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I thought strong induction relied on a proposition being true for all (natural) numbers from a given one up to (say) k, then showing p(k+1) must be true.

For example, I’m trying to understand this proof: https://math.stackexchange.com/a/2096485/442515

In the inductive hypothesis:

Assume $m\in\mathbb{Z}^+$ with $1\le m \le k$ and that our proposition holds for $m$.

If the role of the $m$ there is not to establish that the proposition is true from 1 up to k, then what role does it actually play please? The hypothesis does not seem equivalent to saying "assume the proposition holds for all positive integers from 1 to k inclusive."

At this stage, observations that strong and weak induction are the same are unlikely to help. I feel I need to get my head around this "p(all values from 1 to k inclusive) implies p(k+1) idea first, as it's the only thing I've read about string induction that actually makes sense to me at my current level of understanding.

Robin Andrews
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  • The author assumes that the assertion is true for each $m\in{1,2,3,\ldots,k}$, and uses this assumption to prove that it also holds for $k+1$. – José Carlos Santos Oct 03 '21 at 11:56
  • So what does m have to do with anything? – Robin Andrews Oct 03 '21 at 12:10
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    I don't understand your obsession with the letter $m$. We have to give names to things if we want to talk about them. Here, we have a sequence $(P(n))_{n\in\Bbb N}$ of propositions and we want to prove then by strong induction. So, we state the inductive state as follows: for each $m\in{1,2,\ldots,k}$, $P(m)$ holds. What's strange about it? – José Carlos Santos Oct 03 '21 at 12:14
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    Because what you say is claimed about m doesn't appear to be what is actually claimed, according to the answers here: https://www.quora.com/In-strong-induction-is-Assume-m-in-mathbb-Z-with-1-le-m-le-k-and-that-our-proposition-holds-for-m-equivalent-to-saying-assume-the-proposition-holds-for-all-positive-integers-from-1-to-k-inclusive – Robin Andrews Oct 03 '21 at 12:17

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Say that the statement to be proved is $S$, a statement about $n$, and that $S(m)$ means that the statement is true for $n=m$.

As an example from the proof you have linked to: In the proof the assumption $S(m)$ for $m: 1 \leq m \leq k$ is used to draw the conclusion that the statement, in particular, holds for $\dfrac{k+1}{2}$, that is, $S\Big(\dfrac{k+1}{2} \Big)$. This in turn is used to prove $S(k+1)$.

So while the proof only uses the assumption $S\Big(\dfrac{k+1}{2} \Big)$ in order to prove $S(k+1)$, strong induction provides a straight forward way to let us assume $S\Big(\dfrac{k+1}{2} \Big)$ in the first place, and in particular, weak induction would not have allowed us to do this.

I imagine it would be difficult to come up with some kind of induction proof that as its inductive hypothesis only assumes $S\Big(\dfrac{k+1}{2} \Big)$ in order to prove $S(k+1)$. Even if it was possible, why should one do this when strong induction provides a straight forward way for us to assume $S\Big(\dfrac{k+1}{2} \Big)$, and does not pose any other difficulties?

Edit: After reading your comment and re-reading the linked proof, I am inclined to agree with you. I didn't read your question carefully enough the first time, sorry about that.
I just think that it is a bit of a sloppy formulation in the linked proof. What the author should mean, is indeed this:

"Assume that our proposition is true for all $m$ such that $m \in \mathbb{Z}_+, 1 \leq m \leq k$.

It is probably just a case of being used to how proofs by induction are usually formulated, so I guess most people (including myself) assume that the author really talks about all $m$ such that... not a single $m$ with those properties. And this really should be what the author means, because the proof does not work otherwise.

Paradox
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