Base Phi

     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 

(+/-)(F_n + F_n+1 φ)  or  (+/-)(F_n+2 - F_n+1 φ)

where F_n 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!