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Self-reference And The Sq. Rt. Of -1

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Posted by Alan on August 12, 2002 10:58:09 UTC


I dreamed this up so I'll write it down and see what happens!

I asked myself this question:

Why is it that a 2 dimensional number is represented by the square root of minus one as its one-dimensional representitive?

I draw a square with sides "1 unit".
This is a 2-D unit.

Imagine the square is standing vertically on the desk. Suppose I look down at it so I only see the top edge. Now it looks 1-dimensional; just a 1-unit line.

What is this 1 visible Dimension, this visible top edge to the square; as a 2D viewpoint?

It is a view of the missing side of the unit square! It is the square root of the missing one!


So the square root of minus one is: an alternative 2-D view of one of the alternative perspectives in a 2-D view.

A view of a square that is different from the usual view, by looking at just the top edge and seeing that edge as being a 2-D view so that edge is the square root of the missing remainder of the square.

So an alternative view that something has of itself via a different view of itself (in this case, via looking at itself as a top edge of a vertical square).

So the square root of minus one involves seeing a different dimension of itself via self-reference?

Complex number: a + ib

An ordinary number 6 can be seen as:
a vertical rectangle with vertical side 3 units and top side 2 units.

Look at this from above: you see 2 units.

A 2-D view of that above-view:

(1D) x2 + (square root minus 1D x 3units)

2 + i3

Complex conjugate:

a - ib so 2 - i3

Real number: (a + ib)(a - ib) =
(2 + i3) (2 - i3) = 4 - - 9 = 4 + 9 = 13.

So the complex number gives a 2-D view of 13? A 2x2 square plus a 3x3 square; a view of 13 as a sum of squares? A sum of a 2x2 square and a 3x3 square?

A 1-D view take a side from each square and get 2x3 = 6?


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