Find the mistake in the following proof that all dogs are the same color (if there's any):
Let's use induction. Consider groups of 1 dog. All dogs of every group are the same color, of course. So we now that it's true for 1. Suppose it's true for groups of k dogs, i.e. every group of k dogs are the same color. Then let's consider any group A of k+1 dogs. Consider a subgroup of A containing k dogs. Let's call x the dog in A but not in the subgroup. Then by induction, all dogs in the subgroup are the same color. Now consider a subroup of A of k dogs, with x in the subgroup. All dogs except for x are the same color. Then, since every group of k dogs are the same color (by induction), all dogs in A are the same color. So x and every dog are the same color.
The proof actually exhibits the fallacy of composition. See a discussion of this on
Just because the dogs in one group of one dog are all the same color, and all the dogs in another group of one dog are the same color, does not mean that when these two groups are combined they are all the same color.
It also illustrates equivocation. The "same" does not mean the same in different circumstances. In a given group "same" means the same as the others in the group. "Same" is not a quality like "yellow"--it has to be the same as something
, so just because you're not specifying the thing, that does not mean it is out of the meaning entirely. So in two groups of dogs, if in each group the dogs are the same color, that means they are the same color as all the other dogs in that group, not the same color as all the other dogs in a new group you might form by combining that group with another group. In other words you are changing the referent of "same color as".
Posted by Charlie
on 2003-05-30 10:22:18