Note: There's a small mistake in the video. When I write c is positive 0 in property 3, I don't mean c positive 0 as a real number, but c positive 0 with the new ordering squiggly greater than (so it should be c squiggly greater than 0).
When you say that *c* has to be squiggly greater than 0, it's just in order to make sure the inequality sign doesn't switch when multiplying it by *c* , isn't it? I mean, could we actually tell where is *c* (which quadrant) knowing it is squiggly greater than 0?
This is one of my all time favourite proofs, which demonstrates the fact that as you add more dimensions to numbers, they lose neat properties. So ordering is lost when real numbers become complex numbers; multiplication loses commutativity when complex numbers become quaternions; and multiplication loses associativity when quaternions become octonions, etc.
and multiplication even loses alternativity (a weaker form of associativity (xx)y = x(xy) that octonions retain) with sedenions. And beyond that, you cannot have an algebraic structure at all !!
In that spirit I question whether complex numbers should be called "numbers." (The point is purely about terminology, not about substantive math.) Complex numbers as a vector space over the reals with an amazingly useful vector product seems like a more straightforward way to describe what we are supposed to call "complex numbers." I don't t think many people would say that quaternions are "numbers." Why should we insist on according that honor to "complex numbers"? (Once again, this is not some crackpot theory -- I just think it would be better to refer to complex numbers as we refer to quaternions rather than as we refer to real numbers.)
Though complex number lose some properties, it's also interesting how many properties they retain when compared to quaternions, octonians, etc. Nearly all of the arithmetical properties of real numbers are kept with complex numbers, but with further dimensions added, you lose basic properties like associativity. Complex numbers even have some very nice properties that the real numbers alone don't have, like the fundamental theorem of algebra.
My intuitive proof: If we have a function u(r,k) = re^ik, where r and k are both Re. Then u(-r,0) = u(r,π). For simplicity later, let r be bounded on the interval R=-π to -R=π. We can create two distinct continuous paths from (R,0) to (-R,0) by keeping one variable constant and adjusting the other. Keeping r constant and increasing k from 0 to π takes us from (R,0) to (-R,0) thus u(r,k) > u(r,k+ε). Keeping k constant and increasing r from R to -R takes us from (R,0) to (-R,0) thus u(r,k) > u(r+ε,k). But if u(r,k+ε) and u(r+ε,k) are both continuous and intercept at (R,0) and (-R,0) but not at any other point then they cannot be ordered without being made discontinuous somehow. Damn, it was logical in my head but I cannot seem to explain xD!
Oh my goodness that address book explanation for lexicographic ordering was spot on!!! Also my professor did proof by intimidation with this one in real analysis so seeing in spelled out makes me believe in this now. Thank you!
It seems like property 3 is the one that's no good. When C is real and positive, it seems rather obvious that that should hold. But when defining the new inequality, it doesn't seem as reasonable to assume that property 3 is something that we want because if an ordering does exist, we don't know how that would be affected when multiplying by a complex number.
+Dr. Peyam's Show I think he means that it isn't reasonable to assume that c must be squiggly-greater-than 0, since with the real numbers literally nothing changes if we take that arg(c) = 0 (i.e. c is scalar) is the condition for property (3). Of course using that rule instead would _not_ lead to a contradiction...
If that property doesn't hold, it becomes difficult to do algebra. Imagine if you couldn't rely on multiplying both sides of an inequality by something and having the inequality still hold. The same goes for all the other properties -- we want them so that the ordering is algebraically useful.
@@gcewing But that rule still holds if you not only say that c > 0 but also that c is a real number. You could still have two complex numbers that are not real numbers, multiply both of them by a real c value and the inequality will still hold. You can do algebra as usual. That property still holds if you require c to be a positive real number. So such an ordering is still algebraically useful.
I have a perfect ordering of the complex numbers, but the margin is not wide enough to contain it. Also, I wish I had shirts that shine like Dr. Peyam's. Is there a laundry detergent for that? P.S. I watch Dr. P and Dr. B/R with passion ;-)
informally you can prove this in a simpler way by adding the requirement of transitivity. Now take N complex roots of unity (a_1, ..., a_N), which of course form a circle and you arrive at a_1 < a_N and a_N < a_1 which breaks assimetry.
Yes, you can. There is an order relation on C. Suppose a complex number is smaller than another if it is closer to the origin. So, if it is on a circle centered originally and with a smaller radius. Any complex number is well defined by its radius and the angle made by r with the ox axis. If we do angle abstraction and compare r we can find a relation of order in C. Since r is in R, it means that the order relation
Two classic exercises in one video!(the complex field is non-ordered and 1>0 for all ordered fields) Great! Byyy the way, I am still waiting for 3 videos(!!!)
Did you forgot it only because I did not asked for it for a month(-ish, close enough) !? We talked about it! D^n f(x) for all n and f! tch tch tch... Dr. Peyam, do you need to go to a dr. to check your memory?
But does a ordering exist with only finite many exceptions? Like we could claim there exists an ordering for Complex numbers without 0 (and other exceptions)
Maybe in the case of complex numbers, we should define a new comparison operation. For simplicity I'll refer to the real numbers as a line and the complex numbers as a plane. *on the line*: For any two real numbers, there can be only one of this cases: a < b or a = b or a > b. When plotting a and b on the line we could state that the number "closer" to +∞ than the other is greater. *on the plane*: For any two complex numbers, we can compare their real parts and, separately, their imaginary parts. So, for the real parts we could say that the number whose real part is greater ("closer" to +∞ than the other) is, let's say, 'greater on the real part'. Likewise, for the imaginary parts we could say that the number whose imaginary part is greater ("closer" to the vertical +∞ than the other) is 'greater on the imaginary part'. On the line there were only 3 states of comparison: = < > On the plane there would be 9 states of comparison (the operators are overloaded here, one for the real part comparison, one for the imaginary part comparison): == < >> =< => = i.e. (2, 4) (3, 2); (1, 0) (1, -4)... I don't know how it would be useful, but why not think of it? :)
@@drpeyam Well, laws are made to be violated (or at least bent... a little). 😊 There were times when negative numbers or sqrt(-1) were outrageous abominations, but now they're quite OK and nice to have. Apart from law (1) from your clip (about the trichotomy), which might be well suited for real numbers, but not for R^2, R^3 numbers and so on, I didn't find other violations by my "theory" (I'm a lawyer, so when it comes to maths, I might be way off). Anyway, nice clip... made me wonder. Thank you for sharing your work and knowledge with such passion.
Dear Mr. Peyam, Try making a video on (mathematical) Fields. A field is such a fundamental concept in mathematics, that knowledge of what a field is would help lots of viewers getting a basic understanding of the underlying algebraic structure and its consequences, I think. (especially considering the questions I read in response to this video). It also gives a nice example of how known properties (in Q and R e.g.) can be abstracted to a more general setting. Be certain to include the (in my opinion) shortest proof in mathematics: the fact that the neutral element with respect to addition is unique: 0 = 0 + 0' = 0'. (can it get any shorter? ;-)) (it could even be the title of the video: "The shortest proof in mathematics!" ;-)) (probably better start with groups first)
What about comparison of magnitude and phase? Clearly you wouldn't be able to order them in a line. But you can compare the magitudes of two complex numbers and then their phases to order them in the complex plane. Or is the order you mean different from the one I'm thinking about?
You can do that, just be aware that you won't have all the neat properties of the ordering of real numbers. Specifically, property 2 as stated in the video won't hold.
It’s *essentially* (but not exactly) the same as the lexicographic ordering I talked about at the beginning; it’s nice but doesn’t satisfy the three properties we have
@Dr Peyam , if I use polar form and by convention, start from modulus > 0 and increment the argument 0 to max 2(pi), and when I reach argument 2(pi), allow the modulus to increment, I have created a way to order the complex numbers, therefore, a way to compare complex numbers. Any other number form also, need some basic conventions, to be able to compare 2 numbers. Where am I wrong?
@@drpeyam thank you for your reply! I am of the view that properties 2 and 3 can only be applied at a time on either modulus or argument, but, not simultaneously, without altering the result.
Tried searching the video you mentioned at 5:53 where if 0=1 everything will become 0. I unfortunately cannot find it under the complex analysis playlist from title. Can you please help redirect me Dr. Peyam?
I would argue that (3) is too broad. It would be useful to let c be any complex number, but complex multiplication is ... complex. I'd draw an analogue to other types of values like vectors, matrices, quaternions, etc. With these, self-multiplication is equally as complex and non-intuitive, and this is why we limit it to scalar multiplication when talking about multiplicative properties. The lexicographic ordering that you gave works with these properties if c is constrained to real values, but there is the problem of deciding which term has precedence. I wonder then, if there is some logical way to map an unordered pair of real values to a single real value, then we could define the total ordering based on that mapping.
