1. Prove the following statement:If for some n € Z>1, 2" – 1 is prime then n is prime.by both contradiction and contrapositive.Hint. Let x, n € Z>o and let 1 ≤ k ≤n, 1 ≤l≤n with n= kl. Then (x-1) is composite:(x¹ − 1) = (x² − 1) (xl(k-1) + xl(k-2) +...x² +1).

Given statement: Statement 1: For some n∈ℤ+, 2n-1 is prime. Statement 2: n is prime. Prove that…

Answered: Prove the following statement: If for…

Prove the following statement: If for some ne Z>1, 2" - 1 is prime then n is prime. by both contradiction and contrapositive.

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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8 people are accused of robbery. Exactly 3 of the are guilty, the others are innocent. They confessed the following: A: "G is innocent" E: "B is innocent" H: "F is innocent" C: "H is innocent" F: "D is innocent" D: "A is innocent" G: "E is innocent" Innocent people always tell the truth. Who committed the roberry? Mathematicians never accept solutions without proofs. Write up ac onvincing explanation for your solution.You must justify why these are the only three people who could have committed the crime. This includes a reason for why no one else may have been one of the three.
Prove or disprove the following statement, (c):
- A Pythagorean prime number is a prime number of the form 4n + 1, where n ≥ 1. A Mersenne prime number is a prime number of the form 2" – 1. A perfect number is a number that is equal to the sum of all of its divisors, including 1 but excluding the number itself. Consider the following statements: (1) Let m is the smallest Pythagorean prime number. Then, 2¹ – 1 is a Mersenne prime number. (2) If m is a perfect number, then 2 – 1 is a Mersenne prime number. Select one of the following choices: (a) (1) is True and (2) is False. (c) (1) and (2) are True. (b) (1) is False and (2) is True. (d) (1) and (2) are False.
12. Consider the following (true) claim and two possible proofs. Claim. If n is a positive integer, then n² + 3n+2 is not a prime number. Proof 1. Note that n² + 3n+2= (n + 2)(n+1). Since n 1, we have n + 1>1 and n+2 > 1, so n²+3n+2 has at least two divisors other than 1 and is therefore not prime. Proof 2. If n is even then we can write n = 2k for some positive integer k. Then n² + 3n+2 = 4k² + 6k+2 = 2(2k² +3k+1), which is even and greater than 2. Since 2 is the only even prime, it follows that n² + 3n+ 2 is not prime. Which of these proofs are valid (i.e. which of them actually prove the claim)? (a) Both proofs are valid. (b) Proof 1 is valid, but Proof 2 is not. (c) Proof 2 is valid, but Proof 1 is not. (d) Neither proof is valid.
25. If p >1 is an integer and nłp for each integer n for which 2
what steps need more justification or explanation Proof: 1) Proof by contradiction, so I need to show that the negation is false. 2) Negation: There exists a largest prime number N. 3) Suppose there exists a largest prime number N. Then all numbers bigger than N can be factored into some smaller primes. 4) Consider the product N!+1= (1*2*3*4 … *N) +1 5) Any number between 1 and N will have a remainder of 1 when you try to divide it into N! +1. 6) Rephrasing: (N!+1)/x is not a whole number for any integer x in {1, … N}. 7) That means none of those numbers {1, … ,N} can be factors of N!+1 because they don’t divide in evenly. 8) So N!+1 has no factors between 1 and N. But all of the primes are between 1 and N because N is the largest prime. So N!+1 would need to be a prime! 9) But N!+1 is bigger than N. Contradiction
discrete math
If the contradiction method is used to prove the assertion "For any positive integer n > 0, if a" is even, then a is even.", then a valid implementation of that method begins by assuming that the negation of the statement is true. Choose all of the following statements that correctly express the negation of the above assertion. Vn E N – {0}, rem(a?, 2) = 0 → rem(a, 2) = 0 Vn EN – {0}, rem(a², 2) = 0 rem(a, 2) = 1 En e N – {0}, rem(a?, 2) = 0 → rem(a, 2) In eN - {0}, rem(a², 2) = 0 ^ rem(a, 2) = 1 For any positive integer n > 0, if a" is even, then a is odd. There exists a positive integer n > 0 where a" is even implies that a is od. There exists a positive integer n > 0 where a" is even and a is odd.