## Infinite-digit 'integers'

For the discussion of math. Duh.

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### Infinite-digit 'integers'

This has spun out from another thread, but the concept is this: there doesn't seem to be any contradiction that arises from taking the standard rules of arithmetic and extending them to numbers with an infinity of digits left of the decimal point. And if we make this extension, such infinite-digit numbers often equate to standard real numbers rather straightforwardly, so long as the digits eventually fall into a repeating pattern. If they don't, I'm not sure how or if they can be well defined, but let's stick with the repeating patterns for now.

In decimal arithmetic, one obvious number to consider would be ...999. If we want to evaluate this number, we can play with it using only the starting assumption that regular arithmetic still applies, and see what happens. If we add 1, what happens? All the 9s roll over to 0s. For any finite string of 9s, there would be a 1 introduced in a new left-most position, but for an infinite string, there is no leftmost position, so there is never a place for the 1 to exist, there is only ever another 9 that gets rolled over to 0 as the 1 gets carried. So there is never a 1 in any digit position, and the result is simply an infinite string of zeroes, which would most reasonably be interpreted as simply 0, particularly if we still regard digits as coefficients on powers of 10. So ...999 + 1 = 0, therefore ...999 = -1.

But can we play around with this notion and get a contradiction? At least at first glance, it doesn't seem so. For example, starting from ...9999, we can get ...9998 in 2 different simple ways. We can simply subtract 1, or we can multiply by 10 to get ...9990, then add 8. ...9999 - 1 = -1 -1 = -2. (...9999 * 10) + 8 = ((-1) x 10) + 8 = -10 + 8 = -2. We get the same result either way, so there's no contradiction so far. I've tried many such examples, and it's always worked out so far.

But suppose we have a pattern longer than simply a single repeating digit. Let's try, for example, ...037037037. If we multiply by 3, we get ...111. Multiplying by 9 gets us to the known ...999 = -1, so we know that ...037037037 = -1/27. The standard decimal representation of that would be -0.037037037...

More generally, it seems that any repeating pattern of digits to the left corresponds to the exact same pattern on the right side of the decimal for a negative number. We've already seen ...999 = -1 = -0.999..., ...111 = -1/9 = -0.111..., and ...037037 = -1/27 = -0.037037... The pattern always seems to hold. In fact, given that we're saying the standard rules of arithmetic must hold, and all the decimal representations of fractions relate to each other given those rules, we can know for sure that applying those rules to infinite-digit integers will result in the same relative patterns.

From this, we can deduce a general rule for evaluating these numbers. For a repeating infinite integer with pattern of length n, represented as
...an-1an-2...a2a1a0,
the equivalent standard decimal representation is
-0.an-1an-2...a2a1a0...
Or, equivalently, the number is the rational number
-an-1an-2...a2a1a0/(10n - 1)
which is also equal to
an-1an-2...a2a1a0/(1 - 10n)

This result, as it happens, is identical to extending the equation for summing geometric series to allow values of r greater than 1. Namely, the sum of
a * rk for all k from 0 to infinity equals a/(1 - r) In this case, r is the base of your number system to the power of the number of digit positions necessary to establish the pattern, and a is the number represented by a single iteration of that pattern.

So now we know a general rule for evaluating these numbers that applies regardless of the base of your system, so long as the digits fall into a repeating pattern. This means there are a few things we can't do though. A terminating decimal has no equivalent infinite-digit integer representation, and a non-repeating infinite-digit integer has no equivalent standard decimal representation, at least not using the rules I've figured out so far. A terminating decimal can at least be approximated arbitrarily closely by using an arbitrarily long repeating pattern with lots of 0s, or by changing the base of your number system to turn it into a non-terminating decimal. But a non-repeating infinite-digit integer is a bigger problem that I haven't figured out yet.

arbiteroftruth

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### Re: Infinite-digit 'integers'

and the result is simply an infinite string of zeroes, which would most reasonably be interpreted as simply 0

This would be true if the "rolling over" stopped. Which it does not. So this is false.

Think about it like this -- every time you "roll over to 0 and carry the 1", it takes 1 minute. However, a few seconds after the last such operation, you know there's another operation coming, so you're not done yet! You can't determine what the number is until the operation finishes. Since the operation never finishes, you cannot determine what the number is, so your assumption that it's an infinite string of zeros is false --- we just haven't had the time required yet to write down the 1, but it's coming, I promise!
gorcee

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### Re: Infinite-digit 'integers'

gorcee wrote:
and the result is simply an infinite string of zeroes, which would most reasonably be interpreted as simply 0

This would be true if the "rolling over" stopped. Which it does not. So this is false.

Think about it like this -- every time you "roll over to 0 and carry the 1", it takes 1 minute. However, a few seconds after the last such operation, you know there's another operation coming, so you're not done yet! You can't determine what the number is until the operation finishes. Since the operation never finishes, you cannot determine what the number is, so your assumption that it's an infinite string of zeros is false --- we just haven't had the time required yet to write down the 1, but it's coming, I promise!

Your reasoning is equivalent to saying that 1-0.999... must eventually yield a 1 on the far right that we just haven't gotten to yet. Rather, we can determine from the pattern of the computation that the hypothetical 1 will never have a legitimate position to hold, so it doesn't exist. ...999 + 1 = 0 for the same reason that 1 - 0.999... = 0
arbiteroftruth

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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:
gorcee wrote:
and the result is simply an infinite string of zeroes, which would most reasonably be interpreted as simply 0

This would be true if the "rolling over" stopped. Which it does not. So this is false.

