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Loser66

  • one year ago

What is wrong with the following proof by mathematical induction that all horses are the same color: Clearly, in set of 1 horse, all are the same color (basic step is done) Assume that all horses in any set of n horses are the same color.

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  1. Loser66
    • one year ago
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    Consider a set of n+1 horses, labeled with number 1,2,...., n, n+1. By induction hypothesis, any set of n horses are the same color, hence if the set is from 1 to n, we have all the same. if the set is from 2 to n+1, they are the same also. Two set has n members in common, namely 2,3,...., n+1. all horses are the same color.

  2. Loser66
    • one year ago
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    Suppose the set is 2 horses: pink and orange. (hehehe... think of dolls for baby), hence "n" horse is just 1, either pink or orange--> the same color for all "n". but "n+1" = 2 , hence , n+1 can't be the same color for all. Is there other argument for this problem??

  3. ganeshie8
    • one year ago
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    This is classic, check https://en.wikipedia.org/wiki/All_horses_are_the_same_color

  4. anonymous
    • one year ago
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    oh the proof is trying to show for 2 groups of horses one contain N-1 and the other N, that each one has its own common color thus if we sub one horse from N and added to N-1 we would have the second group that all of i'ts elements have same color include the new horse with mean the previous group of the horse have the same color which means "all horses are the same color" the fallacy in this prove is step 1 itself ( show its true for N=1), like group N would have 1 element and group N-1 would have 0 elements and thus are not necessarily the same color as each other, so for N+1 when we add one horse to N and one to N-1 that does not mean the one added to in N-1 is of the same color to the other one so the origin proof failed from first step and everything else falls.

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