alignment

Three and the Golden Ratio

January 19, 202015 min read

Companion paper: Why the Golden Ratio Selects the Prime Three →

Three and the Golden Ratio — The golden ratio's algebra selects the prime 3 from among all primes. The selection is not numerical. It is structural.

Twelve is an accommodating denominator. It gives us quarters, thirds, sixths, and a few decimals that take longer to settle into a pattern. Put all eleven proper fractions over twelve on one page, and the variety resolves into three kinds of behavior.

The interesting part begins after the first two decimal places.

All eleven proper twelfths at a common decimal depth. Four gold rows repeat 3, four gray rows repeat 6, and three teal rows have terminated. Gold and teal rows receive credit, giving seven elevenths. — All eleven rows, with the first two decimal places set aside together. Gold marks a matching tail; teal marks a terminated row. Both receive credit. Open the grid for the full-size view.

Every row gets the same two-place head start. Beyond that line, four rows repeat 3, four repeat 6, and three have finished. Their remaining digits are zeros. The first row, $1/12$ , supplies the reference. Its 3s continue down the gold rows.

This is the kind of view I want from long division. A fraction has its own sequence of digits, but the complete table lets me compare those sequences. At twelve, the repeating rows divide evenly between two complementary digits. Three and six add to nine at every place.

To turn the picture into a number, we need to say exactly what receives credit.

Seven out of eleven

The alignment score in this work counts a row if its repeating tail matches the reference. It also gives a terminating row full credit. That second rule is part of the definition. The zeros in a terminated row do not literally match the reference’s 3s.

At twelve, the accounting is small enough to keep on the page.

Rows	Behavior after two places	Credit
1, 4, 7, 10	Repeat 3, matching the reference	4
3, 6, 9	Terminate, then stay at zero	3
2, 5, 8, 11	Repeat the complementary 6	0

The score is therefore $7/11$ , about $0.636$ . Counting only literal matches would give $4/11$ . Restricting the comparison to the eight repeating rows would give $4/8$ , exactly one half. Those are answers to different questions. I will use the first convention throughout.

The constant tails of all eleven proper twelfths in numerator order. Four gold threes match the reference, three teal zeros terminate, and four gray sixes are complementary. Seven of the eleven rows receive credit. — The settled tail digits, in numerator order from one through eleven. Gold and teal account for the seven credited rows.

There is a useful discipline in this little example. Before admiring a ratio, count the rows that produced it.

The repeating-block notation

In the earlier drawings I wrote the tail of one twelfth as |333|. I repeat whole copies of the shortest pattern until their digit sum is divisible by one less than the base. In decimal that divisor is nine. Three copies of 3 give nine; three copies of 6 give eighteen. Thus |333| and |666| both close under that rule, although their shortest repeating patterns are single digits.

For one seventh, the block |142857| already has digit sum twenty-seven. Its six rotations supply the tails of all six proper sevenths. The twelfths have a different organization, with terminating rows and two constant tails. The digit grid above shows three later positions so the comparison can be made at a glance.

The count behind two thirds

Twelve belongs to a family of decimal denominators $3m$ , where $m$ is made entirely from powers of 2 and 5. Advancing far enough in decimal clears those factors. Use the same depth for every row, and the remaining behavior depends on the numerator’s remainder on division by three.

Among the $3m-1$ proper fractions, there are $m-1$ terminating rows. There are $m$ rows in the same remainder class as the reference numerator 1, and $m$ in the other nonzero class. These two repeating groups carry the reference digit and its nines complement. At twelve, those are the gold and gray rows in the grid.

Credit the terminating and matching groups, and the formula follows directly.

$\alpha_{10}(3m)=\frac{(m-1)+m}{3m-1}=\frac{2m-1}{3m-1}$

Two of the three groups receive credit. As the table grows, the missing zero-numerator row becomes a smaller part of the count, and the score rises toward two thirds. It stays strictly below that limit.

The condition on $m$ does real work. It is what lets us clear the finite prefix in decimal. Nine, for example, is a multiple of three, but its multiplier $m=3$ falls outside this family.

A golden threshold

Now put a horizontal line across the curve at the reciprocal of the golden ratio.

$\varphi=\frac{1+\sqrt5}{2},\qquad \frac1\varphi\approx0.618$

Solving for the point where the score reaches that line gives $m=\varphi^2$ , approximately $2.618$ . This is an equality on the real curve obtained by extending the formula. An actual table still needs an admissible integer multiplier.

The real alignment curve crosses one over phi at multiplier phi squared. Filled points mark admissible multipliers 1, 2, 4, and 5. Multiplier 4, giving denominator 12 and score 7 over 11, is the first actual table above the line. — The hollow point belongs to the real curve. The gold dot belongs to the fraction table at twelve. The admissible multipliers on this part of the curve are 1, 2, 4, and 5.

The allowable multipliers begin 1, 2, 4, 5. Three is absent because it cannot be built from 2s and 5s. Four gives the first table above the line, with denominator twelve and score $7/11$ .

So there are two things to keep distinct in the drawing. The hollow point marks the exact golden crossing at an irrational scale. The gold dot marks an actual fraction table whose rational score has passed it. No table has acquired an irrational number of matching rows.