Think about it like this -- every time you "roll over to 0 and carry the 1", it takes 1 minute. However, a few seconds after the last such operation, you know there's another operation coming, so you're not done yet! You can't determine what the number is until the operation finishes. Since the operation never finishes, you cannot determine what the number is, so your assumption that it's an infinite string of zeros is false --- we just haven't had the time required yet to write down the 1, but it's coming, I promise!

Your reasoning is equivalent to saying that 1-0.999... must eventually yield a 1 on the far right that we just haven't gotten to yet. Rather, we can determine from the pattern of the computation that the hypothetical 1 will never have a legitimate position to hold, so it doesn't exist. ...999 + 1 = 0 for the same reason that 1 - 0.999... = 0

No, it's not. Because we don't add the .1 starting from the left and go right. And we can't start from the rightmost position, because there is no rightmost position. We need to use other intuition to prove it.

In your ..99999 example, we're starting from the right and going left.
gorcee

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### Re: Infinite-digit 'integers'

gorcee wrote:No, it's not. Because we don't add the .1 starting from the left and go right.

In your ..99999 example, we're starting from the right and going left.

Let's compute ...9999 + 1 step by step.

First we get ...9990 with a carried 10.

Carrying the 10 gives us ...9900 with a carried 100.

Carrying against gives ...999000 with a carried 1000, and so on.

Now let's compute 1 - 0.999... step by step.

First we subtract the tenths digit, and get 0.1 with 0.0999... left to subtract.

Then we subtract the hundredths digit and get 0.01 with 0.00999... left to subtract.

Then we subtract the thousandths digit and get 0.001 with 0.000999... left to subtract, and so on.

The fact that the numbers in the two examples extend in opposite directions is taken care of by the fact that computing addition and computing subtraction can be done by dealing with the digits in opposite directions.
arbiteroftruth

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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:
gorcee wrote:No, it's not. Because we don't add the .1 starting from the left and go right.

In your ..99999 example, we're starting from the right and going left.

Let's compute ...9999 + 1 step by step.

First we get ...9990 with a carried 10.

Carrying the 10 gives us ...9900 with a carried 100.

Carrying against gives ...999000 with a carried 1000, and so on.

Now let's compute 1 - 0.999... step by step.

First we subtract the tenths digit, and get 0.1 with 0.0999... left to subtract.

Then we subtract the hundredths digit and get 0.01 with 0.00999... left to subtract.

Then we subtract the thousandths digit and get 0.001 with 0.000999... left to subtract, and so on.

The fact that the numbers in the two examples extend in opposite directions is taken care of by the fact that computing addition and computing subtraction can be done by dealing with the digits in opposite directions.

The thing you're doing with 1- 0.999... isn't a proof that 1 = 0.999.... In fact, it's a commonly used misconception that 1 =/= 0.999.... The argument typically goes just like yours, and then says "since we can never terminate this process, there will always be a 1 in the n+1th decimal place, therefore it can't be zero".

This intuition is wrong, and similarly, your intuition with ......9999999999999 = 0 is wrong.
gorcee

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### Re: Infinite-digit 'integers'

gorcee wrote:The thing you're doing with 1- 0.999... isn't a proof that 1 = 0.999.... In fact, it's a commonly used misconception that 1 =/= 0.999.... The argument typically goes just like yours, and then says "since we can never terminate this process, there will always be a 1 in the n+1th decimal place, therefore it can't be zero".

This intuition is wrong, and similarly, your intuition with ......9999999999999 = 0 is wrong.

You seem to have forgotten that my entire point was that you were looking at ...999 + 1 similarly to a mistaken way of looking at 1 - 0.999 = 0. The way you conclude the inequality of ...999 and -1 is exactly the same mistake as the way people conclude the inequality of 0.999... and 1 because of the non-terminating process.

You literally made the exact same type of argument that "there will always be a 1 in the n+1th position, therefore it can't be 0".

And just to reiterate, I was saying ...999 = -1, not 0.
arbiteroftruth

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### Re: Infinite-digit 'integers'

If you prefer, here's another way to look at it.

...9999 + 1

The ones digit rolls over to 0, leaving ...9990 + 10(carried)

Then the hundreds digit rolls over to 0, leaving ...99900 + 100(carried)

But another way of writing ...9990 + 10 is 10 * (...999 + 1), and another way of writing ...99900 + 100 is 100 * (...999 + 1)

So let's assign ...999 + 1 = x and take advantage of these equalities.

x = 10 * x = 100 *x = ...

9 * x = 0

x = 0

So ...999 + 1 = x = 0, so ...999 = -1.

EDIT: And the same type of algebraic technique can be used to escape the typical error regarding 1 - 0.999... You subtract the tenths first, giving 0.1 - 0.0999..., which is equal to 0.1 * (1- 0.999...), so x = 0.1 * x, 0.9 * x = 0, x = 0.
Last edited by arbiteroftruth on Sat Mar 24, 2012 8:27 pm UTC, edited 1 time in total.
arbiteroftruth

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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:
gorcee wrote:The thing you're doing with 1- 0.999... isn't a proof that 1 = 0.999.... In fact, it's a commonly used misconception that 1 =/= 0.999.... The argument typically goes just like yours, and then says "since we can never terminate this process, there will always be a 1 in the n+1th decimal place, therefore it can't be zero".

This intuition is wrong, and similarly, your intuition with ......9999999999999 = 0 is wrong.

You seem to have forgotten that my entire point was that you were looking at ...999 + 1 similarly to a mistaken way of looking at 1 - 0.999 = 0. The way you conclude the inequality of ...999 and -1 is exactly the same mistake as the way people conclude the inequality of 0.999... and 1 because of the non-terminating process.

