The golden ratio number φ = (1+√5)/2 possesses many interesting properties.
For any even integer n:
φn + 1 /φn is an integer
For any odd integer n:
φn - 1 /φn is an integer
Prove the above.
postings? I’m pasting Word text.
Using phi2 = phi + 1 repeatedly to break down phin to a
linear function of phi shows that phin
= Fn*phi + Fn-1 ,
where Fi are Fibonacci numbers (F1 = 1, F2 =
1, F3 = 2 …).
(for example, phi6 = F6*phi + F5 = 8*phi + 5)
This can be extended to negative values of n by using the
bidirectional Fibonacci series: …5, -3,
2, -1, 1, 0, 1, 1, 2, 3, 5…
Note that for even n, Fn = -F-n
and for odd n, Fn = F-n .
+ 1/phin = (Fn*phi + Fn-1)+ (F-n*phi
(Fn + F-n)*phi + Fn-1 + F-n-1
so for even n the bracket is zero and the result is an integer.
Also phin – 1/phin = (Fn*phi
+ F-n) – (F-n*phi + F-n-1)
– F-n)*phi + F-n – F-n-1
so for odd n the bracket is zero and the result is an integer.
PS Is there a simple way to produce greek letters in these
Posted by Harry
on 2016-10-18 18:44:16