John Cuthber Posted February 8, 2015 Share Posted February 8, 2015 He was just noting (correctly) that [math]\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}[/math]. In any case, again, the OP's number, being infinite, is not a real number. All real numbers are finite. It is, of course, still true that for any finite list of primes, the number formed by concatenating their digits is a natural number. It's true that [math]\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}[/math]. It's just that what Xerxes wrote doesn't parse in English. Link to comment Share on other sites More sharing options...

imatfaal Posted February 8, 2015 Share Posted February 8, 2015 Hm. Let d_{i} be the number of digits of the ith prime p_{i}, with d_{0} = 0. [math]\sum_{i = 1}^{\infty} \left( p_{i} \times 10^{-\left( \sum_{j = 0}^{i} d_{j} \right) - 1} \right)[/math] This gives us 2 * 10^{0} + 3 * 10^{-1} + 5 * 10^{-2} + 7 * 10^{-3} + 11 * 10^{-5} + 13 * 10^{-7} + ... = 2.3571113... Of course, without a good method for determining the next prime (which, as discussed at length in another thread, we don't currently have), we'd need a list of all primes in the first place to populate the terms of our sum, which would mean we might as well just concatenate the digits of all the primes in the list instead. But, eh. Also, I'm more convinced now that the number will be irrational, since Bertrand's postulate (which I'd heard before, but apparently forgotten) guarantees that for any integer n > 1, there exists a prime p such that n < p < 2n. Thus we'll never come across a prime whose digits comprise a list of all primes below it, so my particular idea for how this number might be rational fails. I think this also rules out imatfaal's idea. Yep - with out being thorough - I am pretty sure that rules out my get out from irrationality as well. Bertrands was proved by Chebyshev - of fame/infamy for all probability/statistics students 1 Link to comment Share on other sites More sharing options...

Xerxes Posted February 8, 2015 Share Posted February 8, 2015 In any case, again, the OP's number, being infinite, is not a real number. All real numbers are finite. I disagree - in fact you have not said what exactly you mean by a "finite number" or "infiinite number" Try this. Let [math]S[/math] be set. For every set element [math]x[/math] define the succesor [math]x^+[/math] by [math]x^+ = x \cup \{x\}[/math] Thus, say [math]\O^+ = \O \cup \{\O\}= \{\O\}[/math] I re-label [math]\O = 0,\,\, \{\O\}=1[/math] etc to obtain the natural numbers so [math]0^+=1,\,\,1^+=2[/math] and "so on". In order to give meaning to the "and son on" I need the axiom of infinity which states There exists a set containing 0 and all of its succesors. The first transfinite number so defined is usually written as [math]\omega [/math]. So one may have the seqquence [math]1,2,3,4,5,........,\omega[/math] and then [math]\omega+1,\omega+2,\omega+3,.........,2\omega[/math] and then [math]2\omega+1, 2 \omega+2,.......3 \omega,.........,\omega^2,.......[/math]. Since the Real numbers contain the Natural numbers as a proper subset, there is no difficulty having a Real number the set of whose predecessors has infinite cardinality i.e which can be written as a non-terminating string of natural numbers Link to comment Share on other sites More sharing options...

John Posted February 9, 2015 Share Posted February 9, 2015 (edited) Gotta love broken LaTeX. Hopefully it will be sorted soon or something.Anyway, you are free to disagree, but the fact that no real number is infinitely large is built into the structure. It's why the real numbers have the Archimedean property. It's why the integers with addition form a cyclic group. As far as successors go, what would S(23571113...) be? What natural number n exists such that S(n) = 23571113...?I guess, all in all, what meaningful difference (besides the symbols) is there between 23571113... and infinity?I'm not sure what your point is regarding transfinite numbers. Perhaps if the LaTeX is fixed it will be clearer. Edit: Alright, now that the LaTeX in this thread has been restored, I still don't see your point with the transfinite numbers. I'm talking about the set of real numbers, of which the transfinite ordinals and cardinals are not elements. Rather, transfinite ordinals and cardinals describe the order types and sizes, respectively, of infinite sets. Edited February 9, 2015 by John Link to comment Share on other sites More sharing options...

Xerxes Posted February 9, 2015 Share Posted February 9, 2015 (edited) the fact that no real number is infinitely large is built into the structure. It's why the real numbers have the Archimedean property. It's why the integers with addition form a cyclic group. OK, I assume as you objected to trailing ellipses in a real number, this implies you think every real number can be written as a terminating string of natural numbers. Since the field [math]\mathbb{R}[/math] is a total order, this in turn implies there is a largest real number. Call it [math]x[/math]. Then the Archimedean property claims that there exist some integer [math]n[/math] and some [math]y >0 \in \mathbb{R}[/math] such that [math]ny >x[/math]. Contradiction Moreover, the ingeters [math]\mathbb{Z}[/math] form (under addition) a cyclic group of infinite order, that is, given a generator = 1, then 1 + 1 + 1 +.......= 0 only if we allow this sum to go out to infinity Edited February 9, 2015 by Xerxes Link to comment Share on other sites More sharing options...