You literally made the exact same type of argument that "there will always be a 1 in the n+1th position, therefore it can't be 0".

And just to reiterate, I was saying ...999 = -1, not 0.

No, I'm really not saying that. What I'm saying is that your "conclusions" are equivalent to that. I'm demonstrating by contradiction that you're wrong.

If you think that ...999 = -1, then -9 = 9 * -1 = 9 * (...9999) = ...9991 = (...9999 * 10) + 1

Since -9 \equiv 0 mod 9 and 1 \equiv 1 mod 9, we must have ...9990 \equiv 8 mod 9. Then, ...99982 is divisible by 9. Since ...99982 can be written as 82 + \sum_{i=2}^{\infty} 9 \cdot 10^i, and every term in the sum is divisble by 9, then 82 must be divisble by 9. It is not, therefore the assumption is false.
gorcee

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### Re: Infinite-digit 'integers'

gorcee wrote:No, I'm really not saying that. What I'm saying is that your "conclusions" are equivalent to that. I'm demonstrating by contradiction that you're wrong.

If you think that ...999 = -1, then -9 = 9 * -1 = 9 * (...9999) = ...9991 = (...9999 * 10) + 1

Since -9 \equiv 0 mod 9 and 1 \equiv 1 mod 9, we must have ...9990 \equiv 8 mod 9. Then, ...99982 is divisible by 9. Since ...99982 can be written as 82 + \sum_{i=2}^{\infty} 9 \cdot 10^i, and every term in the sum is divisble by 9, then 82 must be divisble by 9. It is not, therefore the assumption is false.

82 is divisible by 9. It's 82/9.

If you object that divisibility implies divisibility in a whole-numbered format, I can do that too. 82/9 = 81/9 + 1/9 = 9 + 1/9.

In my proposed interpretation of infinite-digited 'integers', 1/9 can be represented as ...88889. So 1/9 + 9 = ...8889 + 9 = ...88898 = 82/9

So it only becomes impossible to make the division work with these numbers if you reject using these numbers in the first place. Thus the contradiction only exists if you reject the system to begin with, and contradiction only goes away if you accept the system to begin with. Meaning the entire issue is axiomatic either way. Given that the entire point of this system is to axiomatically declare numbers of this form to be acceptable, your contradiction ceases to hold.
arbiteroftruth

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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:
gorcee wrote:No, I'm really not saying that. What I'm saying is that your "conclusions" are equivalent to that. I'm demonstrating by contradiction that you're wrong.

If you think that ...999 = -1, then -9 = 9 * -1 = 9 * (...9999) = ...9991 = (...9999 * 10) + 1

Since -9 \equiv 0 mod 9 and 1 \equiv 1 mod 9, we must have ...9990 \equiv 8 mod 9. Then, ...99982 is divisible by 9. Since ...99982 can be written as 82 + \sum_{i=2}^{\infty} 9 \cdot 10^i, and every term in the sum is divisble by 9, then 82 must be divisble by 9. It is not, therefore the assumption is false.

82 is divisible by 9. It's 82/9.

If you object that divisibility implies divisibility in a whole-numbered format, I can do that too. 82/9 = 81/9 + 1/9 = 9 + 1/9.

In my proposed interpretation of infinite-digited 'integers', 1/9 can be represented as ...88889. So 1/9 + 9 = ...8889 + 9 = ...88898 = 82/9

So it only becomes impossible to make the division work with these numbers if you reject using these numbers in the first place. Thus the contradiction only exists if you reject the system to begin with, and contradiction only goes away if you accept the system to begin with. Meaning the entire issue is axiomatic either way. Given that the entire point of this system is to axiomatically declare numbers of this form to be acceptable, your contradiction ceases to hold.

You seriously need to spend more time studying actual number theory, and less time spending inventing new number theory.

The statement that 82 must be divisible by 9 comes from modular arithmetic identities, which presuppose integer relations. ...99990 \equiv 8 mod 9 implies that ...99990 = 9k + 8 for any integer k. This means that ....99990-8 = ....99982 = 9k. You can't just make k not an integer anymore.
gorcee

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### Re: Infinite-digit 'integers'

gorcee wrote:You seriously need to spend more time studying actual number theory, and less time spending inventing new number theory.

The statement that 82 must be divisible by 9 comes from modular arithmetic identities, which presuppose integer relations. ...99990 \equiv 8 mod 9 implies that ...99990 = 9k + 8 for any integer k. This means that ....99990-8 = ....99982 = 9k. You can't just make k not an integer anymore.

Um, yes I can. That is, in fact, a direct consequence of allowing these numbers to have infinite digits. An integer must have a finite number of digits. So in this system, where k would have infinite digits, k is definitionally not an integer, so the identity you cite no longer needs to hold.

EDIT: But thank you for providing a helpful explanation for why your objection isn't a matter of axioms at all. Your objection is based on an identity that only holds for integers, and I'm not talking about integers to begin with, so the objection is simply invalid.
Last edited by arbiteroftruth on Sat Mar 24, 2012 8:51 pm UTC, edited 1 time in total.
arbiteroftruth

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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:
gorcee wrote:You seriously need to spend more time studying actual number theory, and less time spending inventing new number theory.

The statement that 82 must be divisible by 9 comes from modular arithmetic identities, which presuppose integer relations. ...99990 \equiv 8 mod 9 implies that ...99990 = 9k + 8 for any integer k. This means that ....99990-8 = ....99982 = 9k. You can't just make k not an integer anymore.

