Decimal representation

\(\frac{5}{6}\) is in simplest form and \(6=2\times3\). A rational number has a terminating decimal expansion only if the prime factors of its denominator, in simplest form, are only \(2\) and/or \(5\). Since \(3\) is also a factor, \(\frac{5}{6}=0.8333\ldots\) is non-terminating recurring. An even denominator is not sufficient; although \(6\) is even, it contains the factor \(3\). Exam tip: first reduce the fraction to lowest terms and then factorise its denominator.

Let \(x=0.\overline{6}\). Since 6 repeats indefinitely, \(10x=6.\overline{6}\). Subtracting the original equation gives \(10x-x=6.\overline{6}-0.\overline{6}\), so \(9x=6\) and \(x=\frac{6}{9}=\frac{2}{3}\). Therefore, option B is correct. Option C, \(\frac{3}{5}\), equals the terminating decimal \(0.6\), not the recurring decimal \(0.\overline{6}\). Exam tip: a one-digit recurring decimal \(0.\overline{a}\) can be written as \(\frac{a}{9}\) and then reduced.

In \(\frac{13}{40}\), the denominator is \(40=2^3\times5\), so its decimal expansion terminates. The denominator of \(\frac{7}{18}\) also contains 3, so it is non-terminating recurring. Exam tip: in lowest form, only 2 and 5 may occur in the denominator.

\(\frac{13}{2^2\times5\times7}=\frac{13}{140}\) is in lowest terms because 13 and 140 have no common factor. The denominator \(140=2^2\times5\times7\) also contains the prime factor 7. A rational number has a terminating decimal expansion only if, in lowest terms, its denominator has no prime factors other than 2 and/or 5. Therefore, this decimal expansion is non-terminating recurring. A non-terminating non-recurring decimal represents an irrational number. Exam tip: reduce the fraction first, then inspect the prime factors of its denominator.

\(\frac{21}{84}=\frac14\), and \(4=2^2\); hence its decimal form terminates at \(0.25\). In \(\frac{7}{30}\), the factor 3 remains, so it is non-terminating recurring. Exam tip: simplify the fraction before checking the denominator.

The digit block 285714 repeats continuously, so this is a non-terminating recurring decimal. In fact, \(\frac{2}{7}=0.285714285714\ldots\), so both forms represent the same rational number. Option B is wrong because the decimal does not end, and option D is wrong because every recurring decimal is rational. Exam tip: A decimal that does not end but repeats a fixed block of digits is recurring.

The denominator is \(2^3\times5^5\), and 43 has no common factor with 2 or 5. Multiply the numerator and denominator by \(2^2\): \(\frac{43}{2^3\times5^5}=\frac{172}{10^5}\). Hence, the decimal expansion terminates after 5 decimal places. Option 8 is incorrect because the exponents are not added; the larger exponent determines the required number of places. Exam tip: for a reduced denominator of the form \(2^m5^n\), the decimal terminates in at most \(\max(m,n)\) places.

A rational number has a terminating decimal only when, in lowest form, its denominator has prime factors 2 and/or 5 only. Since \(120=2^3\times3\times5\) includes 3, \(\frac{17}{120}\) is non-terminating recurring. Exam tip: always check the denominator after simplification.