Why a Fraction Ends in One Base and Repeats in Another

One half has several exact descriptions

One half in base ten is zero point five. Five first-place units, each one tenth, exactly fill a half.

Now change to base two. One half is zero point one, because a binary first-place unit is already one half.

Change to base three and the ones continue: one third, one ninth, one twenty-seventh, and so on. The fraction did not become less exact in base three. Its notation needs an infinite expansion in that base. Terminating or repeating is a property of the relationship between a fraction and its base, not an intrinsic mark of whether the fraction is good or difficult.

Predict base six before it appears.

Its first place has size one sixth, so three copies make one half. The expansion terminates after one place. We have now seen three terminating descriptions and one repeating description of the same number. Our task is to explain the pattern without trying every possible base one by one.

Ending after k places means a power denominator

One eighth in base two terminates at the third fractional place. The first three place units are one half, one fourth, and one eighth. The last one can express the entire fraction exactly. Any expansion ending after three binary places can be rewritten as a whole number of eighths.

For example, one half becomes four eighths and one fourth becomes two eighths. More generally, an expansion ending after k places in base b is a whole number divided by b to the power k. That gives a necessary condition for termination: the reduced denominator must divide some power of the base.

It is also sufficient. If the denominator divides a power of the base, multiply numerator and denominator to express the fraction using that power denominator, then write the corresponding finite places.

Compare one eighth in base ten. A thousand is divisible by eight, so one hundred twenty-five thousandths gives zero point one two five. Different bases, the same divisibility reason.

The base supplies its prime factors

One sixth does not terminate in base ten. The denominator six contains a factor of three, but every power of ten is made only from twos and fives. No matter how many decimal places we allow, a power of ten cannot become divisible by three. The expansion starts with one and then repeating sixes.

Now change to base six. One sixth is a single first-place unit, so zero point one is exact.

Compare base twelve. Two twelfths make a sixth, giving zero point two. The decimal failure was a mismatch between denominator factors and base factors. Multiplicities determine how many places may be needed, but introducing another power never supplies a new prime factor. For a reduced denominator to terminate, every prime dividing it must also divide the base.

This is the factor version of our power-denominator test. Keep both versions: divisibility explains the relationship directly, and prime factors make it easy to predict examples before exploring.

Reduce before applying the test

Three sixths appears as zero point five. At first that might seem to contradict the claim about a factor of three in a denominator. But three sixths reduces to one half. The common factor of three cancels from numerator and denominator because the same amount can be regrouped in larger equal pieces.

The denominator relevant to termination is two, not six.

Compare one sixth, whose numerator cannot cancel that factor of three. It repeats in base ten.

Return to three sixths.

Then change to two sixths. Two sixths reduces to one third, so it repeats as threes. All three fractions share the written denominator six, yet their decimal behavior differs. The original numerator matters through reduction. This is a good habit with any general rule: check its conditions before applying it.

Here the word reduced is doing essential mathematical work. During exploration, simplify the fraction mentally, list its remaining denominator factors, and only then compare those factors with the chosen base.

Predict one fifth in three bases

One fifth terminates in base ten as zero point two. Now predict base two. The reduced denominator has a factor of five, which no power of two contains, so the expansion must continue.

Inspect the repeating binary block. Its digits should be interpreted with halves, fourths, eighths, and smaller binary places, not as decimal fractions. Predict base five next. The denominator divides the base itself, so one first-place unit is enough.

The expansion is zero point one. Finally consider base six. Powers of six contain only twos and threes, so a factor of five remains unmatched.

The expansion repeats there too. These predictions did not require calculating every digit. The factor test answered whether an ending was possible. Long division then tells us which digits actually occur and what their repeating block is. Pause on a repeating case and distinguish those two questions. Knowing that an expansion repeats does not yet tell you how long its cycle will be.

Shared factors determine whether it ends, powers determine when

Use one eighth in base two. Its digits are zero, zero, one, ending at the third fractional place. The denominator eight equals two cubed, so three binary places supply eighths. The leading zero positions are part of the finite expansion; ending after three places does not mean that all three digits must be nonzero.

Change to base four. The digits are zero, two. Four squared is sixteen, and two sixteenths equal one eighth. We need only two places because the square of the base is already divisible by eight. The base itself need not be divisible by the denominator if a higher power can supply the required subdivision.

Now use base six. The digits are zero, four, three. Four thirty-sixths plus three two-hundred-sixteenths gives twenty-seven two-hundred-sixteenths, exactly one eighth. Six contains a factor two, but three powers of six are needed to provide the three factors of two in the denominator eight.

View that complete base-six expansion. Its additional factor three does not prevent termination. What matters is that every prime factor needed by the reduced denominator can be supplied by powers of the base. The number of positions depends on how many copies of each needed factor those powers supply.

Sharing only part of the denominator is insufficient

Choose one eighteenth in base six. Eighteen is two times three squared, and six squared is thirty-six. The expansion is zero point zero two because two thirty-sixths equal one eighteenth. Both kinds of denominator factor are available in six, and its second power provides enough copies of each.

Change to base twelve. The expansion becomes zero point zero eight: eight one-hundred-forty-fourths equals one eighteenth. Twelve squared contains the needed factors too. The two finite expressions have different final digits because their second-place units are thirty-sixths and one-hundred-forty-fourths, not because the original fraction changed.

Now use base ten. The expansion begins zero point zero five five five, with the fives continuing. Ten supplies factors two and five, but no factor three. Sharing the factor two with eighteen is not enough. The missing factors of three prevent any finite decimal denominator, a power of ten, from representing the fraction exactly.

Return to base six and its terminating second place. State the complete condition: first reduce the fraction, then ask whether its denominator divides some power of the base. A shared factor alone does not settle the question. This divisibility statement explains both our successful finite expansions and the decimal example that continues.

Choose a base that makes the fraction simple

Consider one twelfth in base six. Twelve is two squared times three. A single six does not contain enough factors of two, but thirty-six, the square of six, is divisible by twelve. One twelfth is therefore three thirty-sixths, written zero point zero three in base six. The first place contributes zero sixths;

the second contributes three thirty-sixths.

Compare base twelve, where one twelfth becomes zero point one.

Compare base ten, where the unmatched factor of three forces repetition. The same fraction can have a two-place ending, a one-place ending, or no finite ending at all. A useful exploration is to choose the fraction first and then select bases deliberately. Find one whose factors support its denominator and one whose factors do not.

Predict the outcome, inspect the expansion, and explain the result. The finite geometry shows representative places; the denominator argument establishes the infinite claim. Ending sooner makes a notation convenient for that fraction, without making the underlying number more exact.