Question

What is the value of ${{\left( \dfrac{1}{8} \right)}^{2}}$?

Answer 1

Hint: We first try to explain the proper fraction and the representation in decimal form. Then we apply the long division to express the decimal form. Then we take the square of the decimal to find the final solution.

Complete step-by-step answer:
The given fraction $\dfrac{1}{8}$ is a proper fraction. Proper fractions are those fractions who have lesser value in numerator than the denominator. We need to convert it into decimal.
We express it in regular long division processes. The denominator is the divisor. The numerator is the dividend. The quotient will be the integer of the mixed fraction. The remainder will be the numerator of the proper fraction.
$8\overset{0.125}{\overline{\left){\begin{align}
  & 10 \\
 & \underline{8} \\
 & 20 \\
 & \underline{16} \\
 & 40 \\
 & \underline{40} \\
 & 0 \\
\end{align}}\right.}}$
Therefore, the decimal form of proper $\dfrac{1}{8}$ diffraction is $0.125$.
Now we find the square of $\dfrac{1}{8}$. We have ${{\left( \dfrac{1}{8} \right)}^{2}}={{\left( 0.125 \right)}^{2}}=\text{0}\text{.015625}$.
Therefore, the value of ${{\left( \dfrac{1}{8} \right)}^{2}}$ is $\text{0}\text{.015625}$.

Note: We can also find the square before converting it into the form of decimal.
We know that ${{\left( \dfrac{a}{b} \right)}^{2}}=\dfrac{{{a}^{2}}}{{{b}^{2}}}$. Therefore, ${{\left( \dfrac{1}{8} \right)}^{2}}=\dfrac{1}{{{8}^{2}}}=\dfrac{1}{64}$.
Now we can convert the fraction $\dfrac{1}{64}$ into decimal using same long division process.
We get $\dfrac{1}{64}=\text{0}\text{.015625}$. Therefore, ${{\left( \dfrac{1}{8} \right)}^{2}}=\dfrac{1}{{{8}^{2}}}=\dfrac{1}{64}=\text{0}\text{.015625}$.