Zeckendorf's Theorem

Theorem
Every positive integer has a unique representation as the sum of non-consecutive Fibonacci numbers.

That is: Let $n$ be a positive integer.

Then there exists a unique increasing sequence of integers $\left\langle{c_i}\right\rangle$ such that:
 * $\forall i \in \N: c_i \ge 2$
 * $c_{i + 1} > c_i + 1$
 * $\displaystyle n = \sum_{i \mathop = 0}^k F_{c_i}$

where $F_m$ is the $m$th Fibonacci number.

For any given $n$, such a $\left\langle{c_i}\right\rangle$ is unique.

Proof
Let such an increasing sequence of integers be referred to in the below as a Fibonacci representation.

First note that every Fibonacci number $F_n$ is itself a Fibonacci representation of itself, where the sequence $\left\langle{c_i}\right\rangle$ contains $1$ term.

Existence
We will use strong induction on $n$.

We have that:
 * $F_2 = 1$
 * $F_3 = 2$
 * $F_4 = 3$

so each of the integers $1$, $2$, and $3$ have a Fibonacci representation.

$4$ has the Fibonacci representation:
 * $4 = 1 + 3 = F_2 + F_4$

This is the base case.

Let the induction hypothesis be that each $n \le k$ has a Fibonacci representation.

The induction step remains to be proved: that $k + 1$ has a Fibonacci representation.

If $k + 1$ is a Fibonacci number, then it is itself a Fibonacci representation.

In that case, the induction step holds.

Suppose that $k + 1$ is not a Fibonacci number.

Then:
 * $\exists j \in \Z: F_j < k + 1 < F_{j + 1}$

Consider $a = k + 1 - F_j$.

Since $a \le k$, it must have a Fibonacci representation by the induction hypothesis.

In addition:

Thus the Fibonacci representation of $a$ does not contain $F_{j - 1}$.

Thus $k + 1$ has the Fibonacci representation of $a + F_j$.

It follows by strong induction that for every positive integer there exists a Fibonacci representation.

Uniqueness
Let $n \in \Z_{>0}$.

Let $S$ and $T$ be distinct Fibonacci representation for $n$.

Consider $S' = S \setminus T$ and $T' = T \setminus S$, that is, each of the sets without their common elements.

Since $S$ and $T$ had equal sums:
 * $(1): \quad \displaystyle \sum S' = \sum T'$

, assume that $S'$ is empty.

Then in order for $T'$ to have the same sum using only positive integers, $T'$ must also be empty.

But as $S \ne T$, it follows that neither $S'$ nor $T'$ is empty.

Let the largest element of $S'$ be $F_S$.

Let the largest element of $T'$ be $F_T$.

Since $S'$ and $T'$ contain no common elements, $F_S \ne F_T$.

, let $F_S < F_T$.

From Sum of Non-Consecutive Fibonacci Numbers:
 * $\displaystyle \sum S' < F_{S + 1}$

and so:
 * $\displaystyle \sum S' < F_T$

But from $(1)$:
 * $\displaystyle \sum S' = \sum T'$

From this contradiction, it cannot be the case that $S'$ and $T'$ are non-empty.

So $S' = T' = \varnothing$ and so $S = T$.

Thus the Fibonacci representation is unique as desired.