Number Theory-limitation of Pn

[icystrike]

Homework Statement

This is part of Euclid's proof such that he says:
1)

P_{n+1} smaller or equal to P_{1}P_{2}...P_{n} +1
Such that p_n is to denote some prime
I am wondering if anyone could provide me with an elegant prove to the above inequality as i am new to number theory.

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

Homework Statement

This is part of Euclid's proof such that he says:
1)

p_{n+1} smaller or equal to p_{1}p_{2}...p_{n}+1
Such that p_n is to denote some prime
I am wondering if anyone could provide me with an elegant prove to the above inequality as i am new to number theory.

That is part of Euclid's proof of what? The "p_1p_2\cdot\cdot\cdot p_n+ 1" reminds me of Euclides proof that there are an infinite number of primes but he never asserts that p_{n+1} is smaller than or equal to that.

It does, of course, follow from Euclid's proof. Assuming that p_1, p_2, ..., p_n is a list of all primes less than or equal to p_n, p_1p_2\cdot\cdot\cdot p_n+ 1 is certainly not divisible by any of p_1, p_2, ..., p_n. Therefore it is either a prime itself or is divisible by a prime larger than any of p_1, p_2, ..., p_n. In either case there must be a prime less than or equal to p_1p_2\cdot\cdot\cdot p_n+ 1

 

[SrEstroncio]

Homework Statement

This is part of Euclid's proof such that he says:
1)

p_{n+1} smaller or equal to p_{1}p_{2}...p_{n}+1
Such that p_n is to denote some prime
I am wondering if anyone could provide me with an elegant prove to the above inequality as i am new to number theory.

It is not very elegant but the proof follows from the principles of order on the real number field.

Considering positive primes: given p_{1} < p_{2}, then it follows that p_{1} < p_{1}^2 < p_{1} \cdot p_{2}, by inductive reasoning you can prove that p_{n+1}<p_{n+1} \cdot p_{1} < p_{n+1} \cdot p_{1} \cdot p_{2}, and so on.

 

[Gib Z]

Am I understanding the question right? It seems to come quite simply by noting that the primes are larger than 1. We can even omit the equality option of the inequality at a glance.

 

[icystrike]

Thank you for all your replies (=

I think I'm able to understand the concept.
Please tell me your opinions on my proof below.Let P_{1}P_{2}...P_{n} +1 =N
Such that N is an element of positive integer.
Thus, it suggest N is co-prime to P_{1},P_{2}, ... ,P_{n}Which may be a prime number that is P_{k}
Such that k is a value larger or equal to n+1 OR

It may be a composite number that is larger than P_{n+1} as the smallest possible divisor of N should be P_{n+1} . Noting N is co-prime to P_{1},P_{2}, ... ,P_{n} .