Re: Proof of ordered powerset

From: Jon Slaughter (Jon_Slaughter_at_Hotmail.com)
Date: 03/21/05 

Date: Mon, 21 Mar 2005 13:57:45 -0600

"Robin Chapman" <rjc@ivorynospamtower.freeserve.co.uk> wrote in message
news:d1mqci$4kt$1@south.jnrs.ja.net...
> Jon Slaughter wrote:
>
>> Since, AFAIK, all irrationals come about through some limiting process...
>> while we might write sqrt(2),
>
> Is constructing an isosceles right angled triangle a
> "limiting process"?
>
> --
> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
> "Elegance is an algorithm"
> Iain M. Banks, _The Algebraist_

Well, try to measure it(atleast in general) ;) i.e., if your unit length if
proportional to one of the sides, then there is no way to measure the
hypotenuse in a finite number of steps with that unit of measurement... i.e,
if there is, it isn't irrational.

its like saying that sqrt(2) is finite, but what is sqrt(2)? well, if its
1.414213.... then its infinite, but if its sqrt(2) its not? what if we are
in base sqrt(2)?

To me, an right triangle with legs 1 and 1 is distinctly different from one
with legs 3 and 4. Just because we can express the second was as having
hypotenuse of sqrt(2) doesn't mean it doesn't come about in some "infinite"
way.

Its same stuff as taking a limit. When we take a limit we are carrying out
an infinite process in a finite way... but it is still an infinite process.

I'm not sure how to make the distinction, as I guess everything could be
thought of as an "infinite process", such as 3 + 5 = 3/(1+1/n) + 5/(1+1/n),
but this is not "natural".

What I'm looking for is something that can be carried out in a finite number
of arithmetical steps without some recursive like structure or some
intermediate infinite processes involved.

for example:

sum(k,k=0..n) = n*(n+1)/2

neither the lhs, the rhs, or the derivation involves anything to do with any
infinite process(atleast not in a natural way). Both sides take a finite
number of arthimetical computations, the first is n additions, the second is
1 addition, 1 multiplication, and 1 division.

Now, lets say I had n*(n+1)/sqrt(2), then it would be different because
sqrt(2) cannot be computed exactly(well, mainly because the formula could
only be approximated for integer n).

now, for something like lim x^2/sin(x), x->0 = 0, the lhs is an infinite
process while the right is a finite one. I believe that is the whole point
of things like this. We have the "infinite" realm and the "finite" realm and
we are trying to find a way to move from one to the other, since the finite
realm is very easy to work in... actually, its not possible to do anything
in the infinite realm(since it would take an infinite amount of pencil
lead).

I hope I'm making some sense. I know that I'm not being very precise, but
hopefully its somewhat clear in what I'm trying to do. While its not
necessary that I find a "simple" way to "express" my original sum, I think
it would be nice for simplicities sake(I am ofcourse assuming that it would
simple ;/).

Jon

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