John Posted February 9, 2015 Share Posted February 9, 2015 (edited) OK, I assume as you objected to trailing ellipses in a real number, this implies you think every real number can be written as a terminating string of natural numbers. Er, no, I don't care about "trailing ellipses," and obviously any nonterminating decimal cannot be expressed as a terminating string of natural numbers. My point is that since a number consisting of a list of all primes (without a decimal point placed anywhere in the number, mind) is infinite in magnitude, that number cannot be real, since there are no infinitely large real numbers. I'm not sure why you're hung up on the ellipses, but I'm talking about the number itself, not the notation. Since the field [math]\mathbb{R}[/math] is a total order, this in turn implies there is a largest real number. Call it [math]x[/math]. Then the Archimedean property claims that there exist some integer [math]n[/math] and some [math]y >0 \in \mathbb{R}[/math] such that [math]ny >x[/math]. Contradiction I'm not sure what you're arguing here, but putting a total order on [math]\mathbb{R}[/math] in no way implies there is a largest real number, and if you do believe there is a largest real number, then you have some misunderstanding about how the real numbers work. The real numbers are Archimedean. Edit: Also, you seem to have a misunderstanding about the Archimedean property. As it applies to the real numbers, the property implies that for *any* real numbers x, y such that y < x, there exists an integer n such that ny > x. That is to say, the property is a statement about *all* y < x, not just a particular y < x. Moreover, the ingeters [math]\mathbb{Z}[/math] form (under addition) a cyclic group of infinite order, that is, given a generator = 1, then 1 + 1 + 1 +.......= 0 only if we allow this sum to go out to infinity Well yes, the set of integers has infinite cardinality, but each natural number is itself finite in magnitude. The integers form a cyclic group because any integer can be expressed as a finite sum or difference of 1's or -1's. Adding 1 to itself as many times as we like, we'll never reach 23571113.... In any case, analytic continuation notwithstanding, the sum 1 + 1 + 1 + ... diverges, i.e. it "equals" infinity, not zero. Edited February 9, 2015 by John Link to comment Share on other sites More sharing options...

Xerxes Posted February 10, 2015 Share Posted February 10, 2015 John, you getting close to insulting me, which I (obviously) do not like For the resord, I understand the Arhimedean perperty of the Reals perfectly, and I know exactly what is meant by a cyclic group of infinite order. So stop it. Moreover, my comment about there being "a kargest Real number" was part of a logical argument to show a contradiction in what I assumed to be your point, that no Real number can be written as a non-terminating string of Natural numbers. If that assumption was wrong, I apologize. Let me ask you this, then. Is it your position that "all Real numbers are finite"? So my question was, and is, this. What do you mean by "a finite number"? Is it that no Real number can be written as a non-terminating string of Natural numbers? Or do you mean that, viewed as a set, the "string representation" of a Real number must have finite cardinality? Or do you mean something else that I have missed? Link to comment Share on other sites More sharing options...

Endy0816 Posted February 10, 2015 Author Share Posted February 10, 2015 Looks like more or less my original conjecture is as close to valid as is reasonable for an online forum to prove. Thank you everyone who responded. p.s. Feel free to return to the ongoing debate, just don't kill each other over numbers. Link to comment Share on other sites More sharing options...

John Posted February 10, 2015 Share Posted February 10, 2015 (edited) John, you getting close to insulting me, which I (obviously) do not like For the resord, I understand the Arhimedean perperty of the Reals perfectly, and I know exactly what is meant by a cyclic group of infinite order. So stop it. Moreover, my comment about there being "a kargest Real number" was part of a logical argument to show a contradiction in what I assumed to be your point, that no Real number can be written as a non-terminating string of Natural numbers. If that assumption was wrong, I apologize. Let me ask you this, then. Is it your position that "all Real numbers are finite"? So my question was, and is, this. What do you mean by "a finite number"? Is it that no Real number can be written as a non-terminating string of Natural numbers? Or do you mean that, viewed as a set, the "string representation" of a Real number must have finite cardinality? Or do you mean something else that I have missed? Calm down. I said you seem to be misunderstanding a couple of things, because your statements above were a bit off. That's not an insult, but rather a possible source of our disagreement here. Perhaps you understand everything perfectly but simply misspoke. And yes, all real numbers are finite. That's not simply my position, but rather part of the definition and structure of the real numbers. By finite, as indicated in my previous post, I mean finite in magnitude. I'm not sure how else to explain what I mean here. The notation doesn't enter into it. Pi, for instance, is definitely irrational. Its decimal representation is "3." followed by a non-terminating and non-repeating string of natural numbers, but obviously pi isn't infinite in magnitude. The number in the OP, however, has no decimal point. It would be a natural number if there were only finitely many primes, but since there are infinitely many primes, as we add each new prime to the end of the number, it increases without bound. Thus, in the limit, the number becomes infinite in magnitude. No natural number (and indeed, no real number) is infinite in magnitude. If I seem a bit curt, then please forgive me. "Infinity is just a very large number" (usually as an implied corollary to "we can divide by zero") seems to be the Math forum's version of "relativity is wrong." Edited February 10, 2015 by John Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now