I've Eliminated 'irrational' numbers

r norman rsn_ at _comcast.net
Thu Apr 15 08:11:59 EST 2004


On Thu, 15 Apr 2004 12:10:42 GMT, "ken" <kpaulc@[remove]earthlink.net>
wrote:

>While Working with the other stuff that I've
>posted recently, I Totally-Busted Incommens-
>urability.
>
<snip 12 points>
>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'.
>

Ken, your lengthy musings about neural systems are at least nominally
on topic in this news group and so are tolerated.  However your forays
into mathematics are distinctly very off topic and should be taken
somewhere else.

You really don't have a clue as to the nature of "irrational numbers.
There are integers that are "whole numbers" like 1, 2, 3 and rationals
that can be represented as a ratio of integers like 1/2 or 5/7.  Some
of these are also integers, like 12/4, most are not.  Then there are
algebraic numbers that can be represented as the solution of an
algebraic equation,like sqrt(2) which is a solution to x^2 - 2 = 0.
Some algebraic numbers are rational and some are even integers, but
most are not.  It is easy to prove that sqrt(2) is irrational,
something that was done by the ancient Greeks.  Finally, there are
transcendental  numbers like pi and e which are not even algebraic. 
The "irrational numbers" include most of the algebraic numbers plus
the transcendentals.

The fact that an equation like x^2 + y^2 = z^2 may have solutions
mixing rationals and irrationals is entirely irrelevant.  For example,
if x=1 and y=1 then z must be algebraic but z is still irrational. If
x = y = pi , then z must be transcendental.  It doesn't matter that
the numbers satisfy a particular equation. All you have really done is
rediscover the well-known expression sin^2 x + cos^2 x = 1 where it is
also well known that sin x and cos x are irrational for almost all x.




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