
#1
Nov2810, 09:08 AM

P: 1,400

At the end of this video lecture on the Principle of Induction, the teacher gives the following example of a flawed proof that misuses the principle.
Theorem. All horses are the same colour. Proof. (By induction.) Base case. In a set of one horse {h}, the principle clearly holds. Inductive step. Suppose S = {h_{1},...,h_{n+1}}, S' = {h_{1},...,h_{n}}, S'' = {h_{2},...,h_{n+1}} are sets of horses. Sets S' and S'' each have one horse fewer than S. h_{2} is an element of S' and of S''. Therefore all horses in S will have the same colour, in particular the same colour as h_{2}. End proof. The explanation offered is that this proof fails because the inductive step assumes that n does not equal 2, and so isn't general enough. Is there another flaw, namely that the proof treats S' and S'' as distinct sets of horses but has indexed them with the same natural number as each other? To use the principle of induction correctly, wouldn't we need to use distinct labels for distinct objects? EDIT. Correction: It assumes n + 1 does not equal 2. 



#2
Nov2810, 09:18 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

No, what is done their is perfectly legitimate. If you have, say, four objects, labeled a, b, c, and d, you are justified into breaking the set {a, b, c, d} into the union of {a, b, c} and {b, c, d} and that is all that is done. The indexing is completely irrelevant to the proof.




#3
Nov2810, 09:42 AM

P: 1,400

Is the inductive hypothesis here not only that all the horses in S' are the same colour as each other, but also that all the horses in S'' are the same colour as each other? I'm guessing so, since the sets are indexed just by the number of elements they have. Then what the inductive step has to show is that these statements imply that all the horses in S are the same colour as each other.




#4
Nov2810, 09:53 AM

Sci Advisor
P: 905

A flawed proof by induction 



#5
Nov2810, 10:08 AM

P: 1,400

Okay, I think I get it now. Thank to you both for your help.



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