In the Art Of Programming by Knuth there is the following exercise:
There must be something wrong with the following proof. What is it? "Theorem. Let $a$ be any positive number. For all positive integers $n$ we have $a^{n-1} = 1$. Proof. If $n=1$, $a^{n-1} = a^{1-1} = a^{0} = 1$. And by induction, assuming that the theorem is true for $n=1,...,n$, we have $$a^{(n+1)-1} = a^{n} = \frac{a^{n-1}\times a^{n-1}}{a^{(n-1)-1}}=\frac{1\times 1}{1}=1;$$ So the theorem is true for $n+1$ as well.
And the corresponding answer:
The theorem has not been proved for $n=2$. In the second part of the proof take $n=1$. We assume there that $a^{(n-1)-1}=a^{-1}=1$.[...]
The problem is I don't understand the answer. Why Knuth highlights number 2 here? With the same success one could have said: "the theorem has not been proved for $n=1729$". And this doesn't seem to me as a good way to decline proof by induction because (to me) such a proof doesn't deal with concrete numbers but the base. I think the real mistake with the above proof is simply the third equality. It must be: $$\frac{a^{n-1}\times a^{n-1}}{a^{(n-1)-1}}=\frac{1\times 1}{1\times a^{-1}}=a$$ And from that we cannot conclude that it equals one. I think that the original answer confuses a reader. Am I wrong?
See what happens when you go from $n=1$ to $n=2$. The argument breaks down because you will get $a^{0-1}$ in the denominator. When you go from $n$ to $n+1$ you have to make sure that the argument works for any positive integer $n$. In this case it fails for $n=1$. That is why the case $n=2$ creates a problem.