Here it is. The big leagues. We're finally ready to rock our own inductive proof. Let's do this thing. First, we prove our base case: When n = 1, we get 4(1) + 1 = 5, which is an odd number. Beautiful. Now we assume our general statement is true when n = k: 4k + 1 is an odd number. Next up, we replace k with k + 1: Is 4(k + 1) + 1 an odd number? We've got to prove this new statement is true. For starters, let's distribute that 4: 4(k + 1) + 1 = 4k + 4 + 1 Don't combine those constants just yet, though. We're already assuming that 4k + 1 is an odd number, so let's rearrange our new expression in a way that contains 4k + 1. 4k + 4 + 1 = (4k + 1) + 4 There we go. Since we "know" that 4k + 1 is odd, we can think of our expression like this: (4k + 1) + 4 = (odd number) + 4 What happens when we add 4 to an odd number? The same thing that happens when we add any even number to an odd number: odd + even = odd Go ahead and test this baby out with a few values. 3 + 4 = 5 7 + 4 = 11 11 + 4 = 15 835 + 4 = 839 See? Our sum is always odd. So since 4k + 1 is an odd number and 4 is an even number, we'll always, always get an odd sum when we add 'em. (odd number) + 4 = odd number There aren't any possible counterexamples, so (4k + 1) + 4 is always an odd number. We proved it's true for k + 1, which means our original statement is also true: 4n + 1 always equals an odd number when n is a positive integer. |