On May 31, 9:28 pm, chaja...@mail.com wrote: > > No, by "are they real" I mean much more. As well as an order you also > > need > > the field operations (addition, subtraction, multiplcation and > > division) > > and you need them to preserve order. > > Makes sense to me and I agree.
However, not to the extent of actually poducing a definition.
> > > No. If we use the standard definition of real numbers and > > the standard way of talking about decimals (infinite decimals > > are defined in terms of limits), then > > decimals can express any real number (including > > sqrt(2) and pi). > > I find this quite simply impossible to accept.
Too bad. The statement is still true. Take any real number x in (0,1]. Construct the following approximations
a1=0.d1 a2=0.d1 d2 a3=0.d1 d2 d3 ...
at each stage pick dn to be the largest digit such that an is less than or equal to x. It is easy to see that the sequence (a1,a2,a3,...)=(0.d1, 0.d1 d2, 0.d1 d2 d3,...) has limit x. Using the standard way of talking about decimals 0.d1 d2 d3 ... = x
> For any decimal that > tries to express a number beyond zero, a whole infinity of real > numbers lie in between that and zero.
Correct. And every one of these real numbers can be expressed by another decimal. Given any decimal greater than 0 there are an (uncountable) infinite number of other decimals between this and zero.
> Decimal can never point out a > value beyond zero that has no intervening infinity. To me the > consequence is that is cannot point out them all, for to point one out > only shows there are others. >
And every time you point out a decimal, you show that there are other decimals.
> > It was to avoid nonsense like > > "would be able to comply in infinite completion" > > that the concept of limits was developed. > > > - William Hughes > > You reject the idea of infinite completion seemingly because it is > counter-intuitive to you, and indeed to many people, as infinity is > unbounded it our most general concepts. What about this: if you take > the sets of all real numbers in the intervals [0, 1] and [-1, 0] they > can be paired in cancelling values in infinite completion.
Meaningless until you define infinite completion.
The sets of > all real number in the intervals of [0, 1) and [-1, 0] could not be > paired in cancelling values in infinite completion, we could only > traverse infinity and yield pairs, refusing to complete the operation > and acknowledge that one value will not pair.
There is no way to "complete the operation", so it makes no sense to refuse to do this. There is no value that will not pair.
> > This idea you reject, or my applications of it, and the result of such > a rejection yields another counter-intuitive result, that 0.999... is > exactly in and of itself equal to 1. >
Yep, when you deal with infinity you are stuck with either counterintuitive results, or incoherent undefined nonsense like "infinite completion". You have chosen the latter.