to a different kind of number line. Edited April 22, 2015 by MWresearch 0 Share this post Link to post Share on other sites

imatfaal 2481 Posted April 22, 2015 the Then let's start with 0. What is i compared to 0i? i seems to possess some nonzero distance from the origin, doesn't it have to be greater than or less than 0i? Perhaps it is just that i isn't comparable to real numbers, and thus complex numbers containing both a real and imaginary component are inherently incomparable, but that doesn't answer why imaginary numbers aren't comparable in the first place. So you have the square root of a negative number that's going in the opposite direction of the positive direction on a number line, big deal, why should that create all this anti-symmetry and contradictions of logic? The direction of an axis showing the square roots of a negative number is ORTHOGONAL to the positive number line not in the opposite direction 0 Share this post Link to post Share on other sites

MWresearch 40 Posted April 22, 2015 (edited) the The direction of an axis showing the square roots of a negative number is ORTHOGONAL to the positive number line not in the opposite direction The negative number is what's going the opposite direction, not the output of the square root function over the negative portion of the domain. Edited April 22, 2015 by MWresearch 0 Share this post Link to post Share on other sites

John 165 Posted April 23, 2015 (edited) I had to look up John's terms, and I'm still uncertain. Maybe he can clarify it for the both of us. I'll try. Just let me know which part(s) need clarification beyond what Wikipedia can provide. I see more clearly where the infinite numbers are coming from, since imaginary numbers in a way act like their own dimension that also acts to treat fundamental operations differently, whereas just real numbers on a number line would not create a circle of possibilities. Is there any manner of expressing that one complex number is greater than or less than another at all? Yes. We can induce a total order on the complex numbers by moving up along the imaginary axis and then right along the real axis, i.e. [math]a + bi < c + di \iff [(a < c) \textnormal{ or } (a = c \textnormal{ and } b < d)][/math]. So for example, i < 100i < 1 < (2 - 100i). It's an odd sort of ordering if we're looking at the numbers in terms of their conceptual magnitude, but as far as I can tell it does satisfy the axioms. We can also induce a partial order on the complex numbers using the concentric circles of magnitude mentioned earlier, counting clockwise around each circle before moving to the "next." It's probably easier to think about this in terms of polar coordinates, in which we can say that [math](|z_1|, \theta_1) < (|z_2|, \theta_2) \iff \left(|z_1| < |z_2|\right) \textnormal{ or } \left(|z_1| = |z_2| \textnormal{ and } \theta_1 < \theta_2\right)[/math]. However, this is a bit unsatisfying since it means, among other things, that 1 < -1. Edited April 23, 2015 by John 0 Share this post Link to post Share on other sites

MWresearch 40 Posted April 23, 2015 (edited) I'll try. Just let me know which part(s) need clarification beyond what Wikipedia can provide. Yes. We can induce a total order on the complex numbers by moving up along the imaginary axis and then right along the real axis, i.e. [math]a + bi < c + di \iff [(a < c) \textnormal{ or } (a = c \textnormal{ and } b < d)][/math]. So for example, i < 100i < 1 < (2 - 100i). It's an odd sort of ordering if we're looking at the numbers in terms of their conceptual magnitude, but as far as I can tell it does satisfy the axioms. We can also induce a partial order on the complex numbers using the concentric circles of magnitude mentioned earlier, counting clockwise around each circle before moving to the "next." It's probably easier to think about this in terms of polar coordinates, in which we can say that [math](|z_1|, \theta_1) < (|z_2|, \theta_2) \iff \left(|z_1| < |z_2|\right) \textnormal{ or } \left(|z_1| = |z_2| \textnormal{ and } \theta_1 < \theta_2\right)[/math]. However, this is a bit unsatisfying since it means, among other things, that 1 < -1. Hmm, I think I see where the ambiguity lies. If the imaginary component is bigger than that of another, but the real component is smaller than that of the same other, you cannot determine which is fundamentally greater? That seems to imply imaginary numbers have no precedence over real numbers, its as if they were never related to real numbers at all, completely independent like its own dimension which I guess explains the orthogonality. For practical purposes, lets say I'm dropping a bowling ball from a height of 3 meters, but I solve for 4 meters of a downward parabola with a maximum. Where is that imaginary solution in physical reality? Where is that extra imaginary dimension I'm not seeing? Edited April 23, 2015 by MWresearch 0 Share this post Link to post Share on other sites

