It's All Gold 2.

Level 2

Let $\phi = \dfrac{1 + \sqrt{5}}{2}$ .

Which of the following below is true:

(1) $\:\ \dfrac{3\phi}{2} = (10.001001001001 ... )_{\phi}$ and $4\phi = (1010.101010101010 ...)_{\phi}$ and $\dfrac{\phi - 1}{2} = (0.100100100100 ... )_{\phi}$ .

(2) $\:\ \dfrac{3\phi}{2} = (10.100100100100 ... )_{\phi}$ and $4\phi = (1010.001001001001 ...)_{\phi}$ and $\dfrac{\phi - 1}{2} = (0.100100100100 ... )_{\phi}$ .

(3) $\:\ \dfrac{3\phi}{2} = (10.100100100100 ... )_{\phi}$ and $4\phi = (1010.010101010101 ...)_{\phi}$ and $\dfrac{\phi - 1}{2} = (0.001001001001 ... )_{\phi}$ .

(4) $\:\ \dfrac{3\phi}{2} = (10.001001001001 ... )_{\phi}$ and $4\phi = (1010.101010101010 ...)_{\phi}$ and $\dfrac{\phi - 1}{2} = (0.001001001001 ... )_{\phi}$ .

Bonus: Write a program that inputs a positive real number $x$ and outputs it in a real positive base $b$ .

For $(1 \leq n \leq 50)$ , use the program above to represent $\dfrac{\phi}{n}$ in base $10\phi$ to the desired number of decimal places. For base $10\phi$ you will need to use $A - 10, \:\ B - 11, \:\ C - 12, \:\ D - 13, \:\ E - 14, \:\ F - 15$ and $G - 16$ for digits.

1 2 4 3

1 solution

Rocco Dalto
Feb 13, 2019

$\phi$ is one solution to $x^2 - x - 1 = 0$

For $\phi - 1:$

$\implies \phi^2 = \phi + 1 \implies \phi^2 - 1 = \phi \implies \dfrac{1}{\phi^2 - 1} = \dfrac{1}{\phi} = \dfrac{2}{\sqrt{5} + 1} = \dfrac{\sqrt{5} - 1}{2} = \phi - 1$

$\implies \phi - 1 = \dfrac{1}{\phi^2 - 1} = \dfrac{\dfrac{1}{\phi^2}}{1 - \dfrac{1}{\phi^2}} =$ $\sum_{n = 1}^{\infty} (\dfrac{1}{\phi^2})^{n} = \boxed{(0.010101010101 ...)_{\phi}}$

For $\dfrac{\phi}{2}:$

$\phi^2 = \phi + 1 \implies \implies \phi^3 = \phi^2 + \phi = 2\phi + 1 \implies \phi^3 - 1 = 2\phi \implies$

$1 - \dfrac{1}{\phi^3} = \dfrac{2}{\phi^2} \implies \dfrac{1}{1 - \dfrac{1}{\phi^3}} = \dfrac{\phi^2}{2} \implies \dfrac{\phi^2}{2} = \sum_{n = 0}^{\infty} (\dfrac{1}{\phi^3})^{n} \implies \dfrac{\phi}{2} = \sum_{n = 0}^{\infty} (\dfrac{1}{\phi})^{3n + 1} = \boxed{(0.100100100100 ... })_{\phi}$

For $\dfrac{\phi - 1}{2}:$

$\phi$ is one root of $x^2 - x - 1 = 0 \implies \phi^2 = \phi + 1 \implies$ $\implies \phi^3 = \phi^2 + \phi = 2\phi + 1 \implies \phi^3 - 1 = 2\phi \implies \dfrac{1}{\phi^3 - 1} = \dfrac{1}{2\phi} = \dfrac{(\dfrac{\sqrt{5} - 1}{2})}{2} = \dfrac{\phi - 1}{2} \implies$

$\dfrac{\phi - 1}{2} = \dfrac{1}{\phi^3 - 1} = \dfrac{\dfrac{1}{\phi^3}}{1 - \dfrac{1}{\phi^3}} = \sum_{j = 1}^{\infty} (\dfrac{1}{\phi^3})^{j} = \boxed{(0.001001001001 ...)_{\phi}}$

Using the above $\implies$

$4\phi = (2\phi + 1) + \phi + (\phi - 1) = \phi^{3} + \phi + (\phi - 1) = \boxed{(1010.010101010101 ...)_{\phi}}$

and

$\dfrac{3\phi}{2} = \phi + \dfrac{\phi}{2} = \boxed{(10.100100100100 ...)_{\phi}}$

Note:

$2\phi = \phi + 1 + \phi - 1 = (11.010101010101)_{\phi} = (100.010101010101)_{\phi} = \phi^2 + (\phi - 1)$

$3\phi = (\phi + 1) + \phi + (\phi - 1) = \phi^2 + \phi + (\phi - 1) = (110.010101010101 ...)_{\phi} = (1000.010101010101 ...)$ $= \phi^3 + (\phi - 1) = (2\phi + 1) + (\phi - 1)$ .

