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Consider the Fibonacci number $1,2,3,5,8,13,21,\cdots.$ It is well known that the limit of the ratio of Fibonacci numbers tends to the Golden Ratio $\phi$. Today, I want to show that $1.5$ is the smallest ratio between any two fibonacci numbers (achieved by $\frac{3}{2}$).

It is evident by inspection, but I was looking for a more formal proof of this.

I've tried using an induction argument, but it hasn't worked. Any ideas/suggestions would be greatly appreciated.

mathz2003
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  • There may be an easier approach, but I would start by showing the ratios alternate between being above and below $\phi$. Then I would show that the ratios below $\phi$ are increasing by taking one ratio to be $\phi-x$, doing two iterations to find the next ratio below $\phi$, and finding the error is less than $x$. This doesn't come with a guarantee as I didn't fill in the steps. – Ross Millikan Nov 25 '23 at 23:30
  • The sequence often starts with $1,1$ or $0,1$; in that case the smaller ratio $\frac11=1$ is obtained – J. W. Tanner Nov 25 '23 at 23:35
  • I am ignoring the $\frac{1}{1} = 1$ for now. @RossMillikan, how would I necessarily do what you suggested? For example, let's say we have consecutive Fibonacci numbers $a, a(\phi - x), a(1 + \phi - x), a(2\phi - 2x + 1)$, would I just show that the next ratio has a smaller delta? – mathz2003 Nov 25 '23 at 23:43
  • So the next ratio below $\phi$ is $\frac {2\phi -2x +1}{\phi-x+1}$ You want to show this is greater than $\phi-x$ – Ross Millikan Nov 25 '23 at 23:46
  • Looking at Wolfram Alpha, that ratio is less than $x$ for $1.26236<x<2.61803$. However, in our case $x$ would likely be somewhere from $0$ to $\phi - 1$ because all Fibonacci numbers must be increasing. This doesn't seem to work. – mathz2003 Nov 25 '23 at 23:49

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Hint: Consider the ratio of the ($n+1$)-th and $n$-th Fibonacci numbers: $$ \frac{a_{n+1}}{a_{n}}=\frac{a_{n}+a_{n-1}}{a_{n}}=1+\frac{a_{n-1}}{a_{n}}. $$ This quantity is at least as large as $3 / 2$ if only and if $2a_{n-1}\geq a_{n}$. Or, written in another way, $$ a_{n-1}+a_{n-1}\geq a_{n}. $$ Can you finish?

parsiad
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