Archived Notes on Grading: HW5

Please see our guidelines for induction proofs.

In general, we are very picky when we grade induction proofs.

To get full credit for this exercise, you needed to clearly and explicitly state that the reasoning in the body of the inductive step breaks down when n is 0.

Please note that the proof uses the second principle of mathematical induction (pp. 197-9 in Rosen). This alternative form of induction says that you can prove the statement...

P(n) for all nonnegative integers n

...by showing...

1. Base case: P(0)
2. Inductive step: If P(k) for all 0 <= k <= n, then P(n + 1)

Thus, the hypothesis "a^k = 1 for all nonnegative integers k <= n" in the flawed proof's inductive step is perfectly legitimate; that assumption is just a rewording of "P(k) for all 0 <= k <= n," step II's antecedent.

[By the way, "strong induction" is a (quite common) synonym for "the second principle of mathematical induction."]

Also, simply saying, "it needs two base cases since P(n - 1) is referred to in the inductive step," is insufficient. There's no theorem that says...

If both P(n) and P(n - 1) appear in the inductive step, then you must have two base cases.

[The above is a good heuristic to keep in mind when you're constructing induction proofs, but it's not universally true.]