Bonus - Program below written in Free Pascal:

program irrationalbases;

{$R-}

var phi:extended;

myfile:text;

n:longint;

procedure change(var k:string);

begin{change}

IF K = '10' THEN K:= 'A';

IF K = '11' THEN K:= 'B';

IF K = '12' THEN K:= 'C';

IF K = '13' THEN K:= 'D';

IF K = '14' THEN K:= 'E';

IF K = '15' THEN K:= 'F';

IF K = '16' THEN K:= 'G';

END;{CHANGE}

function power(base:extended; exponent:integer):extended;

var n:longint;

product:extended;

begin

product:= 1;

for n:= 1 to exponent do

product:= product * base;

power:= product;

end;

function convert(x:extended; base:extended):string;

type arraytype = array[1 .. 100] of longint;

 var j,k,m,count:longint;

 a:arraytype;

 R,holdr:extended;

 strvar,strsum,strsum2:string;

begin

R:= x;

j:= 1;

while power(base,j) <= R do

begin

j:= j + 1;

end;

k:= j - 1;

count:= 1;

While k >= 0 do

begin

a[count]:= trunc(R/power(base,k));

R:= R - a[count] * power(base,k);

k:= k - 1;

count:= count + 1;

end;

count:= count - 1;

strsum:= '';

for m:= count downto 1 do

begin

str(a[m],strvar);

if a[m] > 9 then

change(strvar);

strsum:= strvar + strsum;

end;

holdr:= R;

j:= 1;

while R >= power(1/10,20) do

begin

a[j]:= trunc(R/power(1/base,j));

R:= R - a[j] * power(1/base,j);

j:= j + 1;

end;

j:= j - 1;

strsum2:= '';

for m:= j downto 1 do

begin

str(a[m],strvar);

if a[m] > 9 then

change(strvar);

strsum2:= strvar + strsum2;

end;

If holdr = R then

convert:= strsum

else

convert:= strsum + '.' + strsum2;

end;

begin

assign(myfile,'irr.txt');

rewrite(myfile);

phi:= (1 + sqrt(5))/2;

for n:= 1 to 50 do

begin

writeln(myfile,convert(phi/n,10 * phi));

writeln(convert(phi/n,10 * phi));

end;

close(myfile);

readln;

end.

Output:

1.A

0.D176E9295C4A2B813

0.8BC4B21A0FE1D80F

0.68D43FB5B2394F497

0.53D43FB5B2394F497

0.45E3CG0A9D8E572CA

0.3BFC5BA75B5C5C6C5

0.346A373070881E279

0.2EB70B76FE4F3EE2D

0.2A

0.2626A9DB8AB61E5E8

0.22F35D7791CF037F8

0.203A39D8A598DC845

0.1E140E279A182ABD5

0.1C0FBF31B2F86C03

0.1A4C8A120012C50DE

0.18BF2708B66F644BC

0.175B9E9F69BEA9366

0.161DDF8E790E36183

0.15

0.13G0AD87F8284DB9E

0.131354EF520A34904

0.123D9ABDBA7DF03AD

0.1179C788468G134AD

0.10C5D8822668D59A

0.101D3441EBC49531

0.0FB270CBBB66D43D9

0.0F215E37G045B44D

0.0E9D546C22FF18E09

0.0E1F4C9ED61007A6

0.0DAC3ACA6427F61C

0.0D3DB36CG1DEFE77B

0.0CD8AE39579BDC42

0.0C76EC1DFF6C49CEA

0.0C1ACEFDC0198AC3

0.0BC69BCED4285A29

0.0B743F33496B0B53

0.0B26455EBF6C00427

0.0ADF4204063E92F3D

0.0A98E54A2622B10A9

0.0A55FEDEC1D73EDC

0.0A164E6F09C6AA113

0.09DC8A7050G07A5AD

0.09A276F6B0FB7495A

0.096B2C0973497D96

0.093622F7BE262G068

0.0903626AG0263061D

0.08D5B02A762AD692

0.08A711FG1418A153F

0.087A59458301DA43F

It's All Gold 2.

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