Recurring Decimals [H]
Recurring Decimals
A recurring decimal is one in which one or more digits repeat infinitely. Any fraction \( \frac{p}{q} \) (in lowest terms) whose denominator contains a prime factor other than 2 or 5 will give a recurring decimal. Every recurring decimal is rational — it can be expressed exactly as a fraction — and the method for finding that fraction is elegant and systematic.
Dot notation: a dot placed above a digit (or above the first and last digits of a repeating group) indicates the recurring part.
\[ 0.\dot{3} = 0.3333\ldots \qquad 0.\dot{1}\dot{4} = 0.141414\ldots \qquad 0.3\dot{6} = 0.3666\ldots \]The Algebraic Method
Let \( x \) equal the recurring decimal. Multiply \( x \) by a power of 10 chosen so that the recurring digits align. Subtract to eliminate the recurring part, then solve for \( x \).
The key rule: multiply by \( 10^k \) where \( k \) is the number of digits in one repeating cycle. This shifts the decimal exactly one full cycle, so the recurring tails cancel when subtracted.
| Recurring type | Multiply by | Example |
|---|---|---|
| 1-digit cycle | \( 10 \) | \( 0.\dot{4} \to 10x - x = 9x \) |
| 2-digit cycle | \( 100 \) | \( 0.\dot{2}\dot{7} \to 100x - x = 99x \) |
| 3-digit cycle | \( 1000 \) | \( 0.\dot{1}\dot{4}\dot{2} \to 1000x - x = 999x \) |
| Partial (some non-recurring digits) | Two multiplications | \( 0.3\dot{6} \to 100x - 10x = 90x \) |
Worked Examples
Convert \( 0.\dot{4} = 0.4444\ldots \) to a fraction.
Let \( x = 0.\dot{4} \). The cycle length is 1, so multiply by 10:
\[ 10x = 4.\dot{4} \] \[ 10x - x = 4.\dot{4} - 0.\dot{4} = 4 \] \[ 9x = 4 implies x = \frac{4}{9} \]Check: \( 4 \div 9 = 0.4444\ldots \checkmark \)
Convert \( 0.\dot{2}\dot{7} = 0.272727\ldots \) to a fraction in simplest form.
Let \( x = 0.\dot{2}\dot{7} \). The cycle length is 2, so multiply by 100:
\[ 100x = 27.\dot{2}\dot{7} \] \[ 100x - x = 27.\dot{2}\dot{7} - 0.\dot{2}\dot{7} = 27 \] \[ 99x = 27 implies x = \frac{27}{99} = \frac{3}{11} \]Check: \( 3 \div 11 = 0.272727\ldots \checkmark \)
Convert \( 0.3\dot{6} = 0.3666\ldots \) to a fraction. (Only the 6 recurs.)
Let \( x = 0.3\dot{6} \). Multiply by 10 to move past the non-recurring digit, then by 100 to move one full cycle:
\[ 10x = 3.\dot{6} \] \[ 100x = 36.\dot{6} \] \[ 100x - 10x = 36.\dot{6} - 3.\dot{6} = 33 \] \[ 90x = 33 implies x = \frac{33}{90} = \frac{11}{30} \]Check: \( 11 \div 30 = 0.3666\ldots \checkmark \)
Deep Dive: \( 0.\dot{9} = 1 \)
Applying the algebraic method to \( 0.\dot{9} = 0.9999\ldots \):
\[ x = 0.\dot{9}, quad 10x = 9.\dot{9}, quad 10x - x = 9, quad 9x = 9, quad x = 1 \]This result surprises many students, but it is mathematically rigorous. \( 0.\dot{9} \) and \( 1 \) are two representations of the same real number — just as \( \frac{2}{4} \) and \( \frac{1}{2} \) represent the same value. There is no "last 9" in the infinite sequence, so there is no gap between \( 0.999\ldots \) and 1.
Key Takeaways
- Every recurring decimal is rational and can be converted to an exact fraction using the algebraic method.
- The multiplier is \( 10^k \) where \( k \) is the length of the repeating cycle — this ensures the recurring tails cancel when the expressions are subtracted.
- For a mixed recurring decimal (some non-recurring digits, then a cycle), use two multiplications to isolate the recurring part.
- Always simplify the resulting fraction to its lowest terms and verify by dividing numerator by denominator.
- \( 0.\dot{9} = 1 \) — an exact mathematical equality, not an approximation.