Question

Without actual division, show that each of the following rational numbers is a terminating decimal. Express each in decimal form:

(i) $\frac{23}{(2^3\times 5^2)}$

Answer 1

Solution:

Step 1: Showing that the given rational number is a terminating decimal:

When the denominator of a fraction has the factors of the form $2^m\times 5^n$(Where ‘m’ and ‘n’ are positive integers), then the rational number is a terminating decimal.

For the given rational number, the denominator is $2^3\times 5^2$.

Here, we can observe that the denominator is in the form of $2^m\times 5^n$, where $m=3andn=2$, which satisfies the condition of terminating decimal.

Step 2: Expressing in decimal form:

$$\begin{gathered} \frac{23}{(2^3\times 5^2)} \\ =\frac{23}{(2\times 2^2\times 5^2)} \\ =\frac{23}{2\times {(2\times 5)}^2}\left[\because a^n\times b^n={\left(a\times b\right)}^n\right] \\ =\frac{23}{2\times {10}^2} \\ =\frac{11.5}{100} \\ =0.115 \end{gathered}$$

Final answer: Hence, we have shown that the rational number $\frac{23}{(2^3\times 5^2)}$ is a terminating decimal and its decimal form is $\mathbf{0}\mathbf{.}\mathbf{115}$.