Proof. Assume that there are two integers $a$ and $b$ such that $0 < a < b$ , yet they both have the same Zeckendorf representation $Z$ with $k$ elements all 0s or 1s. We compute $$c = \sum_{i = 1}^k Z_iF_i,$$ where $F_x$ is the $x$ th Fibonacci number. We can be assured that there is only one possible value for $c$ since all $F_i$ are distinct for $i > 0$ and each $F_i$ was added only once or not at all, since each $Z_i$ is limited by definition to 0 or 1. Now $c$ holds the value of the Zeckendorf representation $Z$ . If $c = a$ , it follows that $c < b$ , but that would mean that $Z$ is not the Zeckendorf representation of $b$ after all, hence this results in a contradiction of our initial assumption. And if on the other hand $c = b$ and $c > a$ , then this leads to a similar contradiction as to what is the Zeckendorf representation of $a$ really is.