Dear BobbyM,
May I know from where are you getting all these facts?
Dear anonimnystefy,
The other two which we have chosen are Pythagoras and Turing.
BTW Is Turing OK?
.
.
Some of my classmates don't keep track of Mathematics so I need your advice?

ps.

The clue is that his name was not Frankie.

http://en.wikipedia.org/wiki/Fibonacci

Bob

Quick 'google':

http://www.sciencemuseum.org.uk/visitmu … tAodyhTc0w

http://www.turing.org.uk/turing/

http://en.wikipedia.org/wiki/Alan_Turing

Bob

Does it mean you were making it up?
If so, you have awesome creativity

Hi Agnishom;

If so, you have awesome creativity

Trust me it is just a story. I hoped it would make you laugh.

For the Fibonacci proof, I think it is called Zeckendorf's theorem.

http://en.wikipedia.org/wiki/Zeckendorf%27s_theorem

http://www.mathsisfun.com/algebra/mathe … ction.html

Bob

hi

This article is from IT Now, the magazine of the BCS.  See page 51.

If you cannot scroll about you need to switch off 'full screen mode'.  I did this by saving the document and then re-loading it using Adobe.

http://www.bcs.org/upload/pdf/itnow-jun12.pdf

Bob

Hi phanthanhtom;

Welcome to the forum.

For the invention of the Calculus, the binomial theorem, Newton's iteration and the nurturing of Abraham De Moivre is the reason Newton is there.

hi Agnishom,

I've got to log out now, but I'll answer that later today.

Bob

hi Agnishom,

Night night then. 

Here's the proof for when you wake.

Let's call the number N.

If N is a Fibonacci number then there's nothing to do, because you either accept that a 'sum' can have just one number or if that makes you unhappy, you write it as the sum of the two Fibonacci numbers that made it by definition.

So what if it isn't?

The algorithm requires that we find the next F number below N and subtract it.

Can we definitely do this and be sure we never repeat an F ?

Let the next Fibonacci number above N,  be (a + b) where a and b are the two F numbers that made it.  The Fs go on for ever so there will certainly be one that is just bigger than N.

Suppose also that a < b.

Then b is the  next F number going down from N.  ie.  b < N < a + b

For if there was another  F number squeezing between b and N, let's call it 'c',  then b + c is the next F number after N,  not a + b.

The algorithm requires that we compute N - b and write b in the list.  (You always subtract the biggest F that you can.)

So we are left with a remainder of N-b

Now N < a + b    =>    N - b < a.

So if we now replace N with N - b in the algorithm we will never be able to subtract b again because N - b < a < b

So b will only occur in the list once.

So on this next loop our (new N) is now the (old N) - b

So can we carry on finding smaller and smaller Fs to subtract.

Yes, because 1, 2 and 3 are F numbers so we can either find a bigger F between 3 and N or we must be looking at N = 1, 2 or 3.

So eventually we must escape from the loop.

This is essentially the proof in the Wiki article.  I've just put it into words that, hopefully, make it easier to understand.

I found it was best to try some examples.

eg 1) N = 67                  b = 55                            new N = 67 - 55 = 12.

        N = 12        this isn't an F so b = 8                new N = 12 - 8 = 4

        N = 4          this isn't an F so b = 3                new N = 4 - 3 = 1

        N = 1          this is an F so stop.

67 = 55 + 8 + 3 + 1

eg 2)  N = 103               b = 89                            new N = 103 - 89 = 14

         N = 14                  b = 13                            new N = 14 - 13 = 1

103 = 89 + 14 + 1

Try it yourself and you'll find it always works.

Bob

Ok.  I thought I was just proving that a sum of Fs is always possible. (post #34)

Accidentally, my proof seems to cover what you want. 

Bob

Hi all;

I recommended that page in post #35. No one even saw it!