Lamé's Theorem

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $a, b \in \Z_{>0}$ be (strictly) positive integers.

Let $c$ and $d$ be the number of digits in $a$ and $b$ respectively when expressed in decimal notation.

Let the Euclidean Algorithm be employed to find the GCD of $a$ and $b$.


Then it will take less than $5 \times \min \set {c, d}$ cycles around the Euclidean Algorithm to find $\gcd \set {a, b}$.


Lemma

Suppose it takes $n$ cycles around the Euclidean Algorithm to find $\gcd \set {a, b}$.

Then $\min \set {a, b} \ge F_{n + 2}$, where $F_n$ denotes the $n$-th Fibonacci number.


Proof

Without loss of generality suppose $a \ge b$.

Then $\min \set {c, d}$ is the number of digits in $b$.

By Number of Digits in Number, we have:

$\min \set {c, d} = \floor {\log b} + 1$

Aiming for a contradiction, suppose it takes at least $5 \paren {\floor {\log b} + 1}$ cycles around the Euclidean Algorithm to find $\gcd \set {a, b}$.

Then we have:

\(\ds b\) \(\ge\) \(\ds F_{5 \paren {\floor {\log b} + 1} + 2}\) Lemma
\(\ds \) \(\ge\) \(\ds \phi^{5 \paren {\floor {\log b} + 1} }\) Fibonacci Number greater than Golden Section to Power less Two
\(\ds \) \(>\) \(\ds \phi^{5 \log b}\) Definition of Floor Function

For $b = 1$, both sides are equal to $1$, giving $1 > 1$, which is a contradiction.

Hence we consider $b > 1$ and take $\log$ on both sides:

\(\ds \leadsto \ \ \) \(\ds \log b\) \(>\) \(\ds \paren {5 \log b} \log \phi\) Logarithm of Power
\(\ds \leadsto \ \ \) \(\ds \frac 1 {\log \phi}\) \(>\) \(\ds 5\)

However, $\dfrac 1 {\log \phi} \approx 4.785 < 5$.

This is a contradiction.

Hence the result by Proof by Contradiction.

$\blacksquare$


Source of Name

This entry was named for Gabriel Lamé.


Historical Note

This result was demonstrated by Gabriel Lamé in $1844$.


Sources