Why doesnt Bertrand's postulate imply Legendre's conjecture?

  1. I mean, according to my knowledge, Bertrand's postulate has already been proved, I've already read and understood one, but Paul Erdos, and a few other mathematicians have proved it using various methods. I just finished reading Ramanujan's proof. Its amazingly advanced, and really short. The porblem of the inequality that Ramanujan uses to prove Bertrand's postulate is amazingly similar to an alternate formulation of Legendre's conjecture.

    Well, I played a little bit with other representations of the Prime Counting Function and Ramanujan's inequality of the alternate formulation of Legendre's, and just assuming Bertrand's postulate to be true (which it has already been proved as I mentioned), I found a rather elementary (and obvious) proof of Legendre's.

    Can anyone explain why they say that Bertrand's postulate doesn't imply Legendre's conjecture? I read about the supposed error ratio of the Prime Number Theorem, so if anyone knows about this would they be so kind to elaborate why BT doesnt imply LC?
  2. jcsd
  3. CRGreathouse

    CRGreathouse 3,497
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    Yes. Bertrand's Postulate shows that there is a prime between n^2 and 2n^2, not in the far shorter interval n^2 to (n+1)^2. You'd need something just slightly stronger than the Riemann Hypothesis to get that.
  4. Wow, I think you got your definition of Bertrand's incorrect.

  5. So in fact, your argument about the shorter interval is completely incorrect. In the opposite.

    BP says that there is atleast one prime p that n<p<2n. Any by analyzing the interval between n and 2n for all n is simple:

    2n - n = n.

    the interval between any 2n and n will always be n. If n=100, than 2n=200. the interval between 2n and n is 2n-n=200-100=100=n.

    now, the interval between between n^2 and (n+1)^2 is actually bigger than the interval between n and 2n.
  6. CRGreathouse

    CRGreathouse 3,497
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    Yes, but you need to use n^2 not n. If you use the 2n-2 version, this gives you a prime between n^2 and 2n^2 - 2.

    If this doesn't make sense, you don't understand Bertrand's Postulate.
  7. pardon me, you're right I just analyzed the interval substituting n=x2 and saw that the interval is much much tighter. Just compare the graphs of y=x and y=2√(x)+1.

    I apologize, I see what you're talking about now. Thanks.

    And it's regardless of which version you use, the n<p<2n version gave me the same result. It's a theorem, it has to.
  8. CRGreathouse

    CRGreathouse 3,497
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    No problem, I just thought I needed to write something to get you to look over it again yourself. I was tempted to write [itex]2\sqrt x+1[/itex] myself, but it's a lot better when you can find it on your own. :smile:
  9. Bertrand's Postulate states: For n > 1, there is a prime p satisfying n < p < 2n.

    M. El Bachraoui proved in 2006: For n > 2, there is always a prime p satisfying 2n < p < 3n.

    In general, if you were to prove: For all n >= k >= 1, there is always a prime p satisfying kn < p < (k+1)n, then you would have shown Legendre, since if you allow k = n, then you have that prime p satisfies n^2 < p < n^2 + n < (n+1)^2.

    The reason that Bertrand does not imply Legendre is by examining the intervals for Bertrand: [n,2n] and Legendre: [n^2, (n+1)^2] we can see that Legendre is a "tighter" interval.

    e.g. Suppose we want the lower bound to be 100, then we allow each intervals lower bound to be the n value that produces 100, so we have

    Bertrand (n=100) -> [100, 200] and
    Legendre (n=10) -> [100, 121]

    e.g. Again, we want our lower bound to be 25, then

    Bertrand (n=25) -> [25, 50]
    Legendre (n=5) -> [25, 36]

    This can be proven, but I am too lazy to do so right now.

