Irrationality of the Copeland-Erdos constant

[quincyboy7]

Given that for all positive integers n, there exists a prime p such that n<=p<10n, prove the Copeland-Erdos constant (the concatenation of the primes in order i.e. 0.235711131719...) is irrational.

Alright, so what my gut tells me is that I have to prove somehow that the digits do not recur, and the "10n" part is supposed to help me with decimal places (or something)...I'm not sure if I should apply n to a particular prime or the number of the place or anything exactly. I need a little push in the right direction, so any help would be much appreciated.

 

[Dick]

It's not so hard. As you said, if the constant is rational then it ends with an infinitely repeating series of digits. Suppose a sequence of k digits repeats. Is there a prime containing k digits? What can you say about the next prime after that one?

 

[quincyboy7]

It's not so hard. As you said, if the constant is rational then it ends with an infinitely repeating series of digits. Suppose a sequence of k digits repeats. Is there a prime containing k digits? What can you say about the next prime after that one?

But doesn't that line of reasoning assume that there is ONE prime that makes up the k digits instead of two or three or n primes put together?

Running with the idea, though, if such a prime contains k digits, I assume I'm trying to reach a contradiction where the next prime has more than k digits and thus the original k digits cannot repeat...however, how does the n<=p<10n help me?

 

[Dick]

But doesn't that line of reasoning assume that there is ONE prime that makes up the k digits instead of two or three or n primes put together?

Running with the idea, though, if such a prime contains k digits, I assume I'm trying to reach a contradiction where the next prime has more than k digits and thus the original k digits cannot repeat...however, how does the n<=p<10n help me?

I should have added 'and the prime with k digits falls into the repeating region of the decimal'. It that's so, what must the next prime look like? You'll also need to think about what to do if no prime with k digits falls into the repeating region.

 

[quincyboy7]

If a prime p with k digits falls into the repeating region, then let n=p+1. By the given, it is assured that the prime after p will have k or k+1 digits? I intuitively think that the prime won't have the same digits, but how do I rigorously say that?

 

[Dick]

If a prime p with k digits falls into the repeating region, then let n=p+1. By the given, it is assured that the prime after p will have k or k+1 digits? I intuitively think that the prime won't have the same digits, but how do I rigorously say that?

If the prime falls into the repeating region, then the digits in the decimal repeat every k digits, right? What does that tell you about the first k digits of the prime and the following prime?

 

[quincyboy7]

If the prime falls into the repeating region, then the digits in the decimal repeat every k digits, right? What does that tell you about the first k digits of the prime and the following prime?

Ah! I think I get it, then the first k digits of the next prime will be the same so if the prime is 123456789, the next digits are 123456789(more digits)...more digits are possible only if the next prime is the original prime times at least 10...however, we know the next prime is LESS than 10 times the first prime from the assumption.

Quick question though...why can't the expansion end in the same prime being repeated over and over again. This may be trivial, but when given a prime P, what is the guarantee that there exists a greater prime P0 such that P0>P?

And how in the world would you proceed if there are no primes with k digits in the repeating interval? (As a quick check, we are assuming that the repeating interval is k digits long, correct?)

Thanks so much for what you've done so far, Dick

 

[Dick]

Yes, k is the number of digits in the repeating section. There is a theorem that there are an infinite numbers of primes, isn't there? Besides, that's implied by your n<=p<10n assumption. So the 'no next prime' is not something you have to worry about. What you do have to worry about is whether the decimal doesn't start repeating until after all of the primes with k digits have already been entered into the sequence. Hint: is there a prime with 2k digits? Is there a prime with 3k digits?

 

[quincyboy7]

Well, if n is equal to the repeating prime with k digits, doesn't the less than or equal to sign make the no next prime question not trivial?

I have absolutely no clue how to apply anything to the scenario where the repeating happens "mid-prime".

And what if the interval contains multiple primes, i.e. starts repeating before we get to the prime with k digits?

 

[Dick]

Well, if n is equal to the repeating prime with k digits, doesn't the less than or equal to sign make the no next prime question not trivial?

