Number Theory-limitation of Pn

  • Thread starter icystrike
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  • #1
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Homework Statement



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

[tex]P_{n+1}[/tex] smaller or equal to [tex]P_{1}[/tex][tex]P_{2}[/tex]...[tex]P_{n}[/tex] +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

 
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  • #2
HallsofIvy
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Homework Statement



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

[tex]p_{n+1}[/tex] smaller or equal to [tex]p_{1}[/tex][tex]p_{2}[/tex]...[tex]p_{n}[/tex]+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 "[itex]p_1p_2\cdot\cdot\cdot p_n+ 1[/itex]" reminds me of Euclides proof that there are an infinite number of primes but he never asserts that [itex]p_{n+1}[/itex] is smaller than or equal to that.

It does, of course, follow from Euclid's proof. Assuming that [itex]p_1[/itex], [itex]p_2[/itex], ..., [itex]p_n[/itex] is a list of all primes less than or equal to [itex]p_n[/itex], [itex]p_1p_2\cdot\cdot\cdot p_n+ 1[/itex] is certainly not divisible by any of [itex]p_1[/itex], [itex]p_2[/itex], ..., [itex]p_n[/itex]. Therefore it is either a prime itself or is divisible by a prime larger than any of [itex]p_1[/itex], [itex]p_2[/itex], ..., [itex]p_n[/itex]. In either case there must be a prime less than or equal to [itex]p_1p_2\cdot\cdot\cdot p_n+ 1[/itex]
 
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  • #3

Homework Statement



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

[tex]p_{n+1}[/tex] smaller or equal to [tex]p_{1}[/tex][tex]p_{2}[/tex]...[tex]p_{n}[/tex]+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 [tex]p_{1} < p_{2}[/tex], then it follows that [tex]p_{1} < p_{1}^2 < p_{1} \cdot p_{2}[/tex], by inductive reasoning you can prove that [tex]p_{n+1}<p_{n+1} \cdot p_{1} < p_{n+1} \cdot p_{1} \cdot p_{2}[/tex], and so on.
 
  • #4
Gib Z
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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.
 
  • #5
446
1
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 [tex]P_{1}[/tex][tex]P_{2}[/tex]...[tex]P_{n}[/tex] +1 =N
Such that N is an element of positive integer.
Thus, it suggest N is co-prime to [tex]P_{1}[/tex],[tex]P_{2}[/tex], ... ,[tex]P_{n}[/tex]


Which may be a prime number that is [tex]P_{k}[/tex]
Such that k is a value larger or equal to n+1


OR

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