What about this complex angle thing? Isn't that a way to order them? Moreover.. you can take the real and imaginary part, draw a line trough them and compare the same angled complex numbers via the length of a vector from (0,0i) to this line
Only can compare equal or not equal for complex numbers. Or the absolute value of the complex numbers, like a0+b0*i => r0=sqrt(a0^2+b0^2), then comparing with r1 of a1+b1*i is possible... It looks like a useful question though.
Obviously you can compare the absolute values of complex numbers, but this doesn't hold when comparing them as the base of an exponential function. One of my favorite facts is that e^(pi i x) = (-1)^x -1 is just a base of e^pi (23.1406...) turned on its side on the complex plane. It doesn't make sense to think of -1 as larger than e, but it's way "bigger" as a base. The amount of rotation around the unit circle contributes to the "size" of the base as well as its absolute value.
+Maxx Byron If you mean to ask whether it is possible to order C by means of the modulus function, then no: see my replies to unknown360ful's and Calyo Delphi's questions. It is requirement 3 that's bothering ;-) If you restrict requirement 3 to only "real" numbers, then the lexicographical ordering DOES provide an order in C, but with some weird results (see e.g. my reply to Joseph Noonan's question).
+HelloItsMe That would still get you into trouble, though: see my reply to unknown360ful's question. Only by dropping or weakening one of the 3 requirements in the video, one can come up with an ordering for C. (but whether it is a useful one is another question ... ;-))
Yes, but it doesn't mean the same thing as it does in German. In German it means like a "game", but when the word spiel is used in english it means: noun: spiel; a long or fast speech or story, typically one intended as a means of persuasion or as an excuse but regarded with skepticism or contempt by those who hear it.
(3) does not seem to be that good as an axiom. It even forbids an ordering on the reals, where -1>0, but it does not seem that counterintuitive to order numbers from highest to lowest, so -1">"0 and 0">"1. Maybe it would be good to define it in another way that is more intuitive and does not limit one to "ascending order". Edit: It seems ordering isn't even defined the way you explained it in the video, so I guess it was just a little handwavy. I mean who would even come up with a definition of ordering that needs that much algebraic structure (addition, multiplication).
+ Leonard Romano ... "it even forbids an ordering on the reals, where -1 > 0" It better! In every field (see e.g. en.wikipedia.org/wiki/Field_(mathematics)) - of which Q, R and C are samples - we should have 0 < 1 and -1 < 0. (unless the field consists only of the member 0 - which is as boring as it can get) See my answer to John Baker for a proof. ... "but it does not seem that counterintuitive to order numbers from highest to lowest" I take it you mean something like the "magnitude" of a number here. Like the modulus as defined in R and C. See however my comments on Calyo Delphi and unknown360ful. ... "define it in another way that is more intuitive and does not limit one to 'ascending order'" How intuitive would you like it? Requirement 1 states that every two elements should be comparable: is that not in the heart of a comparison relation? Requirement 2 states that the relation a < b remains true under arbitrary addition (or translation if you like): adding something to "an amount" that is smaller than something else does not make it larger (or equal) when adding the same amount to the other thing ... Both requirements seem very intuitive to me (though your thoughts on this may vary, of course). Requirement 3 is a bit more delicate. In fact, it is a direct translation of "products of positive numbers should remain positive". Namely, suppose a < b. Then by requirement 2 it follows that 0 < b - a (add -a to both sides). If we want to keep the property "positive * positive remains positive" this means that for all c > 0 we should have that c * (b - a) > 0, or cb - ca > 0. Which means that cb > ca (add "ca" to both sides, requirement 2), which is equivalent to ca < cb. It is the fact that "
+Leonard Romano I'm not denying that you cannot define an order on C (or any set for that matter). It's just that mathematicians are stongly "convinced" that on a set that has an algebraic structure (such as fields, like Q, R and C), the ordering shoud behave nicely with respect to addition and multiplication. The example you gave with the set M := {banana, 3, %} does not exhibit any algebraic structure at all: it is just a "mere" set. Therefore requirements 2 and 3 (as mentioned in the video) do not make any sense. What remains is requirement 1. As long as any two elements in the set can be compared and don't contradict requirement 1 (which mathematicians think highy of) ... fine! You've got "some kind of ordening" Mathematicians (on the other hand) are interested in orderings that "behave nicely" with respect to the algebraic structure. Such as: if a > 0, then so is a + b > b (for every b) (requirement 2 in the video) if a > 0, b > 0, then so is ab > 0 (can be proved to be equivalent to requirement 3 in the video) If you "just" want a "flat" relationship of "
I think we are thinking the same but expressing it differently. What you are saying is completly correct and of course it makes sense to define orderings that satisfy more nice properties to gain new insights, but still as I was saying the definition is not sufficient for all kinds of sets but only for fields and maybe some rings or ringlike structures but that's it. As you are saying "mathematicians want more: they want to extend the (logical) notion of "
This reminds of something I read once, I forget where. It's an attempted proof of the Riemann hypothesis a math teacher received from their student. How do we know all the nontrivial zeros of the Riemann zeta function lie on a straight line? Well, you can look up a list of zeros on the internet. Putting the zeros in a list means that you've ordered them. But the zeros are complex numbers, and hence can't be ordered! The only way out of this conundrum is if all the zeros lie on a line. QED.
I take it you mean to define that z1 < z2 if mod(z1) < mod (z2) or (if they are equal) that Arg(z1) < Arg (z2) (where Arg(z) denotes the so-called principal value of "an" argument of z). (remember that arg(z) is a multi-valued function and that both arg(0) and Arg(0) are in principle not defined - although that last part is less of a concern) For your definition to make sense, we must therefore first agree on the limits of Arg(z). Common choices are (-π, π] or [0, 2*π). The last choice, together with your proposed definition of "
Well, there is an ordering with those three properties, it just isn’t helpful. Because the reals and the complex numbers have the same cardinality, there is a bijective function between them. Because R has an order relation with those ordering properties, you can just use the bijection to let the complex numbers piggyback on that ordering relation in R. But the problem in practice is that we either don’t have a formula for the bijection or the ordering relation just isn’t meaningful.
Well, there is an ordering with those three properties, and it is helpful, it just lives in R - not in C. The bijection between R and C doesn't help, since in order to C be an ordered field, in (2) and (3) the addition + and multiplication * must be in C - not in R. And any bijection between R and C is not a field homomorphism.
By AM-GM inequality : (-1+1)/2≥sqrt(-1*1) so 0≥i but i≠0 so 0>i. Solved ! : ) Btw, the squiggly symbol looks like the majorize symbol (\succ in LaTeX).
Is it possible to have an ordering if we change rule 3 to use positive real numbers (i.e. c>0) rather than numbers that are "squiggly greater" than zero? I don't think there's any reason it should apply to all njmbers thag are squiggly greater than 0. After all, if the complex numbers are represented as vectors, it makes sense that the ordering wouldn't be affected by adding the same vector to both (rule 2) or by scaling both by the same amount (multiplying by real c>0), but I see no reason that it shouldn't be affected by multiplying by scaling and rotating (which rule 3 would imply in some cases), which is just like multiplying by a negative with real numbers.
+Joseph Noonan Actually, you can. It's straightforward (though a bit tedious) to check that the lexicographical ordering of C as defined in the video will supply an ordering for C with only requirement 3 weakened as you propose. (unless I made a mistake verifying it) This ordering does lead to some strange results, though. It is clear from the definition that the ordering is the same ordering as that on R if we restrict the definition to real numbers (considered as being embedded in C in the usual sense, that is: we identify x, as an element of R, with the complex number z = x + 0i). A result that makes mathematicians happy! However, from the definition, we derive that for every real x > 1 the complex number z = 1 + xi > 0, z is therefore considered positive. However, z * z = (1 - x*x) + 2xi, which is negative according to the definition of the ordering. Hence, the product of two positive numbers can be negative. Mathematicians do not like such outcomes ... ;-)
Based on this logic, couldn’t you say there’s no ordering of absolute values that satisfies these criteria. Is there a way to modify these criteria to allow an ordering of absolute values and could that then be applied to create an ordering of complex numbers?
Simple is that , that we cant compare complex numbers cz they give irregular values and we can also cliam that one imaginary number can't be less,equal or grrater than other
Nice idea, but property 3 seems to be kind of artificial... what if (pr3) was: if i apply a growing function both sides i keep the sign, either way i use another symbol. i.e. define f(x)=-10·x. apply this decreasing function and change sign 2f(5) -10>-50 (ok) next for an imaginary product define g(x)=i·x and change symbol from < to # 2-5) so applying twice a "growing complex funcion" the # transforms into >... we redefine: g(x)=i·x and change original symbol < to # and # to > ¿will i fail if i keep going on? can we force math?