Um, yes I can. That is, in fact, a direct consequence of allowing these numbers to have infinite digits. An integer must have a finite number of digits. So in this system, where k would be ...1110, k is definitionally not an integer, so the identity you cite no longer needs to hold.

The identity absolutely needs to hold if you want your system to jive with anything else in mathematics. Either way, if your argument is that k isn't an integer, then yes, you're right. Under this construction, k is not an integer. But this does not mean the identity no longer holds. Instead, it proves that the assumptions used must be false. This is the essence of proof by contradiction.

If these identities don't hold, then what you have created is incompatible with mathematics. In which case, take this thread over to the fictional science section, because you're now just making up rules as you go along.

In actual mathematics, if I come to the conclusion that 82 is divisible by 9, then I don't all of a sudden get a situation where I have proven modular arithmetic wrong. I get a situation where I am wrong.
gorcee

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### Re: Infinite-digit 'integers'

gorcee wrote:
arbiteroftruth wrote:
gorcee wrote:You seriously need to spend more time studying actual number theory, and less time spending inventing new number theory.

The statement that 82 must be divisible by 9 comes from modular arithmetic identities, which presuppose integer relations. ...99990 \equiv 8 mod 9 implies that ...99990 = 9k + 8 for any integer k. This means that ....99990-8 = ....99982 = 9k. You can't just make k not an integer anymore.

Um, yes I can. That is, in fact, a direct consequence of allowing these numbers to have infinite digits. An integer must have a finite number of digits. So in this system, where k would be ...1110, k is definitionally not an integer, so the identity you cite no longer needs to hold.

The identity absolutely needs to hold if you want your system to jive with anything else in mathematics. Either way, if your argument is that k isn't an integer, then yes, you're right. Under this construction, k is not an integer. But this does not mean the identity no longer holds. Instead, it proves that the assumptions used must be false. This is the essence of proof by contradiction.

If these identities don't hold, then what you have created is incompatible with mathematics. In which case, take this thread over to the fictional science section, because you're now just making up rules as you go along.

In actual mathematics, if I come to the conclusion that 82 is divisible by 9, then I don't all of a sudden get a situation where I have proven modular arithmetic wrong. I get a situation where I am wrong.

Or, horror of horrors, you've discovered the mysterious number 82/9. This number is not particularly relevant to the property of modular arithmetic you're discussing, because the property you're discussing deals with integers, but that doesn't make 82/9 some sort of false result, it just makes 82/9 not an integer. Similarly, numbers with infinite-digits may or may not be able to satisfy this identity, but they're not integers, so it doesn't really matter.
arbiteroftruth

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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:Or, horror of horrors, you've discovered the mysterious number 82/9. This number is not particularly relevant to the property of modular arithmetic you're discussing, because the property you're discussing deals with integers, but that doesn't make 82/9 some sort of false result, it just makes 82/9 not an integer. Similarly, numbers with infinite-digits may or may not be able to satisfy this identity, but they're not integers, so it doesn't really matter.

Look, this is an immutable law of mathematics:

If a\equiv b \mbox{ mod } n and c\equiv d \mbox{ mod } n, then a+c\equiv b+d \mbox{ mod } n.

This is another immutable law of mathematics:

If a\equiv b \mbox{ mod } n, then a = kn + b for integer k, and some 0 <= b < n.

You can't axiomatically declare these false, and expect to be consistent with mathematics. k is an integer. Anything property of mathematics you find must be consistent with these properties. k must always be an integer. Always. It can never not be an integer. This is a fact.

If you get 90 \equiv 8 mod 9, then you have 90 = 9k + 8 for some integer k. Then, 82 = 9k for some integer k. Since we clearly do not have any integer k such that 9k = 82, then something must be false. What could it be? Ah! It's that 90 \equiv 8 mod 9 is false! You can't have both 90 \equiv 8 mod 9 and 82 = 9k for integer k. They can't both be true. Pick one to be true.

If you pick 82=9k for integer k is not true, then your system is incompatible with mathematics.
If you pick 90 \equiv 8 mod 9 is not true, then your system is wrong.
If you try to claim that both are true, then you invent some new integer k with these properties, and your system is incompatible with mathematics.
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### Re: Infinite-digit 'integers'

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 your 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.
arbiteroftruth

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### Re: Infinite-digit 'integers'

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.
gorcee

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### Re: Infinite-digit 'integers'

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.
arbiteroftruth

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### Re: Infinite-digit 'integers'

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.
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### Re: Infinite-digit 'integers'

gorcee wrote: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.

I claimed that the infinite summation does not have to be divisible by 9. You argued that if this were the case, there must exist some term in the summation that is not divisible by 9. Support this claim without using induction.

EDIT: Technically, you didn't claim that, but if that wasn't your implicit argument, then I see no reason whatsoever for having gone through the exercise of assuming there was some term not divisible by 9 and showing how that leads to contradiction.
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### Re: Infinite-digit 'integers'

gorcee, I don't think this is as meaningless as you're claiming. I've seen lots of people trying to do the things you criticize by forgetting to distinguish between these left-infinite things and integers, but I don't really see that here so far. Let me try to make this a little more formal.

Consider the set X of all eventually-periodic infinite sequences of digits, equipped with addition and multiplication which act as with normal integer addition with the sequence read right-to-left.

Assertion 1: (X,+,*) defines a commutative ring with identity: i.e., + and * are closed on X, commutative, distributive, identities and + inverses exist. [I haven't proved any of these assertions, though most of them look trivial. I'm just formalizing what I think arbiteroftruth is trying to get at.]

