Let Go n! + 1, where n is a positive integer. We will generate a sequence...

 

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Let Go n! + 1, where n is a positive integer. We will generate a sequence of primes by performing the following steps. Begin by finding P₁ which is any prime divisor of G₁, then compute P2 by finding any prime divisor of G and in general Pj+1 which is any prime divisor of G (a) Show that p₁ = 2, P₂ = 3, P3 = 7 and P4 = 71. Begin by finding G₁ then p₁, followed by Go then p2 and so on. Hint: G₁ = 1! + 1 = 2 and P₁2, G = G₂ = 21 + 1 = 3 and P2 = 3, find Gz = G3, then P3 and so on (b) Clearly G is an integer bigger than 1 and it follows that it has a prime divisor p. If p ≤n, why must p divide Gan!. Now simplify G-n! and explain why you get a contradiction when p divides G-n!. (c) From part (b) above, it follows that p > n and if we construct the sequence of primes p1, p2.p3,p4..., as outlined above we get that p1 <p2 <p3 <p4 <<pj < Explain why? From this it follows that there are infinitely many primes. Let Go n! + 1, where n is a positive integer. We will generate a sequence of primes by performing the following steps. Begin by finding P₁ which is any prime divisor of G₁, then compute P2 by finding any prime divisor of G and in general Pj+1 which is any prime divisor of G (a) Show that p₁ = 2, P₂ = 3, P3 = 7 and P4 = 71. Begin by finding G₁ then p₁, followed by Go then p2 and so on. Hint: G₁ = 1! + 1 = 2 and P₁2, G = G₂ = 21 + 1 = 3 and P2 = 3, find Gz = G3, then P3 and so on (b) Clearly G is an integer bigger than 1 and it follows that it has a prime divisor p. If p ≤n, why must p divide Gan!. Now simplify G-n! and explain why you get a contradiction when p divides G-n!. (c) From part (b) above, it follows that p > n and if we construct the sequence of primes p1, p2.p3,p4..., as outlined above we get that p1 <p2 <p3 <p4 <<pj < Explain why? From this it follows that there are infinitely many primes.

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a Lets find the values step by step G n 1 G 1 1 2 P1 any prime divisor ... View the full answer