Math Problems

[Numeri]

So the other day I found this awesome math problem, http://nrich.maths.org/6660, and started working on it.

Basically:

This challenge involves building up a set F of fractions using a starting fraction and two operations which you use to generate new fractions from any member of F.

Rule 1: F contains the fraction 1/2.

Rule 2: If p/q is in F then p/(p+q) is also in F.

Rule 3: If p/q is in F then q/(p+q) is also in F.

After thinking about it a bit, you can prove that no fraction exists where p|q (p divides q), and that all fractions are between 0 and 1. Also, you can work your way backward from a fraction to get the list of operations that generated it (from 1/2). I used 'A' to represent Rule 2 and 'B' for Rule 3. It turns out that the pattern ABABABABABAB...B homes in on sqrt(2)-1 while ABABABABABAB...A homes in on sqrt(1/2) as the length of the pattern increases. This is related to continued fractions, and using the operations A and B, you can generate any continued fraction (in between 0 and 1) and thus any rational number in that range.

Anyway, I had fun doing this, and was wondering if anyone else had some problems like it.

Thanks,
Numeri

[helios]

After thinking about it a bit, you can prove that no fraction exists where p|q (p divides q)

???
All fractions derived from only rules 1 and 2 are of the form 1/n. 1|n for all natural n.

that all fractions are between 0 and 1

Actually, the supremum of F is the reciprocal of the golden ratio. See if you can prove that.

[ne555]

> Actually, the supremum of F is the reciprocal of the golden ratio
¿? I don't understand your joke

[helios]

Joke?

1/2 < 2/3 < 3/5 < 5/8 < 8/13 < 13/21 < ... < Fib(i-1)/Fib(i)

Fib(n-1)/Fib(n) --(n ---> inf)--> 1/phi

EDIT: Ah, crap. Actually, 2/3 is the maximum of F, isn't it? Damn...

[ne555]

¿why did you eliminate Rule 2?

[Script Coder]

All fractions derived from only rules 1 and 2 are of the form 1/n. 1|n for all natural n.

Is my maths off?
From (1/2) where p=1 and q=2 then by rule two three (2/3)

EDIT: Corrected. Thank You Ispil.

[Ispil]

No, that's rule 3.

Rule two states that for every value in set F of p/q, there exists a value in F of p/(q+p). So, for 1/2, there would also be 1/3, 1/4, 1/5, and so on (because p never changes there).

Rule three states that for every value in set F of p/q, there exists a value in F of q/(p+q). So for 1/2, there would also be 2/3, 3/5, 5/8, 3/4, 4/5, 4/7, 2/7, and so on.

[Numeri]

Well, I guess I did forget to say 'excluding 1|n,' but I guess I meant no fractions in a reducible form. :D