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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- October 17th 2009, 12:07 PMNivlaPrime 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. - October 17th 2009, 01:25 PMtonio

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 - October 17th 2009, 01:35 PMPaulRS
No need for Bertrand's postulate.

Simply note that is coprime to , 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

(*) for some hence - October 17th 2009, 03:17 PMNivlaPrime 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. - October 17th 2009, 08:55 PMaman_cc
- October 18th 2009, 12:12 AMtonio
- October 18th 2009, 12:26 AMaman_cc
- October 18th 2009, 10:31 AMNivlaDonation 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 (Wink)

Nivla