Prime comparison

**Nivla** · October 17th 2009, 12:07 PM

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.

**tonio** · October 17th 2009, 01:25 PM

Originally Posted by Nivla

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

**PaulRS** · October 17th 2009, 01:35 PM

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 $p_{n+1} \leq p_{n+k}\leq p_1...p_n-1$

**Nivla** · October 17th 2009, 03:17 PM

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.

**aman_cc** · October 17th 2009, 08:55 PM

Originally Posted by PaulRS

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?

**tonio** · October 18th 2009, 12:12 AM

Originally Posted by aman_cc

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

**aman_cc** · October 18th 2009, 12:26 AM

Originally Posted by tonio

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

Tonio

yep - 61 thanks

**Nivla** · October 18th 2009, 10:31 AM

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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