Fibonacci Numbers
Lesson · Intermediate
Algebra
The Fibonacci sequence is defined by
and
for n ≥ 0.
Thus,
Each term is the sum of the previous two terms.
First 20 Fibonacci Numbers
For example,
We prove the identity by induction on n.
Base case: n = 1.
Inductive step: assume the identity holds for n = k. We prove it for n = k + 1.
We prove the identity by induction on n.
Base case: n = 1.
Inductive step: assume the identity holds for n = k. We prove it for n = k + 1.
We prove the identity by induction on n.
Base case: n = 1.
Inductive step: assume the identity holds for n = k. We prove it for n = k + 1.
One of the most useful Fibonacci identities is
We prove the identity by induction on n, with m ≥ 1 fixed.
Base case: n = 0.
Base case: n = 1.
Inductive step: assume the identity holds for n = k and n = k + 1. We prove it for n = k + 2.
Setting m = n (with n ≥ 1),
Therefore,
This quotient is a Lucas number:
where
We prove the identity by induction on n.
Base case: n = 1.
Inductive step: assume the identity holds for n = k. We prove it for n = k + 1.
We prove the identity by induction on n.
Base case: n = 0.
Inductive step: assume the identity holds for n = k. We prove it for n = k + 1.
Suppose
Then for some positive integer k,
We prove by induction on k that
For k = 1, this is immediate.
Now suppose
Using the Fibonacci addition identity,
The first term is divisible by , and by the induction hypothesis, the second term is also divisible by .
Therefore,
We first need the key fact:
Key Step
Using
we get
Since consecutive Fibonacci numbers are coprime,
so
Euclidean Algorithm
Now we prove the main result using the Euclidean algorithm.
Suppose n ≥ m. Write
Applying the key step repeatedly,
Thus, one step of the Euclidean algorithm on the indices
produces exactly the same reduction for the Fibonacci numbers:
Continue the Euclidean algorithm until the indices reach
where
Therefore,
Since F₀ = 0,
Hence
Let
Then
This is the closed form of the Fibonacci sequence.
We prove the formula by induction on n, using two base cases.
Base case: n = 0.
Base case: n = 1.
Inductive step: assume the formula holds for n = k and n = k + 1. We prove it for n = k + 2.
Also,
Therefore,