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In 1.2.1 Mathematical Induction section, Knuth presents mathematical induction as a two steps process to prove that P(n) is true for all positive integers n:

a) Give a proof that P(1) is true;

b) Give a proof that "if all P(1), P(2),..., P(n) are true, then P(n+1) is also true";

I have serious doubt about that. Indeed, I believe that point b) should be:

b) Give a proof that "if P(n) is true, then P(n+1) is also true". The major difference here is that you are only assuming that P(n) is true, not P(n-1), etc.

However, these books are old and have been read by many people (most of them being much more clever than I am^^).

So what is my confusion here?

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Korchkidu
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  • If `n` is arbitrary and P(1) ... P(n) are true, then can't you say `k = n + 1`, so by the previous statement, P(1) ... P(k) = P(1) ... P(n + 1) is also true? – Blender Mar 15 '13 at 08:12
  • Maybe a question for http://math.stackexchange.com/? – Markus Mar 15 '13 at 08:13
  • possibly a question for [math.stackexchange](http://math.stackexchange.com/) ? Although, the reference is from Computer Science. – Srikanth Venugopalan Mar 15 '13 at 08:14
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    There are two typical ways of doing induction. The version you're using is usually called "weak induction" or just "induction," while Knuth is using "complete induction" or "strong induction." The two are equivalent to one another. – templatetypedef Mar 15 '13 at 08:17
  • @blender: yes, this is my point. Knuth is doing it the other way around. Proving P(n) => P(n+1) proves that P(1)..P(n) is true. Stating that P(1)...P(n) is true, without actually proving it seems too strong to me;) – Korchkidu Mar 15 '13 at 08:34
  • @templatetypedef: thanks, I only knew weak induction indeed. I believe the main difference is that in strong induction, you must prove that P(1)...P(n) is true before proving P(n) => P(n+1), right? – Korchkidu Mar 15 '13 at 08:35
  • @Korchkidu- Not quite. In stronginduction, you make more assumptions than just P(n). Speicifcally, you assume P holds for 1, 2, ..., n,then use that to prove P(n + 1). – templatetypedef Mar 15 '13 at 16:13
  • I hope you have understood this by now, because your comments on the question and on the accepted answer sound quite confused. The two are entirely equivalent. – ShreevatsaR Nov 05 '16 at 19:48

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The entire point here is that the choice of n is arbitrary. Since P(n) implies P(n+1) is the conerstone of induction, then all the intermediate values between 1 and n will also hold under the assumption of P(n). You are supposed to show that if P(0) implies P(1) and P(n) implies P(n+1) then all conditions hold by the nature of n being arbitrary.

mathematician1975
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  • Yes, this is my point. But with Knuth's definition, you must prove that P(1)...P(n) holds prior to proving that P(n) => P(n+1). Right? Otherwise, stating that P(1)...P(n) is true already solved the problem because it is easy to state P(n+1) in terms of P(1)..P(n). – Korchkidu Mar 15 '13 at 08:32