Endy0816 Posted February 4, 2015 Share Posted February 4, 2015 Stupid question, would a constant composed of all the primes 23571113... be an irrational number? I've been thinking it would be, but wanted to double check. Link to comment Share on other sites More sharing options...

imatfaal Posted February 4, 2015 Share Posted February 4, 2015 That hurts my head! a. Not definable as there is an infinite number of primes - ie there is always another one to add to the end (multiple all them used so far and add one - that number either divides by a new prime or is a new prime) b. Infinity is not a real number - irrationals are all real c. It would be a ratio of integers - itself and one, ie an integer and thus not irrational d. It might not even be prime 1 Link to comment Share on other sites More sharing options...

Endy0816 Posted February 4, 2015 Author Share Posted February 4, 2015 What if we modified it to 2.3571113... instead? I was just thinking something akin to Pi, to have a look at and make some observations of. Link to comment Share on other sites More sharing options...

John Posted February 4, 2015 Share Posted February 4, 2015 It depends on what specific restrictions there are in the construction. For instance, 29 is prime, but would something like 3.29537... be valid, since 9 is not prime? If not, are only the digits 1 (which isn't usually considered prime anyway), 2, 3, 5 and 7 allowed? If only those five digits are allowed, then we could easily construct either a rational or an irrational number. If all the primes are required to be part of the number, then it seems clear the result would be irrational, since each prime is a unique sequence of digits and thus there would be no way to construct a repeating decimal from them (and of course the decimal would never terminate, since there are infinitely many primes). Link to comment Share on other sites More sharing options...

imatfaal Posted February 4, 2015 Share Posted February 4, 2015 Would the "behind the decimal point" version be transcendental as well? Or might one day we be able to provide an algebraic route to it. It depends on what specific restrictions there are in the construction. For instance, 29 is prime, but would something like 3.29537... be valid, since 9 is not prime? If not, are only the digits 1 (which isn't usually considered prime anyway), 2, 3, 5 and 7 allowed?If only those five digits are allowed, then we could easily construct either a rational or an irrational number. If all the primes are required to be part of the number, then it seems clear the result would be irrational, since each prime is a unique sequence of digits and thus there would be no way to construct a repeating decimal from them (and of course the decimal would never terminate, since there are infinitely many primes). Can we really rule out a repeating decimal? When the numbers get big and beyond our knowledge ( hypothesizing that mersennes etc have run out) then is it not possible that all remaining primes are formed from a repeating unit. It could not be a simple building block as subsequent units would be divisible by 1000....0001 times a previous block - but if the block was split at different places ie can you have a repeating pattern that could be subdivided into smaller units all of which were prime. Seems completely unfeasible - but can we rule it out? [mp][/mp] Come to think of it I think we can rule it out. If it can be shown as a repeating unit then it can be shown to be an integer over an integer and by definition that is not prime Link to comment Share on other sites More sharing options...

John Posted February 4, 2015 Share Posted February 4, 2015 Yeah, I can't entirely rule that sort of thing out. Since gaps between primes can be arbitrarily large, it's also possible that we'll occasionally come across a prime whose digits are some permutation of all the primes below it, which could lend itself to some repetition, though since primes can be arbitrarily large I think we'd still run into issues. As for the number being transcendental, I have no idea, but then I think that, especially since we apparently allow repetition of primes in our number, we can probably construct at least all the irrational numbers, in which case of course some would be transcendental and others not. Link to comment Share on other sites More sharing options...

imatfaal Posted February 4, 2015 Share Posted February 4, 2015 I thought Endy's idea was to concatenate the known primes - 2, 3, 5,7,11,13,17,19 etc gives 2.35711131719... This would be unique Link to comment Share on other sites More sharing options...

Endy0816 Posted February 4, 2015 Author Share Posted February 4, 2015 (edited) ^Yeah, that was the basic idea(known & unknowns). I'm just not sure of what all the properties of such a number would be. I am glad my thought process was on the right track at least. Edited February 4, 2015 by Endy0816 Link to comment Share on other sites More sharing options...

John Posted February 4, 2015 Share Posted February 4, 2015 Ah, yeah, I suppose so, in which case don't mind me. Also, the constructing all irrationals thing is wrong in any case, since something like 0.101001000100001... obviously doesn't make use of all the primes. This is what I get for trying to do math while rushing to get to work. Of course, we can still make different numbers by moving the decimal point, but I don't think that affects much. And yeah, Endy, it's interesting to consider. Link to comment Share on other sites More sharing options...

