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  1. 2 years ago
    Anonymous

    Not a valid metric of distance

    • 2 years ago
      Anonymous

      >Not a valid metric of distance

      correct, a right triangle cannot have a hypotenuse of length zero, the shape implied by the variables in OPs picture is a line, and the length "i" is also 1.

      • 2 years ago
        Anonymous

        >the shape implied by the variables in OPs picture is a line

        correction, a line segment.

  2. 2 years ago
    Anonymous

    For the side to have length i it must have length 1 and be rotated 90°, so it is on the same line as the side of length 1 therefore the hypothenuse is 0.

    • 2 years ago
      Anonymous

      >length i
      that doesn't even make sense do your argument is invalid

      • 2 years ago
        Anonymous

        You are moronic if that doesnt make sense.

        • 2 years ago
          Anonymous

          calling me moronic doesn't make your argument less wrong

        • 2 years ago
          Anonymous

          you just exposed yourself as mentally moronic

          In order for a metric to be well-defined, you need to have that d(x,y) > 0 for any x=/=y. Is i > 0?

          >In order for a metric to be well-defined, you need to have that d(x,y) > 0 for any x=/=y.
          you missed the part where d: MxM->R being M the metric space and R the real numbers
          >Is i > 0
          according to what order relation?

          you guys have to study more before pretending you know math
          https://en.wikipedia.org/wiki/Order_theory#Basic_definitions
          https://en.wikipedia.org/wiki/Metric_space#Definition

      • 2 years ago
        Anonymous

        Complex numbers form a Kolmogorov-Asimov-Rutherford-Eisenstein-Michelin quasisemipseudometricoidal space that lack all the properties you will rise to object.

    • 2 years ago
      Anonymous

      but it is rotated 90 CCW with respect to X. so it is already where it needs to be. why do you want to rotate it back to 1? then it would be 1 not i.

      • 2 years ago
        Anonymous

        No, because if the side were of length 1, then it would stay where it is, but the side is of length i, so it is rotated relative to where it is.

    • 2 years ago
      Anonymous

      Abstractly, yes, however, negative lengths are unfounded in math/physics. The negative sign is useful to indicate direction (opposite of that which is agreed to be the positive side, almost always left), but the length is modulus wherever, whenever.

      That's using the complex numberline, anon, and even so the distances are modulus of whatever quantity is chosen. So far as I have come to understand it's interesting to use complex numbers because with them you can represent rotations algebraically, I'm sure it must be more profound than that, but I'd be lying If I said I knew.

  3. 2 years ago
    Anonymous

    In order for a metric to be well-defined, you need to have that d(x,y) > 0 for any x=/=y. Is i > 0?

    • 2 years ago
      Anonymous

      sqrt(n) is always positive. Thus sqrt(i^2) must be positive.

      • 2 years ago
        Anonymous

        How do you define > on C rigorously?

        If you assume i > 0 then either you get -1 = i^2 > 0 or you consider that a > 0 and b > 0 doesn't imply ab > 0. In both cases you lose some property that you'd like to have.

        • 2 years ago
          Anonymous

          >How do you define > on C rigorously?
          I don't. I'm too busy fricking your dad.

        • 2 years ago
          Anonymous

          Are you assuming my algebraic properties?

        • 2 years ago
          Anonymous

          >How do you define > on C rigorously?
          Define a bijection g from C to R. Then for x, y in C define x>y if g(x)>f(x)

    • 2 years ago
      Anonymous

      plus
      In order for a function to be a metric*

    • 2 years ago
      Anonymous

      Who said that? You can define a metric to be whatever you want. You could have a complex metric

  4. 2 years ago
    Anonymous

    You multiply i by its negative, therefore its 2

  5. 2 years ago
    Anonymous
  6. 2 years ago
    Anonymous

    Refute this

    • 2 years ago
      Anonymous

      Imagine believing in irrational numbers

      • 2 years ago
        Anonymous

        > irrational numbers

      • 2 years ago
        Anonymous

        Painfully moronic

      • 2 years ago
        Anonymous

        > irrational numbers

        Painfully moronic

        The word "irrational" is LITERALLY in "irrational numbers"! Mathematicians have taken us for absolute fools...

    • 2 years ago
      Anonymous

      The image has pixels.
      The length can be computed by counting the number of pixels.
      Labelling two sides as length one is only an assigned value.
      It doesn't represent reality.
      When an object is made up of a easily countable number of real things, approximations are not neccessary.

