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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?

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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
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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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