A Tricky Problem from Harvard University Interview

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  • Опубліковано 16 лис 2024

КОМЕНТАРІ • 27

  • @emrekarayalcn1236
    @emrekarayalcn1236 2 місяці тому +3

    This problem highlights a very important principle: One cannot simply apply the tools of real analysis to complex numbers.
    This is a fun problem with some nice algebra. However, we run into a problem with square roots of complex numbers. You did not define what taking a square root of a complex number is and whereas in the real case it is common convention that the square root function maps a non-negative real number to a non-negative real number, there is no such convention for complex numbers. Historically square roots arise from geometry and measures of length, going (at least) as far back as the ancient Greeks. In that context we are concerned only with real numbers and it makes sense to consider the 'positive square root' as we consider lengths. Technically the square root sqrt(a) of a number a is a solution of the equation X^2 - a = 0, and therefore (unless a = 0) there are two choices for sqrt(a) (two embeddings of the formal number sqrt(a) as a real number). When defining a square root FUNCTION we must choose one of these solutions, as a function associates one element in the codomain to an element in the domain. The historical context gives the motivation for the choice (of the positive embedding) when a >= 0. However in the case of complex values a no such historical argument exists (not to mention that this argument does not imply that the square root of a positive number is positive, but merely an explanation for why mathematicians choose this embedding, it is a choice, not a fact, that the square root of a real number is positive). If you wish to define a 'square root function' for complex numbers we must therefore choose one of the two possible solutions as our image. One can do so by demanding that the function sqrt( ):C --> C maps 0 to 0, and any non-zero complex number z to the unique solution of X^2 -z = 0 whiose argument lies in the interval [0, pi) (note that this agrees with the conventional square root function for real numbers).
    When you evaluate sqrt((2+i)^2) you say it is equal to 2 + i, but using basic algebra one could show that
    sqrt((2+i)^2)=sqrt((-2-i)^2)= - 2 - i
    So you could make the argument that you defined the square root function implicitly using the construction above, but then we would have
    sqrt((2-i)^2) = - 2 + i, not 2 - i.
    You could of course be using a more exotic complex square root function, but if so it would be worth it to define it explicitly. As neither 'natural' definitions (choosing arguments in [0 pi) or [pi, 2*pi)) Would give the solution 2i. Instead, substituting the values of x you found into the LHS of the original equation and evaluating using these definitions would give an answer of 4 and -4 respectively.
    What I am trying to say is that there is no one way to define a square root function, certainly not for complex numbers. It always requires a choice (e.g. positive or negative for real numbers, for the complex case see above) and it is important that if you wish to define complex valued square root functions , one should make explicit the choices they make, otherwise we get non-well defined or seemingly inconsistent mathematics such as in your video.
    (edit 1: I noticed a grammatical error, it said 'This problems' in line 1, I changed it to 'This problem')
    (edit 2: Changed " the square root function maps a non-negative integer to a non-negative integer" to "the square root function maps a non-negative real number to a non-negative real number "

    • @sil1235
      @sil1235 2 місяці тому

      Yea exactly what I thought when I saw sqrt((2+i)^2) being simplified to 2+i. It's for that similar reason why we can't also just use (a^b)^c=a^(b*c) in complex numbers - one of the common misconceptions.

    • @iqtrainer
      @iqtrainer 2 місяці тому +1

      In the complex domain, sqrt has two answers. But, if you check, one of them doesnt with the ORIGINAL equation. I think both of you didnt catch that. So the answer presented in the video is correct.

    • @iqtrainer
      @iqtrainer 2 місяці тому +1

      Plus, this problem is written with square root. So you need to think about how the question was written. In this problem, we are only looking for principal square root. If this question is written with the power of 1/2 instead of sqrt, then what you two said is correct

    • @sil1235
      @sil1235 2 місяці тому

      I don't see anyone claiming the video is incorrect. Instead, it is pointed out one needs to be careful with calculations in complex numbers - the same rules do not apply the same way as they do in real numbers. In this case it wouldn't certainly hurt to remind that square root (sqrt symbol) means we take the principal square root, and which one it is. Otherwise I guarantee you that people will take whatever they see as a in sqrt(a^2), without thinking (and then they are stumped by 0=1 proofs which are basically exploiting this, or for example the other rule I mentioned in previous post).

    • @sil1235
      @sil1235 2 місяці тому +1

      @@broccoli2237 "The square root has 2 solutions" - not really. Yes, there are two solutions to x^2=3+4i, and yes there are two values of (3+4i)^(1/2), but that is not the same as sqrt(3+4i). The sqrt is usually taken to be the principal square root (in this case that is the one solution with positive real part), so it's 2+i and NOT -2-i (generic definition of principal root is slightly more complicated, I suggest to look it up). So under that definition sqrt(3+4i)=2+i and sqrt(3-4i)=2-i, exactly as on the video. (The point of the above comments was more didactical than technical - so students/viewers are aware of this choice)

  • @Min-cv7nt
    @Min-cv7nt Місяць тому +1

    This is one of the best algebraic equation I've seen in my life

  • @mathnerd5647
    @mathnerd5647 Місяць тому +1

    Since this equation is given as square root, one needs to get only principal square root. No confusion at all.

    • @drpkmath12345
      @drpkmath12345  Місяць тому

      Nicely said my friend haha👍👍👍

  • @iqtrainer
    @iqtrainer 2 місяці тому +1

    Very nice problem professor and your explanation is just gorgeous🎉

  • @trustme_emtsurt1846
    @trustme_emtsurt1846 2 місяці тому +2

    Can someone explain the Left hand side of the first step?
    Where the [ (5-x)+(1+x)- 2√(5-x)(1+x) ] came from rather than just (5-x)-(1+x)

    • @sil1235
      @sil1235 2 місяці тому +4

      Square √(5-x)-√(1+x) and use (a-b)^2=a^2+b^2-2ab

  • @domedebali632
    @domedebali632 2 місяці тому +1

    This is so nice problem professor

    • @drpkmath12345
      @drpkmath12345  2 місяці тому

      Thanks a lot my friend👍👍👍

  • @MrGLA-zs8xt
    @MrGLA-zs8xt 2 місяці тому +1

    The best

    • @drpkmath12345
      @drpkmath12345  2 місяці тому

      Thank you very much my friend👍👍👍

  • @mr.robot001mr.robot00
    @mr.robot001mr.robot00 2 місяці тому +1

    🙏🙏🙏