Honsberger's Identity
Theorem
Let $F_k$ be the $k$th Fibonacci number.
Then:
- $\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
Negative Indices
Let $n \in \Z_{< 0}$ be a negative integer.
Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers).
Then Honsberger's Identity:
- $F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
continues to hold, whether $m$ or $n$ are positive or negative.
Proof 1
From the initial definition of Fibonacci numbers, we have:
- $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$
Proof by induction:
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
- $\forall m \in \Z_{>0} : F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
Basis for the Induction
$\map P 1$ is the case:
\(\ds F_{m + 1}\) | \(=\) | \(\ds F_{m - 1} + F_m\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{m - 1} \times 1 + F_m \times 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_{m - 1} F_1 + F_m F_2\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{m - 1} F_n + F_m F_{n + 1}\) | for $n = 1$ |
and so $\map P 1$ is seen to hold.
$\map P 2$ is the case:
\(\ds F_{m + 2}\) | \(=\) | \(\ds F_{m + 1} + F_m\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{m - 1} + F_m + F_m\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{m - 1} \times 1 + F_m \times 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_{m - 1} F_2 + F_m F_3\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{m - 1} F_n + F_m F_{n + 1}\) | for $n = 2$ |
and so $\map P 2$ is seen to hold.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ and $\map P {k - 1}$ are true, where $k > 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $F_{m + k} = F_{m - 1} F_k + F_m F_{k + 1}$
and:
- $F_{m + k - 1} = F_{m - 1} F_{k - 1} + F_m F_k$
from which it is to be shown:
- $F_{m + k + 1} = F_{m - 1} F_{k + 1} + F_m F_{k + 2}$
Induction Step
This is our induction step:
\(\ds F_{m + k + 1}\) | \(=\) | \(\ds F_{m + k} + F_{m + k - 1}\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{m - 1} F_k + F_m F_{k + 1} + F_{m - 1} F_{k - 1} + F_m F_k\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{m - 1} \paren {F_k + F_{k - 1} } + F_m \paren {F_{k + 1} + F_k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_{m - 1} F_{k + 1} + F_m F_{k + 2}\) | Definition of Fibonacci Number |
So $\map P k \land \map P {k - 1} \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall m, n \in \Z_{>0} : F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
$\blacksquare$
Proof 2
\(\ds \) | \(\) | \(\ds F_{m - 1} F_n + F_m F_{n + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\phi^{m - 1} - \hat \phi^{m - 1} } {\sqrt 5} \dfrac {\phi^n - \hat \phi^n} {\sqrt 5} + \dfrac {\phi^m - \hat \phi^m} {\sqrt 5} \dfrac {\phi^{n + 1} - \hat \phi^{n + 1} } {\sqrt 5}\) | Euler-Binet Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\phi^{m + n - 1} - \phi^{m - 1} \hat \phi^n - \phi^n \hat \phi^{m - 1} + \hat \phi^{m + n - 1} + \phi^{m + n + 1} - \phi^m \hat \phi^{n + 1} - \phi^{n + 1} \hat \phi^m + \hat \phi^{m + n + 1} } 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\phi^{m + n - 1} \paren {1 + \phi^2} + \hat \phi^{m + n - 1} \paren {1 + \hat \phi^2} - \phi^{m - 1} \hat \phi^n \paren {1 + \phi \hat \phi} -\phi^n \hat \phi^{m - 1} \paren {1 + \phi \hat \phi} } 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\phi^{m + n - 1} \paren {1 + \phi^2} + \hat \phi^{m + n - 1} \paren {1 + \hat \phi^2} } 5\) | as $\phi \hat \phi = -1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\phi^{m + n - 1} \paren {2 + \phi} + \hat \phi^{m + n - 1} \paren {2 + \hat \phi} } 5\) | as both $\phi$ and $\hat \phi$ satisfy $x^2 = x + 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\phi^{m + n - 1} \paren {2 + \dfrac {1 + \sqrt 5} 2} + \hat \phi^{m + n - 1} \paren {2 + \dfrac {1 - \sqrt 5} 2} } 5\) | Definition of $\phi$ and $\hat \phi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\phi^{m + n - 1} \paren {\dfrac {5 + \sqrt 5} 2} + \hat \phi^{m + n - 1} \paren {\dfrac{ 5 - \sqrt 5} 2} } 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\phi^{m + n - 1} \paren {\dfrac {1 + \sqrt 5} 2} - \hat \phi^{m + n - 1} \paren {\dfrac {1 - \sqrt 5} 2} } {\sqrt 5}\) | dividing numerator and denominator by $\sqrt 5$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\phi^{m + n} - \hat \phi^{m + n} } {\sqrt 5}\) | Definition of $\phi$ and $\hat \phi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds F_{m + n}\) | Euler-Binet Formula |
$\blacksquare$
Source of Name
This entry was named for Ross Honsberger.
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.8$: Fibonacci Numbers: $(6)$