What is wrong with the following “proof” by Mathematical Induction? We will prove statements that all computers are built by the same manufacturer. In particular, we prove statements B(n) with n ∈ N, that “in any collection of n computers, all of the computers are built by the same manufacturer”. First check that B(1) is true, since with only one computer there is only one manufacturer. Now assume B(k); that is, in any collection of k computers, all are built by the same manufacturer. To prove B(k + 1) consider any collection of k + 1 computers. Pull out one of these computers, ‘HAL’ say, leaving just k computers, all of which must have the same manufacturer. Swap the ‘HAL’ computer with one of the others, so again there are k computers, so all must have the same manufacturer. Thus ‘HAL’ must have the same manufacturer as the others, hence all k + 1 computers have the same manufacturer; that is B(k + 1) must be true.
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2Hint: Intuitively, when would the induction proof have to fail. Is the theorem true for $k=2$? What happens when starting with $k=1$ and proving for $k=2$? – Thomas Andrews Oct 11 '15 at 12:35
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https://en.m.wikipedia.org/wiki/All_horses_are_the_same_color – Michael Hoppe Oct 11 '15 at 16:09
2 Answers
If you take out one, you get a set $A$ of $k$ computers, and if you take out another, you get a different set $B$ of $k$ computers.
But the assumption is that $A\cap B$ is not empty - that they have an element in common.
That is not true when $k+1=2$. Then $A$ and $B$ are singleton sets, and they are disjoint. So it is true that $A$ is a set all made by one manufacturer, and $B$ is a set all made by one manufacturer. To conclude that $A\cup B$ was all made by one manufacturer, $A$ and $B$ must have a common element.
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The inductive step actually involves $k-1$, as well as $k$, to prove $B(k+1)$, when you say that HAL and $k-1$ other computers have the same manufacturer, because the same $k-1$ computers are part of a set of $k$ earlier. So, you can go from $k$ to $k+1$ only when $k-1 > 0$, or equivalently, when $k > 1$. This is because your base case is for $k = 1$ (try to prove that a 'collection' of zero computers all have the same manufacturer!) So you can't go from 1 to 2, and the induction fails.
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Technically, the proof isn't using $B(k-1)$ - you don't need to know $B(k-1)$ is true for the induction step to work, but you do need $k-1>0$ for it to work. Indeed $B(0)$ is true - it is "vacuously true." – Thomas Andrews Oct 11 '15 at 12:44
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