Base Phi
Base phi, like base 2, needs only two
digits; 0 and 1. Here I will also use the digit 2 in mid-calculation.
In
base phi, 10 equals φ = (1 + root(5))/2 =
1.618033989…
Phi obeys the equation:
φ 2 = φ +1
so in base phi:
100 = 11
2 = 1+1 = 1.11
= 10.01
3
= 2+1 =
11.01 = 100.01
4
= 3+1 =
101.01
5 = 4+1 =
101.01+1 = 102.01 = 110.02 =
1000.1001
6
= 5+1 =
1001.1001 = 1010.0001
7
= 6+1 =
1011.0001 = 10000.0001
8
= 7+1 =
10001.0001
9
= 8+1 =
10002.0001 = 10010.0101
10 = 9+1 =
10011.0101 = 10100.0101
11 = 10+1 =
10101.0101
12 = 11+1 =
10102.0101 = 10110.0201 = 11000.1002
= 100000.101001
1
= 0.11 =
0.1011 = 0.101011 = … = 0.1010101…
0.11111… =
0.101010…+0.010101… = 1+.1 = 1.1
= 10
10
= 0.11111… = 1.11*(0.100100100…)
= 2*(0.100100100…);
Therefore 0.100100100… = 10/2
= φ/2
Also 0.010010010… = (φ/2)/φ
= ½
Also that 0.001001001… =
1/(2φ)
Consider this multiplication:
10.1*10.1 = 101+1.01 = 102.01 = 110.02 = 1000.1001 = 5
Therefore 10.1 = the square
root of 5.
1
= 0.101010101… =
101*(0.100010001000…)
Therefore 0.100010001000…
= 1/101
= 10/10.1
= φ/root(5)
= (5+root(5)) / ten
Phi and the Fibonacci sequence are
closely related. Consider this Fibonacci-like sequence:
G(0) =
1 ; G(1)
= 3
; G(n+2) = G(n+1) + G(n)
G(0) = 1 = 1.
G(1) = 3 = 100.01
G(2) = 4 = 101.01
G(3) = 7 = 10000.0001
G(4) = 11 = 10101.0101
G(5) = 18 = 100000.00001
G(6) = 29 = 1010101.010101
…
and so on, alternating most prettily between a positive power of phi plus a
negative power of phi and a positive power of phi minus a negative power of phi.
Here
is the Fibonacci sequence in base phi:
F(1) = 1 = 1.
F(2) = 1 = 1.
F(3) = 2 = 10.01
F(4) = 3 = 100.01
F(5) = 5 = 1000.1001
F(6) = 8 = 10001.0001
F(7) = 13 = 100010.001001
F(8) = 21 = 1000100.010001
F(9) = 44 = 100010001.0001001
These all repeat 1000, then end with
0001 or 001. So they tend towards 1000100010001000…, which is related to φ/root(5)
.
Call a number terminating base phi if
it has an expansion base phi with finitely many 1’s. Any integer is terminating
base phi, as is phi and its powers. Here is an exercise for the ambitious student:
Prove
that any number terminating base phi is of the form:
A
+ B φ, where A and B are integers.
This
is Z[φ]; the integers, adjoin phi. Z[φ] is closed under plus, minus, and times,
but not division.
Which numbers terminating base phi are
invertible? Here an exercise for the
ambitious student:
The only invertible elements of Z[φ]
are plus or minus the powers of φ; (+/-)φn and these are of the form
(+/-)(Fn
+ Fn+1 φ) or (+/-)(Fn+2 - Fn+1 φ)
where
Fn is the nth Fibonacci number.
Addendum:
Base Minus One Over Phi
Phi
solves the equation x 2 = x +1 ; so in base phi, 100 = 11. But -1/φ solves the same equation;
so base -1/φ also has 100 = 11.
The expansions for integers derive
from 100=11. Therefore integers have the same expansions in base -1/φ as they do in base φ!
