Question

After how many decimal places the decimal expansion of $\dfrac{51}{150}$ will terminate.
In Euclid’s Division Lemma, when $a=bq+r$ where $a,b$ are positive integers then what values r can take.

Answer 1

Hint : If the factors of denominator of the given rational number is of form ${{2}^{n}}{{5}^{m}}$ ,where $n,m$ are non-negative integers, then the decimal expansion of the rational number is terminating otherwise non terminating recurring.

Complete step-by-step answer :
According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers $q$ and $r$ which satisfies the condition $a=bq+r$ where $0~\le \text{ }r\text{ }<\text{ }b$ .
$\Rightarrow \dfrac{51}{150}$
$=\dfrac{3\times 17}{3\times 5\times 2\times 5}$
$=\dfrac{17}{{{2}^{1}}\times {{5}^{2}}}$
$=\dfrac{17\times 2}{{{2}^{2}}\times {{5}^{2}}}$
$=\dfrac{34}{{{(2\times 5)}^{2}}}$
$=\dfrac{34}{{{10}^{2}}}$
$=\dfrac{34}{100}$
$=0.34$
Hence it will terminate after two places of decimal.
According to Euclid lemma if we have two positive integers a and b, then there exist unique integers $q$ and $r$ which satisfies the condition $a=bq+r$ where $0~\le \text{ }r\text{ }<\text{ }b$ ,
Hence, the value of r will be between zero and b i.e. $0~\le \text{ }r\text{ }<\text{ }b$.

Note : The rational number for which the long division terminates after a finite number of steps is known as terminating decimal. The rational number for which the long division does not terminate after any number of steps is known as non-terminating decimal. A repeating decimal or recurring decimal is a decimal representation of a number whose digits are periodic (repeating its values at regular intervals) and the infinitely repeated portion is not zero. A non-terminating, non-repeating decimal is a decimal number that continues endlessly, with no group of digits repeating endlessly. Decimals of this type cannot be represented as fractions, and as a result are irrational numbers.