Rational numbers — definition, decimal expansion (terminating - repeating): Worked examples

The scenario matrix

Before solving anything, let us lay out the full battlefield. Every rational-decimal problem you meet is one of these case classes:

#	Case class	What makes it special	Covered by
A	Terminating, denom $= 2^{a} 5^{b}$	division just stops	Ex. 1
B	Pure repeating, one prime $\neq = 2, 5$	remainder loops, decimal point right before the loop	Ex. 2
C	Mixed repeating (both a $2/5$ part and another prime)	some digits settle, then a loop starts	Ex. 3
D	Negative numerator / sign handling	where does the minus go?	Ex. 4
E	Denominator $= 1$ (integer / degenerate)	"decimal" that is just a whole number	Ex. 5
F	Zero numerator (degenerate)	$0/ q$ for any $q$	Ex. 5
G	Repeating decimal $\to$ fraction (reverse direction)	the "multiply and subtract" trick	Ex. 6
H	Limiting value: $0. \overline{9}$	the famous edge case where a repeat equals a terminator	Ex. 7
I	Word problem (real-world units)	translate, then classify	Ex. 8
J	Exam-style twist: "for which $n$ does $\frac{1}{n}$ terminate?"	reasoning about the whole family at once	Ex. 9

Two tools do all the work, so let us name them up front.

Figure — Rational numbers — definition, decimal expansion (terminating - repeating)

Look at the figure: the horizontal bar is the number line of possible remainders $0$ to $q - 1$ . Long division walks along it, and because there are only finitely many slots, the walk must eventually step onto a slot it already stamped — that is the moment repetition is born (unless it lands on $0$ first, which is termination).

Case A — Terminating

Worked example Example 1 · cell A

Write $\frac{9}{40}$ as a decimal.

Forecast: Guess now — will this stop or loop? Look only at $40$ .

Factor the denominator. $40 = 2^{3} \cdot 5^{1}$ . Why this step? The rule from the parent note says a fraction terminates iff the denominator (in lowest terms) has only $2$ s and $5$ s. $g cd (9, 40) = 1$ , so it is already lowest terms, and $40$ passes the test — it will stop.
Balance the powers of 2 and 5. We have three $2$ s but only one $5$ . To make a power of ten $1 0^{n} = 2^{n} 5^{n}$ , the exponents must match. The bigger exponent is $3$ , so we need $1 0^{3}$ . Multiply top and bottom by $5^{2} = 25$ : $\frac{9}{40} = \frac{9}{2 ^{3} 5 ^{1}} = \frac{9 \cdot 5 ^{2}}{2 ^{3} 5 ^{3}} = \frac{225}{1000} .$ Why this step? A denominator of $1 0^{3}$ means "move the decimal point 3 places" — that is what base ten is.
Read off the decimal. $\frac{225}{1000} = 0.225$ .

Verify: $0.225 \times 40 = 9$ ✓. And it stopped after exactly $max (a, b) = max (3, 1) = 3$ digits, exactly as the balancing predicted.

Case B — Pure repeating

Worked example Example 2 · cell B

Write $\frac{4}{11}$ as a decimal.

Forecast: $11$ is prime and it is not $2$ or $5$ . Stop or loop?

Classify. $11$ is a prime other than $2$ or $5$ , so this repeats, and since $11$ shares no factor of $2$ or $5$ with the numerator, the repeat starts right after the decimal point (a pure repeater). Why this step? Only a $2$ / $5$ part in the denominator can produce non-repeating "settling" digits; here there is none.
Long division, tracking remainders.
- $40 \div 11 = 3$ , remainder $7$ .
- $70 \div 11 = 6$ , remainder $4$ ← remainder $4$ is the one we started with. Why this step? The instant a remainder reappears, every future step is a copy — that is the loop.
Read the repetend. The digits produced between the first " $4$ " remainder and its return are $3, 6$ : $\frac{4}{11} = 0. \overline{36} .$ Length $2$ , and $ϕ (11) = 10$ ; $2$ divides $10$ ✓ (matches Euler's totient function as an upper bound, per the parent note).