The relation between the two golden quantities is more suggestive. The threshold is $1/\varphi$ ; its crossing scale is $\varphi^2$ . One is the reciprocal square of the other. We will return to that relation after checking how much of the selection is already explained by the count.

Three among the odd primes

Replacing three with another prime requires some care with the base. For this construction, choose a base $b$ that leaves remainder one on division by $p$ , and let $m$ divide a power of $b$ . These conditions give constant repeating tails and the same form of counting argument.

$\alpha_b(pm)=\frac{2m-1}{pm-1}\longrightarrow\frac2p$

Three works in decimal. Five works in base six; seven works in base eight. We are comparing such admissible families. The formula does not describe every prime in decimal.

For an odd prime, the score approaches its limit from below. To reach the golden threshold, that limit must be greater than $1/\varphi$ . Three has room to cross. Five and every larger prime run out of room first.

Limiting alignment scores for primes 3, 5, 7, and 11 are two thirds, two fifths, two sevenths, and two elevenths. Only three has a limit above the reciprocal golden threshold. The interval between two fifths and two thirds is shaded. — Each prime is considered in a base that leaves remainder one on division by that prime. The shaded interval contains every threshold strictly between two fifths and two thirds. All of them separate three from the larger odd primes.

The arithmetic is $p<2\varphi\approx3.236$ . Only one odd prime fits. Two is a separate case, with score 1 throughout its admissible families.

But the shaded interval also shows how much freedom we have. Every threshold strictly between two fifths and two thirds separates the same odd primes. Choosing $0.6$ would do it. The golden ratio needs a more specific reason to stand out.

The reciprocal-square condition

Return to the relation between the golden threshold and its crossing scale. Write $\tau$ for a threshold between zero and one. Ask for the scale where the score reaches $\tau$ to equal $1/\tau^2$ .

This is an additional condition I choose to study. The digit count supplies the curve; it does not require this relation. Once the condition is stated, however, we can solve it across the entire prime family.

Substitute $m=1/\tau^2$ into the score. The resulting equation is

$\frac{2-\tau^2}{p-\tau^2}=\tau.$

The next drawing makes the test visible. Each colored curve gives the score at that prescribed scale. The diagonal gives the threshold itself. An intersection is a place where they agree.

Scores evaluated at multiplier one over tau squared for primes 3, 5, and 7 intersect the diagonal score equals tau at approximately 0.618, 0.382, and 0.278. Only the first intersection lies between two fifths and two thirds. — The diagonal asks for the score to equal the threshold used to choose its scale. Three meets it at one over phi; five at one over phi squared. The shaded interval is the same separating interval as in the preceding figure.

Multiplying out turns the intersection problem into a cubic.

$f_p(\tau)=\tau^3-\tau^2-p\tau+2=0$

For every prime $p\geq3$ , there is exactly one root between zero and one. At three the cubic factors, and the golden relation appears explicitly.

$f_3(\tau)=(\tau-2)(\tau^2+\tau-1)$

The root in our interval is $1/\varphi$ . Five also factors, giving $1/\varphi^2\approx0.382$ . That second golden value lies below the separating interval. From seven onward, the threshold roots are cubic irrationals. None belongs to a quadratic field, including the field built from rational numbers and $\sqrt5$ .

The factorization in a few lines

A monic cubic with integer coefficients can factor over the rationals only if it has a rational root. Here the possible roots are $1,-1,2,-2$ .

Candidate	Value of the cubic	Prime giving zero
$1$	$2-p$	$2$
$-1$	$p$	None
$2$	$6-2p$	$3$
$-2$	$2p-10$	$5$

The factorizations at three and five are

$f_3(\tau)=(\tau-2)(\tau^2+\tau-1),$ $f_5(\tau)=(\tau+2)(\tau^2-3\tau+1).$

Both quadratic factors have discriminant five. Their roots in $(0,1)$ are $1/\varphi$ and $1/\varphi^2$ . At two the factorization is $(\tau-1)(\tau^2-2)$ , with no root strictly between zero and one. For primes at least seven, the cubic is irreducible over the rationals.

Uniqueness of the root is equally short. For $p\geq3$ , the cubic is positive at zero, negative at one, and strictly decreasing between them. Its root also lies below $2/p$ . Consequently the roots for primes at least five all fall below the separating interval.

Three in the remainder

The cleanest expression of the selection comes from polynomial division. Divide the prime-indexed cubic by the golden quadratic $\tau^2+\tau-1$ . The quotient is $\tau-2$ . The remainder is

$(3-p)\tau.$

At the positive golden threshold, this vanishes exactly when $p=3$ . The same three that organized the fraction table now appears in the condition for the golden quadratic to divide its cubic. Every step is available for inspection, from the eleven rows to this single remainder.

The ingredients are classical. Repeating expansions, complementary digit sums, rational-root tests, and golden-ratio identities have long histories. The paper Why the Golden Ratio Selects the Prime Three gives the proofs and references. Its particular result belongs to this alignment score and the added reciprocal-square condition.

What I find useful here is the route from looking to counting. Keep the whole fraction table, state how each part contributes, and follow the resulting formula. Two thirds comes from the three classes. The golden ratio enters through a specified threshold and scale relation. The final remainder tells us exactly where those choices meet.

The earlier plots

These are the three plots from the earlier presentation. Open any image to inspect it at full resolution.