Question

You know that \[\dfrac{1}{7} = 0.\overline {142857} \]. Can you predict what the decimal expansions of \[\dfrac{2}{7},\dfrac{3}{7},\dfrac{4}{7},\dfrac{5}{7},\dfrac{6}{7}\] are, without actually doing the long division? If so, how?

Answer 1

Hint: We will first consider the given value that is \[\dfrac{1}{7} = 0.\overline {142857} \]. We need to find the decimal expansions of \[\dfrac{2}{7},\dfrac{3}{7},\dfrac{4}{7},\dfrac{5}{7},\dfrac{6}{7}\] without using the long division method. So, we will take the numerator outside the bracket from each term and as we have the value of \[\dfrac{1}{7}\] we can directly multiply the numerator taken outside with the given value of \[\dfrac{1}{7}\]. Thus, we will get the desired result.

Complete step-by-step answer:
We are given that \[\dfrac{1}{7} = 0.\overline {142857} \].
Now, we need to find the decimal expansion of \[\dfrac{2}{7},\dfrac{3}{7},\dfrac{4}{7},\dfrac{5}{7},\dfrac{6}{7}\] without using the long division method.
We will start with \[\dfrac{2}{7}\],
Thus, we have,
\[
   \Rightarrow \dfrac{2}{7} = 2\left( {0.142857} \right) \\
   \Rightarrow \dfrac{2}{7} = 0.285714 \\
 \]
Next, we have \[\dfrac{3}{7}\],
Thus, we have,
\[
   \Rightarrow \dfrac{3}{7} = 3\left( {0.142857} \right) \\
   \Rightarrow \dfrac{3}{7} = 0.428571 \\
 \]
Next, we have \[\dfrac{4}{7}\],
Thus, we have,
\[
   \Rightarrow \dfrac{4}{7} = 4\left( {0.142857} \right) \\
   \Rightarrow \dfrac{4}{7} = 0.571428 \\
 \]
Next, we have \[\dfrac{5}{7}\],
Thus, we have,
\[
   \Rightarrow \dfrac{5}{7} = 5\left( {0.142857} \right) \\
   \Rightarrow \dfrac{5}{7} = 0.714285 \\
 \]
Next, we have \[\dfrac{6}{7}\],
Thus, we have,
\[
   \Rightarrow \dfrac{6}{7} = 6\left( {0.142857} \right) \\
   \Rightarrow \dfrac{6}{7} = 0.857142 \\
 \]
Hence, we get all the decimal expansions without actually performing the long division method.

Note: We can first observe the decimal part of \[\dfrac{1}{7}\]. To find the decimal expansions of the numbers let say of \[\dfrac{2}{7}\], we just have to shift the decimal point two places to the right similarly for \[\dfrac{3}{7}\], we will shift the decimal point three places to the right. And the same can be done for the other numbers. This we can do because \[\dfrac{1}{7}\] is recurring and hence we will get the same remainders again and again after a certain pattern. This one can be done as an alternative method.