Question

Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.
i. $\frac{13}{5}$ ii. $\frac{2}{11}$ iii. $\frac{29}{16}$ iv. $\frac{17}{125}$ v. $\frac{11}{6}$

Answer 1

i) Given $\frac{13}{5}$
Since, $5=2^0\times 5^1$
⇒ The denominator is in the form of$2^m\times 5^n$,
where m and n are non-negative integers.
So, the decimal form of $\frac{13}{5}$ will be terminating type.

ii) Given $\frac{2}{11}$
Since, $11=2^0\times 5^0\times {11}^1$
⇒ The denominator is not in the form of $2^m\times 5^n$,
where m and n are non-negative integers.
So, the decimal form of $\frac{2}{11}$ will be non-terminating recurring type.
iii) Given $\frac{29}{16}$
Since, $16=2^4\times 5^0$
⇒ The denominator is in the form of $2^m\times 5^n$, where m and n are non-negative integers.
So, the decimal form of $\frac{29}{16}$ will be terminating type.

iv) Given $\frac{17}{125}$
Since, $125=2^0\times 5^3$
⇒ The denominator is in the form of $2^m\times 5^n$, where m and n are non-negative integers.
So, the decimal form of $\frac{17}{125}$ will be terminating type.
v) Given $\frac{11}{6}$
Since, $6=2^1\times 5^0\times 3^1$
⇒ The denominator is not in the form of $2^m\times 5^n$, where m and n are non-negative integers.
So, the decimal form of $\frac{11}{6}$ will be non-terminating recurring type.