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1996 Paper 1 Q8
D: 1500.0 B: 1500.0
geometric series recurring decimal fraction conversion sum of geometric series rational numbers proof

  1. By using the formula for the sum of a geometric series, or otherwise, express the number \(0.38383838\ldots\) as a fraction in its lowest terms.
  2. Let \(x\) be a real number which has a recurring decimal expansion \[ x=0\cdot a_{1}a_{2}a_{2}\cdots, \] so that there exists positive integers \(N\) and \(k\) such that \(a_{n+k}=a_{n}\) for all \(n>N.\) Show that \[ x=\frac{b}{10^{N}}+\frac{c}{10^{N}(10^{k}-1)}\,, \] where \(b\) and \(c\) are integers to be found. Deduce that \(x\) is rational.

Solution:

  1. \(\,\) \begin{align*} && 0.383838\ldots &= \frac{3}{10} + \frac{8}{100} + \frac{3}{1000} + \cdots \\ &&&= \frac{38}{100} + \frac{38}{10000} + \cdots \\ &&&= \frac{38}{100} \left (1 + \frac{1}{100} + \frac{1}{100^2} + \cdots \right) \\ &&&= \frac{\frac{38}{100}}{1 - \frac{1}{100}} \\ &&&= \frac{38}{99} \end{align*}
  2. Let \(x = 0\cdot a_{1}a_{2}a_{2}\cdots\) such that there exists \(N\), \(k\) such that \(a_{n+k} = a_n\) for \(n > N\). First notice that \begin{align*} x &= \frac{a_1a_2\cdots a_N}{10\ldots000} + \frac{a_{N+1}}{10^{N+1}} + \cdots \\ &= \frac{b}{10^N} + \frac{a_{N+1}a_{N+2}\cdots a_{N+k}}{10^{N+k}} + \frac{a_{N+1}a_{N+2}\cdots a_{N+2k}}{10^{N+k}} + \cdots \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \left (1 + \frac{1}{10^k} + \cdots \right) \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{1}{1- \frac{1}{10^k}} \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{10^k}{10^k-1} \\ &= \frac{b}{10^N} + \frac{c}{10^N(10^k-1)} \end{align*} where \(b = a_1a_2\cdots a_N\) and \(c = a_{N+1}a_{N+2}\cdots a_{N+k}\). Clearly as the sum of two rational numbers, \(x\) is rational.

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