Mathematical Induction vs Strong Induction

Question

In Rosen's book Discrete Mathematics and Its Applications, 8th Edition it is mentioned that:

You may be surprised that mathematical induction and strong induction are equivalent. That is, each can be shown to be a valid proof technique assuming that the other is valid.

One of the examples given for strong induction in the book is the following:

Suppose we can reach the first and second rungs of an infinite ladder, and we know that if we can reach a rung, then we can reach two rungs higher … prove that we can reach every rung using strong induction

If the two proof techniques are "equivalent", how can I prove the above example using mathematical induction (as opposed to strong induction)?

Does this answer you? https://math.stackexchange.com/questions/108297/strong-induction-proofs-done-with-weak-induction — ancient mathematician, Apr 14 '22 at 06:42
Does this answer your question? Strong Induction proofs done with Weak Induction — ancient mathematician, Apr 14 '22 at 06:43
The statements of the strong and weak induction do not come out of thin air. They have proofs. If you know how to prove something using one of them, to see how to prove it using he other, you follow the proof of the other with that example in mind. It's actually a good exercise to integrate your understanding of the proof. The goal, however, is to not have to do this. The proof abstracts away the details and you can then just use it. — Git Gud, Apr 14 '22 at 06:46
@ancientmathematician it would be nice if the answer could be explained in the context of the "infinite ladder" example given in the textbook. — Sandeep, Apr 14 '22 at 06:46
@Sandeep Then include that requirement in your question. Also, do you know how to prove it using strong induction? — Git Gud, Apr 14 '22 at 06:47
@GitGud yes I know how to prove it using strong induction because the textbook explains it {basically when you are on the (n-1)th rung and you reach out for two rungs higher, you get to the (n+1)th rung. With classical mathematical induction you would know how to get to the (n+2)th rung from the nth rung, but you would not know how to get to the (n+1)th rung}. — Sandeep, Apr 14 '22 at 06:52
@Sandeep FYI, using this search, I found the fairly closely related Demonstration of strong induction using ladder rungs and Question about why a certain element in a strong induction is part of the inductive hypothesis and not the base cases. — John Omielan, Apr 14 '22 at 07:02

score 0 · Answer 1 · answered Apr 14 '22 at 06:47

Just because they're equivalent doesn't mean you can easily swap between them - the point of strong induction is that it lets you skip a lot of steps in your proof.

However, in this case you could do something like this:

Let $P(n)$ be the statement "I can reach the $2n$-th and $2n+1$-th rungs of the ladder." Then we shall prove that $P(n)$ is true for all integer $n \geq 0$.

As the base case, note that $P(n)$ = "I can reach the 1st and 2nd rungs of the ladder", which is given.

Then, assume that $P(n)$ is true for some arbitrary $n$. Then knowing I can always reach two rungs higher, I can reach the $2n+2$-th and $2n+3$-th rungs, but since $2n+2 = 2(n+1)$ and $2n+3 = 2(n+1) + 1$, this means that $P(n+1)$ is true, and our proof is complete.

Mathematical Induction vs Strong Induction

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