Could you have a < sign for complex number, but only when the two number are parallel? Like when Z1= C* Z2. But c is a real number. Like if it was an eigen vector or a basis? The real number would be a ( 1 0 ) basis. [2+2 i] is bigger then [ 1+ 1i ] (With a basis of ( 1 1) ). I know in the video you talk about it, its prop (3) but you kinda created a new "positive" number definition.
Is it valid to assume that i [or -i for that matter] is on (0, infinity) just because we found i > 0 ? It does hold for real numbers but does it hold for complex numbers?
The first required property (not sure how you spell it), we used to call it total comparability, meaning any two elements are comparable and thus can be ordered (The idea is to get total ordering on the entire set). Any way, can someone please spell it out for me?
I would like to posit that it is possible to order the complex, and even the quaternion, numbers in some way, but not in the simplistic intuitive ordering that works on the reals. Let's instead define the curly greater/less than signs to represent the following: z ≺ w ⇒ |z| < |w| z ≻ w ⇒ |z| > |w| Instead of comparing complex or quaternion numbers at face value, instead compare their magnitudes. What we're doing in this case is transforming a simple comparison of position on a one-dimensional number line into comparison of radii of circles of n dimensions where n = 2, 4, or 8 (let's throw octonions into the mix just for fun). By consequence, this then raises the interesting issue where ±3±4i = ±4±3i = ±5 = ±5i, since all eight complex numbers in this equivalence all have the same magnitude. In effect, by ordering the complex numbers in this way, you are ordering complex numbers by solutions to the Pythagorean Theorem of dimension n for n in {2,4,8}. Therefore, for a given magnitude Z, all complex numbers that form a Pythagorean Set with Z are equal to each other, and all other complex numbers are either strictly less than or strictly greater than Z. It then follows that ALL complex/quaternion/octonian numbers are strictly greater than 0, unless you defined some means of taking into account a complex number's direction with respect to, say, the positive Real axis. A problem that would be simple to solve for the complex numbers, but considerably more difficult for quaternions and octonions. Or it may even be such that this property of discrete directionality is lost for the quaternions and octonions, and ordering by this method becomes increasingly opaque and ill-defined. But this then satisfies (2) and (3) quite nicely in that you can add or multiply a constant to both sides of the comparison, and it does not change the ordering in the comparison! Although it does lose the ability to compare an entirely "negative" complex number with an entirely "positive" complex number (again unless the aforementioned method of determining directionality is defined), it does define a very clear means of ordering the complex numbers. Albeit not as simple as simply observing position on a number line. This means of comparison could also work on the reals, but you get weird results like -1 = 1 and -1 > 0 and 1 < -2 and so on. I can't thoroughly explore this idea right now since I'm in a hotel room without a whiteboard to doodle on, so if this half-baked idea is flawed in any significant ways, please do let me know and provide a clear example of such.
+Calyo Delphi Actually, you can NOT order C in that way, unless you are willing to drop requirement 1 in the video, a requirement that mathematicians think highly of: every two numbers z and w in an ordered set should be comparable in the sense that either z < w, w < z or z = w. And one and exactly one of these outcomes should hold. As you mentioned yourself, the ordering "
@Mathematician It was a half-baked idea that I kinda typed out and formulated in my head on-the-fly, so I wasn't really expecting it to be rigorous to any serious degree. :P Just thought of it as an idea of ordering C in a way that kind-of "works" by making all complex numbers with the same modulus "equal" to each other, in that they are all the same absolute distance away from the origin. Which is where that weird behaviour on scaling back a dimension to the reals arises where -1 = 1, et cetera. It definitely requires one to redefine the behaviour of the comparison operators when moving up a dimension from the reals to the complex, which causes problems when you try to go back down by simply setting the additional dimensions to be equal to 0 so that only the real part has any defined nonzero value. And that was a thought that was running in the back of my head when I wrote all this out. I just didn't want to take the time to explore those thoughts to their illogical conclusions at the time that I had them. :P
+Calyo Delphi There's nothing wrong with thinking about a "weird" idea in mathematics ... by all means, state it and explore it! Most great mathematicians have challenged things, only by either finding out they were wrong (which can help the next time ;)) or by proving "the establishment" wrong! Cantor reasoning about different kinds of infinity is a nice example! (to be honest ... I'm not that creative myself ;-)) Now that I come to think of it ... your idea (apart from the clash with the equality definition already existing in C) is not that weird at all! I have no idea about your knowledge of foundations of mathematics. Maybe you know something about topology, maybe you don't. But topological spaces try to "mimic" the idea of vicinity, being in the neighborhood of something, trying to generalize limits and continuity even! (see e.g. metric spaces, a special kind of topological spaces: en.wikipedia.org/wiki/Metric_space)). In sets with a metric defined (mimicking our notion of distance), it turns out that (about) the simplest metric you can define (the so-called discrete metric) which defines that the distance between two different points is always 1 (and the distance between equal points is 0, of course), turns out to be quite useful! Not that much as a topological space in itself, but to (dis)prove theorems! We learn most by the mistakes we make! (and trust me, I have made a few myself). Keep thinking about (the foundations of) mathematical concepts! Keep challenging yourself! ;-) And keep loving math, it's beautiful!
@Mathematician: Thinking about a "weird" idea in mathematics is how I figured out on my own the Leibniz Formula that generalizes the power rule to higher-order derivatives, and then I went a step further to figure out the product rule generalized to higher-order derivatives and connected that with Pascal's triangle and binomial coefficients. ^^ And I spent a while trying to do the same to the chain rule, and I got as far as making connections to partitions of sets, Bell numbers, and Stirling numbers of the second kind, but a whole lot of that dives deep into number and set theory mathematics that's waaaay over my head. And then I discovered Faa di Bruno's formula and felt deflated that two hundred years ago someone had already discovered a generalization. XD Oh well. I didn't really memorize the formula and I plan to eventually dive back into it to try and figure it all out on my own again. And maybe figure out my own generalization that might be different from Faa di Bruno's formula in its own ways! Or it might end up being exactly the same by happenstance. And just recently (last night and today recently) I explored the idea of the parallel operator as a multi-input operation (instead of just a||b = 1/(1/a + 1/b), a1||a2||...||aN = 1/(1/a1 + 1/a2 + ... + 1/aN); I don't know how prevalent this operator is in EE textbooks, but this exact definition of the parallel operator is used in EE to compute parallel resistances/inductances and series capacitances, so I use it as a personal shorthand) and determined (with the help of said Ph.D. friend correcting my terminology) that it is symmetric, but non-associative, just by throwing a few parentheses into the mix and playing with the numbers 'til I got something that I definitively could not make equal to the other thing through algebraic manipulation. And I've also in the recent past played with the idea of using slightly different notation to indicate the inverse of a function, instead of using the negative-one-th power, since it REALLY niggles me that that is the *one* special case where suddenly raising *any* defined function to a specific power means anything other than multiplying/composing that function with itself that many times over. E.g. sin^-1 x != (sin x)^-1, even though sin^2 x == (sin x)^2. Whyyyyy?? It's always been a point of contention and confusion for me because it makes no freaking sense, so I went and made my own far less ambiguous notation for it that I use in my own mathematics. c: (FYI my formal mathematical education stops about most of the way through Calc I. Anything above that that I know is self-taught, largely through blackpenredpen, Dr. Peyam, Flammable Maths, and other maths youtubers, as well as looking stuff up on Wikipedia and asking questions to an online friend of mine who's a maths Ph.D.)
+Calyo Delphi Wowww ... what a long reply! Where do I begin to reply? ... "and I got as far as making connections to partitions of sets, Bell numbers, and Stirling numbers of the second kind" I hope you didn't skip the numbers of the first kind ... ;-) (apart from my numbers ... which haven't been given an official name yet ;-)) ... "And just recently (last night and today recently) I explored the idea of the parallel operator as a multi-input operation" Symmetric, but non-associative ... that would make mathematicians "mad/crazy"! ;-) ..."slightly different notation to indicate the inverse of a function" Actually, there is a reason for denoting the inverse of a function f as f^-1 (didn't mean that as a power, sorry) I'm not going to explain it to you here: you (most likely) have already figured it out for yourself! ... (FYI) my formal mathematical education stops about most of the way through Calc I What stops you from reaching level Calc II (or even higher?) Apparently your "constantly wondering why" doesn't! Keep challenging yourself! And eventually you will reach level ... (wel, your grades will fill it in! ;-))
How will you compare, for example, 3+4i and 5? If you add the argument as a second condition, you'll have to compare cis(4π/3) and cis(10π/3). They are the same number, and have a abs of 1, but 10π/3 > 4π/3 (by radius).
why not you can define a good function f, such that C->R for ordering. f(x,y)=R? if you don't do such artificial function defining in some problem, you probably lose the good chance of using complex number
As complex numbers used to represent dots in a plan the question to compare them is without logic. We may calculate the modules and to express the distance from origin but to what use??? The distances may vary from 0 to infinite. So what, we may say that infinite is grated than 10!!! Difficult to understand. If there are real numbers the comparison is odd.
