Catalan's Identity/Proof 2
Theorem
- ${F_n}^2 - F_{n - r} F_{n + r} = \left({-1}\right)^{n - r} {F_r}^2$
Proof
Proof by induction:
For all $n, r \in \N_{>0}$ where $n > r$, let $\map P {n, r}$ be the proposition:
- ${F_n}^2 - F_{n - r} F_{n + r} = \paren {-1}^{n - r} {F_r}^2$
Basis for the Induction
$n = 1$ yields no suitable $r$, so we look at $n = 2$ instead, which only gives us $r = 1$.
$\map P {2, 1}$ is true:
- ${F_2}^2 - F_3 F_1 = 1^2 - 2 \times 1 = -1 = -1 \times {F_1}^2$
$n = 3$ gives us only $r = 1$ and $r = 2$.
$\map P {3, 1}$ is true:
- ${F_3}^2 - F_2 F_4 = 2^2 - 1 \times 3 = 1 = 1 \times {F_1}^2$
$\map P {3, 2}$ is true:
- ${F_3}^2 - F_1 F_5 = 2^2 - 1 \times 5 = -1 = -1 \times {F_2}^2$
This is our basis for the induction.
First Induction Hypothesis
Now we need to show that, if $\map P {n, r}$ is true for all $r$, where $n > 3$, then it logically follows that $\map P {n + 1, r}$ is true for all $r$.
So this is our induction hypothesis:
- $\forall r < n : {F_n}^2 - F_{n - r} F_{n + r} = \paren {-1}^{n - r} {F_r}^2$
Then we need to show:
- $\forall r < n : {F_{n + 1} }^2 - F_{n - r + 1} F_{n + r + 1} = \paren {-1}^{n - r + 1} {F_r}^2$
Induction Step
This is our induction step:
It will again be a proof by induction.
Basis for the Induction
When $r = 1$:
\(\ds {F_{n + 1} }^2 - F_n F_{n + 2}\) | \(=\) | \(\ds {F_{n + 1} }^2 - F_n \paren {F_{n + 1} + F_n}\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds {F_{n + 1} }^2 - F_n F_{n + 1} - {F_n}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_{n + 1} \paren {F_{n + 1} - F_n} - {F_n}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_{n + 1} F_{n - 1} - {F_n}^2\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1} \paren {F_n^2 - F_{n - 1} F_{n + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1} \paren {-1}^{n - 1} {F_1}^2\) | First induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^n {F_1}^2\) |
So $\map P {n + 1, 1}$ holds.
This is our basis for the induction.
Second Induction Hypothesis
Now we need to show that, if $\map P {n + 1, r}$ is true, where $2 < r < n$, then it logically follows that $\map P {n + 1, r + 1}$ is true.
So this is our second induction hypothesis:
- ${F_{n + 1} }^2 - F_{n - r + 1} F_{n + r + 1} = \paren {-1}^{n - r + 1} {F_r}^2$
Then we need to show:
- ${F_{n + 1} }^2 - F_{n - r} F_{n + r + 2} = \paren {-1}^{n - r} {F_{r + 1} }^2$
Induction Step
This is our induction step:
\(\ds {F_{n + 1} }^2 - F_{n - r} F_{n + r + 2}\) | \(=\) | \(\ds {F_{n + 1} }^2 - F_{n - r + 1} F_{n + r + 1} + F_{n - r + 1} F_{n + r + 1} - F_{n - r} F_{n + r + 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n - r + 1} {F_r}^2 + F_{n - r + 1} F_{n + r + 1} - F_{n - r} F_{n + r + 2}\) | Second induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n - r + 1} {F_r}^2 + F_{n - r + 1} F_{n + r + 1} - F_{n - r} \paren {F_{n + r} + F_{n + r + 1} }\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n - r + 1} {F_r}^2 + \paren {F_{n - r + 1} - F_{n - r} } F_{n + r + 1} - F_{n - r} F_{n + r}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n - r + 1} {F_r}^2 + F_{n - r - 1} F_{n + r + 1} - F_{n - r} F_{n + r}\) | Definition of Fibonacci Number | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n - r + 1} {F_r}^2 - F_{n - r} F_{n + r} + F_{n - r - 1} F_{n + r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n - r + 1} {F_r}^2 + {F_n}^2 - F_{n - r} F_{n + r} - {F_n}^2 + F_{n - r - 1} F_{n + r + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n - r + 1} {F_r}^2 + \paren {-1}^{n - r} {F_r}^2 - \paren {-1}^{n - r + 1} {F_{r + 1} }^2\) | First induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n - r + 1} {F_r}^2 + \paren {-1}^{n - r} {F_r}^2 + \paren {-1}^{n - r} {F_{r + 1} }^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n - r} \paren {- {F_r}^2 + {F_r}^2} + \paren {-1}^{n - r} {F_{r + 1} }^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{n - r} {F_{r + 1} }^2\) |
So $\map P {n + 1, r} \implies \map P {n + 1, r + 1}$ and the result follows by the Principle of Mathematical Induction.
So $\map P {n, r} \implies \map P {n + 1, r}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $F_n^2 - F_{n - r} F_{n + r} = \paren {-1}^{n - r} {F_r}^2$
$\blacksquare$
Source of Name
This entry was named for Eugène Charles Catalan.