Xerxes Posted February 5, 2015 Share Posted February 5, 2015 (edited) Stupid question, would a constant composed of all the primes 23571113... be an irrational number? I've been thinking it would be, but wanted to double check. No it is not irrational. Since your "list" is a natural nuber, it cannot possibly be irrational. The fact that the ellipses imply it "goes on forever" is neither here nor there, since the natural numbers have a countably infinite number of elements. Interestingly, it is one of the quirks of modulo 9 arithmetic i.e. base 10, that once you introduce the decimal point, everything changes. So no real number, with trailing ellipses to right of the point can be rational (unless of course the ellipses denotes a repeating pattern) Edited February 5, 2015 by Xerxes Link to comment Share on other sites More sharing options...

John Cuthber Posted February 5, 2015 Share Posted February 5, 2015 23571113... is a rational number because it's the ratio of two numbers, an example of two such numbers is twice 23571113... and two 2 X 23571113... 2 =23571113... Consideration of that makes it clear that any integer is rational On the other hand 0.23571113... is not rational. All rational fractions, when expressed as a decimal, terminate like 1/4 =0.25 or recur like 1/3 = .333333... Similarly, 2.3571113... is irrational Link to comment Share on other sites More sharing options...

imatfaal Posted February 5, 2015 Share Posted February 5, 2015 snip All rational fractions, when expressed as a decimal, terminate like 1/4 =0.25 or recur like 1/3 = .333333... Similarly, 2.3571113... is irrational Or recur like 1/7 = 0.142857(142857) Or eventually settle into a recurring pattern like 7009/63000 =.111253968(253968) Link to comment Share on other sites More sharing options...

Xerxes Posted February 5, 2015 Share Posted February 5, 2015 23571113... is a rational number because it's the ratio of two numbers, an example of two such numbers is twice 23571113... and two 2 X 23571113... 2 =23571113... Consideration of that makes it clear that any integer is rational On the other hand 0.23571113... is not rational. All rational fractions, when expressed as a decimal, terminate like 1/4 =0.25 or recur like 1/3 = .333333... Similarly, 2.3571113... is irrational This is silly. If 2 X 23571113..... 2 proves that 23571113..... is rational then why is it NOT the case that 2 X 0.23571113..... 2 is also rational? Link to comment Share on other sites More sharing options...

imatfaal Posted February 5, 2015 Share Posted February 5, 2015 This is silly. If 2 X 23571113..... 2 proves that 23571113..... is rational then why is it NOT the case that 2 X 0.23571113..... 2 is also rational? Because the product of the top of the second expression is not an integer - whereas the product at the top of the first expression must be an integer. [mp][/mp] I am still not convinced that the OP's decimal (2.35711...) must be irrational. Can we say for sure (not just in all likelihood) that the primes do not settle into the following organisation: there is a large run of numbers and each prime is a concantenation of a number of instances of that large run of numbers ending at some point of that large run and the next prime number starting with the continuation and then a larger multiple number of concatenations etc? That would be a repeating decimal. But is that sort of repeat necessarily going to generate composite numbers - I think it will but I am unsure. Eventually you will end up with two numbers that would need to be prime Number 1 = concatenate {(large number x), (repeating unit n times), (large number y)} and and Number 2 = concatenate {(large number x), (repeating unit m times), (large number y)} and Will at some point the repetition get to a stage that one of these must be a composite of the other. Link to comment Share on other sites More sharing options...

John Posted February 6, 2015 Share Posted February 6, 2015 (edited) The decimal is probably irrational, but yeah, I don't think we can entirely rule out the possibility of some sort of repetition. I also wouldn't say 23571113... is a real number, though, let alone rational, since the fact that there are infinitely many primes means the number must be infinitely large. Of course, imatfaal covered this point earlier, but in light of more recent discussion, it bears repeating. Edited February 6, 2015 by John Link to comment Share on other sites More sharing options...

Xerxes Posted February 6, 2015 Share Posted February 6, 2015 The decimal is probably irrational, but yeah, I don't think we can entirely rule out the possibility of some sort of repetition. I also wouldn't say 23571113... is a real number, though, let alone rational, since the fact that there are infinitely many primes means the number must be infinitely large. Of course, imatfaal covered this point earlier, but in light of more recent discussion, it bears repeating. Then maybe I misunderstood the OP. Was I wrong to assume it was saying, for example, that if [math]p_1=2,\,p_2=3,\, p_3=5[/math] etc then we are talking about a string of the form [math]p_1p_2p_3,p_4........[/math] which is clearly a natural number, and as such is a subset of the Real numbers and therefore real. Is it your contention that this number is, say, complex? Do you doubt that a natural number can be as "large" as we like? And is it your contention that, given the OP, we may have [math]p_1. p_2 p_3.....p_2 p_3.....p_2 p_3.....[/math] as a rational real number? If so I have misunderstood the OP and I apologize. If not, the above is gibberish Link to comment Share on other sites More sharing options...

