What is the purpose of replacing P(n) with P(k) during the inductive step? Isn't n and k representing the same thing? Why can't I just assume that if P(n) is true, then P(n+1) is true?
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- Convenience, 2. Yes, 3. You can.
– zkutch Feb 13 '23 at 11:45 -
1For ordinary induction you only assume $P(n)$ for the induction step, but for strong induction you assume $P(k)$ for all $k<n$. So this is the reason for replacing $P(n)$ then by $P(k)$. By the way, please use MathJax. Here is a tutorial. – Dietrich Burde Feb 13 '23 at 11:49
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2You can't "just assume that if $P(n)$ is true, then $P(n+1)$ is true" - you have to prove it. – Henry Feb 13 '23 at 12:06
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2When teaching the principle of induction it is often easier to explain things by using k and then to work with cases k=n and k=n+1. Otherwise people get confused when substituting n+1 into n and write nonsense like n=n+1. – Jaap Scherphuis Feb 13 '23 at 12:20
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Firstly, you assume that $n=k$. Then you have to prove that it is also true for $n=k+1$. – ThankYouForFlyingRyanair Mar 03 '25 at 07:03
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Given that $P(n)$ denotes the proposition $n^2>7n+1,$ its mathematical-induction proof has the structure
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Take an arbitrary integer greater than $7,$ denote it by $\boldsymbol k,$ and suppose that $P(\boldsymbol k)$ is true.
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$P(\boldsymbol k+1)$ is true.
Hence, by induction, for every integer greater than $7,\;P(n)$ is true.
By not recycling the variable name $n$ for the induction step, this systematic and reader-friendly presentation avoids giving the impression of a circular proof (that $P(n)$ is being proved by supposing $P(n)\,).$
P.S. Insisting on recycling the name $n$ for the induction step is technically still correct, since all occurences of $n$ within the inductive step would be bound within it.
ryang
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