Isn't the order where every complex number is equal to every other complex number an order which satisfies all the criteria? It is of course a totally trivial ordering that you could apply to any set, but isn't it a more precise statement to say that the only total order on the complex numbers respecting both addition and multiplication is the order in which every pair of elements coincides?
Daniel Darroch I think that ordering would lead to the same contradiction as 0=i. EDIT: Maybe not, because in that "ordering" saying 0=i is completely valid and does not contradict.
It is an order, but it is not compatible with the algebraic structure of C. Basically, it's just an order on the "symbols" that has no connection to the actual arithmetic of C. We can always arrange the elements of a set in some order.
Technically you didn't violate the trichotomy law because you never specified that there are no two numbers a and b such a is both "squiggly less than" and "squiggly greater than" b.
John Baker No, because this ordering it's not the same as the real ordering, and in the video he ended up with *0 squiggly less than -1* . That's not the same as *0 < -1* so there's no contradiction yet; but when multiplying by -1 we have to remember what we had: *0 squiggly less than -1* , and that means -1 is positive here, so you don't switch the inequality, since -1 is no longer negative (in this situation).
+John Baker In a sense, you are correct: IF C were to have a nice ordering (adhering to the three requirements mentioned in the video), then the following MUST be true: a < b implies -b < -a (hence, -a > -b) (the "
CAUTION! Wrong! if you want to compare 2 with -4 in complex plane (2 vs -4) you will say |2| < |-4|=4 (True) So, the answer to the question (2 vs -4) is 2 < -4 (Wrong) !!! (we already know that -4 < 2) Cause you aren't comparing numbers you are comparing functions!
PrinzMajdazari No, because this ordering it's not the same as the real ordering, and in the video he ended up with *0 squiggly less than -1* . That's not the same as *0 < -1* so there's no contradiction yet; but when multiplying by -1 we have to remember what we had: *0 squiggly less than -1* , and that means -1 is positive here, so you don't switch the inequality, since -1 is no longer negative (in this situation).
Aaron Quitta We need a bijective function that takes a complex number and outputs a real number and its inverse function. Thats the same as a function which takes two real numbers and returns one. We can construct a new number by filling all odd decimal positions with the first number and all even decimal positions with the second number. The inverse would be to take all the odd decimal positions to construct the first number and all the even for the second. You can put every number inside the function, so it has to be bijective
Aaron Quitta The problem is that it would imply that there are not equal in size therefore there wouldn't be a bijective function between these two sets. But there is -> contradiction.
This video got me thinking about a question my foundations professor posed to our class, namely, how would one go about well ordering the real numbers. Curious to see if you have any thoughts are on this. Cheers!
I'm pretty sure the real numbers cannot be well ordered without the Axiom of Choice, which means there's no way to explicitly define a well ordering of the real numbers. This has in fact been proved. If we assume the reals can be well ordered, then it is possible to construct non-measurable subsets of the reals (eg. the Vitali set), and it has been proved that the existence of non-measurable subsets of R is independent of the axioms of Zermelo-Fraenkel set theory without the Axiom of Choice. That is, there are models of ZF in which every subset of R is measurable. All this means that you must assume the axiom of choice (or a weaker form of it that is strictly stronger than ZF) in order to demonstrate the fact that the real numbers can be well ordered. If it were possible to explicitly demonstrate a well-ordering of R, then one would have proved the Axiom of Choice (or a weaker form) which is impossible in set theory as we usually understand it.
+Samuel Roberts Smart professor you have there: it is always good to think about foundations of mathematics! en.wikipedia.org/wiki/Well-order might provide you with some insight (though it depends heavily on rigorous foundations of set theory).
You give up too easily. We can say that two complex numbers are equal if the modulus of each is equal and the argument of each is equal. From there we can say that one is less than the other if the modulus is less, or if the modulus is equal but the argument is less. Done!
Note: There's a small mistake in the video. When I write c is positive 0 in property 3, I don't mean c positive 0 as a real number, but c positive 0 with the new ordering squiggly greater than (so it should be c squiggly greater than 0).
c strictly squiggly greater than 0
(was enjoying mysekf greatly with intentionally confusing 'squiggly' and 'strictly' in the video)
Yeah, I was confused by that. I thought "wait, isn't any countably generated real vector space orderable?"
When you say that *c* has to be squiggly greater than 0, it's just in order to make sure the inequality sign doesn't switch when multiplying it by *c* , isn't it? I mean, could we actually tell where is *c* (which quadrant) knowing it is squiggly greater than 0?
Andy Arteaga Yes for your first question. As for your second question, we don’t know where c is in the quadrants
Dr. Peyam's Show Thank you! And great video btw!
This is one of my all time favourite proofs, which demonstrates the fact that as you add more dimensions to numbers, they lose neat properties. So ordering is lost when real numbers become complex numbers; multiplication loses commutativity when complex numbers become quaternions; and multiplication loses associativity when quaternions become octonions, etc.
and multiplication even loses alternativity (a weaker form of associativity (xx)y = x(xy) that octonions retain) with sedenions. And beyond that, you cannot have an algebraic structure at all !!
hence why it's better to work in it as a vector space
In that spirit I question whether complex numbers should be called "numbers." (The point is purely about terminology, not about substantive math.) Complex numbers as a vector space over the reals with an amazingly useful vector product seems like a more straightforward way to describe what we are supposed to call "complex numbers."
I don't t think many people would say that quaternions are "numbers." Why should we insist on according that honor to "complex numbers"? (Once again, this is not some crackpot theory -- I just think it would be better to refer to complex numbers as we refer to quaternions rather than as we refer to real numbers.)
Though complex number lose some properties, it's also interesting how many properties they retain when compared to quaternions, octonians, etc. Nearly all of the arithmetical properties of real numbers are kept with complex numbers, but with further dimensions added, you lose basic properties like associativity. Complex numbers even have some very nice properties that the real numbers alone don't have, like the fundamental theorem of algebra.
And what did you conclude? Are there infinitely many properties?
Since I am here earlier than flammy,
Squiggly
Duncan Schaafsma now that's better lol
Flammable Maths I shall henceforth use your version of theta in rememberance of you getting rekt. No more ugly circular fuckbois.
Hu
Short answer: No
Long Answer: nO
I will teach complex numbers also in Autumn, so I plan to use this in my class. This is not a "traditional" complex number problem.
Hey Dr. Payem! I love your videos!
Kedves Sándor! Hol tanít Ön? Sok örömet a tanításhoz!
@@matekurucz2 Köszönöm :-) Ilyen remek videókkal könnyű :-)
i
❤️❤️❤️
i
@@drpeyam sqrt(-1)
My intuitive proof:
If we have a function u(r,k) = re^ik, where r and k are both Re.
Then u(-r,0) = u(r,π).
For simplicity later, let r be bounded on the interval R=-π to -R=π.
We can create two distinct continuous paths from (R,0) to (-R,0) by keeping one variable constant and adjusting the other.
Keeping r constant and increasing k from 0 to π takes us from (R,0) to (-R,0) thus u(r,k) > u(r,k+ε).
Keeping k constant and increasing r from R to -R takes us from (R,0) to (-R,0) thus u(r,k) > u(r+ε,k).
But if u(r,k+ε) and u(r+ε,k) are both continuous and intercept at (R,0) and (-R,0) but not at any other point then they cannot be ordered without being made discontinuous somehow.
Damn, it was logical in my head but I cannot seem to explain xD!
Oh my goodness that address book explanation for lexicographic ordering was spot on!!! Also my professor did proof by intimidation with this one in real analysis so seeing in spelled out makes me believe in this now. Thank you!
See remark above please
I think the problem comes from the fact that when you multiply both sides by -1,the > sign becomes
It seems like property 3 is the one that's no good. When C is real and positive, it seems rather obvious that that should hold. But when defining the new inequality, it doesn't seem as reasonable to assume that property 3 is something that we want because if an ordering does exist, we don't know how that would be affected when multiplying by a complex number.
It’s a typo, btw, it’s supposed to be c squiggly greater than 0, not real positive c
+Dr. Peyam's Show
I think he means that it isn't reasonable to assume that c must be squiggly-greater-than 0, since with the real numbers literally nothing changes if we take that arg(c) = 0 (i.e. c is scalar) is the condition for property (3). Of course using that rule instead would _not_ lead to a contradiction...
