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Re: NonEuclidean Arithmetic
Posted:
Sep 14, 2012 12:16 PM


On Thu, Sep 13, 2012 at 11:22 PM, Joe Niederberger <niederberger@comcast.net> wrote: > Kirby says: >>OK, so repeated addition is OK up to the point where instead of 1/2 of something, you want 1/pi of it, and that something happens to be irrational already. That's when repeated addition breaks down, > > Well, some people like to hold on up to the bitter end. One way to approach irrational numbers is as the limits of certain sequences  and define the product of two such limits as the limit of the products of corresponding terms of rationals. That way, you can keep your repeated addition going infinitely. Or indefinitely. Or maybe indinfinitely. > > Joe N >
I explicitly have repeatedly said that we can use an infinite sum or series  but we don't need to appeal to infinity to represent any real number as a sum. I already demonstrated repeatedly that any real number can be represented as a finite sum of any natural number of addends in which all the addends are the same real number. For every real number x and every natural number n, just do x =(x/n)n = (x/n)(1_1 + ... + 1_n) = (x/n)_1 + ... + (x/n)_n. This works generally for any field that has characteristic 0 and that contains the natural numbers.
But I also keep saying that this does not prove the repeated addition crowd claim that repeated addition is what real number multiplication *is*. It proves the opposite  it proves what Devlin in
"It's Still Not Repeated Addition" http://www.maa.org/devlin/devlin_0708_08.html
implied in the last part of the article, which is this:
Repeated addition is merely a derivable property of the two operations of such a field like the real numbers, derived in such a field from just some of the algebraic properties (here they would be the identity, inverse, and distributive properties) as well as the property of having characteristic 0.
I have repeatedly said that this is about the need to constantly redefine what the whole "repeated addition" computation process is each time we move from the naturals to the rationals to the reals. I have said again and again that a finite computation process works as a representation for some sort of repeated addition process for naturals and rationals, but that is no finite computation process that can work for the multiplication of two irrationals without redefining things more to the point to where we are now talking about a new function altogether, only approximations of the result of the multiplication, not the original multiplication which is a function that takes us from the two inputs to the product, the output  it does not take us to roughly the output.
And note that repeated addition has been touted as a process by which we can always compute the product of any two real numbers.
But there is still no finite or infinite process of repeated addition that computes a noncomputable irrational as a product. Note that that subset of the reals that is the set of all noncomputable irrationals is what makes the set of all reals an uncountable set, the number of elements in the set of all reals being a greater cardinal number than the cardinal number of either the set of all naturals or the set of all rationals, which are countable sets. The union of the set of all rationals (which are algebraic and computable numbers) and the set of all computable irrationals (which covers the set of all algebraic irrationals and covers a subset of the transcendental irrationals) is still only countable. This means that the multiplication of two noncomputable irrationals yielding a noncomputable irrational as a product makes up almost all of real number multiplication.
This is the proof that repeated addition cannot model as a computational process almost all of real number multiplication. It is the proof of the falsity of the repeated addition crowd claim that repeated addition is what real number multiplication *is*.
 End of Forwarded Message
Message was edited by: Paul A. Tanner III