John Posted February 6, 2015 Share Posted February 6, 2015 The OP is asking about a number whose digits are a list of all the prime numbers. There are infinitely many primes, thus there are infinitely many digits in the number, thus the number is infinitely large, thus not a real number.As for the rationality of the decimal, I don't think we can absolutely rule out the idea that, every so often, we'll arrive at a prime number whose digits are a list of all the prime numbers below it, since gaps between primes can be arbitrarily large. Someone better versed in number theory could probably say something more definite, but in my mind, while the number is almost certainly irrational, I don't know if we can prove that it is irrational. Link to comment Share on other sites More sharing options...

Xerxes Posted February 7, 2015 Share Posted February 7, 2015 (edited) This is silly. If 2 X 23571113..... 2 proves that 23571113..... is rational then why is it NOT the case that 2 X 0.23571113..... 2 is also rational? Because the product of the top of the second expression is not an integer - whereas the product at the top of the first expression must be an integer So you multipy by 2 then divide by 2 and get back the same number. Not very surprisimg, is it? So in order to assert that 23571113 is rational, you start by assumiing that it is ratiomal. Likewise for 2.357,,,,,, being irratiomal This we call asssuming the truth of the proposition to be proved Edited February 7, 2015 by Xerxes Link to comment Share on other sites More sharing options...

John Cuthber Posted February 7, 2015 Share Posted February 7, 2015 No, we don't start by assuming it's rational. We start by assuming it's an integer What do you think is the definition of a rational number? Do you think it is that a rational number is the ratio of two integers? Do you think that 2.357,,,,,, is an integer? Do you think that 2 times 2.357,,,,,, is an integer? Link to comment Share on other sites More sharing options...

Sensei Posted February 7, 2015 Share Posted February 7, 2015 Stupid question, would a constant composed of all the primes Are you claiming that there is limited quantity ("all") of prime numbers? Link to comment Share on other sites More sharing options...

Xerxes Posted February 7, 2015 Share Posted February 7, 2015 (edited) No, we don't start by assuming it's rational. We start by assuming it's an integer What do you think is the definition of a rational number? Do you think it is that a rational number is the ratio of two integers? Do you think that 2.357,,,,,, is an integer? Do you think that 2 times 2.357,,,,,, is an integer? Please don't patronize me Look every natural number is an integer is a rational number is real number. The reverse "chain" is false however No non-terminating decimal without a repeating pattern can possibly be rational Edited February 7, 2015 by Xerxes Link to comment Share on other sites More sharing options...

John Cuthber Posted February 7, 2015 Share Posted February 7, 2015 (edited) Look every natural number is an integer is a rational number is real number That does not make sense. The point is that any integer is a rational number Since the OP has no indication of a decimal point, it's an integer- albeit, an infinitely big one. So, it is rational. I can prove that any integer is rational because it's the ratio of twice the integer to 2. Pointing out that you are wrong isn't patronising. For what it's worth, I already said that "No non-terminating decimal without a repeating pattern can possibly be rational" However, someone made the perfectly reasonable point that they don't see it that way, and they would like proof. Can you offer any such proof? [Edit Sorry, my mistake there; they said that they couldn't be sure the list of primes didn't repeat itself, and asked for proof. John seems to have offered that proof} Edited February 8, 2015 by John Cuthber Link to comment Share on other sites More sharing options...

John Posted February 7, 2015 Share Posted February 7, 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. Link to comment Share on other sites More sharing options...

Endy0816 Posted February 8, 2015 Author Share Posted February 8, 2015 (edited) Are you claiming that there is limited quantity ("all") of prime numbers? No, I know that there are an infinite number. Wouldn't have even bothered anyone if I thought the result would be of finite length. At this point based on imatfaal's helpful post, I'm amending my question to be regarding the decimal form instead. Also trying to figure out how/if it can be expressed as the sum of an infinite series. I was able to make a sort of programming loop version. Can't exactly define the infinite array it utilizes, nor provide it with the infinite memory and time needed; but were those things possible, it would work beautifully. Edited February 8, 2015 by Endy0816 Link to comment Share on other sites More sharing options...

John Posted February 8, 2015 Share Posted February 8, 2015 (edited) 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. Edited February 9, 2015 by hypervalent_iodine 1 Link to comment Share on other sites More sharing options...

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