I have absolutely no clue how to apply anything to the scenario where the repeating happens "mid-prime".

And what if the interval contains multiple primes, i.e. starts repeating before we get to the prime with k digits?

I'm not sure what you mean by 'repeating prime'. But you do want to make sure that you can choose a prime entirely contained in the repeating part of the decimal. The hint I'm trying to give you is that the argument you have at the beginning of post 7 works if the prime has k digits, but it also works if the prime has 2k, 3k, 4k etc digits. Can you prove one of those must fall into the repeating part? Suppose the decimal starts repeating at decimal point number N and at that point you are listing a prime with L digits...

 

[Dick]

In fact, if the prime has 2k digits and falls into the repeating part, is it even a prime?

 

[quincyboy7]

Hmmm...still stumped. If a prime has 2k digits for example, if the first k digits make up the repeating part, then the last k digits of the prime are just the same as the first k digits. Why can't those k digits repeat forever?

And what if the repeating interval with k digits is made up of digits from more than one prime. Argh this is frustrating

 

[quincyboy7]

How can I exhaust cases where the digits of the prime in question are greater than k?

How can I exhaust cases where the repeating interval contains digits from more than one prime?

 

[Dick]

Look. You are worrying about stuff you don't have to worry about. If the decimal sequence starts repeating while you are spelling out the digits of a prime with L digits then you still have to have to spell out digits of the primes with more digits than L in the sequence, right? Many of them will have numbers of digits that are a multiple of k, also right? What does that mean? Derive a contradiction from this. See your comments in post 7 again.

 

[quincyboy7]

But my comments in #7 are completely reliant on the prime being k digits and being in the interval. If the prime is, say, 2k digits, then I have no clue where the interval might start. Even if I did, it wouldn't matter. If the interval starts at the first digit of the 2k-digit prime, then the prime is not a prime (it is a multiple of 1000000...(k-1 zeros)1). If the interval starts at the first digit of the 3k-digit prime, then the prime is not a prime (it is a multiple of 10000...(k-1 zeros)10000000(k-1 zeros)1). But if the interval starts ANYWHERE else, or contains more than one prime, then there's no telling what will happen. Eventually, sure, you'll run into a prime with 2k, 3k, 4k digits, but there is no telling where that's going to happen.

 

[Dick]

But my comments in #7 are completely reliant on the prime being k digits and being in the interval. If the prime is, say, 2k digits, then I have no clue where the interval might start. Even if I did, it wouldn't matter. If the interval starts at the first digit of the 2k-digit prime, then the prime is not a prime (it is a multiple of 1000000...(k-1 zeros)1). If the interval starts at the first digit of the 3k-digit prime, then the prime is not a prime (it is a multiple of 10000...(k-1 zeros)10000000(k-1 zeros)1). But if the interval starts ANYWHERE else, or contains more than one prime, then there's no telling what will happen. Eventually, sure, you'll run into a prime with 2k, 3k, 4k digits, but there is no telling where that's going to happen.

You still aren't thinking clearly. If you are SURE you will eventually run into a prime with 2k, 3k, 4k etc digits (however high the constant in front of the k has to go), in the repeating zone then you are DONE. You've already said that if a sequence of nk digits repeats in blocks of k digits, then you know it's NOT prime. That's a contradiction. It means the Copeland Erdos constant is NOT rational.

 

[quincyboy7]

If you are SURE you will eventually run into a prime with 2k, 3k, 4k etc digits (however high the constant in front of the k has to go), in the repeating zone then you are DONE. You've already said that if a sequence of nk digits repeats in blocks of k digits, then you know it's NOT prime.

Well, the argument I made about the nk digits repetition implying that the number is not a prime completely rests on the interval starting at the first digit of the "prime" with nk digits. I completely agree with you that the postulate guarantees a prime with 2k, 3k, 4k digits. But I'm not sure how it guarantees that eventually such a prime will have its first k digits in the repeating k digits of the interval.

