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My textbook illustrates strong induction using a ladder analogy as follows: Suppose we can reach the first and second rungs of an infinite ladder, and we know that if we can reach a rung, we can reach two rungs higher. Prove we can reach every rung.

Proof: “Assume we can reach the first $k$ rungs (inductive hypothesis). From the given information, we can reach the $(k+1)$st rung from the $(k-1)$st rung for all $k\ge2$, and, because $k-1 \le k$, by the inductive hypothesis we can reach the $(k-1)$st rung. This completes the inductive step and finishes the proof.”

But I thought the inductive step required us to prove that if we can reach the first $k$ rungs, we can reach the $(k+1)$st rung, not if we can reach the first $(k-1)$st rungs, we can reach the $(k+1)$st rung. In other words, what happened to $k$? I just don't see how this proof fits in with the definition of strong induction...

yroc
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1 Answers1

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If we had used weak induction, then we would only be allowed to assume that the induction hypothesis holds true for $k' = k$, where $k \geq 2$. In other words, we would only be allowed to assume that the $k^\text{th}$ rung is reachable.

Strong induction, on the other hand, lets us assume that the induction hypothesis holds true for all $k' \in \{1, 2, \ldots, k\}$, where $k \geq 2$, so that any of the first $k$ rungs are reachable. Thus, since $1 \leq k - 1 \leq k$, we know in particular that the induction hypothesis holds true for $k' = k - 1$. This lets us assume that the $(k - 1)^\text{th}$ rung is reachable.

Adriano
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  • Yes, but the induction hypothesis is supposed to be that all of the first $k$ rungs are reachable, but the proof only seems to use the fact that the first $k-1$ rungs are reachable. This is what's confusing me. – yroc Nov 02 '14 at 00:30
  • Indeed, the induction hypothesis is very powerful, and we end up not using its full power $-$ only some of it. All we need is the fact that the $(k - 1)^\text{th}$ rung is reachable. We don't even use the fact that the first $k - 1$ rungs are reachable. – Adriano Nov 02 '14 at 00:32
  • OK, but doesn't the definition of strong induction specify that you have to prove that if the first $k$ statements are true, then the $k+1$st statement is true? It doesn't allow (by my reading) that one can pick and choose which statements (from 1 to $k$) to assume as the induction hypothesis? – yroc Nov 02 '14 at 00:46
  • The induction hypothesis lets us assume that the first $k$ statements are true. Since all $k$ statements are true, we may pick and choose any one of the first $k$ statements if it helps us prove that the $(k + 1)^\text{th}$ statement is true. – Adriano Nov 02 '14 at 00:50
  • I'm sorry, but there must be something very basic that I'm misunderstanding. If you're required prove the conditional "The first $k$ statements are true implies the $(k+1)$th statement is true," doesn't this mean you have to use the fact that all of the first $k$ are true in your proof? Suppose, as you suggest, you only use the $(k-1)$th statement in your proof. But you're not using the $k$th statement, the $(k-2)$th statement, the $(k-3)$th statement etc. So how have you really proved the conditional? – yroc Nov 02 '14 at 01:00
  • Here's an analogy. Suppose that we want to prove the conditional: "If you live in Canada and you like chocolate, then you live in North America." Indeed, suppose that you live in Canada and you like chocolate. Then since you live in Canada, we know that you live in North America, as desired. The hypothesis here involved an extra statement that wasn't actually required to prove the conclusion. Nevertheless, we've proven the conditional. – Adriano Nov 02 '14 at 02:03
  • Let us continue this discussion in chat. – yroc Nov 02 '14 at 02:28