Note that this is pure abstract algebra; I haven't yet tried to claim anything about Z or Q. There's nothing particularly weird here. Now (if assertion 1 holds) define a ring homomorphism from X into Q. There are a few things we need: additive identity to 0, multiplicative identity to 1.

Assertion 2: The evaluation in the first post defines such a homomorphism. [Again, I haven't tried to prove this, I'm just trying to formalize the assertions.]

Assertion 3: This homomorphism is in fact uniquely specified by the ring-homomorphism requirements.

These constructions, by the way, are basically the 10-adic numbers; in base p they're the p-adic numbers.

--

As long as arbiteroftruth doesn't forget what ring he's working in, this just seems like playing around with algebra.

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### Re: Infinite-digit 'integers'

Edit: ninja'd
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### Re: Infinite-digit 'integers'

eta oin shrdlu wrote:gorcee, I don't think this is as meaningless as you're claiming. I've seen lots of people trying to do the things you criticize by forgetting to distinguish between these left-infinite things and integers, but I don't really see that here so far. Let me try to make this a little more formal.

Consider the set X of all eventually-periodic infinite sequences of digits, equipped with addition and multiplication which act as with normal integer addition with the sequence read right-to-left.

Assertion 1: (X,+,*) defines a commutative ring with identity: i.e., + and * are closed on X, commutative, distributive, identities and + inverses exist. [I haven't proved any of these assertions, though most of them look trivial. I'm just formalizing what I think arbiteroftruth is trying to get at.]

Note that this is pure abstract algebra; I haven't yet tried to claim anything about Z or Q. There's nothing particularly weird here. Now (if assertion 1 holds) define a ring homomorphism from X into Q. There are a few things we need: additive identity to 0, multiplicative identity to 1.

Assertion 2: The evaluation in the first post defines such a homomorphism. [Again, I haven't tried to prove this, I'm just trying to formalize the assertions.]

Assertion 3: This homomorphism is in fact uniquely specified by the ring-homomorphism requirements.

These constructions, by the way, are basically the 10-adic numbers; in base p they're the p-adic numbers.

--

As long as arbiteroftruth doesn't forget what ring he's working in, this just seems like playing around with algebra.

I can't claim to be well-versed enough in number theory to know what all of your assertions really mean, but as far as I do understand it, that looks like what I'm saying.

And though there are definite similarites to p-adics, there are also differences insofar as that wikipedia article is accurate. Just to give one example, there would be no equivalent leftward representation of a terminating decimal in my system. Only repeating decimals have an equivalent leftward representation.(meaning that although the implicit 0s after a terminating decimal can be represented as leftward 0s, or just 0, the non-repeating decimals before that cannot)
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### Re: Infinite-digit 'integers'

I got what you're saying for the p-adic numbers from his previous thread. However, I still think there's some inconsistency with how he's choosing to define it. Basically, he's attempting the construction as in the wikipedia article, while simultaneously rejecting the multiplicity of integers.

Edit #2: From Wikipedia:
However, the ring of 10-adic numbers have one major drawback. It is possible to find pairs of non-zero 10-adic numbers whose product is 0. In other words, the ring of 10-adic numbers is not an integral domain because they contain zero divisors.

This issue addresses the concern I raised earlier with the modular arithmetic. Just replace the addition operation with the multiplication operation, and you'll see where/how those such issues arise.
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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:I don't follow. How does it remove well-ordering? Perhaps I'm not understanding something, but for any 2 numbers in this system, you can easily subtract them, see whether the result is greater than, equal to, or less than 0, and conclusively define which number is larger, if either.
That's not what well-ordering means. Well-ordering means that any nonempty set has a least element. (So yeah, granted, the integers as a whole don't have this property either, but the natural numbers do.)

arbiteroftruth wrote: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.
Hold on. Isn't this the very "paradox" you used to start your other thread? Perceived "discontinuities" where something that held true for every finite example suddenly ceased in the limit at infinity?

Except, there actually was no problem before, since you were simply confusing sets with their elements, but there really is a problem now, unless you're just declaring by fiat that your new numbers are never congruent to anything modulo something else. (Either that, or you're denying that continuing to add 0 (mod 9) will eventually get you something which is not 0 (mod 9).)
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gmalivuk
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### Re: Infinite-digit 'integers'

I think that arbiter is attempting to construct the 10-adic numbers and claim that they are isomorphic to the field domain of integers, or perhaps attempt a "correction" of the 10-adic numbers that attempts to bypass the issue of zero divisors. However, his argument in doing so is "bwaha! I declare those multiplicative properties to not hold, because infinity!"

His realization of the 10-adic numbers is clever enough, but by stretching too hard to make them something they're not, opened up to many fallacies.
Last edited by gorcee on Sun Mar 25, 2012 12:36 am UTC, edited 1 time in total.
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### Re: Infinite-digit 'integers'

I got what you're saying for the p-adic numbers from his previous thread. However, I still think there's some inconsistency with how he's choosing to define it. Basically, he's attempting the construction as in the wikipedia article, while simultaneously rejecting the multiplicity of integers.

Edit #2: From Wikipedia:
However, the ring of 10-adic numbers have one major drawback. It is possible to find pairs of non-zero 10-adic numbers whose product is 0. In other words, the ring of 10-adic numbers is not an integral domain because they contain zero divisors.

This issue addresses the concern I raised earlier with the modular arithmetic. Just replace the addition operation with the multiplication operation, and you'll see where/how those such issues arise.

Can you clarify these objections?

