Question

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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Answer 1

(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 $(2^m\times 5^n)$.
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 $(2^m\times 5^n)$.
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 $(2^m\times 5^n)$.
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 $(2^m\times 5^n)$.
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 $(2^m\times 5^n)$.
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 $(2^m\times 5^n)$.
Hence, the given rational is terminating.