I've Eliminated 'irrational' numbers

ken kpaulc at [remove]earthlink.net
Thu Apr 15 07:10:42 EST 2004

While Working with the other stuff that I've
posted recently, I Totally-Busted Incommens-

And it's Easy to see.

Given a Circle, having =any= radius, construct
a right-Triangle,

1. having the Circle's radius as it's hypotenuse,
call it C,

2. one side as a semi-chord, with one end shar-
ing a point with the circle-end of the hypotenuse,
call it A,

3. and the other side drawn from the center of
the Circle, to meet the semi-chord, forming a
right-angle, and sharing a point with the 'dangl-
ing'-end of the semi-chord, call it B.

4. The Pythagorean Theorem applies, and

A^2 + B^2 - C^2 = 0,

5. and is =ALWAYS= Equals Zero.

6. As one varies the angle that the hypotenuse
make with the X axis, A and B adjust their
lengths continuously, and

A^2 + B^2 - C^2 = 0, always,

7. even when A or B becomes Equal to

8. at which point, either the semi-chord, A,
or the center-pinned line, B, becomes Equal
to the hypotenuse, C.

9. This seems, to me, to be quite "astounding",
because "Incommensurability" is nowhere to
be seen.

10. Since the radius can be of =any= length,
this Circle operation scales from zero to
Infinity, which covers =any= plane, and does
so, Continuously, without a hint of "Incom-

11. It does this be-cause when A and B go-
Irrational, they balence each other =Exactly=!

12. Which means that Irrational Numbers
are no longer a "Problem" - just construce
a Circle with the necessary radious, and, voilà,
there's your "Conjugated"-pair of Irrationals,
ready to do anything that you want to do with

13. Of course, you cannot measure these
Numbers, but you can "address" them, Exactly,
through the Circle-Geometry, given above. You
know, pick your favorite Greek symbol, and
its, "Bye-bye, 'Irrationals'.

Q. E. D.

K. P. Collins
(c) 2004-04-15, by K. P. Collins

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