Find the demonstration error for the statement "All positive integers are equal" 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? 
 A: 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 $1$).
A: 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?


*

*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.

*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...

*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.
