The Three-Tier Theorem

Write out the five proper sixths. Then the eleven proper twelfths. A small difference between these two tables marks an interval that no denominator’s alignment can enter.

For the sixths, the reference is $1/6=0.1666\ldots$. Once the first two decimal places have passed, two rows repeat its digit 6, two repeat 3, and one has terminated. For the twelfths, the reference is $1/12=0.08333\ldots$. After those same two places, four rows repeat its digit 3, four repeat 6, and three have terminated.

Give each repeating row its proportion of agreements with the reference. Include the reference row itself, which of course agrees with itself. Give a terminating row score 1, as the alignment convention requires, even if its zero tail differs from the reference. Average over all the proper fractions.

The sixths receive three credits among five rows. The twelfths receive seven among eleven.

$$\alpha_{10}(6)=\frac35, \qquad \alpha_{10}(12)=\frac7{11}.$$

The two scores are about $0.6000$ and $0.6364$. No denominator gives an alignment strictly between them. Not a small denominator overlooked in the count, and not an enormous one whose repeating block we have yet to compute. Six alone attains the lower edge. Twelve alone attains the upper edge.

Here are all the denominators from 2 through 500. The top row of points sits at 1. A second family rises toward $2/3$. The remaining points lie below, with 6 at their highest position.

A plot can show us the empty strip. To know that it stays empty, we need a way to control denominators of every size, including those with composite repeating parts. The useful observation comes from subtracting a fraction from one. Each repeating digit acquires a partner.

The part that stops and the part that repeats

Forty is made entirely of twos and fives. Enough multiplications by ten absorb every factor of its denominator, so all thirty-nine proper fortieths terminate. Every row receives score 1. Their average is 1.

At 60, the base can absorb a factor of 20, but a factor of 3 remains. At 84, it can absorb a factor of 4, leaving 21. That remaining factor need not be prime.

Write this separation as $n=tm$. The supported part $m$ contains every factor supplied by the base. In decimal, those are the twos and fives. The rough part $t$ is what remains, relatively prime to ten.

A row $k/n$ terminates exactly when its numerator cancels the whole rough part. Its numerator must therefore be one of

$$t,\ 2t,\ \ldots,\ (m-1)t.$$

There are $m-1$ such rows. Of the $tm-1$ proper fractions, the other $(t-1)m$ repeat.

We compare those repeating rows on a common clock. Choose enough initial decimal places to clear the supported part from every row, then compare over a full period of the reference. A shorter tail repeats as many times as needed to fill that period. Starting each row at a separately chosen rotation would change the question.

The alignment formula for prime rough parts counts agreements through the remainder cycle selected by that clock. With a composite rough part, the reduced fractions can have different periods. We can still bound their total without finding a separate formula for each one.

A partner for every repeating row

At denominator 21, take $3/21$ and $18/21$. Their sum is one. Their repeating blocks are

 3/21 = 0.142857142857…
18/21 = 0.857142857142…

Each column adds to nine. Now set the reference $1/21=0.047619\ldots$ above them. The first row matches its 4. The second matches its 7. Both rows can contribute, but never in the same column. Two equal integer digits cannot add to nine.

The shorter rows obey the same rule. Seven twenty-firsts repeats 3; fourteen twenty-firsts repeats 6. Across the reference’s six places, the first has no match and the second has one. A pair may leave many positions unmatched. All we need is that it cannot supply two matches at any one position.

The argument works for every nonterminating row $k/n$ and its partner $(n-k)/n$. Their tail digits add to nine, so their alignment scores add to at most 1. No repeating row is left without a partner. The only fraction that could be its own partner is $1/2$, and that one terminates in decimal.

Consider 84 again. Three of its eighty-three proper fractions terminate. The other eighty form forty pairs. Even if each pair supplied a whole credit, the total could not exceed forty-three.

$$\alpha_{10}(84)\le\frac{3+40}{83}=\frac{43}{83}.$$

Its actual alignment is $11/83$, well below the bound. Pairing deliberately throws away detail. In exchange, it gives an estimate that does not care how long or complicated the repeating blocks become.

For any rough part $t>1$, the same count gives

$$\begin{aligned} \alpha_{10}(tm) &\le\frac{m-1+(t-1)m/2}{tm-1}\\[4pt] &<\frac{t+1}{2t}. \end{aligned}$$

The first numerator counts the terminating rows individually and the repeating rows in pairs. The last inequality is strict for every finite $m$.