 

[Dick]

Well, the argument I made about the nk digits repetition implying that the number is not a prime completely rests on the interval starting at the first digit of the "prime" with nk digits. I completely agree with you that the postulate guarantees a prime with 2k, 3k, 4k digits. But I'm not sure how it guarantees that eventually such a prime will have its first k digits in the repeating k digits of the interval.

Think about this. If you are in the region of the decimal that repeats with period k, then ANY group of nk digits is NOT prime if n>1. You sketched out why. It DOESN'T MATTER where it starts. Take 123123123123123123123123123123123123... Find me a seqence of 6 digits in there that doesn't factor in an obvious way.

 

[quincyboy7]

But that's kind of hand-wavy...just "sketching out" why leaves me unsatisfied, I guess.

And even if the prime is k digits and falls into the repeating interval, does the n<=p<10n really cause a contradiction? Let's assume that the prime of k digits is the highest prime with k digits...then our previous contradiction is not a contradiction, since "n"=the prime and we can move on to k+1 digits.

 

[Dick]

But that's kind of hand-wavy...just "sketching out" why leaves me unsatisfied, I guess.

And even if the prime is k digits and falls into the repeating interval, does the n<=p<10n really cause a contradiction? Let's assume that the prime of k digits is the highest prime with k digits...then our previous contradiction is not a contradiction, since "n"=the prime and we can move on to k+1 digits.

n<=p<10n tells you there is a prime of ANY digit length whatsoever, right? If you don't agree with this then tell me. Now totally ignore ANYTHING you or I have said before. Sit back and clear your mind. Try and fit a prime that has 2k, 3k, 4k etc digits into a sequence of digits that repeats with period k. It CAN'T be prime, can it?

 

[quincyboy7]

Try and fit a prime that has 2k, 3k, 4k etc digits into a sequence of digits that repeats with period k. It CAN'T be prime, can it?

I agree that there must be a prime with x digits, whatever x is. Let's say a prime has nk digits. Then let's say m of those digits appear in the first repeating interval containing the prime. So it is structured so that we have m digits/k digits/k digits/k digits/...(n-1 times)/k-m digits. I "see" that. But what does that have to do with its being prime.

 

[Dick]

I agree that there must be a prime with x digits, whatever x is. Let's say a prime has nk digits. Then let's say m of those digits appear in the first repeating interval containing the prime. So it is structured so that we have m digits/k digits/k digits/k digits/...(n-1 times)/k-m digits. I "see" that. But what does that have to do with its being prime.

You are still fixated on where it starts. It doesn't matter where it starts. It's k digits/k digits/k digits .. /k digits. The 'k digits' might be a permutation of the ORIGINAL 'k digits' you identified. But it still repeats that way. 231231 is an obvious non-prime in 123123123... See? I didn't start at the 1. It still repeats in blocks of 3.

 

[quincyboy7]

So what you're saying that is if we have k digits ANYWHERE in a repeating interval of k digits, then those k digits repeat? Then we're home free, since we can state that the repeating interval will eventually contain a prime with nk digits with a repeating interval of k digits n times, which is not a prime since it is a multiple of 100000(k-1 zeros)100(k-1zeros...and so on n times. Contradiction.

This is such a "punch-the-wall" problem where half the frustration is not getting it and the other half is knowing that people smarter than you easily get it.

 

[Dick]

So what you're saying that is if we have k digits ANYWHERE in a repeating interval of k digits, then those k digits repeat? Then we're home free, since we can state that the repeating interval will eventually contain a prime with nk digits with a repeating interval of k digits n times, which is not a prime since it is a multiple of 100000(k-1 zeros)100(k-1zeros...and so on n times. Contradiction.

This is such a "punch-the-wall" problem where half the frustration is not getting it and the other half is knowing that people smarter than you easily get it.

Yes, that's it. I wouldn't exaggerate the smarter people 'easily' get it thing. I had to think about it for a while. Pick a smarter friend of yours who is not in this class and spring it on them. Tell me how long it takes them to get it. Notice there is no 'advanced math' involved in proving it. I didn't notice a ton of Physics Forum people jumping into help with this one. Might mean it's not that easy. At least until you've seen the solution.