Prove, using induction, that for any odd integer n > 1, the sum of all positive odd integers less than n is less thanthe sum of all positive even integers less than n

Proof by induction involves two steps: the base case and the inductive step.

Base Case: Let's start with the smallest odd integer greater than 1, which is 3. The sum of all positive odd integers less than 3 is 1. The sum of all positive even integers less than 3 is 2. So, the base case holds.

Inductive Step: Assume the statement is true for some odd integer n = 2k + 1, where k is an integer. That is, assume that the sum of all positive odd integers less than 2k + 1 is less than the sum of all positive even integers less than 2k + 1.

We need to prove that the statement is true for the next odd integer n = 2k + 3.

The sum of all positive odd integers less than 2k + 3 is the sum of all positive odd integers less than 2k + 1 plus 2k + 1.

The sum of all positive even integers less than 2k + 3 is the sum of all positive even integers less than 2k + 1 plus 2k + 2.

By the inductive hypothesis, the sum of all positive odd integers less than 2k + 1 is less than the sum of all positive even integers less than 2k + 1.

Therefore, the sum of all positive odd integers less than 2k + 3 is less than the sum of all positive even integers less than 2k + 3.

Since the base case is true and the inductive step has been proven, we can conclude by induction that for any odd integer n > 1, the sum of all positive odd integers less than n is less than the sum of all positive even integers less than n.