As for multiplying to get 0, I can tell pretty quickly that a real proof would need to be rather extensive, but after looking at it some it looks like any algorithm for producing two sequences of digits that multiply to return 0 in all digit positions would fail to fall into a repeating pattern, and thus not be a valid number within this system. Each time you move one more digit to the left, the number of digit combinations that multiply to that position increases, and additionally you must always consider the carried value from the previous multiplication. Some recursive algorithm could ensure that you always get digits that work, but I doubt any such algorithm could result in a repeating pattern of digits.
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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:

I got what you're saying for the p-adic numbers from his previous thread. However, I still think there's some inconsistency with how he's choosing to define it. Basically, he's attempting the construction as in the wikipedia article, while simultaneously rejecting the multiplicity of integers.

Edit #2: From Wikipedia:
However, the ring of 10-adic numbers have one major drawback. It is possible to find pairs of non-zero 10-adic numbers whose product is 0. In other words, the ring of 10-adic numbers is not an integral domain because they contain zero divisors.

This issue addresses the concern I raised earlier with the modular arithmetic. Just replace the addition operation with the multiplication operation, and you'll see where/how those such issues arise.

Can you clarify these objections?

As for multiplying to get 0, I can tell pretty quickly that a real proof would need to be rather extensive, but after looking at it some it looks like any algorithm for producing two sequences of digits that multiply to return 0 in all digit positions would fail to fall into a repeating pattern, and thus not be a valid number within this system. Each time you move one more digit to the left, the number of digit combinations that multiply to that position increases, and additionally you must always consider the carried value from the previous multiplication. Some recursive algorithm could ensure that you always get digits that work, but I doubt any such algorithm could result in a repeating pattern of digits.

If a \equiv b \mbox{ mod } n and c \equiv d \mbox{ mod } n, then ac \equiv bd \mbox{ mod } n. If you have some nonzero a,c such that ac = 0, then 0 \equiv bd \mbox{ mod } n, implying that bd = 0.

However, a = k_1n + b, c = k_2n+d, so ac = (k_1n+b)(k_2n+d)= k_1k_2n^2+k_1dn+k_2bn+bd = 0. Since bd = 0, we can restrict our attention to k_1k_2n^2+k_1dn+k_2bn = 0. If bd = 0, then we can have b=0, d=0, or both, leaving three cases to investigate.

If b,d = 0, then k_1k_2n = 0, implying that one of k_1, k_2 is zero, meaning that either a or c are zero (since b and d are both 0), which contradicts the assumption of nonzero a, c.

If b = 0, then k_1k_2n+k_1d = k_1(k_2n+d) = 0. Since k_1 can't be zero (because b is zero), then k_2n + d = 0 \rightarrow k_2 = -d/n. However, n > d > 0, so |d/n| < 1, which contradicts that k_2 must be an integer.

A similar argument is made for d=0.
Last edited by gorcee on Sat Mar 24, 2012 11:16 pm UTC, edited 1 time in total.
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### Re: Infinite-digit 'integers'

gmalivuk wrote:Hold on. Isn't this the very "paradox" you used to start your other thread? Perceived "discontinuities" where something that held true for every finite example suddenly ceased in the limit at infinity?

Yeah, but that's why I made a separate thread. I've reconciled that "paradox" with my intuition and moved to a separate idea that came out of that thought process.

gmalivuk wrote:but there really is a problem now, unless you're just declaring by fiat that your new numbers are never congruent to anything modulo something else. (Either that, or you're denying that continuing to add 0 (mod 9) will eventually get you something which is not 0 (mod 9).)

I think this issue is closely related to the way this system changes depending on your base. A decimal number system is a summation of powers of 10, a binary system is a summation of powers of 2, etc. When talking about simply incrementing a constant value, what you are doing is (essentially) using a unary system, just with 9s(in this case) instead of 1s.

The thing is, unary is a particularly unique case in this type of system. There is no distinction between digit positions, so there's no situation where all the digits roll over to zeroes or any equivalent situation. There is no equivalent object to a decimal point with less significant digits on the right of it, so no way to define that type of interpretation of the left-infinite numbers. For all other bases, the arithmetic works out to be an extension of the sum of geometric series, which when the base is 1, results in division by 0 and is undefined.

For all these reasons, there is no meaningful way in this system to define the behavior of incrementation as it becomes infinite.

In the scenario gorcee brings up though, we're still not just incrementing by 9s. ...999900 is not a sum of incremented 9s, but a sum of 9 * 10k for incremented values of k, so it would be more sensible to say that it is the nature of powers of 10 that changes "at infinity", and the nature and reasoning of that change is described exactly by applying decimal arithmetic as I have done.

As for the issue of modular congruence: a given left-infinite number can still be evaluated as congruent to something modulo n, it's just that certain methods of computing that value cease to work. One method that should still work is to divide by the modulus and take the remainder, which is easily done by dividing from right to left. For example ...99999 (mod 9). Divide by 9 from right to left, using the highest possible value for each digit. The rightmost digits are 99, so we use 9 * 9 = 81 to get rid of as much as possible, leaving ...99918. From The next pair of digits is 91, so we use another 9 to subtract 810 and leave ...99108. The next digit lets us subtract 8100 and leave ...9991008. Continuing to infinity, all digits of the quotient are 9s, the 0s get pushed out to infinity, and only the 8 in the ones place is left, and that is the remainder. So the division result is ...999 r8, meaning ...999 is congruent to 8 (mod 9), and is found on the ...999th cycle(which is the -1st cycle).

