Difference between revisions of "Fallacious proof/all horses are the same color"

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Our base case is not the appropriate base case: if one could show that every pair of horses has the same color (the result for <math>n = 2</math>), the fact that all horses have the same color would follow.  Unfortunately, the case <math>n = 2</math> does not follow from the case <math>n = 1</math>. The first horse is the same color as itself, and so is the second horse, but there is no overlap.
 
Our base case is not the appropriate base case: if one could show that every pair of horses has the same color (the result for <math>n = 2</math>), the fact that all horses have the same color would follow.  Unfortunately, the case <math>n = 2</math> does not follow from the case <math>n = 1</math>. The first horse is the same color as itself, and so is the second horse, but there is no overlap.
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Revision as of 22:45, 7 August 2006

Our base case is not the appropriate base case: if one could show that every pair of horses has the same color (the result for $n = 2$), the fact that all horses have the same color would follow. Unfortunately, the case $n = 2$ does not follow from the case $n = 1$. The first horse is the same color as itself, and so is the second horse, but there is no overlap.

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