MATH Section 1.7

Use the following building blocks in the right column to assemble a direct proof that the product of two odd numbers is odd. Not all blocks belong in the proof.

Correct Blocks and order (going down): Suppose that n and m are odd numbers. Then, n = 2k + 1 for some integer k and m = 2q + 1 for some integer q. By multiplying these equations, we obtain nm = (2k + 1)(2q + 1). Distributing, we get nm = 2(2kq + k + q) + 1. By definition of an odd integer, that means that nm is odd.

Show that at least ten of any 64 days chosen must fall on the same day of the week.

Correct blocks and order (going down): We prove this using the method of contradiction. Suppose there are nine or fewer days that fall on the same day of the week. This would account for at most 9 x 7 = 63 days, but we have chosen 64 days. This contradiction shows that at least 10 of the days must be on the same day of the week.

Use the following building blocks to assemble a proof that if m and n are integers, and mn is even, then m is even or n is even. Not all blocks belong in the proof. Before you start, you might want to write out such a proof on paper, and carefully consider whether a direct proof, a proof by contraposition or a proof by contradiction is the most appropriate.

Correct blocks and order( going down): We will give a proof by contraposition. Suppose that m is odd and n is odd. By definition of odd number, m = 2p + 1 for some integer p, and n = 2q + 1 for some integer q. Therefore, nm = (2q + 1)(2p + 1) = 4pq + 2q + 2p + 1 = 2(2pq + q + p) + 1. By definition of odd number, this means that nm is odd. Therefore, we have proved the contrapositive of the desired statement and thus completed the proof by contraposition.

Let P(n) be the proposition "If a and b are positive real numbers, then (a + b)n ≥ an + bn." What is P(1)? (You must provide an answer before moving to the next part.)

If a and b are positive real numbers, then (a + b)1 ≥ a1 + b1 = a + b.

Consider a positive integer n. Select which of the following biconditionals is true. (Check all that apply.) (You must provide an answer before moving to the next part.) Check All That Apply If n is even, then 7n + 4 is even; if 7n + 4 is even, then n is even. If n is odd, then 7n + 4 is odd; if 7n + 4 is odd, then n is odd. If n is even, then 7n + 4 is even; if 7n + 4 is odd, then n is odd. If n is odd, then 7n + 4 is even; if 7n + 4 is even, then n is odd.

If n is even, then 7n + 4 is even; if 7n + 4 is even, then n is even. If n is odd, then 7n + 4 is odd; if 7n + 4 is odd, then n is odd.

Prove that if n is an integer and 3n + 2 is even, then n is even using a proof by contradiction. Explanation: Suppose that 3n + 2 is even and that n is odd. Since 3n + 2 is even, so is 3n. We know that if we subtract an odd number from an even number, we get an odd number. As 3n is even and n is odd, 3n - n should be odd, but 3n - n = 2n is even. This is a contradiction.. Therefore, our supposition was wrong; hence n is even.

Steps: Suppose that 3n + 2 is even and that n is odd. Since 3n + 2 is even, so is 3n. We know that if we subtract an odd number from an even number, we get an odd number. As 3n is even and n is odd, 3n - n should be odd, but 3n - n = 2n is even. This is a contradiction. Therefore, our supposition was wrong; hence n is even.

Let P(n) be the proposition "If a and b are positive real numbers, then (a + b)n ≥ an + bn." The statement P(1) is true because (a + b)1 ≥ a1 + b1 = a + b. (You must provide an answer before moving to the next part.) True or False

True

Let P(n) be the proposition "If a and b are positive real numbers, then (a + b)n ≥ an + bn." What type of proof was used to prove that P(1) is true? (You must provide an answer before moving to the next part.)

direct proof

Prove that if n is an integer and 3n + 2 is even, then n is even using a proof by contraposition. Explanation: In order to prove the contrapositive, assume that n is odd. Then, we can write n = 2k + 1 for some integer k. Then, 3n + 2 = 3(2k + 1) + 2 = 6k + 5 = 2(3k + 2) + 1. This number is again of the form 2p + 1 for the integer p = 3k + 2; hence, it is odd. Thus, if n is odd, then 3n + 2 is odd.

steps: Assume that n is odd. Then, we can write n = 2k + 1 for some integer k. Then, 3n + 2 = 3(2k + 1) + 2 = 6k + 5 = 2(3k + 2) + 1. This number is again of the form 2p + 1 for the integer p = 3k + 2; hence, it is odd. Thus, if n is odd, then 3n + 2 is odd.

Show that if n is an integer and n3 + 5 is odd, then n is even using a proof by contraposition. Explanation: In order to prove the contrapositive, assume that n is odd. We can write n = 2k + 1 for some integer k. Then, n3 + 5 = (2k + 1)3 + 5 = 8k3 + 12k2 + 6k + 6 = 2(4k3 + 6k2 + 3k + 3). As n3 + 5 is two times an integer, it is even. Thus, if n is odd, then n3 + 5 is even.

steps: Assume that n is odd. We can write n = 2k + 1 for some integer k. Then, n3 + 5 = (2k + 1)3 + 5 = 8k3 + 12k2 + 6k + 6 = 2(4k3 + 6k2 + 3k + 3). As n3 + 5 is two times an integer, it is even. Thus, if n is odd, then n^3 + 5 is even.

Consider a positive integer n. Put the steps in correct order to prove that if n is even, then 7n + 4 is even (You must provide an answer before moving to the next part.) Rank the options below.

steps: Since n is even, it can be written as 2k for some integer k. Then 7n + 4 = 14k + 4 = 2(7k + 2). This is 2 times an integer, so it is even.

Show that if n is an integer and n3 + 5 is odd, then n is even using a proof by contradiction. Explanation: Suppose that n3 + 5 is odd and that n is odd. As n is odd, n3 is odd. We know that the sum of two odd numbers is even. As n3 and 5 are odd, their sum n3 + 5 should be even, but it is given to be odd. This contradicts that n3 + 5 is odd and that n is odd. Therefore, our supposition was wrong; hence n is even.

steps: Suppose that n3 + 5 is odd and that n is odd. As n is odd, n3 is odd. We know that the sum of two odd numbers is even. As n3 and 5 are odd, their sum n3 + 5 should be even, but it is given to be odd. This is a contradiction. Therefore, our supposition was wrong; hence n is even.

Consider a positive integer n. Put the steps in correct order to prove that if 7n + 4 is even, then n is even using contraposition.

steps: Assume that n is odd. n can be written as 2k +1 for some integer k. Then 7n + 4 = 14k +11 = 2(7k + 5) + 1. The obtained integer is 1 more than 2 times an integer, so it is odd.