Question

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

(i) \(\frac{23}{2^3\times5^2}\)

(ii) \(\frac{24}{125}\)

(iii) \(\frac{171}{800}\)

(iv) \(\frac{15}{1600}\)

(v) \(\frac{17}{320}\)

(vi) \(\frac{19}{3125}\)

Answer 1

(i) \(\frac{23}{2^3\times5^2}\) = \(\frac{23\times5}{2^3\times5^3}\) = \(\frac{115}{1000}\) = 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 × 5ⁿ). 

Hence, the given rational is terminating

(ii) \(\frac{24}{125}\) = \(\frac{24}{5^3}\) = \(\frac{24\times2^3}{5^3\times2^3}\) = \(\frac{192}{1000}\) = 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 × 5ⁿ ). 

Hence, the given rational is terminating.

(iii) \(\frac{171}{800}\) = \(\frac{171}{2^5\times5^2}\) = \(\frac{171\times5^3}{2^5\times5^5}\) = \(\frac{21375}{100000}\) = 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 × 5ⁿ ). 

Hence, the given rational is terminating.

(iv) \(\frac{15}{1600}\) = \(\frac{15}{2^6\times5^2}\) = \(\frac{15\times5^4}{2^6\times5^6}\) = \(\frac{9375}{1000000}\) = 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 × 5ⁿ). 

Hence, the given rational is terminating.

(v) \(\frac{17}{320}\) = \(\frac{17}{2^6\times5}\) = \(\frac{17\times5^5}{2^6\times5^6}\) = \(\frac{53125}{1000000}\) = 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 × 5ⁿ). 

Hence, the given rational is terminating

(vi) \(\frac{19}{3125}\) = \(\frac{19}{5^5}\) = \(\frac{19\times2^5}{5^5\times2^5}\) = \(\frac{608}{100000}\) = 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 × 5ⁿ). 

Hence, the given rational is terminating.