CBSE 10th Mathematics | Decimal Representation Of Rational Numbers

Decimal Representation Of Rational Numbers

Theorem:

Let $\displaystyle x=\frac{p}{q}$ be a rational number such that $\displaystyle q\ne 0$ and prime factorization of q is of the form $\displaystyle {{2}^{n}}\times {{5}^{m}}$ where m, n are non-negative integers then x has a decimal representation which terminates.

For example :     $\displaystyle 0.275=\frac{275}{{{10}^{3}}}=\frac{{{5}^{2}}\times 11}{{{2}^{3}}\times {{5}^{3}}}=\frac{11}{{{2}^{3}}\times 5}=\frac{11}{40}$

Theorem:

Let $\displaystyle x=\frac{p}{q}$ be a rational number such that $\displaystyle q\ne 0$ and prime factorization of q is not of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$, where m, n are non-negative integers, then x has a decimal expansion which is non-terminating repeating.

For example :     $\displaystyle \frac{5}{3}=1.66666...$

Solved Examples Based on Decimal Representation Of Rational Numbers

Question:

Without actually calculating, state whether the following rational numbers have a terminating or non-terminating repeating decimal expansion.

(i)     $\displaystyle \frac{27}{343}$            (ii)    $\displaystyle \frac{19}{1600}$        (iii)    $\displaystyle \frac{129}{{{2}^{2}}\times {{5}^{5}}\times {{3}^{2}}}$

Hint: If the denominator is of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$ for some non negative integer m and n, then rational number has terminating decimal otherwise non terminating.

Solution:

(i)    $\displaystyle \frac{27}{343}=\frac{27}{{{7}^{3}}}$

Since $\displaystyle q={{7}^{3}}$ which is not of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$.

\    It has non terminating decimal representation.

(ii)    $\displaystyle \frac{19}{1600}=\frac{19}{{{2}^{6}}\times {{5}^{2}}}$

Since q = $\displaystyle {{2}^{6}}\times {{5}^{2}}$ which is of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$.

\    It has a terminating decimal representation.

(iii)    $\displaystyle \frac{129}{{{2}^{2}}\times {{5}^{5}}\times {{3}^{2}}}$

Since $\displaystyle q={{2}^{2}}\times {{5}^{5}}\times {{3}^{2}}$ is not of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$. It has a non-terminating decimal representation.

Question:

What can you say about the prime factorization of the denominators of the following rationales:

(i)     36.12345            (ii)    $\displaystyle 36.\overline{5678}$

Solution:

(i)    Since 36.12345 has terminating decimal expansion. So, its denominator is of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$ where m, n are non-negative integers.

(ii)    Since $\displaystyle 36.\overline{5678}$has non terminating repeating decimal expansion. So, its denominator has factors other than 2 or 5.