• HakFoo@lemmy.sdf.org
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    5 months ago

    I like it.

    Read the “1” unit side as “move left 1 unit” and the “i” side as “move up i units”, and the hypotrnuse is the net distance travelled.

    The imaginary line is perpendicular to the real line, so “up i unit” is equivalent to “right 1 unit”. The two movements cancel out giving a net distance of zero.

    • mozz@mbin.grits.dev
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      5 months ago

      Yep. A vertical line segment above A with length 𝑖 is a horizontal line segment to the left that’s 1 unit long. So, the diagram needs a “not to scale” caveat like a map projection, but there’s nothing actually wrong with it, and the triangle’s BC side is 0 units long.

      • MachineFab812@discuss.tchncs.de
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        5 months ago

        i= √(-1) = imaginary number (1^2) + (√(-1))^2 = 1 - 1 = 0 7

        At least, I thought that was the idea in the OP.

        Also, for your version, on a number line or Cartesian plane, the distance from -1 to 1 is 2, not 0

        • mozz@mbin.grits.dev
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          5 months ago

          Also, for your version, on a number line or Cartesian plane, the distance from -1 to 1 is 2, not 0

          Yeah. I cheated. You have to either deliberately misunderstand how to measure vectors or else drop a minus sign for it to work my way.

          (Or, from my previous example, you could just frame it as you’re getting the hypotenuse by measuring between |AB| and -|AC|𝑖 instead of the way I framed it – but that makes it more obvious that you’re fishing for a particular answer.)