Verify: $0. \overline{36} = \frac{36}{99} = \frac{4}{11}$ ✓ (we reverse this move in Example 6).

Case C — Mixed repeating

Worked example Example 3 · cell C

Write $\frac{7}{60}$ as a decimal.

Forecast: $60$ has $2$ s, $5$ s and a $3$ . What happens when a denominator carries both kinds of prime?

Factor. $60 = 2^{2} \cdot 3 \cdot 5$ . Lowest terms already ( $g cd (7, 60) = 1$ ). Why this step? The presence of the extra prime $3$ guarantees a repeat; the $2^{2} \cdot 5$ part will produce a short non-repeating lead-in.
Predict the shape. Split off the $2/5$ part. The largest $2$ / $5$ exponent is $2$ (from $2^{2}$ ), so 2 digits settle before the loop begins. Why this step? Multiplying by $1 0^{2}$ absorbs the $2^{2} \cdot 5$ influence, leaving a pure repeater over the leftover $3$ .
Long division.
- $70 \div 60 = 1$ , remainder $10$ .
- $100 \div 60 = 1$ , remainder $40$ .
- $400 \div 60 = 6$ , remainder $40$ ← remainder $40$ repeats. $\frac{7}{60} = 0.11 \overline{6} .$ The " $11$ " is the settling part, " $\overline{6}$ " is the loop.

Verify: $0.11 \overline{6} = 0.11 + 0.00 \overline{6}$ . Now $0.00 \overline{6} = \frac{6}{900} = \frac{1}{150}$ , and $0.11 = \frac{11}{100} = \frac{16.5}{150}$ … cleaner: $\frac{7}{60} = 0.1166 \dots$ ; check $0.11666 \times 60 \approx 7.0$ ✓. Exact check in VERIFY.

Case D — Negatives

Cases E & F — Degenerate inputs

Worked example Example 5 · cells E and F

(a) Write $\frac{6}{1}$ as a decimal. (b) Write $\frac{0}{17}$ as a decimal.

Forecast: These look silly — but the definition $\frac{p}{q}$ still applies. Are they rational? Terminating?

(a) Denominator $= 1$ . $1 = 2^{0} \cdot 5^{0}$ — an empty product of $2$ s and $5$ s, which still counts. Why this step? $a = b = 0$ satisfies $q = 2^{a} 5^{b}$ , so by the rule it terminates. Indeed $\frac{6}{1} = 6 = 6.0$ : it "terminates" after zero decimal digits. Every integer is a terminating rational — this is why Real numbers contain $Z$ as terminating decimals.
(b) Zero numerator. $\frac{0}{17} = 0$ for any nonzero denominator. Why this step? $0$ divided by anything (nonzero) is $0$ ; the denominator never gets to matter because the division ends immediately with remainder $0$ . So $\frac{0}{17} = 0.0$ , terminating — even though $17$ is a "bad" prime! The lesson: the denominator rule assumes lowest terms, and $\frac{0}{17}$ reduces to $\frac{0}{1}$ .

Verify: (a) $6.0 \times 1 = 6$ ✓. (b) $0 \times 17 = 0$ ✓, and $g cd (0, 17) = 17$ so lowest terms is $\frac{0}{1}$ ✓.

Case G — Decimal back to fraction

Case H — The limiting edge case

Case I — Word problem

Case J — Exam-style twist (the whole family)

Recall Quick self-check

Does $\frac{7}{28}$ terminate? ::: Reduce first: $\frac{7}{28} = \frac{1}{4} = 0.25$ . Yes — $4 = 2^{2}$ . Why must a long division of $\frac{p}{q}$ eventually repeat or stop? ::: Only $q$ possible remainders ( $0$ to $q - 1$ ); by pigeonhole one must recur (or hit $0$ ). What number does $0. \overline{9}$ equal? ::: Exactly $1$ . Does the length of the repetend of $\frac{1}{n}$ have to be $n - 1$ ? ::: No — it divides $ϕ (n)$ ; it can be much shorter.