Question

Express the following in the form \[\dfrac{p}{q}\], where \[p{\text{ and }}q\] are integers and \[q \ne 0\].
(i)\[0.\overline 6 \]
(ii)\[0.4\overline 7 \]
(iii)\[0.\overline {001} \]

Answer 1

Hint: We will first let the given number equal to \[x\]. We can write the right-hand side in expanded form as the bar over the number shows that the digit is repeating and mark it as equation 1. Then we will multiply both sides of the equation with 10 as in the first part the bar is over 6 only so, we will multiply 10 on both sides of the equation and name it as equation 2. After this we will subtract equation 2 from equation 1 and thus evaluate the value of \[x\] from it.

Complete step-by-step answer:
(i)We will first consider the given number that is \[0.\overline 6 \].
Now, we will let the number equal to \[x\].
Thus, we get,
\[ \Rightarrow x = 0.\overline 6 \]
We can also write the above equation by expanding the right-hand side as the bar over the digit 6 shows that the number is repeating itself again and again.
Thus, we have,
\[ \Rightarrow x = 0.66666.... - - - - \left( 1 \right)\]
Next, we will multiply equation (1) with 10 since there is a bar over 6 only that is the one digit.
Hence, we have,
\[
   \Rightarrow 10x = 10\left( {0.66666....} \right) \\
   \Rightarrow 10x = 6.6666..... - - - - \left( 2 \right) \\
 \]
Next, we will subtract equation (1) from equation (2),
\[
   \Rightarrow 10x - x = 6.6666..... - 0.6666.... \\
   \Rightarrow 9x = 6 \\
   \Rightarrow x = \dfrac{6}{9} \\
   \Rightarrow x = \dfrac{2}{3} \\
 \]
Thus, we can conclude that the \[\dfrac{p}{q}\] form of \[0.\overline 6 \] is \[\dfrac{2}{3}\].

(ii)We will first consider the given number that is \[0.4\overline 7 \].
Now, we will let the number equal to \[x\].
Thus, we get,
\[ \Rightarrow x = 0.4\overline 7 \]
We can also write the above equation by expanding the right-hand side as the bar over the digit 7 shows that the number is repeating itself again and again.
Thus, we have,
\[ \Rightarrow x = 0.477777.... - - - - \left( 1 \right)\]
Next, we will multiply equation (1) with 10 since there is a bar over 7 only that is the one digit.
Hence, we have,
\[
   \Rightarrow 10x = 10\left( {0.47777....} \right) \\
   \Rightarrow 10x = 4.7777..... - - - - \left( 2 \right) \\
 \]
Next, we will subtract equation (1) from equation (2),
\[
   \Rightarrow 10x - x = 4.7777..... - 0.47777.... \\
   \Rightarrow 9x = 4.300 \\
   \Rightarrow x = \dfrac{{4.3}}{9} \\
   \Rightarrow x = \dfrac{{43}}{{9 * 10}} \\
   \Rightarrow x = \dfrac{{43}}{{90}} \\
 \]
Thus, we can conclude that the \[\dfrac{p}{q}\] form of \[0.4\overline 7 \] is \[\dfrac{{43}}{{90}}\].

(iii)We will first consider the given number that is \[0.\overline {001} \].
Now, we will let the number equal to \[x\].
Thus, we get,
\[ \Rightarrow x = 0.\overline {001} \]
We can also write the above equation by expanding the right-hand side as the bar over the digits 001 shows that the numbers are repeating themselves again and again.
Thus, we have,
\[ \Rightarrow x = 0.001001.... - - - - \left( 1 \right)\]
Next, we will multiply equation (1) with 1000 since there are bars over 3 digits.
Hence, we have,
\[
   \Rightarrow 1000x = 1000\left( {0.001001....} \right) \\
   \Rightarrow 1000x = 1.001001..... - - - - \left( 2 \right) \\
 \]
Next, we will subtract equation (1) from equation (2),
\[
   \Rightarrow 1000x - x = 1.001001..... - 0.001001.... \\
   \Rightarrow 999x = 1.000... \\
   \Rightarrow x = \dfrac{1}{{999}} \\
 \]
Thus, we can conclude that the \[\dfrac{p}{q}\] form of \[0.\overline {001} \] is \[\dfrac{1}{{999}}\].

Note: Remember that the bar means the digit will repeat itself again and again. Do multiply the equation with 10 or 1000 accordingly to the bar over the digits. Simplify the equation carefully and do not make any calculation mistakes in that. When multiplied the expression with 10 or 1000, do not get confused with the decimal point if multiplied by 10 then the decimal will shift to one place to the right.