Order 7 of the following sentences so that they prove the following statement by contrapositive: For any integers m and n, it 3 + mn, then 3 + m. Proof by contrapositive of the statement (in order): Choose from this list of sentences Since 31m, there exists an integer k such that m = 3k Since k and n are integers and integers are closed under multiplication kn must be an integer 3|m Since 31mn, there exists an integer k such that mn = 3k It follows that, mn = 3kn = 3(kn) Let m and n be integers and 3 mn

Order 7 of the following sentences so that they prove the following statement by contrapositive: For any integers m and n, it 3 + mn, then 3 + m. Proof by contrapositive of the statement (in order): Choose from this list of sentences Since 31m, there exists an integer k such that m = 3k Since k and n are integers and integers are closed under multiplication kn must be an integer 3|m Since 31mn, there exists an integer k such that mn = 3k It follows that, mn = 3kn = 3(kn) Let m and n be integers and 3 mn

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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IV. Let P(n):n and n + 2 are primes. be an open sentence over the domain N. Find six positive integers n for which P(n) is true. If n E N such that P(n) is true, then the two integers n ,n + 2 are called twin primes. It has been conjectured that there are infinitely many twin primes.
Note: For all integers k,n it is true that kn, k+n, and k-n are integers. An integer k is even if and only if there exists an integer r such that k=2r. An integer k is odd if and only if there exists an integer r such that k=2r+1. For every integer k it is true that if k is even then k is not odd. For every integer k it is true that if k is odd then k is not even. For every integer k it is true that if k is not even then k is odd. For every integer k it is true that if k is not odd then k is even. If P then Q means the same thing as P → Q (P implies Q). 3. Consider the argument form pvr .. p→r Is this argument form valid? Prove that your answer is correct. 4. Prove that for every integer d, if d³ is odd then d is odd.
7. For n 2 1, in how many out of the n! permutations T = (T(1), 7(2),..., 7 (n)) of the numbers {1, 2, ..., n} the value of 7(i) is either i – 1, or i, or i +1 for all 1 < i < n? Example: The permutation (21354) follows the rules while the permutation (21534) does not because 7(3) = 5. Hint: Find the answer for small n by checking all the permutations and then find the recursive formula depending on the possible values for 1(n).
Show that the following argument is valid for any natural number n ∈N. p1 →p2 p2 →p3 ... pn−1 →pn ¬pn ∴¬(p1 ∨···∨pn−1................
es remaining 8. Consider the function f:NxN-N defined recursively by: 1) Base case: Let meN and define (0,m) = 0 2) Recursive case: For any x,meN, x>0, define f(x,m) = (x-1,m) + (m+m) Prove the following theorem holds using proof by induction: Thereom: For any n,meN, m>0 we have (n.m) I m = n+n Fill in your answer here 9 Help BIU X, x L - ɔE =N E X Format
I need answer of all sentences p
Write the following as proposition: For every integer n, there exists an integerm such that m > n?
A student wants to prove by induction that a predicate P holds for certain nonnegative integers. They have proven that for all integers n ≥ 0 that P(n) → P(3n). Suppose the student has proven P(3). Which of the following propositions can they infer? (The domain for any quantifiers appearing in the answer choices is the natural numbers.) O Vn, P(3+3) O P(n) does not hold for ŉ < 0 P(n) for n = 6, 9, 12,... Vn ≥ 1, P(3¹)
5. Prove by simple induction onn that 2" > n
= For integer n ≥ 1 let P(n) be the predicate that 9" - 5n CEZ 4c for some For the induction hypothesis, consider k ≥ 1, and suppose that P(k) is true. For the inductive step, we want to show that P(k+1) is true. True or false: The following proof correctly proves P(k+ 1) true, where every step other than the one labelled IH follows by algebra. (I'm asking: is this a valid algebraic proof? Is the algebra correct? Did I use the IH correctly? Did I get the correct final result?) 9k+15k+19.9k - 5.5k True False = 9(4c – 5k) – 5 ⋅ 5% for CE Z by the IH = 4.9c-9.5k - 5.5k = 4.9c-4.5k = 4d for d = Z
Suppose that the equation ax b .mod n/ is solvable (that is, d j b, whered D gcd.a; n/) and that x0 is any solution to this equation. Then, this equation has exactly d distinct solutions, modulo n, given by xi D x0 C i.n=d / fori D 0; 1; : : : ; d 1
The Modular Operation x mod m = r denotes that r is the remainder of the division of x by m. For example, 27 mod 4 = 3. If two integers have the same remainder, then they are equivalent. For example, 27 ≡ 55 mod 4. An integer x is called prime if the only two positive integers that evenly divide it are 1 and x. Using these defifinitions, rewrite each of the following theorems using quantififiers and predicates. Note that the theorems are not precisely stated. You are allowed to use only the predicate Prime(x) that is True if x is a prime, and False otherwise. No other predicates can be used. You can also use either | or the mod defifinition to indicate that a number is divisible by another. Consider all numbers as positive integers greater than 0. 1. Lagrange’s four-square theorem: Every natural number can be expressed as a sum of four integer squares. 2. Kaplansky’s theorem on quadratic forms (partial): Any prime number p equivalent to 1 mod 16 can be represented by both or neither…