I just recently found out this fake proof by induction that all positive integers are equal from The Mathematical Gazette:
Let $P(n)$ be the propositition:
"If the maximum of two positive integers is $n$ then the integers are equal."
Clearly $P(1)$ is true. Assuming that $P(n)$ is true, assume that $u$ and $v$ are positive integers such that the maximum of $u$ and $v$ is $n + 1$. Then the maximum of $u - 1$ and $v - 1$ is $n$, forcing $u - 1 = v - 1$ by the validity of $P(n)$. Therefore, $u = v$.
I see this, almost a duplicate: Find the fallacy in the following treatment, and I understand it, but I got into an argument with someone. They say that the base case $P(1)$ is in fact, not true, because, either the two integers are already the same, or they are different, and only case where $P(1)$ is true is where they must be already the same, in which case we haven't proved anything.
I say, that the special case $n = 1$ forces the numbers to be the same, which makes $P(1)$ true.
Who is correct?
The base case is correct. The fallacy is in the induction step when you assume that $u-1$ and $v-1$ are positive integers.