If that property doesn't hold, it becomes difficult to do algebra. Imagine if you couldn't rely on multiplying both sides of an inequality by something and having the inequality still hold. The same goes for all the other properties -- we want them so that the ordering is algebraically useful.
@@gcewing But that rule still holds if you not only say that c > 0 but also that c is a real number. You could still have two complex numbers that are not real numbers, multiply both of them by a real c value and the inequality will still hold. You can do algebra as usual. That property still holds if you require c to be a positive real number. So such an ordering is still algebraically useful.
I have a perfect ordering of the complex numbers, but the margin is not wide enough to contain it.
Also, I wish I had shirts that shine like Dr. Peyam's. Is there a laundry detergent for that?
P.S. I watch Dr. P and Dr. B/R with passion ;-)
1:20 We compare complex numbers like that when we are finding 2D convex hull ,
Graham scan compares complex numbers in that way but in polar form
informally you can prove this in a simpler way by adding the requirement of transitivity. Now take N complex roots of unity (a_1, ..., a_N), which of course form a circle and you arrive at a_1 < a_N and a_N < a_1 which breaks assimetry.
I like that!!
Well, you lose something after each addition of structure. No free lunch.
AndDiracisHisProphet Until you get to sednions, then reapplying doesn't make things much worse.
what do you mean. Quaternions already are not commutative anymore
AndDiracisHisProphet sedenions are so bad already you don't lose a lot anymore
oh yeah, I should read properly
@@sugarfrosted2005 i know octernions, are sednions 16 or 32?
Yes, you can.
There is an order relation on C. Suppose a complex number is smaller than another if it is closer to the origin. So, if it is on a circle centered originally and with a smaller radius. Any complex number is well defined by its radius and the angle made by r with the ox axis. If we do angle abstraction and compare r we can find a relation of order in C. Since r is in R, it means that the order relation
Yooo thanks for this! I linked to it in my video for today so that I didn't have to do it myself.
Equivalent to comparing points on 2D or 3D grid. You can't say one point is "bigger" than another.
Two classic exercises in one video!(the complex field is non-ordered and 1>0 for all ordered fields) Great!
Byyy the way, I am still waiting for 3 videos(!!!)
OMG, 3?!? So far I have transcendence of e and pi, what’s the third one?
Did you forgot it only because I did not asked for it for a month(-ish, close enough) !? We talked about it! D^n f(x) for all n and f!
tch tch tch... Dr. Peyam, do you need to go to a dr. to check your memory?
But does a ordering exist with only finite many exceptions? Like we could claim there exists an ordering for Complex numbers without 0 (and other exceptions)
You always have lexicographical and polar lexicographical orders.
Maybe in the case of complex numbers, we should define a new comparison operation.
For simplicity I'll refer to the real numbers as a line and the complex numbers as a plane.
*on the line*: For any two real numbers, there can be only one of this cases: a < b or a = b or a > b. When plotting a and b on the line we could state that the number "closer" to +∞ than the other is greater.
*on the plane*: For any two complex numbers, we can compare their real parts and, separately, their imaginary parts. So, for the real parts we could say that the number whose real part is greater ("closer" to +∞ than the other) is, let's say, 'greater on the real part'. Likewise, for the imaginary parts we could say that the number whose imaginary part is greater ("closer" to the vertical +∞ than the other) is 'greater on the imaginary part'.
On the line there were only 3 states of comparison: = < >
On the plane there would be 9 states of comparison (the operators are overloaded here, one for the real part comparison, one for the imaginary part comparison): == < >> =< => =
i.e. (2, 4) (3, 2); (1, 0) (1, -4)...
I don't know how it would be useful, but why not think of it? :)
I was thinking that too, but it would violate one of the natural laws of comparison, don’t quite remember which one :)
@@drpeyam Well, laws are made to be violated (or at least bent... a little). 😊
There were times when negative numbers or sqrt(-1) were outrageous abominations, but now they're quite OK and nice to have.
Apart from law (1) from your clip (about the trichotomy), which might be well suited for real numbers, but not for R^2, R^3 numbers and so on, I didn't find other violations by my "theory" (I'm a lawyer, so when it comes to maths, I might be way off).
Anyway, nice clip... made me wonder. Thank you for sharing your work and knowledge with such passion.
Dear Mr. Peyam,
Try making a video on (mathematical) Fields.
A field is such a fundamental concept in mathematics, that knowledge of what a field is would help lots of viewers getting a basic understanding of the underlying algebraic structure and its consequences, I think.
(especially considering the questions I read in response to this video).
It also gives a nice example of how known properties (in Q and R e.g.) can be abstracted to a more general setting.
Be certain to include the (in my opinion) shortest proof in mathematics: the fact that the neutral element with respect to addition is unique:
0 = 0 + 0' = 0'.
(can it get any shorter? ;-))
(it could even be the title of the video: "The shortest proof in mathematics!" ;-))
(probably better start with groups first)
What about comparison of magnitude and phase? Clearly you wouldn't be able to order them in a line. But you can compare the magitudes of two complex numbers and then their phases to order them in the complex plane. Or is the order you mean different from the one I'm thinking about?
Kiritsu You can order them, but that ordering wouldn‘t have very intuitive properties associated with ordering.
You can do that, just be aware that you won't have all the neat properties of the ordering of real numbers. Specifically, property 2 as stated in the video won't hold.
It’s *essentially* (but not exactly) the same as the lexicographic ordering I talked about at the beginning; it’s nice but doesn’t satisfy the three properties we have
+Kiritsu
See my reply on unknown360ful's question.
@Dr Peyam , if I use polar form and by convention, start from modulus > 0 and increment the argument 0 to max 2(pi), and when I reach argument 2(pi), allow the modulus to increment, I have created a way to order the complex numbers, therefore, a way to compare complex numbers. Any other number form also, need some basic conventions, to be able to compare 2 numbers. Where am I wrong?
basically, test 1 - compare modulus, and stop if different, if same to go to test 2 - compare argument (0
Yeah but this is similar to lexicographic ordering, you still lose all the natural properties you want
@@drpeyam thank you for your reply! I am of the view that properties 2 and 3 can only be applied at a time on either modulus or argument, but, not simultaneously, without altering the result.
Another cool ordering on the Complex could be by Lexicographic order of the modolus and the argoment (being in the interval [0,2pi))
It works if you don't assume there exists a concept as positive, does it not?
+Avi Mehra
Actually, the concept of "being positive" is already implied to exist, due to requirement 1 in the video.
A (total) ordering "
Tried searching the video you mentioned at 5:53 where if 0=1 everything will become 0. I unfortunately cannot find it under the complex analysis playlist from title. Can you please help redirect me Dr. Peyam?
It’s called 1/0 doesn’t exist
Very clear and useful lesson on a matter which is often done aproximately in many courses, thanks for putting these things clear👌
I would argue that (3) is too broad. It would be useful to let c be any complex number, but complex multiplication is ... complex. I'd draw an analogue to other types of values like vectors, matrices, quaternions, etc. With these, self-multiplication is equally as complex and non-intuitive, and this is why we limit it to scalar multiplication when talking about multiplicative properties.
The lexicographic ordering that you gave works with these properties if c is constrained to real values, but there is the problem of deciding which term has precedence. I wonder then, if there is some logical way to map an unordered pair of real values to a single real value, then we could define the total ordering based on that mapping.
What about this complex angle thing?
Isn't that a way to order them?
Moreover.. you can take the real and imaginary part, draw a line trough them and compare the same angled complex numbers via the length of a vector from (0,0i) to this line
7:20, what if I just do,
1>0
Multiply both sides by i
i×1 > i×0 or, i>0
:P
Hah! It's easier to escape the jaws of determined crocodile, than the jaws of logic. Thanks for the demo!
After 0
Only can compare equal or not equal for complex numbers. Or the absolute value of the complex numbers, like a0+b0*i => r0=sqrt(a0^2+b0^2), then comparing with r1 of a1+b1*i is possible...
It looks like a useful question though.
Can you mesure complex numbers by the length of the number from the origin on a complex plane?
+Boypig24
See my answer to unknown360ful's question.
Yrs but its a partial order. No trichotmy
omg! a clean board ! I like it! :)
Distance to the origin? Pythagoras with a correction for the quadrant the points are in?
Obviously you can compare the absolute values of complex numbers, but this doesn't hold when comparing them as the base of an exponential function.
One of my favorite facts is that e^(pi i x) = (-1)^x
-1 is just a base of e^pi (23.1406...) turned on its side on the complex plane. It doesn't make sense to think of -1 as larger than e, but it's way "bigger" as a base. The amount of rotation around the unit circle contributes to the "size" of the base as well as its absolute value.
