Bob was having fun with his, why this Zeckendorf chap is not even a member!

Here is another six:

Doron Zeilberger;
Herbert S. Wilf
Abraham De Moivre of course
Newton of course
Freeman Dyson ( little known )
Pierre De Fermat of course

That is a few more than six.

Hypatia
Ada Lovelace

I just get lost in it. I will have to sit down one day and just read it carefully and thoroughly.

I'm supposed to be a teacher.  Eeeekkk!

I'll have another go if you ask me to.

Bob

hi Stefy,

That wasn't the answer I was hoping for.  Perhaps I should have said "I will enjoy trying to make a better explanation".

Bob

Because you said you didn't understand it => it wasn't clear.

B

Actually you said your tiny mind couldn't comprehend it.

Since (as a teacher) I have the patience of a saint, I was prepared to spell it out in more simple terms, specifically taking account of your tiny mind.        But if you are rejecting my offer, I'll just seek out someone who appreciates my talent.

Bob

Oh thank you!    I'd love to.

The algorithm shows how you can reduce any counting number to a sum of Fibonacci numbers (without repetition).

A number N has an F number subtracted which is saved in a list.

Then the remainder (ie from the subtraction, no division involved) is treated as the new N and the algorithm repeated.

This continues until the remainder is an F; at which point the process stops and the list of subtracted Fs  gives the required sum.

There are three things to prove:

(i) Will it always be possible to find an F to subtract?

(ii)  How do I know that no F is ever subtracted again in the process?

(iii) Will this ever stop?

The key to this is to consider the number N and find the largest F below N and subtract that.

If N is an F already then just stop.

If N isn't an F how do I know I can find an F that is lower?

Well the first three counting numbers {1,2,3} are all Fs, so there is certainly an F that is lower.  If there are several, then pick the biggest!

Let's call that F, 'b'.  And while we are at it, call the one below that, 'a'.

Consider a + b.

It is the next F after b.  Where does it come in relation to N?

Is it equal to N?  If so, then N is an F and we can stop.

Is it below N?  No, because we were told to pick the biggest F below N and call that one 'b'.  So a + b cannot be under N.

So N < a + b     =>   N - b < a

So the remainder from the subtraction is below 'a' and so there is no chance of ever repeating a subtraction of 'b'.  The remainder is just too small.  In fact, we can go further.  Since N - b < a, that means that even 'a' cannot be subtracted from the remainder.  So, having subtracted 'b',  we can never subtract 'a' as well.

So that proves I can find an F and that I'll never repeat an F.

But will I ever stop?

Well the remainders get smaller and smaller whilst remaining positive.  Eventually I'll have to reach an F because the last possible remainders, if no other F presents itself, will be 3 or 2 or 1 and they are all Fs.

So it is guaranteed that I will eventually end in an F.

Example:

N =150

What's the largest Fibonacci number under 150?

144 I think.  That is 'b'.  And 'a' = 89

Calculate N - b = 150 - 144 = 6.  Notice this is under 89.

Replace 150 by 6.

N = 6.  What is the largest F under this?  Answer 5

Calculate N - b = 6 - 5 = 1

This is the new N.  But it is an F so I can stop.

150 = 144 + 5 + 1

Note:  The representation by a sum of Fs is not always unique (150 = 144 + 3 + 2 + 1) but choosing the largest each time is what guarantees that we don't repeat and we don't get two consecutive Fs.

Bob

Sure it does.  But I wasn't trying to prove that theorem, just show the representation is possible.  And that statement emphasises why the largest is needed at each stage.  (challenge:  Find a number with more than one representation that does satisfy the theorem.)

Thanks for reading it and allowing me the chance to show off some more. 

Bob

Yes, if you can.

Bob

Thank You,
Bob

I cannon believe no one suggested Gottfried Wilhelm Leibniz, possibly the greatest mathematician of all time and the original creator of Calculus.