Technique explanation
Auxiliary fractions simplify the calculation of recurring decimals by using a 'one more' divisor. For a denominator ending in 9, like 29, we use the auxiliary divisor 3 (from 30). We divide by 3 and prefix the remainder to the next quotient digit. Denominators ending in 1, 3, or 7 are first multiplied to end in 9 (e.g., 1/7 becomes 7/49, using divisor 5). Standard long division remains the fallback.
Standard fallback: Standard long division of the fraction to verify the quotient digits.
Worked examples
Example 1: Find the first few digits of 1/19 using an auxiliary fraction.
1. 19 is near 20. The auxiliary divisor is 2.
2. 1 ÷ 2 = 0 remainder 1. (Digit 0, next is 10).
3. 10 ÷ 2 = 5 remainder 0. (Digit 5, next is 05).
4. 05 ÷ 2 = 2 remainder 1. (Digit 2, next is 12).
5. 12 ÷ 2 = 6 remainder 0. (Digit 6, next is 06).
Check: Multiply the divisor by the approximation: 19 × 0.0526 = 0.9994 (close to 1).
Result: 1/19 ≈ 0.0526...
Example 2: Find the first few digits of 1/13 using an auxiliary fraction.
1. Multiply by 3/3 to get 3/39. 39 is near 40. The auxiliary divisor is 4.
2. 3 ÷ 4 = 0 remainder 3. (Digit 0, next is 30).
3. 30 ÷ 4 = 7 remainder 2. (Digit 7, next is 27).
4. 27 ÷ 4 = 6 remainder 3. (Digit 6, next is 36).
5. 36 ÷ 4 = 9 remainder 0. (Digit 9, next is 09).
Check: Verify the first four digits via standard division: 1 ÷ 13 = 0.0769...
Result: 1/13 ≈ 0.0769...
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