Is it fair to compare r values when converting to polar?
+Maxx Byron
If you mean to ask whether it is possible to order C by means of the modulus function, then no: see my replies to unknown360ful's and Calyo Delphi's questions.
It is requirement 3 that's bothering ;-)
If you restrict requirement 3 to only "real" numbers, then the lexicographical ordering DOES provide an order in C, but with some weird results (see e.g. my reply to Joseph Noonan's question).
If I would order the complex numbers, I would firstly order them by length and then if they're the same length order them by angle
It’s almost the same ordering as in the beginning of the video
+HelloItsMe
That would still get you into trouble, though: see my reply to unknown360ful's question.
Only by dropping or weakening one of the 3 requirements in the video, one can come up with an ordering for C.
(but whether it is a useful one is another question ... ;-))
12:59 wait, is "same spiel" a thing you can say in english?
Yeah
Yes, but it doesn't mean the same thing as it does in German. In German it means like a "game", but when the word spiel is used in english it means:
noun: spiel;
a long or fast speech or story, typically one intended as a means of persuasion or as an excuse but regarded with skepticism or contempt by those who hear it.
@@Googahgee yeah, i am german xD
In german you can say "gleiches Spiel" too, so for me it seemed like a half-translated phrase xD
(3) does not seem to be that good as an axiom. It even forbids an ordering on the reals, where -1>0, but it does not seem that counterintuitive to order numbers from highest to lowest, so -1">"0 and 0">"1.
Maybe it would be good to define it in another way that is more intuitive and does not limit one to "ascending order".
Edit: It seems ordering isn't even defined the way you explained it in the video, so I guess it was just a little handwavy. I mean who would even come up with a definition of ordering that needs that much algebraic structure (addition, multiplication).
+ Leonard Romano
... "it even forbids an ordering on the reals, where -1 > 0"
It better! In every field (see e.g. en.wikipedia.org/wiki/Field_(mathematics)) - of which Q, R and C are samples - we should have 0 < 1 and -1 < 0.
(unless the field consists only of the member 0 - which is as boring as it can get)
See my answer to John Baker for a proof.
... "but it does not seem that counterintuitive to order numbers from highest to lowest"
I take it you mean something like the "magnitude" of a number here. Like the modulus as defined in R and C.
See however my comments on Calyo Delphi and unknown360ful.
... "define it in another way that is more intuitive and does not limit one to 'ascending order'"
How intuitive would you like it?
Requirement 1 states that every two elements should be comparable: is that not in the heart of a comparison relation?
Requirement 2 states that the relation a < b remains true under arbitrary addition (or translation if you like): adding something to "an amount" that is smaller than something else does not make it larger (or equal) when adding the same amount to the other thing ...
Both requirements seem very intuitive to me (though your thoughts on this may vary, of course).
Requirement 3 is a bit more delicate. In fact, it is a direct translation of "products of positive numbers should remain positive".
Namely, suppose a < b. Then by requirement 2 it follows that 0 < b - a (add -a to both sides). If we want to keep the property "positive * positive remains positive" this means that for all c > 0 we should have that c * (b - a) > 0, or cb - ca > 0. Which means that cb > ca (add "ca" to both sides, requirement 2), which is equivalent to ca < cb.
It is the fact that "
Well all I need for a good "strict" total ordering is the following:
1) x
+Leonard Romano
I'm not denying that you cannot define an order on C (or any set for that matter).
It's just that mathematicians are stongly "convinced" that on a set that has an algebraic structure (such as fields, like Q, R and C), the ordering shoud behave nicely with respect to addition and multiplication.
The example you gave with the set M := {banana, 3, %} does not exhibit any algebraic structure at all: it is just a "mere" set. Therefore requirements 2 and 3 (as mentioned in the video) do not make any sense.
What remains is requirement 1. As long as any two elements in the set can be compared and don't contradict requirement 1 (which mathematicians think highy of) ... fine! You've got "some kind of ordening"
Mathematicians (on the other hand) are interested in orderings that "behave nicely" with respect to the algebraic structure. Such as:
if a > 0, then so is a + b > b (for every b) (requirement 2 in the video)
if a > 0, b > 0, then so is ab > 0 (can be proved to be equivalent to requirement 3 in the video)
If you "just" want a "flat" relationship of "
I think we are thinking the same but expressing it differently. What you are saying is completly correct and of course it makes sense to define orderings that satisfy more nice properties to gain new insights, but still as I was saying the definition is not sufficient for all kinds of sets but only for fields and maybe some rings or ringlike structures but that's it.
As you are saying "mathematicians want more: they want to extend the (logical) notion of "
This reminds of something I read once, I forget where. It's an attempted proof of the Riemann hypothesis a math teacher received from their student. How do we know all the nontrivial zeros of the Riemann zeta function lie on a straight line? Well, you can look up a list of zeros on the internet. Putting the zeros in a list means that you've ordered them. But the zeros are complex numbers, and hence can't be ordered! The only way out of this conundrum is if all the zeros lie on a line. QED.
7:43 wait, multiplication property was about real numbers, and there was other sign (common >), and i is not from range (0,infinity)
Dr. Peyam, what if we compare mod(z) and arg(z)? You know to see which one is farther from the origin and who is rotated by a greater amount?
unknown360ful exactly my thought, as I said in my comment
Oh honestly I hadn't seen your comment, I only saw somebody talk about comparing the norm but thought the angle approach is also important...
'Cuz the argument is modular. 3/2pi>1/2pi, but 3/2pi +pi
so -2>1?
I take it you mean to define that z1 < z2 if mod(z1) < mod (z2) or (if they are equal) that Arg(z1) < Arg (z2) (where Arg(z) denotes the so-called principal value of "an" argument of z).
(remember that arg(z) is a multi-valued function and that both arg(0) and Arg(0) are in principle not defined - although that last part is less of a concern)
For your definition to make sense, we must therefore first agree on the limits of Arg(z). Common choices are (-π, π] or [0, 2*π). The last choice, together with your proposed definition of "
Well, there is an ordering with those three properties, it just isn’t helpful. Because the reals and the complex numbers have the same cardinality, there is a bijective function between them. Because R has an order relation with those ordering properties, you can just use the bijection to let the complex numbers piggyback on that ordering relation in R. But the problem in practice is that we either don’t have a formula for the bijection or the ordering relation just isn’t meaningful.
Well, there is an ordering with those three properties, and it is helpful, it just lives in R - not in C. The bijection between R and C doesn't help, since in order to C be an ordered field, in (2) and (3) the addition + and multiplication * must be in C - not in R. And any bijection between R and C is not a field homomorphism.
In part (2) when you write "for all y", do you mean all complex y, all real y, or all y complex, real, and otherwise?
All y. Complex, real or otherwise
By AM-GM inequality : (-1+1)/2≥sqrt(-1*1) so 0≥i but i≠0 so 0>i. Solved ! : ) Btw, the squiggly symbol looks like the majorize symbol (\succ in LaTeX).
If for whatever reason, we changed Nr3 to only allow c Element of the positive real numbers, would you still be able to produce a contradiction?
+Kuratius
See my answer to Joseph Noonan's question.
Is it also the case for dual numbers ? It seems so as epsilon > 0 and epsilon square = 0. Is the strict ordering a requirement or is
Is it possible to have an ordering if we change rule 3 to use positive real numbers (i.e. c>0) rather than numbers that are "squiggly greater" than zero? I don't think there's any reason it should apply to all njmbers thag are squiggly greater than 0. After all, if the complex numbers are represented as vectors, it makes sense that the ordering wouldn't be affected by adding the same vector to both (rule 2) or by scaling both by the same amount (multiplying by real c>0), but I see no reason that it shouldn't be affected by multiplying by scaling and rotating (which rule 3 would imply in some cases), which is just like multiplying by a negative with real numbers.
+Joseph Noonan
Actually, you can.
It's straightforward (though a bit tedious) to check that the lexicographical ordering of C as defined in the video will supply an ordering for C with only requirement 3 weakened as you propose.
(unless I made a mistake verifying it)
This ordering does lead to some strange results, though.
It is clear from the definition that the ordering is the same ordering as that on R if we restrict the definition to real numbers (considered as being embedded in C in the usual sense, that is: we identify x, as an element of R, with the complex number z = x + 0i). A result that makes mathematicians happy!
However, from the definition, we derive that for every real x > 1 the complex number z = 1 + xi > 0, z is therefore considered positive. However, z * z = (1 - x*x) + 2xi, which is negative according to the definition of the ordering. Hence, the product of two positive numbers can be negative.