    I have actually spent the last 8 months on proving that there is always a prime in the interval [kn, (k+1)n] for n >= k >= 1, and it is under review for a journal as we speak... err, I guess as we type. HeH.
    Last edited: Aug 13, 2009
  10. CRGreathouse

    CRGreathouse 3,497
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    That sounds like a cute result. Weak (like Bertrand's Postulate), but perhaps sometimes useful -- and probably hard to prove.
    Last edited: Aug 13, 2009
  11. It's interesting to note that Chebyshev was the first to show Bertrand in 1850, and Erdos stated it elementarily in 1932 although it wasn't until 2006 when Bachraoui showed [itex][2n, 3n][/itex].

    Another interesting thing is that Erdos checked the values for [itex]n = 1, 2, ..., 96[/itex], and Bachraoui had to check from 2 until 945. It is believed that in order to show each case you need to check exponentially many beginning cases.

    You can find Bachraoui's proof via this URL: http://www.m-hikari.com/ijcms-password/ijcms-password13-16-2006/elbachraouiIJCMS13-16-2006.pdf

    The worst thing is, both Erdos and Bachraoui's proofs are simplistic by most proof techniques, yet Legendre's conjecture has been around for over 150 years.

    Another interesting conjecture is by Dorin Andrica, who stated the following: For all [itex]n>0[/itex], let [itex]p_{n}[/itex] denote the n-th prime number, then [itex]\sqrt{p_{n+1}} - \sqrt{p_{n}} < 1[/itex]. In fact, if Andrica's Conjecture is proven, then Legendre is a direct corollary. I believe the converse is true as well, though I have been unable to find a proof on the internet.
    Last edited: Aug 13, 2009
  12. CRGreathouse

    CRGreathouse 3,497
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    Any chance you'd send me a preprint? Or is this on the arXiv?
  13. Throw me an E-mail: my user name on here @gmail.com and I'll be happy to send a pre-print to you.
  14. Regarding Ramanujan's proof of the Bertrand postulate , I am unable to understand the 2nd equation that he uses ......

    He starts the proof like this , let v(x) be the sum of logarithms of all primes less than or equal to x, Now consider :
    [tex]\Psi[/tex](x) = v(x) + v (x^[1/2]) + v(x^[1/3]) + .... - eq.1

    then he writes :
    log ( [x] ! ) = [tex]\Psi[/tex](x) + [tex]\Psi[/tex](x/2) + [tex]\Psi[/tex](x/3) + .... -eq.2

    where [x] is the greatest integer <= x

    Why is equation 2 valid ...... it certainly does not seem to be an obvious statement to me ...... Can someone please explain how to derive eq 2 , given eq 1 . I was able to follow the steps of Ramanujan's proof after the 2nd equation ( assuming that there exists some Sterlings approx of which he talks about ... nd that the result of the approx is what he mentions it to be ) ... but plz help me with the 2nd step .. Is there something that is very clear to everyone else that I am missing here ??
  15. Ok ..... I was able to show that eq 2 ..i.e :

    log ( [x] ! ) = [tex]\Psi[/tex](x) + [tex]\Psi[/tex](x/2) + [tex]\Psi[/tex](x/3) + ....

    is indeed valid.

    But still this was not at all obvious to me , and only after reading the lemma 2 part in the wikipedia proof of Bertrands postulate ( http://en.wikipedia.org/wiki/Proof_of_Bertrand's_postulate) ... and then thinking for quite some time was i able to see that eq 2 is indeed valid.
    In order to prove lemma 2 , wikipedia states that

    The exponent of p in n! is
    [tex]\sum[/tex] [n/p^j]
    where summation is from j=1 to infinity , and [ ] is the greatest integer function.

    I see that the above statement which is easily proved (but even to see that this was true ,again took me some time ) , does in fact lead to equation 2 that Ramanujan has used .
    I guess I certainly do not have the mental faculty to comprehend such things that are quite obvious to others.
  16. I was able to see that Andrica's Conjecture does indeed lead to Legendre's Conjecture .

