>Prove that Fib(n) is the closest integer to (φ^n)/√5,

I know that phi (φ) represents a number that is the result of (1 + √5)/2, by way of using that algebraic formula.
The other possible number, (1 - √5)/2, is represented by psi (ψ).

I also know the ratio, fib(n+1) / fib(n) approaches the value of φ as the fibonnaci sequence grows (as n increments).

How can I show that by placing n as the exponent of φ, the value of (φ^n)/√5 will be the fib (n)?

From wikipedia....
>any power of φ is equal to the sum of the two immediately preceding powers

Which is just another way of connecting the Fibonacci sequence and the golden ratio.

Other websites created base cases by using the equations in terms of psi and phi to calculate Fib (1) and Fib (0),

What can I do with this information? I know that the problem makes use of something called Binet's formula (or something similar).

...does anyone know SICP?

I'd love a hint, pls

there are probably other anons who will be able to help you more, but this is how i think you would do it

proof by induction should be fairly straightforward with the hint they gave you. if not, various proofs for the closed form of fibonacci numbers are probably widely available online. then since ψ^n = (-1/φ)^n, you could prove |ψ^n/sqrt(5)| < 1/2 for all n >= 0, which i found here

https://en.wikipedia.org/wiki/Fibonacci_number#Computation_by_rounding

which would complete the proof

>>58332447
Thanks, sorry for the freakout post. I'm trying to go solo and it's my first language.

How did you come upon this problem, OP?