After twos and fives have been removed, the decimal rough parts begin $1,3,7,9,11,13,\ldots$. Apart from 1 and 3, they are all at least 7. Consequently every such denominator satisfies

$$\alpha_{10}(n)<\frac{t+1}{2t}\le\frac47<\frac35.$$

They cannot reach the lower edge of the empty strip. Only the family with rough part 3 remains to be counted.

The missing multiplier

When $n=3m$ and $m$ contains only twos and fives, clearing the prefixes leaves thirds. There are $m$ rows that match the reference tail, $m$ with the opposing tail, and $m-1$ that terminate. The sixths and twelfths were the first two tables large enough to show all three groups.

Their count extends without change.

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

At $m=1$, this gives $1/2$. At $m=2$, it gives $3/5$. The next permitted multiplier is 4, giving $7/11$.

Three is missing because it is not built from twos and fives. Substituting $m=3$ into the formula would produce $5/8$, right inside the gap, but the substitution would violate the condition under which we obtained the formula. Denominator 9 has rough part 9 and supported part 1. Its actual alignment is $1/8$.

For the permitted values of $m$, the scores increase toward $2/3$. One rearrangement shows both the direction and the distance remaining.

$$\frac{2m-1}{3m-1} =\frac23-\frac{1}{3(3m-1)}.$$

As $m$ grows, the subtracted amount shrinks, but it never vanishes. From $m=4$ onward, the family stays at or above $7/11$ and strictly below $2/3$.

We can now account for every point in the plot.

Tier	Denominators in decimal	Alignment
I	Only factors of 2 and 5	$1$
II	$3m$, with $m$ supported on ten and $m\ge4$	From $7/11$ up toward $2/3$, never reaching it
III	Every other denominator	At most $3/5$, attained only at 6

Three and six belong to the lowest tier, although their rough part is 3. The supported multiplier decides which side of the gap that family occupies. For instance, 75, 96, and 120 all leave 3 after their twos and fives are removed, and their supported multipliers are all at least 4. They belong to the middle tier. The number 45 leaves 9. It does not.

An even base has no middle digit

Nothing in the pairing depends on the names of the decimal digits. In base $b$, complementary tail digits add to $b-1$.

If the base is even, that sum is odd. Two identical integer digits still cannot produce it. Also, $1/2$ still terminates. The pairing and its finite bound survive intact.

Now suppose that three does not divide the base. The rough part 3 remains available, and its family has the same exact count. The supported multiplier 2 is available because the base is even. So is 4. The multiplier 3 is unavailable because three does not divide the base. The jump from 2 to 4 survives too.

Every other rough part is an odd integer at least 5. Its alignment is therefore strictly below $(5+1)/(2\cdot5)=3/5$. Decimal gives the stronger bound $4/7$ because five has already been removed, but the stronger bound is not needed to keep the strip empty.

Thus every even base not divisible by three has the same forbidden interval

$$\left(\frac35,\frac7{11}\right),$$

with the same unique edge denominators 6 and 12. In the three-tier classification, “supported on ten” simply becomes “supported on the chosen base.”

An odd base changes the arithmetic at exactly the point the proof uses. It has a middle digit, $(b-1)/2$, equal to its own complement. In base three, $1/2=0.111\ldots$. The row is its own partner and agrees with itself everywhere. The two contributions can no longer be limited to one. The even-base hypothesis is doing real work.

A threshold inside the gap

The reciprocal golden ratio lies strictly between the two edges.

$$\frac35<\frac1\varphi<\frac7{11}.$$

It separates the lowest tier from the other two. In each of the stated bases, a denominator reaches that threshold exactly when all its factors are supported on the base, or when its rough part is 3 and its supported part is at least 4.

Any threshold strictly inside the same interval makes the same selection. The special role of the golden ratio comes from the three-core profile, not from the empty interval alone.

The plot can now extend as far as we like. Longer periods may rearrange the partial agreements, but each repeating row still has its complement. Larger supported factors add terminating rows in the same exact count. Neither change can insert a multiplier between 2 and 4. The strip remains empty for reasons already visible in the sixths and twelfths.

Earlier plots

The three earlier illustrations remain available here at their original resolution.

Complementary decimals and further reading

Complementary repeating decimals have a long history. Shrader-Frechette surveys their patterns. Kak and Chatterjee study complement structure and digit-sequence correlations, while Lewittes and Armstrong and Armstrong develop related period and repetend arithmetic. The pairing here applies across all proper fractions of a denominator, including those with composite rough parts, to obtain the finite bound and the complete classification.