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Is $1234567891011121314151617181920212223......$ an integer?

This question came from that one and from that talk where it's noted that "integers have a finite count of digits", so that the "number" in the title is not at all a number (not integer nor rational or real) . This statement seems to me justified because I suspect that if we admit an infinite count of digits we build a set of "numbers" that is not countable (but I've not a proof). I'm wrong?


Reading comments and answers I'm a bit confused. So I add a more specific question: Is the string in the title a number in some model of Peano Axioms?

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    $\begingroup$ No.​​​​​​​​​​​​ $\endgroup$ Jan 2, 2015 at 20:12
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    $\begingroup$ $p$-adic integers have infinitely many digits. $\endgroup$ Jan 2, 2015 at 20:12
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    $\begingroup$ It is not really a number unless you give more context, but rather a sequence of numbers. $\endgroup$
    – Pedro
    Jan 2, 2015 at 20:13
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    $\begingroup$ @Matt Samuel, which despite the name, aren't integers. $\endgroup$ Jan 2, 2015 at 20:13
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    $\begingroup$ Suppose your thing is an integer and call it $S$. What is $S+1$? $\endgroup$
    – MJD
    Jan 2, 2015 at 20:21

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First of all, before we can callthis an integer, we can only call it a blob of ink or a bunch of pixels. How are some buches of pixels integers and others are not? Well, the are not integers, they represent integers. For example MMXV, and $2015$, and 0x7DF are distinct representations of the same integer, and all given with enough "context" to allow us to conclude (more or less easily) what representation scheme was used and that integers "behind" these notations are in fact the same.

I won't go into details of the other two representation, but concentrate on the decimal one. How do we know which number it represents? While the answer seems to be "two thousands and fifteen" obviously, it is not really that obvious. The convention(!) behind the notation is that a finite sequence of digit symbols $d_1d_2\ldots d_k$ represents the number $\sum_{i=0}^{k-1}10^{i}d_{k-i}$ (with a bit of ignoring the distinction between a digit symbol and its numerical value). Your candidate does not match this convention because it is not a finite sequence. We cannot even read the value of $k$ from it. Admittedly, there are other ways to formulate the convention of deimal representation, but they all fail in some way or other with an infinite digit string (it may be appropriate to consider left-infinite sequences of digits and view the natural numbers as a subset of the "numbers" obtained this way (by prependingthem with infinitely many zeroes), but beware - the realm of Archimedes-Plutoniumism lurks behind that; and this does not apply to your right-infinite string).

To put it differently: Asking what integer an ill-formed (because infinite) sequence of decimal digits represents is not much better than asking what integer the ill-formed "roman numeral" MQLXXRT represents.

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This is an infinite string of digits. But this string, when we try to interpret it in decimals, does not encode a number, i.e., an element of ${\mathbb N}$ or ${\mathbb R}$. Natural numbers, when written in decimals, appear as finite strings, and have a last digit. Noninteger real numbers, when written in decimals, require a decimal point, after which an infinity of digits may appear. Therefore, if you decide to insert a decimal point in front or between any two digits in your sequence, you obtain a well defined positive real number.

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To say that it is an integer would be to say that it is a natural number, since it is positive. Then, Peano's $n$th axiom is ill-defined for $ n \ge 6 $, debating the fact that it can even be called a natural number. So I would say no.

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    $\begingroup$ Sorry, I'm not an expert on axiomatic, so can you explain me what means "Peano's nth axiom is ill-defined for n≥6"? $\endgroup$ Jan 2, 2015 at 20:45
  • $\begingroup$ Peano's 6th, 7th, and 8th axioms are ill-defined. $\endgroup$ Jan 2, 2015 at 21:07
  • $\begingroup$ Why is it ill-defined? $\endgroup$
    – quid
    Jan 2, 2015 at 21:16
  • $\begingroup$ Because the successor is ill-defined. $\endgroup$ Jan 2, 2015 at 22:05
  • $\begingroup$ I would not say it is ill-defined; obviously, it is undefined in the question, but why is it ill defined? I could say it is obtained by incrementing the first digit checking if there is a carry if so incrementing the next and so on. I think this would actually be a well behaved operation; it should essentially correspond to arithmetic on the $p$-adic integers, where the string is just written in the sense other than usual. Of course it would not satisfy all axioms but I am not convinced by the argument that the successor is ill defined. $\endgroup$
    – quid
    Jan 2, 2015 at 23:22
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Your number is not an integer. Suppose that you can have an integer with an infinite number of digits. Then by Cantor's diagonal argument you can show that the set of all integers in uncountable, which is a contradiction if you accept the Peano axioms as suggested by Ahaan S. Rungta. Thus your number cannot be an integer.

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  • $\begingroup$ I don't understand what can be a real number with a null decimal expansion that is not an integer. $\endgroup$ Jan 2, 2015 at 20:39
  • $\begingroup$ There isn't a decimal point in this number. So, while you can't, for example, find the last digit of $\pi$, you should be able to find the "ones" digit, which is what asking for the last digit of this number would be. $\endgroup$
    – KSmarts
    Jan 2, 2015 at 20:48
  • $\begingroup$ Ahh.. Sorry I was reading the number backwards my misunderstanding. I've edited my answer. $\endgroup$
    – Wintermute
    Jan 2, 2015 at 20:50
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    $\begingroup$ AFAIK, which is to say not too well, there are models that satisfy the Peano axioms that are in fact not countable. $\endgroup$
    – quid
    Jan 2, 2015 at 21:20
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    $\begingroup$ @quid: By Löwenheim-Skolem, we have that any first-order theory has models of higher cardinality. However, this is not applicable for second-order theories, and indeed Dedekind showed in 1888 that there is only one model satisfying the Peano axioms, up to isomorphism. $\endgroup$ Jan 2, 2015 at 21:43
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It is exactly the same thing as asking whether infinity an integer. But integer MUST have a finite number of digits. (the number in that title is "BOGUS")

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