# Infinite integer

1. Aug 7, 2014

### Vincent_

Mathematicians have long held that infinite integers do NOT exist, but here is a very simple argument that shows that they do exist.

A list of positive integers Z+ can be formed in a base-1 numeral system as...

1
11
111
1111
.
.
.
1111111...

Since the set of integers is infinite, the list will be of an infinite length.

If the length of the list is infinite, the last term will also be infinite in length since it is the same length as the list is long.

If the term is infinite in the base-1 numeral system, then it will be infinite in base 2 or 10.

Therefore infinite integers do exist.

2. Aug 7, 2014

### gopher_p

Your proof contains an error. The list of finite strings of $1$s, written in increasing order of length, does not have a last term.

I would encourage you to give up your quest to prove the existence of infinite integers. All integers are finite according to the definitions of "integer" and "finite".

Edit: I should've maybe said something along the lines "reasonable definitions of 'integer' and 'finite'" given the difficulty in finding definitions that aren't in some way a bit circular.

Last edited: Aug 7, 2014
3. Aug 8, 2014

### Vincent_

Then it must have an infinite number of last terms.

Sir, you are welcome to define 'integer' any way you wish. I, on the other hand, choose to go where logic and reason dictates.

For someone to say that an infinite set can be placed into a one-to-one pairing with only finite numbers is simply not logical.

4. Aug 8, 2014

### phinds

Since your conventions of "logic and reason" do not conform to standard math, this amounts to just making up your own definitions such that no one else is going to agree with you. This is the REASON for standard terminology ... so that we can talk to each other with some sense of what the other is saying. Making up your own definitions is a bad idea.

5. Aug 8, 2014

### PeroK

If there is a last term, what happens when you add 1 to it?

6. Aug 8, 2014

### Vincent_

If you wish to conform to a standard of math that does not conform to logic and reason, you are welcome to do it. Just don't expect me to join you there.

In my world, one can't make two mutually exclusive concepts true by simply defining them to be true.

7. Aug 8, 2014

### Vincent_

You will get another one of the infinite number of last terms.

8. Aug 8, 2014

### PeroK

Is there a first of these last terms?

9. Aug 8, 2014

"Then it must have an infinite number of last terms."

You have a meaningless definition of "last terms". But yes, as asked above, out of curiosity, where is the demarcation point for these last terms? Is "11" the first of them?

Note that if you have infinitely many distinct integers, as you claim your last terms to be, you cannot have a largest one, since all that come after are larger.

It isn't clear what you think you're on about, but it isn't grounded in mathematics.

10. Aug 8, 2014

### disregardthat

There already exists notion of "infinite integers", called p-adic numbers, for prime numbers p. Written in base p, p-adic numbers are on the form $.....a_3a_2a_1$ for infinite sequences of integers $a_i$ such that $0 \leq a_i < p$. For example, in base 2, the number ......111111, which you seem to be interested, is a 2-adic number. It exists in any base p.

The set of p-adic numbers are written as $\mathbb{Z}_p$, and contains the formal series $\pm \sum_{n=0}^{\infty}a_ip^n$, where $0 \leq a_i < p$. The usual integers $\mathbb{Z}$ are contained as a subset in $\mathbb{Z}_p$ (we just say that the terms are eventually 0). These formal infinite series may be added, subtracted, multiplied and even divided by (except for by 0). Unlike $\mathbb{Z}$, it is a field (you cant in general divide integers in $\mathbb{Z}$ and end up with integers), just like $\mathbb{Q}$ and $\mathbb{R}$, and is of particular interest in number theory.

Note that these infinite series are purely formal, and not convergent as infinite series in $\mathbb{R}$. So don't confuse the two notions. There is a notion of convergence in the picture here too, but it is not analytic convergence. For the algebraically inclined, $\mathbb{Z}_p$ may be defined as an inverse limit of rings $\mathbb{Z} / p \gets \mathbb{Z} / p^2 \gets \mathbb{Z} / p^3 \gets ...$, and the p-adic numbers are the "convergent" elements of this sequence.

As a concrete example of what goes on here, let $p = 5$. Then we have a convergent element
$(1,1+5,1+5 + 2 \cdot 5^2,1+5+2 \cdot 5^2+2 \cdot 5^3,1+5+2 \cdot 5^2+2\cdot 5^3+2\cdot 5^4,...)$ in the sequence $\mathbb{Z} / 5 \gets \mathbb{Z} / 5^2 \gets \mathbb{Z} / 5^3 \gets ...$.
In $\mathbb{Z} / 5^{n+1}$ our element is $1+ 5 + 2 \cdot 5^2 + ... + 2 \cdot 5^n$, written as 22....2211 in base 5. It is clear that when we move from $\mathbb{Z} / 5^{n+1} \to \mathbb{Z} / 5^{n}$ (modulo $5^n$), we end up with $1+5 + 2 \cdot 5^2 + ... + 2 \cdot 5^{n-1}$, so this sequence is, in techincal terms, stable under these (modulo) maps. In this way we make sense of the "infinite integer" ....2222222211 in $\mathbb{Z}_5$, equivalently written as $1 +5+ \sum^{\infty}_{n=2}2 \cdot 5^n$.

