# Why is the fraction of a Fibonacci pair always the best estimation of the golden ratio in its range?

It seems, that fib(n)/fib(n-1) is always the best approximation for the golden ratio you can get, when the numerator is smaller than fib(n) and denominator can be chosen arbitrarily.

fib(n) is the n'th Fibonacci number.

I tested this algorithmically, and also the statement seems intuitively right, but I lack the skills to prove it.

Is there any prove to this?

Edit: (because it was marked as duplicate) I know, that the limit of these Fibonacci pairs is converging to the golden ratio, and I also know the proof to it. But why are these pairs the best convergents?

Edit Nr2: There are "para-Fibonacci"-sequences like 1,3,4,7,11,18 that have the same convergent property, but converge slower than the original series. So the only fact, that it converges, says nothing about the speed of convergence.

• These are the convergents to the continued fraction of the golden ratio. Commented Dec 24, 2017 at 11:58
• Possible duplicate of Limit of the ratio of consecutive Fibonacci numbers
– user371838
Commented Dec 24, 2017 at 11:59
• This is absolutely clear. But why are these the best convergent? Commented Dec 24, 2017 at 12:00
• Those are general facts about continued fractions. I recommend Khinchin's short book "Continued Fractions" for a reference for all such results. Commented Dec 24, 2017 at 12:05
• @Wojowu There is also a good account in Hardy and Wright "An Introduction to the Theory of Numbers" Commented Dec 24, 2017 at 12:34

The Fibonacci ratios are the continued fraction convergents to $(1+\sqrt{5})/2$. For any irrational number $\alpha$, each of its continued fraction convergents $p/q$ is the best rational approximation to $\alpha$ among all fractions (in reduced form) with denominator at most $q$. But for a positive integer $d$ that is not a denominator of a continued fraction convergent to $\alpha$, the best rational approximation to $\alpha$ with denominator at most $d$ need not be among the continued fraction convergents to $\alpha$ with denominator at most $d$. This point is often misunderstood.
Here is an example. The continued fraction convergents to $\sqrt{3}$ with denominator at most $10$ are 1, 2, 5/3, and 7/4, but the best rational approximation to $\sqrt{3}$ with denominator at most $10$ is not any of those fractions. It is 12/7.
It turns out that the best rational approximations to an irrational $\alpha$ up to an arbitrary bound on the denominator are found among the convergents and intermediate convergents to the continued fraction of $\alpha$. Find the definition of intermediate convergents on the Wikipedia page for continued fractions. The intermediate convergents to $\sqrt{3}$ include 12/7.
To build intermediate convergents to a standard continued fraction $[a_1,a_2,a_3,\ldots]$, where $a_i$ is a positive integer for $i>1$, you use positive integers less than $a_i$. The special thing about $(1+\sqrt{5})/2$ is that its continued fraction is $[1,1,1,1,\ldots]$, with each $a_i$ equal to $1$, so there are no intermediate convergents. Therefore the best rational approximations to $(1+\sqrt{5})/2$ are always its continued fraction convergents, which are the Fibonacci ratios.