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Insights The Case for Learning Complex Math - Comments

  1. Feb 3, 2016 #1

    anorlunda

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    Last edited: Jun 28, 2016
  2. jcsd
  3. Feb 3, 2016 #2
    Great article. I do have to disagree with the contention that the resistance of many to complex numbers is due to the term imaginary. People don't have a problem accepting all kinds of imaginary things when viewing Star Wars, so I don't think it bogs them down that much in math class. Rather, I think the aversion is quite natural. As was pointed out in the article, the reaction to ## n = 4 - 5 ## was, "Who ever heard of -1 goats?!" That is not an irrational reaction. It is perfectly natural because, well, nobody has ever been able to spot their -1 goat out browsing on the mountain. It is only when -1 is seen as an accounting tool that we can understand what -1 goats means.

    I think a similar reaction is provoked when one encounters ## i = \sqrt{-1} ## for the first time. It is not because it is called the imaginary unit, but because people naturally read ## A = \sqrt{B} ## as, "Some number A, which when multiplied by itself, equals B." They then make the immediate deduction that there is no such number A, which when multiplied by itself, equals -1 or any other negative number! This is one of the reasons I personally think complex numbers should first be taught as ordered pairs with specific operations simply defined. Then the fact that ## (0,1)*(0,1) = (-1,0) ##, does not arouse natural suspicion at all. Later, the convenient notation can be introduced. IMO, this would go a long way towards eliminating the aversion to complex numbers.
     
  4. Feb 4, 2016 #3

    kith

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    There are a number of little errors with your Schrödinger equation: the [itex]\hbar[/itex] needs to be squared, the derivative with respect to [itex]x[/itex] should be of second order and neither the [itex]\bar \psi[/itex] nor the [itex]\nabla[/itex] should be there in the bracket.

    Nice insight!
     
  5. Feb 4, 2016 #4

    anorlunda

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    Thanks for your sharp eye. Corrected.
     
  6. Feb 4, 2016 #5

    QuantumQuest

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    A really good insight. What everyone resisting to the very idea of complex numbers has to be aware of, is that the concept is not something ad-hoc which after some work, made its way and conquered the world of math, but rather it was somewhere hidden and finally came into light. That explains its use as rotation as described in the article and it definitely has its important place in Physics.
     
  7. Feb 5, 2016 #6
    And I was going with, "Learn complex math because it is fun."
     
  8. Feb 5, 2016 #7

    anorlunda

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    When writing that Insights article, I came upon a curiosity.

    Rather than (add,subtract,multiply,divide) as the basic arithmetic operations, students could be taught (sum,negate,multiply,invert). Where sum adds signed numbers, while add implies unsigned positives such as "3 goats plus 2 goats". Subtraction could be defined as negate, then sum. Division is invert, then multiply.

    But then someone could note that negation is just multiplication by -1, so that the operators could be reduced to three (sum, multiply, invert.)

    If we introduce complex numbers, then we could have (sum, negate, multiply, invert, rotate). But negate and rotate are both special cases of multiply, so we are back to three once again (sum, multiply, invert).

    But with complex, there is an additional basic operation so we are back to four (sum, multiply, invert, conjugate)

    I'm curious. Has anyone else been down this path before of redefining the basic arithmetic operators?
     
  9. Feb 5, 2016 #8

    jbriggs444

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    From a abstract algebra perspective, I am not comfortable making multiplication part of the definition of a group under addition.
     
  10. Feb 5, 2016 #9

    anorlunda

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    I don't know what you mean by that.

    Start at the beginning. Are you comfortable with (add, subtract, multiply, divide) being the four basic operators taught in grade school?
     
  11. Feb 5, 2016 #10

    jbriggs444

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    Yes. That is fine. And I would also have no problem with (add, negate), (multiply, invert).
    I have a problem with using multiplication by -1 as part of the definition of the additive inverse. It's superfluous and introduces a multiplication operation that may not even exist.
     
  12. Feb 5, 2016 #11

    anorlunda

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    OK, then I agree. I hadn't thought so deeply into the implications.

    But would you consider negate different than multiply by -1? It is hard for me to think of negate except as a special case of multiply.
     
  13. Feb 5, 2016 #12

    jbriggs444

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    It is hard for me as well. All that training starting from elementary school dealing with integers and rational numbers makes it difficult to get away from that mode of thought and to start thinking of addition and multiplication as abstract operations on other sets.

    Yes, putting my mathematician's hat on, negation is different from multiplication by -1 since the latter operation may not exist.

    That said, if multiplication exists and additive inverses exist and if the distributive law for multiplication over addition holds, then I am pretty sure that one can demonstrate that negation and multiplication by -1 must be equivalent.
     
  14. Feb 6, 2016 #13
    Good that you mentioned a link to Better Explained. That would help a lot of people.

    Nice insight by the way.
     
  15. Feb 19, 2016 #14
    I agree about using ordered pairs. If one considers a vector space of two dimensions equipped with a specific multiplication operation (vector * vector -> vector) corresponding with complex multiplication, the concept of an "imaginary" something goes away, while leaving notation and computation basically unchanged.
     
    Last edited: Feb 19, 2016
  16. Jul 13, 2016 #15
    "Can We Achieve the Same Thing Without Complex? Yes you can, but it is more difficult."

    I have the feeling that this is a somewhat arguable statement and that the "mistery" behind complex numbers is often overemphasized. In my opinion all these "misteries" about imaginary numbers and the square root of -1 disappear when you simply see complex numbers as the even-grade Clifford algebra of R^2. The "misterious" imaginary unit is just the bivector e1*e2 (or a unit pseudoscalar of R^2, in general). Multiplication of complex numbers is equivalent to the Clifford product of scalar+bivector quantities in R^2, with the difference that such quantities can even be multiplied by ordinary vectors in R^2, and they have the effect of rotating (and scaling) the vector.
    I think Clifford algebras, Geometric Algebra, Clifford products and rotors should have been mentioned in this article.



    Reference https://www.physicsforums.com/insights/case-learning-complex-math/
     
  17. Feb 5, 2017 #16
    I've just joined Allan. Great paper - very useful to be able to answer questions like why bother!
     
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