I am working on a problem in my textbook where I am given this proof dealing with Fibonacci numbers. The function $f$ is defined by $f(0) = f(1) = 1$ and for all $n\geq 2$, and $f(n) = f(n-1) + f(n-2)$. The following proof is trying to prove $f(4) = 5$:
\begin{align*} f(4) &= 5\\ f(3)+f(2) &= 5\\ [f(2)+f(1)]+f(2) &= 5\\ 2f(2) + 1 &= 5\\ 2f(2) &= 4\\ 2(f(1) + f(0)) &= 4\\ 2(1+1) &= 4\\ 4 &= 4 \end{align*}
I know that this proof is incorrect, but I'm having a hard time finding how it is incorrect and coming up with sufficient reasoning. Every time I look at it, I can't seem to find a noticeable error. Can anyone give me some pointers and/or suggestions as to how this proof is incorrect? Any help is appreciated.
The proof is going in the wrong direction. It starts with what it wants to prove and arrives at a truth. As all the steps are reversible, that is not a problem this time. It is bad practice because you have to commit to every step being reversible and people who check the proof have to be more careful. That is not the root of the problem, which you do not state clearly. The usual Fibonacci sequence has $f(4)=3$, which conflicts with the claimed result.
The real problem is in the definition of the Fibonacci sequence. The usual definition is $f(0)=0, f(1)=1$. The proof quotes a start that is off by $1$ from the usual definition. Using the starting values given in the proof, the statement that $f(4)=5$ is correct. It is just not the usual Fibonacci sequence. I find it a good fake proof because people tend to overlook the starting values and concentrate on the algebra in the derivation.