Talk:Zeckendorf's theorem

message: * Oppose merging. The theorem article gives the mathematical underpinning, while the coding article gives the practical persperctive. CompositeFan 17:15, 23 January 2007 (UTC)
replies: text: In view of the above comments and the length of time that has passed, I have removed the "merge" tag from the article. Gandalf61 10:16, 18 March 2007 (UTC)

message: I removed the following new section, headed Proof of Zeckendorf's Theorem, from the article:
:Zeckendorf's Theorem can be proved by induction.
:For n=1,2,3 it is clearly true, for n=4 we have 4=3+1.
:Now let each n \leq k have a Zeckendorf's representation. If k+1 is a Fibonacci Number then we're done, else there exists j such that F_j<k+1<F_{j+1}. Now consider a=k+1-F_j. Since it is a< k, a has a Zeckendorf representation; moreover F_j+a<F_{j+1} then a<F_{j-1} so the Zeckendorf representation of a does not contain F_{j-1}. Then k+1 can be represented as a+F_j and we're done.
I removed this because it is not a full proof of Zeckendorf's theorem. It shows that every positive integer has a Zeckendorf representation, but in addition Zecekendorf's theorem also says that this representation is unique. This "proof" does not show uniqueness, so it only proves one half of Zeckendorf's theorem. Gandalf61 20:13, 14 August 2007 (UTC)
replies: text: I copied a proof of uniqueness from PlanetMath. I hope that satisifies the complaint that the proof was not complete. Please feel free to reword it to be more in line with Wikipedia style. CompositeFan 17:20, 15 August 2007 (UTC)
replies: text: I think thats only half of the uniqueness proof, the other half is proving that given any number there is only one Zeckendorf representation. --Ray andrew 18:04, 15 August 2007 (UTC)

text: As Ray Andrew says, the PlanetMath proof shows that two different integers cannot have the same Zeckendorf representation - but this is trivially obvious, anyway. What it does not do is show that one integer cannot have two different Zeckendorf representations - this is the difficult bit. As the "proof" is still incomplete, I have removed the section. Gandalf61 18:15, 15 August 2007 (UTC)
replies: text: So if it's missing one bit, we delete the whole goddamn thing, right? Does that make sense for Wikipedia? Do we also go around deleting stubs because they're incomplete? Knodeltheory 19:26, 15 August 2007 (UTC)

text: Yes, I see where you go from there. If there is a source for this uniqueness proof then it could go into the article. Does anyone have a source ? Gandalf61 11:21, 16 August 2007 (UTC)
thread_title: Proof of Zeckendorf's Theorem

message: I'm concerned about the bit that reads
The first part of Zeckendorf's theorem (existence) can be proved by induction. ..... Then k + 1 can be represented as a + F_j. Moreover, it is obvious that each Zeckendorf representation corresponds to only one integer.
It is possible that a and F_j may have '1' in the same place or in adjacent places. The proof can be patched along the following lines: do people agree?
* If they are in the same place i.e. there is repeated F_l in the representation for some l, then this number itself has a Z-representation. (*)
** If they are in adjacent places, the representation ....0110... can be replaced by .....0001.. because the sum of two Fibs is the next Fib. (**)
However, the result of doing (*) or (**) may create a further problem, in which case (*) or (**) can be repeated ... (I suppose)
Am I right?
Would anybody like to write this up in full? Johnbibby (talk) 07:55, 18 June 2008 (UTC)
replies: text: a and F_j cannot have '1' in the same place or in adjacent places because a<F_{j-1}. so the highest Fibonacci number that can possibly be in the Zeckendorf representation of a is F_{j-2}. Gandalf61 (talk) 07:48, 21 June 2008 (UTC)
thread_title: New concerns about proof

message: Note: F-series: f1=0, f2=1, f3=1, f4=2, f5=3,...; 3 is a F-number, but also 3=1+2=f2+f4, the sum of 2 not succeeding F-numbers.??Nijdam (talk) 21:29, 16 October 2011 (UTC)
thread_title:  unique? 

message: I am considering providing the following proof by induction of the existence part of the theorem:-
Base step: clearly, 1 has a representation (or, alternatively, we could start at 0).
Induction:-
Suppose k has a representation; clearly it must be of the form 1...10 or 1...01 or 1...00. Clearly, if we temporarily relax the requirement that there be no consecutive 1s in the representation of k+1, k+1 can be represented as 1...11 or 1...10 or 1...01 respectively. Now consider the effect of applying the following repair algorithm to restore the requirement: halt if there are no consecutive 1s; find the leftmost consecutive pair of 1s; clearly this has a 0 to the left of it (possibly implied, if the representation of k+1 currently begins with a pair of 1s), so the leftmost consecutive pair of 1s falls within a triple 011; replace this triple with the triple 100; and repeat. Clearly, each iteration maintains the sum of the representation, by the definition of the Fibonacci numbers, so it remains a representation of k+1. Clearly, each iteration reduces the total number of 1s by 1, and so the iteration must terminate as there are only finitely many 1s to begin with. But clearly the iteration can only terminate on a representation with no consecutive 1s (which must be, as shown, a representation of k+1), so there must be a representation of k+1 with no consecutive 1s (and we have constructed it from that of k). Q.E.D.
Would this be an acceptable contribution, perhaps after editing to tidy it up? Maybe it would read more easily if it were rewritten as a proof by infinite descent? PMLawrence (talk) 08:08, 26 December 2013 (UTC)
replies: text: I don't think this is as clear or as concise as the existence proof that is already in the article. I think it is only worth including a second existence proof if thus one has a reliable source. Gandalf61 (talk) 09:31, 26 December 2013 (UTC)
replies: text: On the first point, that's why I was wondering if this would read better if presented in its proof by infinite descent version. Also, judging by such proofs as the irrationality of the square root of 2, there is merit in presenting variant proofs using different approaches, and not just presenting "best" proofs, precisely because the approaches are different - readers can learn different "meta" stuff from each approach. As for sourcing this proof, isn't it of the nature of the beast that a mathematical proof provides its own authority, unlike (say) an assertion about current events? After all, there isn't any such source given for any of the current proof (yes, that has been flagged here - though not for those proofs at the square root of 2 that are unsourced - but nobody has followed that up in over a year, and at least having a different variant would buttress the current situation). PMLawrence (talk) 11:28, 26 December 2013 (UTC)
thread_title:  Alternative existence proof