Consecutive Fibonacci Numbers are Coprime

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Theorem

Let $F_k$ be the $k$th Fibonacci number.


Then:

$\forall n \ge 2: \gcd \set {F_n, F_{n + 1} } = 1$

where $\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$.


That is, a Fibonacci number and the one next to it are coprime.


Proof

From the definition of Fibonacci numbers:

$F_1 = 1, F_2 = 1, F_3 = 2$


Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

$\gcd \set {F_n, F_{n + 1} } = 1$


Basis for the Induction

$\map P 2$ is the case:

$\gcd \set {F_2, F_3} = \gcd \set {1, 2} = 1$

Thus $\map P 2$ is seen to hold.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\gcd \set {F_k, F_{k + 1} } = 1$


Then we need to show:

$\gcd \set {F_{k + 1}, F_{k + 2} } = 1$


Induction Step

This is our induction step:

\(\ds \gcd \set {F_{k + 1}, F_{k + 2} }\) \(=\) \(\ds \gcd \set {F_{k + 1}, F_{k + 2} - F_{k + 1} }\) Common Divisor Divides Integer Combination
\(\ds \) \(=\) \(\ds \gcd \set {F_{k + 1}, F_k}\) Definition of Fibonacci Number
\(\ds \) \(=\) \(\ds \gcd \set {F_k, F_{k + 1} }\)
\(\ds \) \(=\) \(\ds 1\) Induction Hypothesis

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \ge 2: \gcd \set {F_n, F_{n + 1} } = 1$

$\blacksquare$


Sources

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