You can just as well speak of 10-adic numbers, infinite formal sums written in base 10. But the big difference when the base is not prime is not only that you do not end up with a field (so you may not divide in general), but not even an integral domain, which means that two non-zero integers in $\mathbb{Z}_{10}$ multiplied might yield 0. So it behaves very differently from $\mathbb{Z}$. Your sequence 1, 11, 111, 1111, ... is indeed a "convergent" element of the sequence $\mathbb{Z} / 10 \gets \mathbb{Z} / 10^2 \gets \mathbb{Z} / 10^3 \gets ...$, and "converges" to ......11111111. It isn't quite the last element in the sequence (it has no last element), but just like for analytical limits, it is the "limiting" value.

Last edited: Aug 8, 2014
11. Aug 8, 2014

### phinds

That's been clear all along and it caused this thread to be locked. I can't imagine why it got UNlocked, since it STILL isn't about actual math, just some personal theory using personal definitions.

12. Aug 8, 2014

### micromass

Are you sure the 2-adic number is 1111111....?? I think it is ....1111111. For example, in base 2, we have ...1111111 = -1.

I unlocked it because some reputable members thought they could add something to the discussion. The attitude of the OP will determine whether it remains open or not.

13. Aug 8, 2014

### disregardthat

You're right, decimal expansions go the other way. I'll correct that.

14. Aug 8, 2014

### phinds

Seems reasonable. Thanks for the explanation.

15. Dec 9, 2014

### Justin Ponze

I don't understand why you think using an undefined sequence with abstract values, and self predicated validity, especially in explaining the termination of the sequence gives validity to any claim or theory. You can't clearly define what an infinite quantity is because you can't count forever, much less can you live forever and nothing, but matter and energy in it's baser forms lasts forever. You and I and every other human and sentient lifeform(s) in the universe are subject to change and entropy just like anything else, so why do you justify your "sequences" continence with an abstract value (i.e. 11111... or ...), more over your parameters for sustained continence of your number sequence is abstract in concept and self-predicated in validity, (i.e. Your opinion that infinite quantities are a given and the inability to question eternally finite and infinite integers was the very basis for your argument for it.) if only for the reason that humans define everything we perceive as real it would be a given that since humanity will not last forever in any case that we nor any other sentient lifeforms will not be there to count that long nor have we thus far.
.

16. Dec 9, 2014

"Since the set of integers is infinite, the list will be of an infinite length."
Yes, no argument there.

"If the length of the list is infinite, the last term will also be infinite in length since it is the same length as the list is long."
Full stop: if the list is infinite there is not a "last term": no matter how far into the list you go there are more terms.

"You will get another one of the infinite number of last terms."
Begin banging head on table. Here your notion is completely undefined since (as has been pointed out) you have not given, and you cannot give, a mathematically precise (more importantly, consistent) for what constitutes the beginning of the "infinite number of lat terms". (I can say it begins with the second term and you would have no reason to say I'm incorrect).
But even if you could - the fact that there are infinitely many "last terms" means the list has no ending, so there is no final term: each term after the previous is simply 1 digit longer, still finite in length. (Infinity in your discussion is not a place, it is a concept - there is no "there there").

The original post contains some of the highest proof twaddle I've ever seen.

17. Dec 11, 2014

### MayCaesar

Let's denote one of these "last terms" as $x$. For any two integers, the operation of summation is defined. Now add 1 to this number. There are two cases to consider (all other cases can be expressed in terms of these two cases):
1. $x+1$ equals $x$. Then, apparently, since the number is the same, should be $x-1=x$, $x-2=x$ and so on. This means that, no matter what number you deduct from $x$, you will still get $x$. Now take the limit of deducting $x$ from $x$ - you should get 0, according to integer numbers properties, while you are getting $x$. This means that $x=0$, and, apparently, $0+1=0$ is incorrect. From this contradiction we conclude that $x+1=x$ is an incorrect assumption. So is $x+i=x$ for any non-zero $i$.
2. $x+1$ is different from $x$. This contradicts your assumption of "one of the infinite number of last terms".

This is all just a juggling though. There is no such integer as "infinity". Infinity is a mathematical abstraction, IT IS NOT A NUMBER. So you can't talk about "infinite last term", it cannot exist in principle. Infinity can serve as a number in generalized coordinate space, but that space does not equal space of integers.