Mathematicians do not like such outcomes ... ;-)
Based on this logic, couldn’t you say there’s no ordering of absolute values that satisfies these criteria.
Is there a way to modify these criteria to allow an ordering of absolute values and could that then be applied to create an ordering of complex numbers?
Simple is that , that we cant compare complex numbers cz they give irregular values and we can also cliam that one imaginary number can't be less,equal or grrater than other
7:50 Why is I>0?
Beautiful proof!
Nice idea, but property 3 seems to be kind of artificial...
what if (pr3) was:
if i apply a growing function both sides i keep the sign, either way i use another symbol.
i.e. define f(x)=-10·x. apply this decreasing function and change sign
2f(5) -10>-50 (ok)
next for an imaginary product define
g(x)=i·x and change symbol from < to #
2-5)
so applying twice a "growing complex funcion" the # transforms into >...
we redefine:
g(x)=i·x and
change original symbol < to # and # to >
¿will i fail if i keep going on? can we force math?
Cannot we define I=√-1 and still i^2=1 |a+ib|=√|a^2-b^2| instead of √a^2+b^2.then -i
No I don’t think that work’s
Could you have a < sign for complex number, but only when the two number are parallel? Like when Z1= C* Z2. But c is a real number. Like if it was an eigen vector or a basis?
The real number would be a ( 1 0 ) basis. [2+2 i] is bigger then [ 1+ 1i ] (With a basis of ( 1 1) ).
I know in the video you talk about it, its prop (3) but you kinda created a new "positive" number definition.
That's what I was wondering. I wondered if you can compare imaginary number with real number.
Is it valid to assume that i [or -i for that matter] is on (0, infinity) just because we found i > 0 ?
It does hold for real numbers but does it hold for complex numbers?
The first required property (not sure how you spell it), we used to call it total comparability, meaning any two elements are comparable and thus can be ordered (The idea is to get total ordering on the entire set). Any way, can someone please spell it out for me?
ZeT 47 Trichotomy
I would like to posit that it is possible to order the complex, and even the quaternion, numbers in some way, but not in the simplistic intuitive ordering that works on the reals. Let's instead define the curly greater/less than signs to represent the following:
z ≺ w ⇒ |z| < |w|
z ≻ w ⇒ |z| > |w|
Instead of comparing complex or quaternion numbers at face value, instead compare their magnitudes. What we're doing in this case is transforming a simple comparison of position on a one-dimensional number line into comparison of radii of circles of n dimensions where n = 2, 4, or 8 (let's throw octonions into the mix just for fun).
By consequence, this then raises the interesting issue where ±3±4i = ±4±3i = ±5 = ±5i, since all eight complex numbers in this equivalence all have the same magnitude. In effect, by ordering the complex numbers in this way, you are ordering complex numbers by solutions to the Pythagorean Theorem of dimension n for n in {2,4,8}. Therefore, for a given magnitude Z, all complex numbers that form a Pythagorean Set with Z are equal to each other, and all other complex numbers are either strictly less than or strictly greater than Z.
It then follows that ALL complex/quaternion/octonian numbers are strictly greater than 0, unless you defined some means of taking into account a complex number's direction with respect to, say, the positive Real axis. A problem that would be simple to solve for the complex numbers, but considerably more difficult for quaternions and octonions. Or it may even be such that this property of discrete directionality is lost for the quaternions and octonions, and ordering by this method becomes increasingly opaque and ill-defined.
But this then satisfies (2) and (3) quite nicely in that you can add or multiply a constant to both sides of the comparison, and it does not change the ordering in the comparison! Although it does lose the ability to compare an entirely "negative" complex number with an entirely "positive" complex number (again unless the aforementioned method of determining directionality is defined), it does define a very clear means of ordering the complex numbers. Albeit not as simple as simply observing position on a number line.
This means of comparison could also work on the reals, but you get weird results like -1 = 1 and -1 > 0 and 1 < -2 and so on.
I can't thoroughly explore this idea right now since I'm in a hotel room without a whiteboard to doodle on, so if this half-baked idea is flawed in any significant ways, please do let me know and provide a clear example of such.
+Calyo Delphi
Actually, you can NOT order C in that way, unless you are willing to drop requirement 1 in the video, a requirement that mathematicians think highly of: every two numbers z and w in an ordered set should be comparable in the sense that either z < w, w < z or z = w. And one and exactly one of these outcomes should hold.
As you mentioned yourself, the ordering "
@Mathematician
It was a half-baked idea that I kinda typed out and formulated in my head on-the-fly, so I wasn't really expecting it to be rigorous to any serious degree. :P Just thought of it as an idea of ordering C in a way that kind-of "works" by making all complex numbers with the same modulus "equal" to each other, in that they are all the same absolute distance away from the origin. Which is where that weird behaviour on scaling back a dimension to the reals arises where -1 = 1, et cetera.
It definitely requires one to redefine the behaviour of the comparison operators when moving up a dimension from the reals to the complex, which causes problems when you try to go back down by simply setting the additional dimensions to be equal to 0 so that only the real part has any defined nonzero value. And that was a thought that was running in the back of my head when I wrote all this out. I just didn't want to take the time to explore those thoughts to their illogical conclusions at the time that I had them. :P
+Calyo Delphi
There's nothing wrong with thinking about a "weird" idea in mathematics ... by all means, state it and explore it! Most great mathematicians have challenged things, only by either finding out they were wrong (which can help the next time ;)) or by proving "the establishment" wrong! Cantor reasoning about different kinds of infinity is a nice example!
(to be honest ... I'm not that creative myself ;-))
Now that I come to think of it ... your idea (apart from the clash with the equality definition already existing in C) is not that weird at all!
I have no idea about your knowledge of foundations of mathematics. Maybe you know something about topology, maybe you don't. But topological spaces try to "mimic" the idea of vicinity, being in the neighborhood of something, trying to generalize limits and continuity even!
(see e.g. metric spaces, a special kind of topological spaces: en.wikipedia.org/wiki/Metric_space)).
In sets with a metric defined (mimicking our notion of distance), it turns out that (about) the simplest metric you can define (the so-called discrete metric) which defines that the distance between two different points is always 1 (and the distance between equal points is 0, of course), turns out to be quite useful! Not that much as a topological space in itself, but to (dis)prove theorems!
We learn most by the mistakes we make! (and trust me, I have made a few myself).
Keep thinking about (the foundations of) mathematical concepts! Keep challenging yourself! ;-)
And keep loving math, it's beautiful!
@Mathematician:
Thinking about a "weird" idea in mathematics is how I figured out on my own the Leibniz Formula that generalizes the power rule to higher-order derivatives, and then I went a step further to figure out the product rule generalized to higher-order derivatives and connected that with Pascal's triangle and binomial coefficients. ^^ And I spent a while trying to do the same to the chain rule, and I got as far as making connections to partitions of sets, Bell numbers, and Stirling numbers of the second kind, but a whole lot of that dives deep into number and set theory mathematics that's waaaay over my head.
And then I discovered Faa di Bruno's formula and felt deflated that two hundred years ago someone had already discovered a generalization. XD Oh well. I didn't really memorize the formula and I plan to eventually dive back into it to try and figure it all out on my own again. And maybe figure out my own generalization that might be different from Faa di Bruno's formula in its own ways! Or it might end up being exactly the same by happenstance.
And just recently (last night and today recently) I explored the idea of the parallel operator as a multi-input operation (instead of just a||b = 1/(1/a + 1/b), a1||a2||...||aN = 1/(1/a1 + 1/a2 + ... + 1/aN); I don't know how prevalent this operator is in EE textbooks, but this exact definition of the parallel operator is used in EE to compute parallel resistances/inductances and series capacitances, so I use it as a personal shorthand) and determined (with the help of said Ph.D. friend correcting my terminology) that it is symmetric, but non-associative, just by throwing a few parentheses into the mix and playing with the numbers 'til I got something that I definitively could not make equal to the other thing through algebraic manipulation.
And I've also in the recent past played with the idea of using slightly different notation to indicate the inverse of a function, instead of using the negative-one-th power, since it REALLY niggles me that that is the *one* special case where suddenly raising *any* defined function to a specific power means anything other than multiplying/composing that function with itself that many times over. E.g. sin^-1 x != (sin x)^-1, even though sin^2 x == (sin x)^2. Whyyyyy?? It's always been a point of contention and confusion for me because it makes no freaking sense, so I went and made my own far less ambiguous notation for it that I use in my own mathematics. c:
(FYI my formal mathematical education stops about most of the way through Calc I. Anything above that that I know is self-taught, largely through blackpenredpen, Dr. Peyam, Flammable Maths, and other maths youtubers, as well as looking stuff up on Wikipedia and asking questions to an online friend of mine who's a maths Ph.D.)
