Each one of the following is an attempted proof of the statement For every integer n, there is an odd number k such that n < k < n+3. Only one of the proofs is correct. Match each proof with a correct analysis of its merits. Let the integer n be given. If n is even, let k be n+1. If n is odd, let k be n+2. Either way, k is odd and n < k < n+3. That proves that for any integer n, an odd k such that n < k < n+3 exists. Let n be given. Then k = 2n+1 is odd by definition, and greater than n. Since also k < n+3, we have shown the existence of an odd k between n and n+3 for all n. Given the integer n, pick k - n + 2. Then n < k < k+3. Thus, for every integer n, an odd k with n < k < n+3 exists. Let the odd integer k be given. Pick n = k-1. Then n < k < n+3. We have shown that for every integer n, an odd integer k with n < k < n+3 exists. A. The proof is correct. By starting with the assumption that n is an arbitrary integer, it sets up universal generalization. Then it makes a case distinction, so that no matter whether n is even or odd, k comes out to be odd, and between n and n+3. By universal generalization, that proves that such k exists for all integers n.
B. This proof shows that for every odd k, an integer n with n < k < n+3 exists, which is a different statement than what was supposed to be proved, and not logically equivalent.
C. The proof starts with the proper assumption but constructs the wrong k. While the k is between n and n+3, it may not be odd. The k chosen Is even when n is even.
D. The proof starts with the proper assumption, but then constructs the wrong k. 2n+1 while odd, is generally not between n and n+3.