Question

Prove that for all integers n, there exists an even integer k such that

n < k + 3 ≤ n + 2
(You can use that facts without proof that even plus even is even or/and even plus odd is odd.)

Answer 1

To prove that for all integers n, there exists an even integer k such that n < k + 3 ≤ n + 2, we can consider two cases. In both cases, we have found an even integer k that satisfies the given condition for any integer n.

Step-by-step explanation:

To prove that for all integers n, there exists an even integer k such that n < k + 3 ≤ n + 2, we can consider two cases:

Case 1: If n is even, let k = n. Then, k + 3 = n + 3, which is an even integer and n + 3 ≤ n + 2.

Case 2: If n is odd, let k = n + 1. Then, k + 3 = n + 4, which is an even integer and n + 4 ≤ n + 2.

In both cases, we have found an even integer k that satisfies the given condition for any integer n.