## Infinite-digit 'integers'

For the discussion of math. Duh.

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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:I guess my question then would be why p-adics are regarded as an alternate construction at all. Given everything I've looked at here, aren't the p-adics just a particular category of real numbers?

Well, the algebraic completion of any p-adic field is isomorphic (as a field only) to the complex numbers. So there is an injection of the p-adics into the complex numbers, but it's not canonical and can't even be discribed (it's existence requires the axiom of choice). Since the 2-adics don't contain a 4th root of 1, the 2 adics can be embedded in the reals.

The problem is, while you can do 'tricks' with some subsets of them, eg repeating expansions, there's no canonical way to relate an arbitrarily left-infinite number to a real number in a way that respects multiplication and addition.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
mike-l

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### Re: Infinite-digit 'integers'

Since you're doing this work on your own without any formal training, it might be worthwhile to explain some concepts that might be unfamiliar to you.

First, you have designed a system of numbers, which is great. And you've made a bit of an attempt to describe operations on those numbers, which you wish to consider identical to multiplication and addition. Rather than trying to explain the mechanics of these operations in gritty detail in limits to infinity, it might be more worthwhile to take the operations for granted, and then show that your set of numbers, together with your versions of multiplication and addition (call them M and A), are related in some manner to some known structure.

In other words, you have a set, we'll call them T. Then, you've got two operations, M and A. What you want is to let them be similar to some set of numbers (reals, complex, rationals, whatever) with the known analogues of multiplication and addition. So, you want to draw a relationship between (T,A,M) and (S,+,*).

In algebra, a set with two operations, that satisfies a few properties I won't go into (because they're not complicated; you can look them up) are called "rings". You can pile on some additional properties to rings to get domains, fields, etc. However, rings are more general and it makes sense to talk about them at first.

When we talk about two rings being "equivalent", we're talking about isomorphism. Mathematically, two rings are isomorphic if there is a bijective homomorphism between the respective sets. What that means is that if you put your two sets side by side, you can define some function that maps something in your set to something in the Reals (complex, rationals, integers, etc) and vice versa. In addition, this function has to have some properties, namely that multiplication and addition are respected. So if you had f : T -> R, then you need to have f(x A y) = f(x) + f(y) and f(x M y) = f(x)*f(y). In other words, adding two numbers in your set, then taking the map over to the reals is the same thing as taking the map to the reals for both numbers independently, then adding them together.

Isomorphism is very powerful, because if you can show that two rings are isomorphic, then any property that holds in one set automatically holds in the other.

This way, if you can show that your number system is isomorphic to some known number system, then you know that your number system is consistent up to the details of your operations M and A. Once you've shown that your systems are isomorphic, then you can mop up the details between the difference in the operations. This makes it a lot easier to develop the constructions you want to do, particularly when dealing with things that go out to infinity. It also eliminates the need to try to bandy about awkward arguments about notation and places, etc.
gorcee

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### Re: Infinite-digit 'integers'

gorcee wrote:
arbiteroftruth wrote:
gorcee wrote:
arbiteroftruth wrote:Hold on, I found the real problem with your objection. But first, let's solve ...99982 = 9k for k. In my system, ...99982 evaluates to -18, so k must be something that evaluates to -2. This would be ...9998. Now let's check the solution.

...9998 * 9 = 72+810+8100+... = ...99982

How do we reconcile this with your analysis of the contradiction? Well, I've found the fallacy in you original analysis. You say that every term in the sum \sum_{i=2}^{\infty} 9 \cdot 10^i is divisible by 9. This is true, however, your objection is only valid if divisibility by 9 also holds for the entire infinite sum. Your observation about the individual terms, combined with induction, demonstrates that divisibility by 9 holds true for any finite sum, but induction does not extend to infinity, only to integers. Thus the entire infinite summation does not necessarily have to give a result divisible by 9, and in my proposed system it in fact doesn't, as the result is -100.

You're trying to prove my standard mathematics, which is definitely right, with your nonstandard mathematics, which is definitely not. Using this, you can derive anything you please.

Assume that there is a term in the sequence \left\{ 9\cdot 10^i\right\}_{i=2,3,\ldots} that is not divisble by 9. However, every term in the sequence is a multiple of 9. Then, there must be a term in the sequence that is a multiple of 9 and not divisible by 9. Since every multiple of 9 must be divisible by 9, then this is a contradiction, and every term in the sequence is divisble by 9.

This is how we prove things in actual mathematics. This proof holds. Without fail. If you have constructed something that is contradictory to this proof, you have either done something wrong, or you are working with a system that is incompatible with mathematics. You can't handwave this away.

There does not have to be a term in the sequence that is not divisible by 9. Every single one of the terms in the sequence is divisible by 9. Any number of these terms summed together yields a result divisible by 9. But the infinite sum does not yield a result divisible by 9.

Your proof does not hold without fail, because your proof is based on mathematical induction, which explicitly only applies for finite iterations. I don't have to handwave it away, because your proof is flawed, because you're using induction beyond the scope in which it is valid.

My proof is absolutely not based on induction, because I did not use any induction step. All I did was applied modular arithmetic, which can be done without loss of generality, to any elements in a countably infinite set. This is the similar to the way that we show, for instance, that the group of integers is cyclic with generator 1.

This is not a countably infinite set, mirror all infinite digits over the decimal point and we receive the reals numbers that are infinitely long, as the set of all finite real numbers is surely countable, the union of the set of finite real numbers with the union of the set of infinite real numbers, MUST yield an uncountable set, Cantor claims that the union of countable sets is again a countable sets, shows us that the set of infinite real numbers must be uncountable, this carries over to infinite strands of digits
jstapleton93

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### Re: Infinite-digit 'integers'

jstapleton93 wrote:This is not a countably infinite set, mirror all infinite digits over the decimal point and we receive the reals numbers that are infinitely long, as the set of all finite real numbers is surely countable, the union of the set of finite real numbers with the union of the set of infinite real numbers, MUST yield an uncountable set, Cantor claims that the union of countable sets is again a countable sets, shows us that the set of infinite real numbers must be uncountable, this carries over to infinite strands of digits

I assume you meant to say that the set of all finite length real numbers is surely countable.

PM 2Ring

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