I suspect that computing Fibonacci numbers with Binet formula is slower than with matrix form - if mul(n) is the number of multiplications required to compute x

I suspect that computing Fibonacci numbers with Binet formula is slower than with matrix form - if mul(n) is the number of multiplications required to compute x^n, then matrix form will use 8 * mul(n) integer multiplications, while Binet's formula requires 2 * mul(n) floating point operations. I suspect the latter is slower.