  7. 2 years ago
    Anonymous

    Pythagorean's theorem holds for lengths. Lengths are always nonnegative real numbers, so there's no such thing as a side of a triangle having length $i$

  8. 2 years ago
    Anonymous

    I see no problem here.

  9. 2 years ago
    Anonymous

    You're treating an imaginary number as a real number. Guess what, you can't.

  10. 2 years ago
    Anonymous

    This is actually correct. If you plot this in a 2d cartesian space then it makes no sense but if you plotted this out in a Riemann surface it would make perfect sense.
    Imaginary numbers are not imaginary, they are very real. The need for the complex plane arises as basically a work around for fundamental flaws in standardized math (ie. taking the square root of a negative number). This is because negative number themselves dont exist (you cant have a shape with a "negative" side) but again were introduced (after some controversy) because they can be useful in some cases and it allows math to be explored further without the need to rethink the rest of mathematics. It was only a matter of time before this workaround would cause issues and imaginary numbers were created as another workaround.

  11. 2 years ago
    Anonymous

    I refute your image

  12. 2 years ago
    Anonymous

    Define the objects that you are measuring, explicictly define the distance and prove that it satisfies the definition of distance

  13. 2 years ago
    Anonymous

    >uses imaginary numbers in real geometery >gets surprised when shit breaks

  14. 2 years ago
    Anonymous

    The arithmetics is correct.
    i^2 + 1^2 = 0^2
    -1 + 1 = 0
    0 = 0
    But it is senseless to assign a nonpostive number to a distance. Geometrically this is nonsensical.

    Don't mind me now, just trying to relearn how to use latex and edit my text (haven't been here in a LONG while).

    [math]sqrt{-1}^{2} + 1^{2} = 0^{2}[/math]

    • 2 years ago
      Anonymous

      >But it is senseless to assign a nonpostive number to a distance. Geometrically this is nonsensical.
      It makes sense in the theory of relativity.

      • 2 years ago
        Anonymous

        where in "relativity" is a nonpositive number assigned to a positive distance

      • 2 years ago
        Anonymous

        Have never heard that, and still I'm more inclined to think that the negative sign does not represent a value less than zero but rather direction (like vectors) or something else, perhaps time dialation in this case, but again, I got no clue on that.

        Does at any point in high level physics or math a negative distance (not indicating direction) is axiomatically accepted?

        • 2 years ago
          Anonymous

          It's imaginary numbers, not negative numbers. Compute the distance between two points in spacetime that are actually separated by a time, and you'll get an imaginary distance. Similarly, compute the time between two points in spacetime that are actually separated by a distance, and you'll get an imaginary time.

          • 2 years ago
            Anonymous

            which has nothing to do with assigning a nonpositive number to a positive distance

          • 2 years ago
            Anonymous

            >It's imaginary numbers, not negative numbers
            Lowkey they seem negative. I mean, it pretty much revolves around the fact that there's a number, convetionally called "i", that when squared equals negative one. And I gotta agree with the other anon, I still see no relation between this and assigning nonpositive numbers to positive distances.

  15. 2 years ago
    Anonymous

    I refute it with a ruler.

  16. 2 years ago
    Anonymous

    Fixed it.

    • 2 years ago
      Einstein was a thieving jew

      lol I don't know if this is a joke or not but this is actually a better visual representation of what's actually going on. Better yet would be to make this a 3D cartesian space and have the side of length i on the Z-axis.

      • 2 years ago
        Anonymous

        homosexual beat me to it

  17. 2 years ago
    Anonymous

    You are on your way to discovering hyperbolas, OP.
    Unit hyperbola:
    [math](x)^2+(iy)^2 = 1[/math]

    • 2 years ago
      Einstein was a thieving jew

      You are on your way to discover hyperbolas, I am on my way to discover hyperborea. We are not the same.

  18. 2 years ago
    Anonymous
    • 2 years ago
      Anonymous

      alternative angle

    • 2 years ago
      Anonymous

      no it's the other way around
      the green hypotenuse is the 45 degree angle, which is supposed to be precisely 0 no matter the quadrant, while the i length is the Y axis

  19. 2 years ago
    Anonymous

    Yes (i)^2 + (1)^2 equals (0)^2, but you can't have side with length zero or length i. It's inbreds like you who can't think beyond basic shapes who held back math for thousands of years

    • 2 years ago
      Anonymous

      Geometric Algebra expresses blades as essentially hypercomplex numbers

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