# Can you find the fallacy in this inductive argument that all things are the same?

In honor of the start of the school year, here is what I think is a fairly interesting use, or rather misuse, of mathematical induction. When I first saw this problem, it took me a while to catch the flaw.

To prove: Everything is the same as everything else.
Base case: n =1. Obviously everything is the same as itself.

Inductive step: Assume that any n things are the same as each other. Choose any group of n+1 objects. The first n are all the same by the inductive assumption. So are the last n. Since the two groups overlap, all n+1 objects must be the same.

We conclude that any number of objects are all the same, that is, everything is the same as everything else.

Observing members: 0 Composing members: 0

Here’s my attempt (no guarantees): This is a generalization of n, not everything.

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I appreciate your response. Here is a hint. There is one crucial case where the assumption of an overlap between the first n and the last n is not valid.

Here is the answer. The fallacy lies in the case of n=2. From the case n=1, each of the two objects are the same as themselves, but there is no overlap, so you can’t say that the two objects are the same.

If you could say that every two objects are the same then it would indeed be true that everything is the same. Just fix one object and compare it to everything else. Every other object would be the same as the one and therefore everything would be the same as everything else.

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