44 thoughts on “Can you compare complex numbers?”

1. 

   Yuval Paz says:

   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(!!!)

2. 

   blackpenredpen says:

   Since I am here earlier than flammy,

   <3

3. 

   Daniel Darroch says:

   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?

4. 

   The Major says:

   Of course there is no linear ordering of the complex numbers…
   Since when was the complex plane linear?

5. 

   Mi Les says:

   Just use the norm!

   |i| < |3|

6. 

   Kiritsu says:

   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?

7. 

   Bibin Muttappillil says:

   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)

8. 

   Aaron Quitta says:

   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?

9. 

   therealAQ says:

   i <3 Peyam

10. 

    cexploreful says:

    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
    2<5, f(2)>f(5) <=> -10>-50 (ok)
    next for an imaginary product define
    g(x)=i·x and change symbol from < to #
    2<5, g(2)#g(5) <=> 2i#5i
    so i apply it again
    2i#5i, g(2i)%g(5i) <=> -2%-5. (i want -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?

11. 

    ZeT 47 says:

    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?

12. 

    unknown360ful says:

    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?

13. 

    John Baker says:

    But when multiplying an inequality by -1 the less-than should be changed into a greater-than: (-1) [ 2 < 3] = -2 > -3

14. 

    Leonard Romano says:

    (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).

15. 

    PrinzMajdazari says:

    shouldn't the multiplication by -1 always change the "direction" if the order sign?

16. 

    Flammable Maths says:

    I see what you did there in the title <3

17. 

    l'islam La vérité says:

    Nice !

18. 

    Calyo Delphi says:

    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.

19. 

    hari kishan says:

    nice!

20. 

    Ralph Dratman says:

    That was beautiful.

21. 

    Timothy Augustine says:

    That's what I was wondering. I wondered if you can compare imaginary number with real number.

22. 

    Avi Mehra says:

    It works if you don't assume there exists a concept as positive, does it not?

23. 

    Samuel Roberts says:

    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!

24. 

    Kuratius says:

    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?

25. 

    Boypig24 says:

    Can you mesure complex numbers by the length of the number from the origin on a complex plane?

26. 

    david wright says:

    Hah! It's easier to escape the jaws of determined crocodile, than the jaws of logic. Thanks for the demo!

27. 

    HelloItsMe says:

    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

28. 

    Stydras says:

    Squigglely awesome :3

29. 

    Elias Caeiro says:

    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).

30. 

    Maxx Byron says:

    Is it fair to compare r values when converting to polar?

31. 

    Jürgen Lochner says:

    omg! a clean board ! I like it! 🙂

32. 

    Jacek Soplica says:

    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

33. 

    Joseph Noonan says:

    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.

34. 

    Joseph Noonan says:

    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.

35. 

    r75shell says:

    7:43 wait, multiplication property was about real numbers, and there was other sign (common >), and i is not from range (0,infinity)

36. 

    Mathematician says:

    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)

37. 

    Seth Person says:

    I think absolute value is a fair way to do it.

38. 

    Seth Person says:

    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.

39. 

    PoweredDragon says:

    After 0<-i You could just use 0(-i)<(-i)(-i), and later from 0<-1 you get (by the same multiplication) 0<i, which gives contradiction in terms of first condition

40. 

    John Baker says:

    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 😉

41. 

    Phoenix Fire says:

    If you replace the second requirement with multiplication, you end up with absolute value.

42. 

    nikulaye says:

    some complex numbers can be compared those with 0i and maybe those with the same i

43. 

    Bottle Of Fresh Squeezed Ass Juice says:

    You always have lexicographical and polar lexicographical orders.

44. 

    Dr. Peyam's Show says:

    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).