I'm trying to find a flaw in the following proof, but I am unsure if I am correct or not:
Identify the flaw in the proof that $2n = 0$ for all $n \ge 0$.
Base case: If $n=0$ then $2\cdot n = 2\cdot 0 = 0$ Inductive step: Assume $n \gt 0$ and $2m=0$ for all integers $m$ where $0 \le m \lt n$. Then we have
$2n = 2(a + b)$ for some integers a and b where $0 \le a, b \lt n$
$= 2a + 2b$
$= 0 + 0$ $= 0$
The part of the proof that seems incorrect to me is the fact that $b \lt n$, but the proof says $b=0$. For b to be less than n, it must be less than 0 according to the initial theorem.
Is this the only flaw in the proof?