John 165 Posted April 23, 2015 (edited) Hmm, I think I see where the ambiguity lies. If the imaginary component is bigger than that of another, but the real component is smaller than that of the same other, you cannot determine which is fundamentally greater? Well, we can always just go by the magnitude to get a sense of relative sizes. We just have to accept that in that context, certain apparently unequal numbers are equal in magnitude, e.g. 1 has the same magnitude as i and -1. The difficulty arises when we try to put the complex numbers into some order such that two unequal numbers have unequal positions in the order, e.g. even though 1 and i have the same magnitude, if we want to order the complex numbers, we'd like either i < 1 or 1 < i (but not both), and fulfilling either requirement can be tricky, especially once we start adding operations into the mix. For practical purposes, lets say I'm dropping a bowling ball from a height of 3 meters, but I solve for 4 meters of a downward parabola with a maximum. Where is that imaginary solution in physical reality? Where is that extra imaginary dimension I'm not seeing? Hm, I'm not sure what you're asking here. The parabola has real solutions for all t, but in practice the Earth will keep the bowling ball from following the parabola past 3 meters, so in practice (disregarding wind and other such adjustments we'd need to make), the bowling ball will follow the parabola until it bounces, then follow another until it bounces again, and so on until it comes to rest. The parabola only describes the ball's path from the moment you throw/drop it until the moment it collides with some other object, so at the risk of upsetting any philosophers of science lurking around, the fact that the graph of the parabola continues down past h_{0} - 3 meters doesn't have a particularly profound meaning. Edit: To answer your question, the solution is real not imaginary, and it lies one meter under the ground. Edited April 23, 2015 by John 0 Share this post Link to post Share on other sites

MWresearch 40 Posted April 23, 2015 This is what I mean with the physical problem http://www.wolframalpha.com/input/?i=solve+-x%5E2%2B3%3D4+for+x 0 Share this post Link to post Share on other sites

John 165 Posted April 23, 2015 (edited) Alright, I see what you're saying. The situation then is that you're standing at a height of three meters, dropping a bowling ball, and then trying to determine at what time the bowling ball will be at a height of four meters. But a height of four meters doesn't lie on this parabola, so there's no solution. Of course, in practice, you could drop the bowling ball onto a machine that will catch the ball and launch it with enough force to send the ball four meters or higher into the air, but then, mathematically, you're dealing with a different parabola.Edit: Just randomly decided to search for "imaginary time" and this came up: http://en.wikipedia.org/wiki/Imaginary_time. But it's something you'd probably want to discuss in the Physics forum. Edited April 23, 2015 by John 0 Share this post Link to post Share on other sites

MWresearch 40 Posted April 23, 2015 Alright, I see what you're saying. The situation then is that you're standing at a height of three meters, dropping a bowling ball, and then trying to determine at what time the bowling ball will be at a height of four meters. But a height of four meters doesn't lie on this parabola, so there's no solution. Of course, in practice, you could drop the bowling ball onto a machine that will catch the ball and launch it with enough force to send the ball four meters or higher into the air, but then, mathematically, you're dealing with a different parabola. Edit: Just randomly decided to search for "imaginary time" and this came up: http://en.wikipedia.org/wiki/Imaginary_time. But it's something you'd probably want to discuss in the Physics forum. No imaginary time makes sense in terms of treating imaginary numbers as this extra dimension and it is used in line-integrals to define probabilistic paths of particles, like if there's extra actions that happen in each particular moment in time to allow particles to move as they do. But still, treating this one special square root of negative one as its own entire dimension just be multiplying it by different real numbers, like 1i, 2i, 3i, 4i just seems like too much for one number to account for, and we can't even see it. But with that physical situation, I'm confused as to when you say there is no solution. Is there not an imaginary solution? On which coordinate in time is that imaginary dimension perpendicular too? Can I say at y=0, there's some imaginary time where the ball was at 4 meters? How would I go about making sense of the solution i? It seems imaginary numbers encompass all of nature, they must be around somewhere... It kind of seems like wolfram is saying that at an imaginary perpendicular axis in time where t is t=+/- i when real time = 0. A lot functions also make sense when you overlap their complex and real components, like the symmetry of the logarithm function that you'd expect from being the integral of 1/x, except with an addition +pi or the symmetry of the square root function 0 Share this post Link to post Share on other sites

John 165 Posted April 23, 2015 There is an imaginary solution mathematically, but what I was referring to was the physical interpretation. That is to say, physically, there is no time at which the ball is at a height of four meters so long as it's following the parabolic trajectory defined by -t^{2} + 3 = h. We mustn't confuse the mathematical model of a system with the system itself.As for applications of complex numbers in physics, I'll have to leave that for those who know more about physics and applied mathematics. My current knowledge there doesn't extend much beyond knowing that there are applications. 0 Share this post Link to post Share on other sites

MWresearch 40 Posted April 23, 2015 Well, if real numbers or values exist, mustn't imaginary values as well? Like I said, they appear everywhere in nature, yet they are no where to be found with our eyes. 0 Share this post Link to post Share on other sites