Fibonacci series is an infinite sequence of integers, starting with $1$ and $2$ and defined recursively after that, for the $n$th term in the array, as $F(n) = F(n-1) + F(n-2)$. How is the countability of Fibonacci sequence proven?


closed as unclear what you're asking by JMoravitz, GNUSupporter 8964民主女神 地下教會, Lord Shark the Unknown, Arnaud Mortier, kingW3 Feb 15 '18 at 19:24

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    $\begingroup$ The Fibonacci series has an index of a natural number. That’s the definition of countability. $\endgroup$ – Jack Moody Feb 15 '18 at 16:40
  • $\begingroup$ There is a bijection between the natural numbers and F(n) $\endgroup$ – Jack Moody Feb 15 '18 at 16:41
  • $\begingroup$ To check that its values are an infinite set, just notice that it is strictly increasing ($F(n-2)> 0$). $\endgroup$ – user530891 Feb 15 '18 at 16:53
  • $\begingroup$ Ask yourself: how could it NOT be countable, given a standard definition of "countable"? $\endgroup$ – Peter Smith Feb 15 '18 at 17:09

A set $S$ is countable iff there exists a bijection between $\mathbb{N}$ and $S$

This means that we have to find that bijection, and as the Fibonacci Numbers is a subset of $\mathbb{N}$, it must be countable. Or stated differently, we could just map a number $n$ to the $n$th fibonacci number.


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