Is there a prime $p$ whose successor is greater than $2p$?

Question

Toying with Goldbach's Conjecture, I encountered myself in a situation where the following question arose.

Is there a prime $p$ whose successor is greater than $2p$?

You see. If the answer to this question is true, then Goldbach's Conjecture's disproven.

Answer 1

The answer is no. Actually, you have Bertrand-Chebyshev's theorem:

For any natural number $n>1$, there exists a prime number $p$ such that $$n<p<2n.$$

As a consequence, if we denote $p_n$ the $n$-th prime number, we have $$p_{n+1}<2p_n.$$

Answer 2

Among the positive primes, no, there is not. In 1845, Bertrand conjectured that there is always a prime between $n$ and $2n$. Five years later, Chebyshev proved it "using non-elementary methods," according to MathWorld.

So, if $n$ is prime greater than 3, then the next prime is at most $2n - 3$. And even that is an overestimate for $n > 7$. See Sloane's A062234.

If I wanted to be a smart-aleck, I would insist that "successor" of a prime $p$ means a prime that is greater than $p$. Then, if $p = -2$, we'd have $2p = -4$ but the successor of $p$ is 2. But I do understand that you're only thinking about positive primes.

Of course this doesn't resolve Goldbach's conjecture one way or the other, no matter how strongly it suggests it to be true.

Answer 3

By the prime number theorem, $$\pi(p) \approx \frac{p}{\log p},$$ so $$\pi(2p) \approx \frac{2p}{\log 2p},$$ which suggests there are about $$\frac{p \log \frac{p}{2}}{\log p \log 2p}$$ primes between $p$ and $2p$. Note that this is an increasing quantity above $p = 2$.

You might be saying that the primes need to be rather large for these formulas to have anything resembling accuracy. But the bright side of this is that it means that if a counterexample exists, it would be a small prime. And the small primes are generally packed more closely together than the large primes!

Furthermore, the twin prime conjecture remains unproven.

Of course I wouldn't have thought about this if everyone else hadn't already brought up Bertrand's postulate, which still lacked an elementary proof when I was born (plus back then kids were taught that $1$ is prime, which doesn't really affect any of the foregoing).

Answer 4

Just invoking Bertrand's postulate might be enough for the OP, but I also expect something towards actually starting to address the Goldbach conjecture, which of the existing answers at the time I posted this bounty, there was only one.

Okay. Consider then even number $2p + 2$. Where $p$ is prime so that there are no primes between $p$ and $2p$.

This is a counter example to the gold conjecture.

Let $2p + 2 = q + r$ where $q$ and $r$ are prime. Wolog assume $q \ge r$.

If $q > p$ then $q \ge 2p+1$ so $r \le 1$. That can't work. If $q \le p$ then $q + r \le 2p < 2p+2$.

So $2p + 2$ can not be written as sum of two primes.

So the goldbach conjecture would be disproven if we could find such a prime $p$. But Bertrand's postulate which has beed proven directly says there is no such $p$.

So the goldbach conjecture has not been disproven.

If the number $2p + 2 = q + r$ where $p,q,r; q \ge r$ are primes it would have to be that $p < q < 2p$ but as there will always be some prime between $p$ and $2p$, this is not a problem.

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If the goldbach conjecture is true, a (maybe) interesting conesequence is that for any prime $p$, there is a prime $q$ so that $p < q < 2p$ and $2p-q+2$ is also prime.

Ex. $2 < 3 < 2*2; 2*2+2-3 = 3$. $3< 5< 6$ and $8-5 = 3$. $5 < 7 < 10; 12-7 = 5$. and ... $13 < 17,\langle19\rangle,23 < 26$ and $28 -17,\langle19\rangle,23 = 11,\langle9\rangle, 5$ etc.

Answer 5

As many of the other answers have noted, there is no prime $p$ such that its successor is greater than $2p$. This is a trivial consequence of Bertrand's postulate which states that

For any natural number $n>1$, there exists some prime $p$ such that $$n<p<2n$$

This ensures that for all primes $p_n$, its successor, $p_{n+1}$, must fall within the range $p_n<p_{n+1}<2p_n$. If you do not understand why that is, think about $p_n$ as a natural number. By Bertrand's postulate there must exist a prime greater than $p_n$ but less than $2p_n$. This garuntees the existence of a prime $p_{n+k}$ such that $p_n<p_{n+k}<2p_n$. As such, $p_{n+1} \leq p_{n+k}$ and hence must lie in this range as well.

The proof of Bertrand's postulate is quite long. If you wish to see one, I would recommend Erdös' elementary proof. Particularly, the following paper by Michael Tang presents the proof in a fashion particularly easy to follow if you have little background in number theory. http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNy8yLzcyZTgyOGRhZDgxMmQ5MTY0ODIwOTJjZTUyZWQ0OWI4ZjIzYWVmLnBkZg==&rn=QmVydHJhbmQucGRm

Although I do not believe that this was your original intention, I will now address the link between this problem and the Goldbach Conjecture. You are correct when you say that the existence of a prime $p$ whose successor is larger than $2p$ would disprove the Goldbach conjecture. In fact, there exists a deeper link between Bertrand's postulate and the strong Goldbach conjecture. The Goldbach conjecture implies Bertrand's postulate. A proof of this can be found here. https://proofwiki.org/wiki/Goldbach_implies_Bertrand

However, although Goldbach's conjecture implies Bertrand's postulate, this does not work the other way around. That is, Bertrand's postulate does not imply the Goldbach conjecture (otherwise the Goldbach conjecture would be proven).

Answer 6

From the book "ELEMENTARY THEORY OF NUMBERS" ;
written by W. SIERPINSKI ,
Edited by A. SCH1NZEL ,
Chapter III. Prime Numbers
section 10. Proof of Bertrand's Postulate (Theorem of Tchebycheff)
page 144

Theorem7. If $n$ is a natural number $> 5$, then between $n$ and $2n$ there are at least two different prime numbers.
i.e. there are primes $p_1 \neq p_2$ such that $n < p_1 , p_2 < 2n$.

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Now let $p$ be any arbitrary prime number.

• If $p \in \{ 2, 3, 5 \}$, then we are done!
• If $7 \leq p$, this theorem implies that there exists one prime number $q$ such that $p < q < 2p;$
  so the successor is less than $2p$.

Answer 7

This is in fact never the case. The French mathematician Joseph Bertrand postulated in the mid 19th century that there is always a prime between $n$ and $2n$ ($n\in\mathbb{N}$). This was proven a few years later by the Russian mathematician Pafnuty Chebyshev using a rather lengthy and cumbersome proof.

In 1932, the then 19 year old Hungarian Paul Erdős published a far neater and shorter proof of the same statement utilizing binomial coefficients, which prompted some humorous fellow to compose the following rhyme:

Chebyshev said it and I say it again.

There's always a prime between $n$ and $2n$.