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Mathematics

Try to find the decimal expansions of 103\dfrac{10}{3} and 1112\dfrac{11}{12}. What do you observe about the repetition of the digits after the decimal point?

Whole Numbers

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Answer

Decimal expansion of 103\dfrac{10}{3} :

By long division :

3)13.3333)10.0003))193)))103),93)1))103)1093)))10103)10093))10001\begin{array}{l} \phantom{3)\quad1}{3.333\ldots} \ 3\overline{\smash{\big)}\quad 10.000\ldots} \ \phantom{3)}\phantom{)1}\underline{-9} \ \phantom{3)\quad))}10 \ \phantom{3)\quad,}\underline{-9} \ \phantom{3)\quad1))}10 \ \phantom{3)\quad10}\underline{-9} \ \phantom{3)\quad))10}10 \ \phantom{3)\quad100}\underline{-9} \ \phantom{3)\quad)1000}1\ldots \end{array}

103=3.3333=3.3\dfrac{10}{3} = 3.3333\ldots = 3.\overline{3}

The digit 3 repeats infinitely after the decimal point.

Decimal expansion of 1112\dfrac{11}{12} :

By long division:

12)10.916612))11.00001210812)111200012)0120012)111180012)0072012)111118012)0007212)1111118\begin{array}{l} \phantom{12\overline{)}\,1}0.9166\ldots \ 12\overline{\smash{\big)}\phantom{)}11.0000\ldots} \ \phantom{12\overline{}}\underline{-108} \ \phantom{12\overline{)}111}20\phantom{00\ldots} \ \phantom{12\overline{)}\,0}\underline{-12}\phantom{00\ldots} \ \phantom{12\overline{)}1111}80\phantom{0\ldots} \ \phantom{12\overline{)}\,00}\underline{-72}\phantom{0\ldots} \ \phantom{12\overline{)}11111}80\phantom{\ldots} \ \phantom{12\overline{)}\,000}\underline{-72}\phantom{\ldots} \ \phantom{12\overline{)}111111}8\ldots \end{array}

1112=0.91666=0.916\dfrac{11}{12} = 0.91666\ldots = 0.91\overline{6}

After the decimal point, "91" appears once and then the digit 6 repeats infinitely.

Observations :

1. Both decimal expansions are non-terminating but repeating.

2. In 103\dfrac{10}{3}, the repetition starts immediately after the decimal point (pure repeating decimal).

3. In 1112\dfrac{11}{12}, there are some non-repeating digits before the repeating block begins (general/mixed repeating decimal).

This happens because the denominator 3 (in 103\dfrac{10}{3}) has only the prime factor 3, and the denominator 12 = 22 × 3 (in 1112\dfrac{11}{12}) has both 2 and 3 as factors. The presence of a prime factor other than 2 and 5 in the denominator makes the decimal repeating.

Hence, 103=3.3\dfrac{10}{3} = 3.\overline{3} and 1112=0.916\dfrac{11}{12} = 0.91\overline{6}. Both have repeating digits after the decimal point.

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