Question

Express $0.12\overline{3}$ in the form of $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$.

Answer 1

Step 1: Express the given decimal number in infinite decimal form:

Given decimal: $0.12\overline{3}$.

$$\begin{gathered} 0.12\overline{3}=0.1233333.... \\ \Rightarrow 0.12\overline{3}=0.12+\left(0.003+0.0003+0.00003+...\right) \\ \Rightarrow 0.12\overline{3}=0.12+3\left(0.001+0.0001+0.00001+...\right) \\ \Rightarrow 0.12\overline{3}=0.12+3\left({10}^{-3}+{10}^{-4}+{10}^{-5}+...\right) \end{gathered}$$

Step 2: Simplify the equation

It is clear that, $\left({10}^{-3}+{10}^{-4}+{10}^{-5}+...\right)$ is an infinite geometric progression.

$$\begin{gathered} \left({10}^{-3}+{10}^{-4}+{10}^{-5}+...\right)=\frac{{10}^{-3}}{1-{10}^{-1}} \\ \Rightarrow \left({10}^{-3}+{10}^{-4}+{10}^{-5}+...\right)=\frac{0.001}{1-0.1} \\ \Rightarrow \left({10}^{-3}+{10}^{-4}+{10}^{-5}+...\right)=\frac{0.001}{0.9} \\ \Rightarrow \left({10}^{-3}+{10}^{-4}+{10}^{-5}+...\right)=\frac{1}{900} \end{gathered}$$

Thus, $0.12\overline{3}=0.12+3\left({10}^{-3}+{10}^{-4}+{10}^{-5}+...\right)=0.12+\left(3\times \frac{1}{900}\right)$

$$\begin{gathered} \Rightarrow 0.12\overline{3}=\frac{12}{100}+\frac{1}{300} \\ \Rightarrow 0.12\overline{3}=\frac{\left(12\times 3\right)+1}{300} \\ \Rightarrow 0.12\overline{3}=\frac{36+1}{300} \\ \Rightarrow 0.12\overline{3}=\frac{37}{300} \end{gathered}$$

Therefore, the fraction form of the given decimal number is $\frac{\mathbf{37}}{\mathbf{300}}$.