Another algebra problem about prime and induction

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SUMMARY

The discussion focuses on proving by induction that the n-th prime number is less than 2 raised to the power of 2 raised to n. The approach involves assuming the statement is true for all n less than or equal to k and then comparing the (k+1)-th prime, p_{k+1}, with the product of the first k primes plus one. The use of the Euclidean algorithm is suggested to analyze the relationship between p_{k+1} and the product of the first k primes, leading to a contradiction if p_{k+1} exceeds the proposed upper bound.

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Homework Statement



prove by induction that the [itex]n^{\text th}[/itex] prime is less than [itex]2^{2^{\text n}}[/itex]

Homework Equations


hint:assume it is correct for all [itex]n \leq k[/itex], and then compare [itex]p_{k+1}[/itex] with [itex]p_{1}p_{2}...p_{k}+1[/itex]

The Attempt at a Solution


is [itex]p_{k+1}[/itex] smaller/greater than [itex]p_{1}p_{2}...p_{k}+1[/itex] so that i can extend the use of inequality?
I attempt to use euclidean algorithm but do not know where to use.
Thank you.
 
Physics news on Phys.org
If [tex]p_{k+1}[/tex] is larger than the suggested number, you should be able to prove that it has no prime divisors which is a contradiction
 

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