 # Without actual division, show that each of the following rational numbers is a terminating decimal. Question:

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

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

(ii) $\frac{24}{125}$

(iii) $\frac{171}{800}$

(iv) $\frac{15}{1600}$

(v) $\frac{17}{320}$

(vi) $\frac{19}{3125}$

Solution:

(i) $\frac{23}{2^{3} \times 5^{2}}=\frac{23 \times 5}{2^{3} \times 5^{3}}=\frac{115}{1000}=0.115$

We know either 2 or 5 is not a factor of 23, so it is in its simplest form.

Moreover, it is in the form of $\left(2^{m} \times 5^{n}\right)$.

Hence, the given rational is terminating.

(ii) $\frac{24}{125}=\frac{24}{5^{3}}=\frac{24 \times 2^{3}}{5^{3} \times 2^{3}}=\frac{192}{1000}=0.192$

We know 5 is not a factor of 23, so it is in its simplest form.

Moreover, it is in the form of $\left(2^{m} \times 5^{n}\right)$.

Hence, the given rational is terminating.

(iii) $\frac{171}{800}=\frac{171}{2^{5} \times 5^{2}}=\frac{171 \times 5^{3}}{2^{5} \times 5^{5}}=\frac{21375}{100000}=0.21375$

We know either 2 or 5 is not a factor of 171, so it is in its simplest form.

Moreover, it is in the form of $\left(2^{m} \times 5^{n}\right)$.

Hence, the given rational is terminating.

(iv) $\frac{15}{1600}=\frac{15}{2^{6} \times 5^{2}}=\frac{15 \times 5^{4}}{2^{6} \times 5^{6}}=\frac{9375}{1000000}=0.009375$

We know either 2 or 5 is not a factor of 15, so it is in its simplest form.

Moreover, it is in the form of $\left(2^{m} \times 5^{n}\right)$.

Hence, the given rational is terminating.

(v) $\frac{17}{320}=\frac{17}{2^{6} \times 5}=\frac{17 \times 5^{5}}{2^{6} \times 5^{6}}=\frac{53125}{1000000}=0.053125$

We know either 2 or 5 is not a factor of 17, so it is in its simplest form.

Moreover, it is in the form of $\left(2^{m} \times 5^{n}\right)$.

Hence, the given rational is terminating.

(vi) $\frac{19}{3125}=\frac{19}{5^{5}}=\frac{19 \times 2^{5}}{5^{5} \times 2^{5}}=\frac{608}{100000}=0.00608$

We know either 2 or 5 is not a factor of 19, so it is in its simplest form.

Moreover, it is in the form of $\left(2^{m} \times 5^{n}\right)$.

Hence, the given rational is terminating.