The only things you can't do are to figure out modular congruence based on continued incrementation of the modulus(because that's essentially a unary system and will conflict with the nature of whatever base you're using), and you can't apply concepts of divisibility to infinite digits (because such numbers cease to be integers, so divisibility is not well-defined).
Last edited by arbiteroftruth on Sun Mar 25, 2012 12:03 am UTC, edited 2 times in total.
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### Re: Infinite-digit 'integers'

I don't see how you can simultaneously claim that ...999 is not an integer, because it has an infinite representation, and that ...9999 = -1.

Either you are claiming that representation by infinite "places" yields something that is not an integer, or representation by infinite places can yield something like -1, which is an integer.

Again, to claim one thing, you have to admit the other. So if you claim that ...9999 = -1, then it is an integer, and all the other happy integer things such as the division algorithm and congruence relations must hold.
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### Re: Infinite-digit 'integers'

gorcee wrote:I don't see how you can simultaneously claim that ...999 is not an integer, because it has an infinite representation, and that ...9999 = -1.

Either you are claiming that representation by infinite "places" yields something that is not an integer, or representation by infinite places can yield something like -1, which is an integer.

Again, to claim one thing, you have to admit the other. So if you claim that ...9999 = -1, then it is an integer, and all the other happy integer things such as the division algorithm and congruence relations must hold.

Well, I was speaking about the general case of left-infinite numbers. It's like saying that certain integer rules don't hold for decimals because the decimals aren't integers, but in the special cases of repeating 9s and repeating 0s, decimals are integers. Same thing. A left-infinite repeating decimal is not an integer unless the repeating pattern is a string of infinite 9s or a string of infinite 0s.

As it pertains to modular math, the point is that rules for divisibility that depend on the values of digits in various positions do not hold. Because then you're not just looking at the properties of the number itself, but also the properties of the decimal representation. In this case, the decimal representation is shorthand for the sum an infinite series of 9 * 10k. Although the result of this series is still an integer, the effective coefficient on 9 is not. That is, the 9 can be pulled outside the summation, meaning the value is 9 * sum(k,0->inf,10k). 9 is an integer, and the final value is an integer, but the coefficient applied to 9 ceases to be an integer, as it is -1/9.

In general, any left-repeating number can be represented similarly.
...a1a0an-1an-2...a1a0 =
an-1an-2...a1a0 * sum(k,0->inf,10n * k)

The infinite summation in that equation is never an integer. That is why rules of divisibility do not hold. There is, essentially, an implied denominator that already divides out all the 9s that the number looks like it should be divisible by. (as the summation evaluates to 1/(1-10n) ).
arbiteroftruth

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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:
gorcee wrote:I don't see how you can simultaneously claim that ...999 is not an integer, because it has an infinite representation, and that ...9999 = -1.

Either you are claiming that representation by infinite "places" yields something that is not an integer, or representation by infinite places can yield something like -1, which is an integer.

Again, to claim one thing, you have to admit the other. So if you claim that ...9999 = -1, then it is an integer, and all the other happy integer things such as the division algorithm and congruence relations must hold.

Well, I was speaking about the general case of left-infinite numbers. It's like saying that certain integer rules don't hold for decimals because the decimals aren't integers, but in the special cases of repeating 9s and repeating 0s, decimals are integers. Same thing. A left-infinite repeating decimal is not an integer unless the repeating pattern is a string of infinite 9s or a string of infinite 0s.

The integers you're describing are just the 10-adic numbers. Let's ignore, for a moment, the rest of your infinite expansion representation.

The integers form a domain, which is well-known. What you need to do to show that your representation is consistent, at least for integers, is show that your representation also forms a domain. Your numbers form a commutative ring with unity, which is also fairly straightforward. What you need to show is that there are no zero divisors in your arithmetic.
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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:a non-repeating infinite-digit integer has no equivalent standard decimal representation, at least not using the rules I've figured out so far.
But why not just admit you're essentially describing the 10-adics, and use the rules people already figured out for them?
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gmalivuk
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### Re: Infinite-digit 'integers'

gmalivuk wrote:
arbiteroftruth wrote:a non-repeating infinite-digit integer has no equivalent standard decimal representation, at least not using the rules I've figured out so far.
But why not just admit you're essentially describing the 10-adics, and use the rules people already figured out for them?

Right, in which case many of the things you've said become a lot more clear. For instance, rather than talking about places and positions and handwaving around, you can claim that if you assume that ...999 = -1, then you could, for instance generate the group of integers with ...999. Then you can do things like show isomorphism under addition. Now that I look at your description in this light, and I filter out all the stuff you've talking about positions and repeating representations and whatnot, it becomes much more clear. I'm still not sold on isomorphism between them and the domain of integers, however.
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### Re: Infinite-digit 'integers'

gorcee wrote:The integers you're describing are just the 10-adic numbers. Let's ignore, for a moment, the rest of your infinite expansion representation.

The integers form a domain, which is well-known. What you need to do to show that your representation is consistent, at least for integers, is show that your representation also forms a domain. Your numbers form a commutative ring with unity, which is also fairly straightforward. What you need to show is that there are no zero divisors in your arithmetic.

As I said before, though there are definite similarities to the p-adics, there are also definite differences, at least if wikipedia is to be trusted. A terminating standard decimal has an equivalent p-adic representation, but no representation in my system of left-infinite 'integers'. Also, non-repeating p-adic numbers have equivalent decimal representations, but non-repeating left-infinite 'integers' are undefined.

