-A Pythagorean prime number is a prime number of the form 4n + 1, where n ≥ 1. A Mersenne primenumber is a prime number of the form 2" – 1. A perfect number is a number that is equal to the sum ofall 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.

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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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3. Consider the statement: "for all integers n, if n is even then 6n is even." (a) Prove the statement. (b) Is the converse true? Prove or disprove.
Do n.9
Define an integer k to be odd if k − 1 is even. Write up a formal proof of the following, using an indirect proof: Claim: For any natural numbers m and n, if m is odd and n is odd, then m + n is even.
1- Following statement is given to you "If m and n are even integers, then mn is divisible by 4." The student used following steps to prove the given statement. If m and n are even integers then there exists integers a and b such that m = 2a andn= 2b Therefore mn can be written mn= (2a)(2b). mn=4ab Since ab is an integer, mn is 4 times an integer so it is divisible by 4. What is the name of the proof method which is used in the solution? a) O Contrapositive proof b) O Direct Proof c) O Mathematical Induction d) O Proof by Contradiction
Only need part(d) and part(f). Thank you for the help!
Let n be an integer. Then n3 – n + 3 is an odd number is an odd number for some values of n and an even number for some values. O is an even number O may not be an integer None of these
Let n be an integer. Then n3 – n + 5 O is an odd number O None of these is an odd number for some values of n and an even number for some values. O is an even number O may not be an integer
1. Induction and prime factorisation of natural numbers: Let n E N be a natural number. A prime factorisation of n is an equation n = prime number and d; e {0}UN. Using Mathematical Induction, prove that every natural number has a prime factorisation. di d2 P1 P2 de ..Pk where for i = 1, . . . , k, The following statements inay be useful in your proof: (Q1) Vm, n E N (m 2) →
n > 1 is an integer. Which of the following statements is false? Select one: a. If 10" – 1 is prime then n is prime. b. If n is even then 2" + 1 is prime. c. If 2" – 1 is prime then n is prime. d. If n is composite then 10" – 1 is composite. e. If n is composite then 2" – 1 is composite.
Every positive integer is always divisible by 1 and by itself. That is, for every posi- tive integer n we always have 1|n and n|n. We say an integer p ≥ 2 is a prime number if its only divisors are 1 and p. (By convention, 1 is not a prime number.) (a) List the first ten prime numbers. (b) If a positive integer n is not prime, then it is called a composite number. Show that every positive integer n ≥ 2 can be written as a product of prime numbers. (This is called a prime factorization of n. Notice that some primes may need to be repeated. For example, 180 = 2² × 3² × 5 and the primes 2 and 3 are repeated.) Hint: Consider two separate cases based on whether a given number n is prime or composite. For composite numbers, suppose by strong induction that all smaller integers have a prime factorization.
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
4. Explain why the following statements are true or False. (a) There is an even number that is a multiple of 3. (b) All prime numbers are odd. (c) Vm,neN, m •nzm+ n

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