For Question 16 consider the theorem and proof below then answer the question:Theorem. There are infinitely many prime numbers.Proof. Suppose to the contrary that there is only a finite number of prime numbers, viz, p1 , P2, ..., Pk-Let N = (p1 · p2 · ... · Pk) + 1. Since N is greater than each of the p1, p2, ..., Pk it must be acomposite number. Also, since N > 1 and N is a natural number, then N must have a primedivisor. Let p E {P1, P2, ... , Pk} be a prime divisor of N. Then p|N and p|p1 · P2 · ... Pr sincep occurs as a factor of p1 · P2 · ... Pr. Now, recall that for any subset {a, b, c, m, n} of the setof integers if alb and alc, then a|(mb + nc). Hence, we have that p| [N – (pı · P2 · ... Pk)], whichimplies that p|1. But p is a prime number, which implies that p 2 2, a contradiction. Therefore,there are infinitely many prime numbers.Question:In the proof above we let “N = (p1 · P2 · ... Pk) + 1." Is it necessary to add precisely 1 in thisline of the proof?If yes, justify your answer clearly; if no, explain why any (positive) integer l would work.

Name: For Question 16 consider the theorem and proof below then answer the
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Description: For Question 16 consider the theorem and proof below then answer the question:Theorem. There are infinitely many prime numbers.Proof. Suppose to the contrary that there is only a finite number of prime numbers, viz, p1 , P2, ..., Pk-Let N = (p1 · p2

Here we need to show whether is it necessary to take in N exactly plus 1.

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For Question 16 consider the theorem and proof below then answer the question: Theorem. There are infinitely many prime numbers. Proof. Suppose to the contrary that there is only a finite number of prime numbers, viz, p1, P2 · . . , Pk- Let N = (p1 · P2 · ... · Pk) + 1. Since N is greater than each of the p1, p2, ... , Pk it must be a composite number. Also, since N > 1 and N is a natural number, then N must have a prime divisor. Let p E {p1, P2,. .. , Pk} be a prime divisor of N. Then p|N and p|p · P2 · ...· Pk since p occurs as a factor of p1 · P2 · ... · Pk. Now, recall that for any subset {a, b, c, m, n} of the set of integers if alb and alc, then a|(mb+ nc). Hence, we have that p| [N – (pı · P2 · ... Pk)], which implies that p|1. But p is a prime number, which implies that p2 2, a contradiction. Therefore, there are infinitely many prime numbers. Question: In the proof above we let "N = (p1 ·P2 · ... Pk) + 1." Is it necessary to add precisely 1 in this line of the proof? If yes, justify your answer clearly; if no, explain why any (positive) integer l would work.

For Question 16 consider the theorem and proof below then answer the question: Theorem. There are infinitely many prime numbers. Proof. Suppose to the contrary that there is only a finite number of prime numbers, viz, p1, P2 · . . , Pk- Let N = (p1 · P2 · ... · Pk) + 1. Since N is greater than each of the p1, p2, ... , Pk it must be a composite number. Also, since N > 1 and N is a natural number, then N must have a prime divisor. Let p E {p1, P2,. .. , Pk} be a prime divisor of N. Then p|N and p|p · P2 · ...· Pk since p occurs as a factor of p1 · P2 · ... · Pk. Now, recall that for any subset {a, b, c, m, n} of the set of integers if alb and alc, then a|(mb+ nc). Hence, we have that p| [N – (pı · P2 · ... Pk)], which implies that p|1. But p is a prime number, which implies that p2 2, a contradiction. Therefore, there are infinitely many prime numbers. Question: In the proof above we let "N = (p1 ·P2 · ... Pk) + 1." Is it necessary to add precisely 1 in this line of the proof? If yes, justify your answer clearly; if no, explain why any (positive) integer l would work.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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