Question 13 Marks
Look at several examples of rational numbers in the form $\frac{p}{q}(q \neq 0),$ where $p$ and $q$ are integers with no common factors other than $1$ and having terminating decimal representations $($expansions$).$ Can you guess what property $q$ must satisfy?
Answer
View full question & answer→Let $\frac{1}{2}$, $\frac{1}{4}$,$\frac{5}{8}$, $\frac{17}{25}$,$\frac{2}{125}$, $\frac{13}{20}$ etc which has terminating decimal expansion . For example:
$\frac{1}{2} = \frac{1}{2^1}=0.5 ,$ denominator $q = 2^1$
$\frac{1}{4}= \frac{1}{2^2} = 0.25$ denominator $q = 2^2$
$\frac{5}{8} = \frac{5}{2^3} = 0.625 $, denominator $q = 2^3$
$\frac{17}{25}= \frac{17}{5^2} = 0.68 ,$ denominator $q = 5^2$
$\frac{2}{125}= \frac{2}{5^3} = 0.016,$ denominator $q = 5^3$
$\frac{13}{20} = \frac{13}{2^2\times 5} = 0.65,$ denominator $q = 2^2\times 5$
It can be observed that the terminating decimal can be obtained in a condition where prime factorization of the denominator$ (i.e. q)$ of the given fractions has $2$ or the power of $2$ or $5$ or power of $5$ or both.
$\frac{1}{2} = \frac{1}{2^1}=0.5 ,$ denominator $q = 2^1$
$\frac{1}{4}= \frac{1}{2^2} = 0.25$ denominator $q = 2^2$
$\frac{5}{8} = \frac{5}{2^3} = 0.625 $, denominator $q = 2^3$
$\frac{17}{25}= \frac{17}{5^2} = 0.68 ,$ denominator $q = 5^2$
$\frac{2}{125}= \frac{2}{5^3} = 0.016,$ denominator $q = 5^3$
$\frac{13}{20} = \frac{13}{2^2\times 5} = 0.65,$ denominator $q = 2^2\times 5$
It can be observed that the terminating decimal can be obtained in a condition where prime factorization of the denominator$ (i.e. q)$ of the given fractions has $2$ or the power of $2$ or $5$ or power of $5$ or both.
