• HakFoo@lemmy.sdf.org
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    8 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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      8 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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        8 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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          8 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.)

        • mozz@mbin.grits.dev
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          8 months ago

          Yeah. We were making a joke about the complex plane – you could say that measuring the hypotenuse of a triangle is equivalent to measuring the distance between points |AB| and |AC|𝑖 on the complex plane. That definition actually makes quite a bit of sense, and I think by sheer coincidence it’s possible to misunderstand how to do it and wind up with a way of looking at it where the hypotenuse of a right triangle with sides 1 and 𝑖 would work out to exactly 0. Which brings it back into concordance with OP’s (also wrong) Pythagorean presentation of it.

          It obviously doesn’t really work that way, but it’s hard to see necessarily anything wrong with it, which makes it a fun math thing.

    • mexicancartel@lemmy.dbzer0.com
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      6 months ago

      If so moving down the imaginary line should be equivalent to miving left but then the answer must be 2 units long

      But (-i)² is also -1 and it still results in 0