Discrete mathematics Proof by contradiction. Assume for the sake of argument that the set P of all primes is finite. Consider the product of all primes, and let r = 1 + ∏p∈P p. Then we have r mod p = 1 for any prime p ∈ P. None of the primes p ∈ P divides r, so r is a prime. This is a contradiction because r not(∈) P.  This proof is not valid. Identify and explain where the error is in the argument.

Prove or disprove the following statement, (c):

Proficiency #7. [Counterexamples] For each of the following two statements, give a counterexample to show that the state- ment is false. a. If x is an integer such that x? = 1 mod 12, then x =1 mod 12 or x = -1 mod 12. b. If A, B, D are sets such that A C B and B Z D, then A Z D.

Give a carefully worded proof for the following statements using direct proof, contrapositive proof or proof by contradiction. No matter which proof method you choose to use, make sure you follow the outline properly. (If-and-Only-If Proof) Given an integer a and b, then b2(a + 3) is even if and only if a is odd or b is even. (Use the basic definitions of even and odd in your proof.)

Assume a, b, c, d ∈ Z. Prove each of the following. (a) If a|(15b−c) and 7c∼=5b (mod2a), then a|20c. (b) If a|b and b|c then a|c. (c) Prove the following deduction ∴ a|0 and if 0|a then a = 0. (d) Prove the deduction ∴(d|(ab+c)&d|b) ⇐⇒ (d|b&d|c)

Find all integers, n, where -15 ≤≤ n ≤≤ 15 which make 15 ≡≡ n mod 4  (4 points) Show all your work: Answer= __________________

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) =…

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