I Problem with infinite decimal numbers?

[rmberwin]

I came across the following argument that attempts to show that the notion of infinite decimal numbers is incoherent. Try adding these two numbers:05.4123482100439884...

16.3482518100560115...
___________________
21.760600020?999999...By the Axiom of Choice, "There exist arbitrary infinite sequences of numbers", the two numbers being added are arbitrary, that is, not computable. In the sum, because of the series of 9's the 10th digit after the decimal is indeterminate. It could be either a 0 or a 1, depending on whether there is eventually a carry-over, which would 'cascade' to the left in the sum. The claim is that by using arithmetic (that is, an algorithm) the sum cannot be derived, and therefore infinite decimal numbers don't exist. I'm not a mathematician, but here's my thought. If by the AoC, "Arbitrary infinite sequences exist", then the sum, even if not calculable, still exists, because the AoC merely asserts the existence of certain infinite sets, not how to derive them. So the argument is faulty.

 

[fresh_42]

There are three points, which don't match up.

1. The decimal representation of numbers, is only one possibility among many. All special effects, which arise from them are restricted to this one representation to the base of ten. E.g. $\frac{1}{3} = 0.\bar{3}$ in this representation, but $\frac{1}{3} = 0.1$ if we'd chosen the base $3$. Even more abstract is the ratio between diameter and circumference of a circle, which we denote by the single letter $\pi$, whereas it has a pretty random sequence of digits to the base $10$. But the letter is as good as any sign, it's only a matter of conventions.
2. The axiom of choice has nothing to do with it at all. There aren't any choices made. The moment we define, which number we're talkig about, this is the moment we have chosen. However, not by the means of the axiom, rather by a convention which allows us to talk about a certain number.
3. What you were actually referring to is the completeness of the real numbers, i.e. the construction from rationals by adding all possible limits to them. This is qua definitionem a limit process, which has nothing to do with the axiom of choice: it is only filling the gaps. So the infinite sequences of digits are only points which can be arbitrarily closely approximated by rationals. If we want to discuss their appearance in a number system as e.g. the usual one to the base $10$, we have to agree on this limit process first.

 

[FactChecker]

The problem with insisting on finite decimal numbers:
There are so many problems where insisting on only finite decimal numbers can not be correct. Specifying any finite stopping place gives a measurable error, so no finite decimal number answer can be correct. The only alternatives are either to say that those problems are unsolvable mathematically, or to allow infinite decimal numbers. Infinite decimal numbers give good answers to some problems that finite decimal numbers can not. So anyone who denies infinite decimal numbers has a lot to justify.

 

[WWGD]

I came across the following argument that attempts to show that the notion of infinite decimal numbers is incoherent. Try adding these two numbers:05.4123482100439884...

16.3482518100560115...
___________________
21.760600020?999999...By the Axiom of Choice, "There exist arbitrary infinite sequences of numbers", the two numbers being added are arbitrary, that is, not computable. In the sum, because of the series of 9's the 10th digit after the decimal is indeterminate. It could be either a 0 or a 1, depending on whether there is eventually a carry-over, which would 'cascade' to the left in the sum. The claim is that by using arithmetic (that is, an algorithm) the sum cannot be derived, and therefore infinite decimal numbers don't exist.I'm not a mathematician, but here's my thought. If by the AoC, "Arbitrary infinite sequences exist", then the sum, even if not calculable, still exists, because the AoC merely asserts the existence of certain infinite sets, not how to derive them. So the argument is faulty.

But sums are "derived" only up to convergence, since an infinite process cannot be carried through. Maybe this is the issue you want addressed?

 

[FactChecker]

I have a question. Is there an example where infinite decimal numbers are used in the proper way and give a demonstrably incorrect answer?

 

[SSequence]

But sums are "derived" only up to convergence, since an infinite process cannot be carried through.

Unless I am missing something, I think for infinite decimal places, sums can also be carried out with high school algorithm (with small extension).

 

[mfb]

You don’t know some digits of your sum because you don’t know the two numbers you add with sufficient precision. So what?

If you add two two-digit integers and I only tell you the first digits are 4 and 5, respectively, you don’t know if the sum starts with 10 or 9 either, and nothing is infinite in this example.

 

[WWGD]

I came across the following argument that attempts to show that the notion of infinite decimal numbers is incoherent. Try adding these two numbers:05.4123482100439884...

16.3482518100560115...
___________________
21.760600020?999999...By the Axiom of Choice, "There exist arbitrary infinite sequences of numbers", the two numbers being added are arbitrary, that is, not computable. In the sum, because of the series of 9's the 10th digit after the decimal is indeterminate. It could be either a 0 or a 1, depending on whether there is eventually a carry-over, which would 'cascade' to the left in the sum. The claim is that by using arithmetic (that is, an algorithm) the sum cannot be derived, and therefore infinite decimal numbers don't exist.I'm not a mathematician, but here's my thought. If by the AoC, "Arbitrary infinite sequences exist", then the sum, even if not calculable, still exists, because the AoC merely asserts the existence of certain infinite sets, not how to derive them. So the argument is faulty.

Just to make the point that what you wrote here are not really " infinite decimal numbers" , since, strictly speaking there are _no_ infinite (standard) Real numbers. I For the sake of accuracy, I would rephrase this as :These are numbers with an infinite decimal expansion _in which there is no last non-zero term_.
Additionally, of course, you cannot add arbitrary numbers, you need to add "actual" numbers, just like you cannot, e.g., run arbitrary computer programs. I don't mean to be facetious; just making the point that in order to address these types of questions it helps a lot, to say the least, to use language that is more precise.

 

[Zafa Pi]

1) Consider all "numbers" of the form: a whole number (include 0), a decimal point, and a sequence of digits.
2) Change a tail of 9s to a tail of 0s in the usual way.
3) Define an order on the "numbers" in the obvious way (look where they 1st disagree).
4) Prove the least upper bound property (easy).
5) If x and y are two "numbers" truncate both to n digits and add.
6) The set of all such sums (n = 1,2,3...) is bounded above.
7) The least upper bound is x + y.