Simple Finite Continued Fraction is Almost Determined by Value

Theorem

Let $n,m \geq 0$ be natural number.

Let $\sequence {a_k}_{0 \mathop \le k \mathop \le m}$ and $\sequence {b_k}_{0 \mathop \le k \mathop \le n}$ be simple finite continued fractions in $\R$.

Let $\sequence {a_k}_{0 \mathop \le k \mathop \le m}$ and $\sequence {b_k}_{0 \mathop \le k \mathop \le n}$ have the same value.

Then either:

• $n = m$, and the sequences are equal.
• $n = m + 1$, $a_k = b_k$ for $k < m$, $a_m = b_m-1$ and $b_{m + 1} = 1$
• $m = n + 1$, $a_k = b_k$ for $k < n$, $b_n = a_n-1$ and $a_{n + 1} = 1$

Proof

This theorem requires a proof. In particular: use Floor of Simple Finite Continued Fraction You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Also see

• Simple Infinite Continued Fraction is Uniquely Determined by Limit