For each of the following statements, either prove that the statement is true, or finda counterexample to show that the statement is false and explain your reasoning.(a) For each integer n, n is even if and only if 6n + 4 is even.(b) For all integers ℓ, m, n, if ℓ + m is odd and ℓ + n is even, then m + n is odd.
Question
For each of the following statements, either prove that the statement is true, or finda counterexample to show that the statement is false and explain your reasoning.(a) For each integer n, n is even if and only if 6n + 4 is even.(b) For all integers ℓ, m, n, if ℓ + m is odd and ℓ + n is even, then m + n is odd.
Solution 1
(a) The statement is true. Here's the proof:
If n is even, then n can be written as 2k for some integer k. Then 6n + 4 = 6(2k) + 4 = 12k + 4 = 2(6k + 2), which is even because it's a multiple of 2.
Conversely, if 6n + 4 is even, then it can be written as 2m for some integer m. Then 6n = 2m - 4 = 2(m - 2), which implies that n = (m - 2) / 3. Since m is an integer, (m - 2) is also an integer, and therefore n is an integer. Because n can be written as a multiple of 2, n is even.
(b) The statement is false. Here's a counterexample:
Let ℓ = 1, m = 2, and n = 1. Then ℓ + m = 1 + 2 = 3, which is odd, and ℓ + n = 1 + 1 = 2, which is even. However, m + n = 2 + 1 = 3, which is odd, not even. Therefore, the statement is not true for all integers ℓ, m, and n.
Solution 2
(a) The statement is true. Here's the proof:
If n is even, then n can be written as 2k for some integer k. Then 6n + 4 = 6(2k) + 4 = 12k + 4 = 2(6k + 2), which is even because it's a multiple of 2.
Conversely, if 6n + 4 is even, then it can be written as 2m for some integer m. Then 6n = 2m - 4 = 2(m - 2), which implies that n = (m - 2) / 3. Since m is an integer, (m - 2) is also an integer, and the division of an integer by 3 is an integer if and only if the numerator is a multiple of 3. Therefore, n is even.
(b) The statement is false. Here's a counterexample:
Let ℓ = 1, m = 2, and n = 1. Then ℓ + m = 1 + 2 = 3, which is odd, and ℓ + n = 1 + 1 = 2, which is even. However, m + n = 2 + 1 = 3, which is odd, not even. Therefore, the statement is not true for all integers ℓ, m, and n.
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