+Calyo Delphi
Wowww ... what a long reply!
Where do I begin to reply?
... "and I got as far as making connections to partitions of sets, Bell numbers, and Stirling numbers of the second kind"
I hope you didn't skip the numbers of the first kind ... ;-)
(apart from my numbers ... which haven't been given an official name yet ;-))
... "And just recently (last night and today recently) I explored the idea of the parallel operator as a multi-input operation"
Symmetric, but non-associative ... that would make mathematicians "mad/crazy"! ;-)
..."slightly different notation to indicate the inverse of a function"
Actually, there is a reason for denoting the inverse of a function f as f^-1 (didn't mean that as a power, sorry)
I'm not going to explain it to you here: you (most likely) have already figured it out for yourself!
... (FYI) my formal mathematical education stops about most of the way through Calc I
What stops you from reaching level Calc II (or even higher?) Apparently your "constantly wondering why" doesn't!
Keep challenging yourself! And eventually you will reach level ... (wel, your grades will fill it in! ;-))
why not ordering by distance from the origin?
But then i = -i in terms of distances, but we wouldn’t have trichotomy
Can there really be negative complex numbers? I thought, "complex" is the third flag though.
Well, strictly speaking no, because you can not compare complex numbers, but you could say the real parts are negative?
I think absolute value is a fair way to do it.
How will you compare, for example, 3+4i and 5?
If you add the argument as a second condition, you'll have to compare cis(4π/3) and cis(10π/3). They are the same number, and have a abs of 1, but 10π/3 > 4π/3 (by radius).
why not you can define a good function f, such that C->R for ordering. f(x,y)=R?
if you don't do such artificial function defining in some problem, you probably lose the good chance of using complex number
Discovering something that doesn’t exist:
An ordering on C
Watching this video i hit upon a question....whats the value of iota in decimals why is that undefined on calculator ??
As complex numbers used to represent dots in a plan the question to compare them is without logic. We may calculate the modules and to express the distance from origin but to what use??? The distances may vary from 0 to infinite. So what, we may say that infinite is grated than 10!!! Difficult to understand. If there are real numbers the comparison is odd.
Isn't the order where every complex number is equal to every other complex number an order which satisfies all the criteria? It is of course a totally trivial ordering that you could apply to any set, but isn't it a more precise statement to say that the only total order on the complex numbers respecting both addition and multiplication is the order in which every pair of elements coincides?
Daniel Darroch I think that ordering would lead to the same contradiction as 0=i. EDIT: Maybe not, because in that "ordering" saying 0=i is completely valid and does not contradict.
Yeah, it would satisfy the three axioms, but would break either the definition of numbers or the definition of equality.
I'm confused then the dictionary order is it an order or not on C?
It is an order, but it is not compatible with the algebraic structure of C.
Basically, it's just an order on the "symbols" that has no connection to the actual arithmetic of C. We can always arrange the elements of a set in some order.
Nice sir...
Thanx for showing properties of zero.question is 1=1?.
Dr. Peyan, answer me this question: how imaginary numbers can quantify some natural laws?
The schrödinger equation is a perfect example for this
Are you trying to argue C is an ordered field? Because I think 1 can argue that -1>0 is a contradiction if we R trying to do that.
If you replace the second requirement with multiplication, you end up with absolute value.
Technically you didn't violate the trichotomy law because you never specified that there are no two numbers a and b such a is both "squiggly less than" and "squiggly greater than" b.
But when multiplying an inequality by -1 the less-than should be changed into a greater-than: (-1) [ 2 < 3] = -2 > -3
John Baker No, because this ordering it's not the same as the real ordering, and in the video he ended up with *0 squiggly less than -1* . That's not the same as *0 < -1* so there's no contradiction yet; but when multiplying by -1 we have to remember what we had: *0 squiggly less than -1* , and that means -1 is positive here, so you don't switch the inequality, since -1 is no longer negative (in this situation).
+John Baker
In a sense, you are correct: IF C were to have a nice ordering (adhering to the three requirements mentioned in the video), then the following MUST be true:
a < b implies -b < -a (hence, -a > -b)
(the "
That was beautiful.
Just use the norm!
|i| < |3|
CAUTION! Wrong!
if you want to compare 2 with -4 in complex plane (2 vs -4)
you will say |2| < |-4|=4 (True)
So, the answer to the question (2 vs -4) is 2 < -4 (Wrong) !!!
(we already know that -4 < 2)
Cause you aren't comparing numbers you are comparing functions!
cexploreful
we’re not in the realm of real numbers
cexploreful |x|=|y| doesn't implie that x=y
Assumption (ii) doesn't hold:
i < 3
i-3 < 3-3
i-3 < 0 BUT |i-3|>|0|
ya he proves it cannot be totally ordered
one of the conditions will always fail
Nice !
At time 8:48 why r you not reversing sign of inequality after multiplying with -1?
You can order the complex numbers; I want to order a complex pizza!!
By these rules you cant order the real numbers either. 0>-1 gives 0*(-1)>-1*(-1) and 0>1. Adding 1 to the first inequality gives a contradiction.
There is a mistake in proving contradiction for proving I>0 as sign of inequality changes on multiplying with negative no.
See pinned comment
Cannot we define I=√-1 and still i^2=1 |a+ib|=√|a^2-b^2| instead of √a^2+b^2.
Proof by intimidation is the antithesis of proof by seduction, which is neither proof by induction nor deduction
So, complex numbers are incomparable as is the teacher!
Indeed!
I still have trouble with being unable to order 0 and i, since they less in the same axis and i should then be measurably greater.
Complex numbers are two dimensional vectors, so ordering is possible by length. In this video you're trying to 'delete' one dimension
shouldn't the multiplication by -1 always change the "direction" if the order sign?
PrinzMajdazari No, because this ordering it's not the same as the real ordering, and in the video he ended up with *0 squiggly less than -1* . That's not the same as *0 < -1* so there's no contradiction yet; but when multiplying by -1 we have to remember what we had: *0 squiggly less than -1* , and that means -1 is positive here, so you don't switch the inequality, since -1 is no longer negative (in this situation).
+PrinzMajdazari
See my reply on John Baker's question.
Squigglely awesome :3
Imagine losing points in your exam bc u didn't prove that 0 isn't i
Of course there is no linear ordering of the complex numbers...
Since when was the complex plane linear?
The Major
There is actually a way to map all the conplex numbers onto a line.
Bibin Muttappillil How?
Aaron Quitta
We need a bijective function that takes a complex number and outputs a real number and its inverse function.
Thats the same as a function which takes two real numbers and returns one.
We can construct a new number by filling all odd decimal positions with the first number and all even decimal positions with the second number.
The inverse would be to take all the odd decimal positions to construct the first number and all the even for the second.
You can put every number inside the function, so it has to be bijective
Bibin Muttappillil Thats a really cool idea! Is there any problem with the set of complex numbers being bigger than the set of real numbers?
Aaron Quitta
The problem is that it would imply that there are not equal in size therefore there wouldn't be a bijective function between these two sets.
But there is -> contradiction.
Magnitude a squared plus magnitude b squared!!! One norm
nice!
This video got me thinking about a question my foundations professor posed to our class, namely, how would one go about well ordering the real numbers. Curious to see if you have any thoughts are on this. Cheers!
I'm pretty sure the real numbers cannot be well ordered without the Axiom of Choice, which means there's no way to explicitly define a well ordering of the real numbers. This has in fact been proved. If we assume the reals can be well ordered, then it is possible to construct non-measurable subsets of the reals (eg. the Vitali set), and it has been proved that the existence of non-measurable subsets of R is independent of the axioms of Zermelo-Fraenkel set theory without the Axiom of Choice. That is, there are models of ZF in which every subset of R is measurable. All this means that you must assume the axiom of choice (or a weaker form of it that is strictly stronger than ZF) in order to demonstrate the fact that the real numbers can be well ordered. If it were possible to explicitly demonstrate a well-ordering of R, then one would have proved the Axiom of Choice (or a weaker form) which is impossible in set theory as we usually understand it.
+Samuel Roberts
Smart professor you have there: it is always good to think about foundations of mathematics!
en.wikipedia.org/wiki/Well-order might provide you with some insight (though it depends heavily on rigorous foundations of set theory).
i
even a UA-cam video thumbnail give me a fake hope 😔
Same Spiel :)
You give up too easily. We can say that two complex numbers are equal if the modulus of each is equal and the argument of each is equal. From there we can say that one is less than the other if the modulus is less, or if the modulus is equal but the argument is less. Done!
But that’s just lexicographic ordering and doesn’t satisfy natural properties of ordering...
@@drpeyam It's a definition- and what's wrong with making a definition in Math?