    Regarding the converse - i.e. given Legendre's Conjecture , then Andrica's Conjecture also holds . -- I can see that if Legendre,s Conjecture is stated for real values of n > 1 , and not just integer values of n , then Legendre does indeed imply Andrica.
    But did Legendre state the Conjecture for only integers or for all real values as well ? I think it was stated for only integers , because if he wanted to state it for real values he could have said that there exists a prime b/w n & n + 2 * sqrt(n) +1 , instead of n^2 and (n+1)^2

    If Legendre only quoted for integers , then I do not see Legendre leading to Andrica ... but I may be wrong.
  17. I believe a proof of Legendre -> Andrica may go something like the attachment.

    Also, if Legendre's Conjecture is proven, it not only guarantees that there's a prime in the interval... it actually gurantees that there are 2 primes satisfying the inequality. This may be easily shown, but I haven't typed up the proof yet.

    Attached Files:

  18. It looks to me like your proof that legendre implies andrica , is implicitly assuming that there are 2 primes between n^2 and (n+1)^2 . But Legendre guarantees only the existence of a minimum of one. If there is only 1 prime bw n^2 and (n+1)^2 , then I am unable to follow how you conclude that sq( n^2 + 2n) - sq( n^2 + 1 ) < 1 implies that sq(p(n+1)) -sq(p(n)) < 1
    because only one of p(n) or p(n + 1) needs to lie in the interval :
    (n^2 + 1 , n^2 + 2n) according to legendre , not both .
  19. CRGreathouse

    CRGreathouse 3,497
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    I have to echo srijithju here. Let me give my own reasoning.

    Suppose that Legendre's conjecture holds, but only barely: there is a huge prime gap near n^2. In particular, the closest primes to n^2 are
    (n-1)^2 + a
    (n+1)^2 - b
    for small a and b. So the prime gap is of length
    (n+1)^2 - b - (n-1)^2 - a = 4n - a - b.

    Now Andrica's conjecture has the convenient form
    p_n+1 - p_n < 2sqrt(p_n) + 1
    (this follows naturally from the fact that sqrt(p_n + 2sqrt p_n + 1) - sqrt(p_n) = 1).

    So in this case we have
    4n - a - b < 2sqrt((n-1)^2 + a) + 1 < 2sqrt((n-1)^2 + a + a^2/4) + 1 = 2(n-1) + a + 1 = 2n + a - 1
    that is
    2n < 2a + b - 1
    which is false for a,b small as assumed.

    Basically, the gaps allowed in Legendre's conjecture are two times too long for it to imply Andrica's conjecture.

    (On an unrelated note, there's a bad error in your pdf relating to the order of quantifiers.)
    Last edited: Feb 17, 2010
  20. i think we can agree that a stronger legendre's conjecture (there is at least 2 primes between consecutive squares) implies andrica's easily.


    if k^2 < pn < p(n+1) <(k+1)^2 than applying sqrt

    k< sqrt(pn) < sqrt(p(n+1)) < k+1

    the smaller interval must fit in the bigger one so

    sqrt(p(n+1)) - sqrt(pn) < k+1 - k = 1

    i realy dont think legendre's and andrica's are equivalent.
  21. CRGreathouse

    CRGreathouse 3,497
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    No, not really.* You showed that that p(n+1) and pn are not a counterexample to Andrica's conjecture, but this does not rule out all possible counterexamples, only ones between two integer squares.

    Just as above: Suppose your strong Legendre conjecture held, but only just barely:
    (n-1)^2 + a = p1
    (n-1)^2 + b = p2
    (n+1)^2 - c = p3
    (n+1)^2 - d = p4
    are the only primes between (n-1)^2 and (n+1)^2, with a,b,c,d small. Then sqrt(p3) - sqrt(p2) is about 2 - (b+c)/2 or roughly 2, a counterexample to Andrica's conjecture.

    * Technical note: I believe that both are true, so that trivially anything implies either one. You can choose to interpret my remark as "you haven't shown that SL implies A" or alternately that a sequence about which we know nothing but that the equivalent of SL holds in it need not have the equivalent of A hold in it. I prefer the latter approach.
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