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Math Help - Prime comparison

  1. #1
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    Prime comparison

    Is this obvious? Starting with the third prime "5",

    a prime is less than the product of all preceding primes plus 1 .

    A few examples seem convincing that the statement is true, but I have not been able to demonstrate it.
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  2. #2
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    Quote Originally Posted by Nivla View Post
    Is this obvious? Starting with the third prime "5",

    a prime is less than the product of all preceding primes plus 1 .

    A few examples seem convincing that the statement is true, but I have not been able to demonstrate it.


    Yes it's true. Google "Bertrand's Postulate", so that if we have the prime p and p' is the prime before it, then p < 2p'...
    And so, in fact, your statement could very seriously be strengthened: every prime >= 3 is less than the prime before it times the first prime.

    No, not obvious, but pretty straigthforward for whoever knows BP.

    Tonio
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  3. #3
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    No need for Bertrand's postulate.

    Simply note that p_1\cdot ...\cdot p_n-1 is coprime to p_1,...,p_n, therefore it must be divisible by another prime (since it's >1) which is not among the first n, since the sequence of primes is increasing (*) we conclude that p_{n+1}\leq p_1\cdot ...\cdot p_n-1<p_1\cdot ...\cdot p_n+1

    (*) p_{n+k}|(p_1...p_n-1) for some k\geq 1 hence p_{n+1} \leq p_{n+k}\leq p_1...p_n-1
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    Prime comparison II

    Thanks, Tonio and PaulRS! I shall look up Bertrand's Postulate, Tonio.
    PaulRS: Yes, now that I have your reply, I see the simplicity I was not able to conger. G.E. Andrews, in his "Number Theory" states the inequality in a hint towards the solution of an exercise, but I could not demonstrate the hint - let alone the entire exercise!

    Thanks to you, both. Now, I can struggle on.

    Nivla.
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  5. #5
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    Quote Originally Posted by PaulRS View Post
    No need for Bertrand's postulate.

    Simply note that p_1\cdot ...\cdot p_n-1 is coprime to p_1,...,p_n, therefore it must be divisible by another prime (since it's >1) which is not among the first n, since the sequence of primes is increasing (*) we conclude that p_{n+1}\leq p_1\cdot ...\cdot p_n-1<p_1\cdot ...\cdot p_n+1

    (*) p_{n+k}|(p_1...p_n-1) for some k\geq 1 hence p_{n+1} \leq p_{n+k}\leq p_1...p_n-1
    All, if I may ask a related question please.

    Is p_1...p_n-1 always a prime, for all n?
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    Quote Originally Posted by aman_cc View Post
    All, if I may ask a related question please.

    Is p_1...p_n-1 always a prime, for all n?

    No: 2*3*5*7*11*13*17 - 1 is divisible by....(look above 53).

    Tonio
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  7. #7
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    Quote Originally Posted by tonio View Post
    No: 2*3*5*7*11*13*17 - 1 is divisible by....(look above 53).

    Tonio
    yep - 61 thanks
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    Donation in thanks!

    Tonio and PaulRS:

    I am making another donation (my second) to MHF in appreciation for your help. Who knows? I might need help again

    Nivla
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