Consider the following proof that the statement "for all integers n, n2 + 2n" is not true. Proof: Let n = 2, then we have n2 = 22 = 4 = 2.2 = 2n. Therefore above statement is not true. Which proof method is used in above proof. Proof by counterexample Direct proof O Proof by contraposition O Proof by contradiction

Only need part(d) and part(f). Thank you for the help!

Hi, the following question is about proof by contradiction, can you please help thanks w working out thanks.

3. Proof by Contrapositive Prove the following statement by contrapositive: If 6n2 +2n+2 is odd, then n is odd.

By dragging statements from the left column to the right column below, construct a valide proof of the statement: For all integers n, if 13n is even, then n is even. The correct proof will use 3 of the statements below. Statements to choose from: Let n be an arbitrary integer and assume n is odd. Let n be an arbitrary integer and assume n is even. Since 13 is odd and the product of an odd number and an even number is even, Let n be an arbitrary integer and assume 13n is even. 13n must be even Since 13 is odd and the product of odd numbers is odd, 13n must be odd. n must be even Since an even number divided by 13 must be even, Your Proof: Put chosen statements in order in this column and press the Submit Answers button.

I need help proving this statement    For all integers a, b and c if aꟾc and bꟾd then abꟾcd   Thank you

Give a counterexample to the statement. The product of 2 primes is always odd.

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

Consider the statement that 3 divides n³ + 2n whenever n is a positive integer. Outline the proof by clicking and dragging to complete each statement. Let P(n) be the proposition that Basis step: P(1) states that Inductive step: Assume that Show that We have completed the basis step and the inductive step. By mathematical induction, we know that vk>0, (P(K)→ P(k+ 1)) is true, that is, vk>0, (3 divides k³ + 2k→ 3 divides (k+ 1)³ + 2(k+ 1)). 3 divided K³ + 2k for an arbitrary integer k> 0. P(n) is true for all integers n ≥ 1. 3 divides 1³ +2 3 divides n³ + 2n. 1, which is true since 1³ +21=3, and 3 divides 3. Reset Click and drag statements to fill in the details of showing that VK(P(K) → P(k+ 1)) is true, thereby completing the induction step. By the inductive hypothesis, 3 divides (k³ + 5k). 3 divides 3(K³ + 1) because it is 3 times an integer. By part (i) of Theorem 1 in Section 4.1, 3 divides the sum (k³ + 2k) + 3(k² + k + 1). This completes the inductive step. (k+ 1)³ + 2(k+ 1) =…

Use the principle of mathematical induction to prove “6? − 1 is divisible by 5 for each integer ? ≥ 0”.You need an introduction, body, and conclusion. Show the Inductive Hypothesis. Points will be givenfor clarity. Hint: ??+1 = ??⋅ ?.