Question

Type your proofs/solutions or handwrite them clearly and legibly. Include all relevant details/justification. Sections: §3.3 (Proof by Contrapositive), §3.4 (Proof by Cases), and §3.5 (Proof Evaluation) 1. Let x∈Z. Prove that if 3x+4 is even, then 7x−5 is odd. 2. Let n∈Z. Prove that 2n²+n is odd if and only if cos(nπ/2​) is even. (Hint: Consider the values of integers that cos(nπ/2​) can assume.) 3. Prove that if n∈Z, then n³−n is even.

Answer 1

1. To prove that if 3x+4 is even, then 7x-5 is odd, we can use proof by contrapositive. 2. To prove that 2n²+n is odd if and only if cos(nπ/2) is even, we can use proof by cases.

Step-by-step explanation:

1. To prove that if 3x+4 is even, then 7x-5 is odd, we can use proof by contrapositive. The contrapositive of the given statement is: if 7x-5 is even, then 3x+4 is odd. Let's assume 7x-5 is even, then we can write it as 7x-5 = 2k, where k is an integer. Solving for x, we get x = (2k+5)/7. Substituting this value of x in the expression 3x+4, we get 3((2k+5)/7) + 4 = 6k/7 + 6 + 4 = 6k/7 + 10. This expression is not an integer, which means 3x+4 is not an even number. Therefore, the contrapositive is true and the original statement is proved.

2. To prove that 2n²+n is odd if and only if cos(nπ/2) is even, we can use proof by cases. We need to consider two cases:
Case 1: Assume 2n²+n is odd. We can write this as 2n²+n = 2k+1, where k is an integer. Solving for n, we get n = (2k+1)/2 = k + 1/2. Substituting this value of n in cos(nπ/2), we get cos((k + 1/2)π/2) = cos((2k+1)π/4). Since k is an integer, (2k+1)π/4 is an odd multiple of π/2, and cos(odd multiple of π/2) = 0, which means cos((2k+1)π/4) is even.

Case 2: Assume cos(nπ/2) is even. We can write this as cos(nπ/2) = 2k, where k is an integer. Taking the arccosine of both sides, we get nπ/2 = arccos(2k). Since the range of arccosine is [0,π], nπ/2 can be written as nπ/2 = mπ, where m is an integer.