As for zero divisors, as I said before, I haven't gone through a proof yet, but looking at it so far makes me suspect that any potential zero divisor would be non-repeating, and thus undefined. Suppose we're multiplying a pair of left-infinite numbers
...an-1an-2...a1a0 and ...bm-1bm-2...b1b0

The first digit of the result, d0, is determined entirely by a0 and b0, which must multiply to a non-zero positive integer multiple of the base of your number system. It must be non-zero because if it is zero, then you either end up with one of your factors being 0, or you simply start using non-zero multiples at a later digit position, a problem which reduces down to this one. Because once you use a single non-zero multiple of your base, you have a carried digit, and all later multiplications have to get rid of the carry digit by adding up with the carry digit to some other non-zero multiple of your base. It must be positive, because I'm assuming a standard number system(not covering cases like balanced ternary just yet). It must be an integer, because we're dealing with finitely many digits at a time, and all digits have values at most n-1 for a base n number system.

If the base of your number system if prime, then the situation is simple. Any non-zero multiple of a prime must include the prime itself as one of its factors, but the highest valid digit is 1 less than the base, so the digits a0 and b0 cannot possibly multiply to a multiple of the base.

A non-prime base is more complex. The first digit of the result, d0, is determined entirely by a0 and b0, which must cause d0 to be 0, and create a carry digit for the next position.

d1 = a0*b1 + a1*b0 + k1(the carry digit for this position) (mod n)

And in general:

di = ai*b0 + ai-1*b1 + ... + a1*bi-1 + a0*bi + ki (mod n)

As i becomes arbitrarily large, so does the unmodulated value of the above summation, thus so does ki+1 (the carry digit for the next position). As the values of k become multi-digit, they start affecting multiple future digit positions.

This is the part where I get stuck, as far as real proof goes. But my suspicion is that the value of k will follow some Fibonacci-esque sequence, and this will force the digits necessary to achieve di = 0 to follow a non-repeating algorithm, thus the string of digits is not a defined number in my system, and there are no numbers that act as zero-divisors.
Last edited by arbiteroftruth on Sun Mar 25, 2012 2:24 am UTC, edited 2 times in total.
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### Re: Infinite-digit 'integers'

gmalivuk wrote:But why not just admit you're essentially describing the 10-adics, and use the rules people already figured out for them?

There's something about in-depth study of the 10-adics that leads people to the realization that they are the first people in the history of mathematics who aren't a complete tool. It's like something out of Lovecraft. This thread takes me back to Archimedes Plutonium on sci.math arguing that Peano and Godel were wrong and that the Riemann Hypothesis and the Goldbach Conjecture were easy when you worked in infinite 10-adic integers.
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### Re: Infinite-digit 'integers'

After looking more closely, I can't believe I overlooked the fact that when the article says a terminating decimal has a terminating p-adic expansion, they're still allowing the p-adic expansion to go right of the decimal. In that case, my numbers are just (at least a subset of) the p-adics.

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?
arbiteroftruth

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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:After looking more closely, I can't believe I overlooked the fact that when the article says a terminating decimal has a terminating p-adic expansion, they're still allowing the p-adic expansion to go right of the decimal. In that case, my numbers are just (at least a subset of) the p-adics.

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?

Kind of, sort of.

The reals form a field. The 10-adics do not. p-adics are a more complete if p is prime. They at least form an integral domain. I'm not sure about multiplicative inverse (which would be required for a field).

I mean it, though, when I mentioned that you should study number theory in more detail if this sort of thing interests you. For instance:

If the base of your number system if prime, then the situation is simple. Any non-zero multiple of a prime must include the prime itself as one of its factors, but the highest valid digit is 1 less than the base, so the digits a0 and b0 cannot possibly multiply to a multiple of the base.

A non-prime base is more complex. The first digit of the result, d0, is determined entirely by a0 and b0, which must cause d0 to be 0, and create a carry digit for the next position.

can be stated much more cleanly if you knew the relations between residues and gcd. This will make it much easier for people to understand what you're saying.
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### Re: Infinite-digit 'integers'

gorcee wrote:
arbiteroftruth wrote:After looking more closely, I can't believe I overlooked the fact that when the article says a terminating decimal has a terminating p-adic expansion, they're still allowing the p-adic expansion to go right of the decimal. In that case, my numbers are just (at least a subset of) the p-adics.

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?

Kind of, sort of.

The reals form a field. The 10-adics do not. p-adics are a more complete if p is prime. They at least form an integral domain. I'm not sure about multiplicative inverse (which would be required for a field).

Well, at least for my particular subset where the p-adic number has to eventually fall into a repeating pattern, every such number corresponds to a rational number, and every rational number corresponds to such a p-adic number, and every rational number has a rational multiplicative inverse.

gorcee wrote:I mean it, though, when I mentioned that you should study number theory in more detail if this sort of thing interests you.

That's basically what I'm doing here. Given that I'm not going to spend the money to go back to college as a math major just because I find it kind of interesting, and given that my best method of learning has always been to question things and see what reasoning falls apart when I do, this is pretty much exactly my preferred method of study.
arbiteroftruth

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### Re: Infinite-digit 'integers'

arbiteroftruth wrote:After looking more closely, I can't believe I overlooked the fact that when the article says a terminating decimal has a terminating p-adic expansion, they're still allowing the p-adic expansion to go right of the decimal. In that case, my numbers are just (at least a subset of) the p-adics.
Yeah, yours are exactly the subset that has only 0's to the right of the decimal point and an eventually repeating sequence of digits to the left.

The question then becomes whether this is a particularly "interesting" subset, in the sense of having any useful properties that aren't shared by some much easier to work with set like the rational numbers.
In the future, there will be a global network of billions of adding machines.... One of the primary uses of this network will be to transport moving pictures of lesbian sex by pretending they are made out of numbers.
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