Actually here is a "

proof" that all ravens are black, which uses a minor

logical flaw in

inductive reasoning to prove it.

`We intend to prove that all ravens are black. To do this, we will reason inductively, and prove that for n ravens, all n ravens are black. `
First, we prove P(1). To do this, we find a raven and confirm that he is black.

Now, we must show that P(n)->P(n+1), so we assume that for n ravens, all n ravens are black.

By our assumption, the first n ravens in the line of n+1 ravens are all black. Also, by the same assumption, the last n ravens in our line of n+1 ravens are black. Therefore, the 1st and the (n+1)th raven must be the same color as the middle (n-2), and therefore all n+1 ravens are black.

`
By the same measure, since P(n)->P(n+1), we have proven that all ravens are black.`

Of course, the error in this proof can be found by examining the basis of induction, which is to show that P(1)->P(2)->P(3)->P(4)...

If you'll note, if we try to use the above argument for P(2), no overlap is created. Since we have no overlap, we have no basis to show that raven 1 is the same color as raven 2, since we can only assume that P(n) ravens are the same color, and P(n)=1 in this case.