Find the demonstration error for the statement "All positive integers are equal"

Question

All positive integers are equal, that is, for each $n \in \mathbb{N}$ the assertion $P(N): 1 = \cdots = n$ is true.

(i) $P(1)$ is true because $1 = 1$

(ii) Suppose that $P(n)$ is true, then $1 = \cdots = n - 1 = n$. Summing $1$ on each member of the equality, it follows that $n = n + 1$, hence $1 = \cdots = n - 1 = n = n + 1$ therefore $P(n+1)$ is true.

By the principle of mathematical induction it follows that $P(n)$ is true for all n $\in \mathbb{N}$

Can I say that n = n + 1 is a contradiction to show that this demonstration is wrong? If not, what is wrong in it?

Answer 1

The problem is, in your second step, there may be only one number, i.e. $1=1$. Therefore, you cannot add both sides by $1$ and get $1=1+1$. In the other words, in your induction hypothesis $n-1=n$, the $n-1$ term may not exist (when $n=1$ because your base case start at $1$).

Answer 2

There is a meta-argument that most fallacious induction tricks have to work the same way that this one does.

How can an induction proof go wrong but not transparently wrong?

1. The base case is defective. As the smallest case, it is usually easy to verify and hard to disguise. Sometimes, using the empty set can create enough confusion.

2. The step $n \to n+1$ is defective for most $n$. Bad algebra and statements that are wrong for almost all $n$ are hard to disguise. Maybe it can be argued based on examples of the first few $n$ that the induction step works, where the next case would fail. But this is less likely than...

3. The step from $n$ to $n+1$ is correct for almost all $n$, but wrong at one (or a very small number of) special value. Here you have only to conceal one error, for a value at least slightly larger than the base case.

The last is the easiest to obfuscate, and so the most efficient disproof technique when confronted with a false induction is to look for a small case where the induction step breaks.