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I am working on a proof, and to do it, I think it would be optimally to use induction backwards.

Show that 1 doesn't work. Assume n doesn't work. Prove that n+1 doesn't work.

Is this valid?

VividD
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Jimmy360
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    Won't a proof by contradiction work just fine? – cgo Apr 17 '15 at 06:45
  • @cgo In the proof that I am working on, this would be easier. – Jimmy360 Apr 17 '15 at 06:46
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    It would help if you include the problem. – Andrey Kaipov Apr 17 '15 at 06:47
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    I should certainly think so. I wouldn't call it inverted induction, though, rather induction of a negative statement. – Archaick Apr 17 '15 at 06:47
  • https://www.wolframalpha.com/input/?i=+sum_%28i%3D1%29%5E%28n-2%29i%28Floor%5Bn%2Fi%5D%29%3D1%2Bsum_%28i%3D1%29%5E%28n-1%29i%28Floor%5B%28n-1%29%2Fi%5D%29 – Jimmy360 Apr 17 '15 at 06:48
  • @AndreyKaipov I am trying to isolate n the best that I can, then use the definition of odds and this reverse induction to proof that it cannot be satisfied by an odd number (n is a natural number). – Jimmy360 Apr 17 '15 at 06:49
  • Your approach is equivalent to "Assume N+1 does work. Show that N works." Depending on your problem, that may be easier. – DanielV Apr 29 '15 at 20:27

2 Answers2

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What you are describing is still a proof by induction. A proof by induction is:

  • Take a statement $P(n)$ about integers.
  • Prove $P(1)$
  • Prove $\forall n:P(n) \implies P(n+1)$

What you are doing is:

  • Take a statement $A(n)$
  • Prove $\neg A(1)$
  • Prove $\forall n: \neg A(n)\implies \neg A(n+1)$

So what you are doing si simply performing induction on the statement $\neg A$, i.e., you are performing standard induction, but your statement $P$ is actually a negation of some other statement. It is still a statement, so there is nothing truly "inverted" happening.

5xum
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  • I would add however, that most proofs constructed this way are likely to be wrong. Unless you really truly know exactly what you're doing, it is very error-prone to try to prove that something is wrong or impossible based on something else being wrong or impossible. – CaptainCodeman Apr 17 '15 at 11:10
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    @CaptainCodeman True, but maybe this is where the induction does become some sort of "inverse" induction, since, instead of $\neg A(n)\implies \neg A(n+1)$, you can prove $A(n+1)\implies A(n)$... – 5xum Apr 17 '15 at 11:28
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Indeed, it can be viewed as "inverted" induction, i.e. as a special case of Fermat's method of infinite descent, since the contrapositive of your induction step is: $\ n$ works $\,\Rightarrow\, n\!-\!1\,$ works. This descent form of induction is a very natural way to present many inductive proofs.

Bill Dubuque
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