
Classify the decimal form of the given rational numbers into termination and non-terminating recurring type.
i) \[\dfrac{{13}}{5}\]
ii) \[\dfrac{2}{{11}}\]
iii) \[\dfrac{{29}}{{16}}\]
iv) \[\dfrac{{17}}{{125}}\]
v) \[\dfrac{{11}}{6}\]
Answer
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Hint: A rational number is any number that can be written in the form of a fraction. It involves integers, fractions, terminating decimals, and repeating decimals or non-terminating decimals. So, every decimal termination is a rational number. Now, classify the decimal form of the given rational numbers into termination and non-terminating recurring type by using the rule $\dfrac{p}{q}$. For terminating decimal, the value of denominator q is ${2^m} \times {5^n}$ where m and n are integers and for non-terminating recurring, the value of denominator q is ${2^m} \times {5^n} \times k$ where m, n are integers, and k is a prime factor other than 2 or 5.
Complete step-by-step answer:
i) Classify the decimal form of the \[\dfrac{{13}}{5}\] rational numbers into termination and non-terminating recurring types.
Now, use the denominator value of the given rational number:
\[ \Rightarrow 5 = {2^0} \times {5^1}\]
\[\therefore \] The above value of the denominator is in the form of ${2^m} \times {5^n}$ which is given in the hint portion, where m and n are non-negative integers.
So, by the definition of the rational number, the decimal form of the \[\dfrac{{13}}{5}\] will be the terminating type.
ii) Classify the decimal form of the \[\dfrac{2}{{11}}\] rational numbers into termination and non-terminating recurring types.
Now, use the denominator value of the given rational number:
\[ \Rightarrow 11 = {2^0} \times {5^0} \times {11^0}\]
\[\therefore \]The above value of the denominator is not in the form of ${2^m} \times {5^n}$ which is given in the hint portion, where m and n are non-negative integers.
So, by the definition of the rational number, the decimal form of the \[\dfrac{2}{{11}}\] will be non-terminating recurring type.
iii) Classify the decimal form of the \[\dfrac{{29}}{{16}}\] rational numbers into termination and non-terminating recurring types.
Now, use the denominator value of the given rational number:
\[ \Rightarrow 16 = {2^4} \times {5^0}\]
\[\therefore \] The above value of the denominator is in the form of ${2^m} \times {5^n}$ which is given in the hint portion, where m and n are non-negative integers.
So, by the definition of the rational number, the decimal form of the \[\dfrac{{29}}{{16}}\] will be the terminating type.
iv) Classify the decimal form of the \[\dfrac{{17}}{{125}}\] rational numbers into termination and non-terminating recurring types.
Now, use the denominator value of the given rational number:
\[ \Rightarrow 125 = {2^0} \times {5^3}\]
\[\therefore \] The above value of the denominator is in the form of ${2^m} \times {5^n}$ which is given in the hint portion, where m and n are non-negative integers.
So, by the definition of the rational number, the decimal form of the \[\dfrac{{17}}{{125}}\] will be the terminating type.
v) Classify the decimal form of the \[\dfrac{{11}}{6}\] rational numbers into termination and non-terminating recurring types.
Now, use the denominator value of the given rational number:
\[ \Rightarrow 6 = {2^1} \times {5^0} \times {3^1}\]
\[\therefore \] The above value of the denominator is not in the form of ${2^m} \times {5^n}$ which is given in the hint portion, where m and n are non-negative integers.
So, by the definition of the rational number, the decimal form of the \[\dfrac{{11}}{6}\] will be non-terminating recurring type.
Option (ii) and (v) are the correct answers.
Note: Simply, it is possible to express rational numbers in the form of decimal fractions. The meaning of terminating decimals are the numbers which come to an end after some decimal points and the non-terminating decimals define a decimal number that persists indefinitely, with no group of digits repeated indefinitely, is a non-terminating or non-repeating decimal. The denominator of the terminating factors, the prime factors of the denominators are either 2 or 5 or both. The denominator of the non-terminating recurring decimals, they have numbers other than 2 and 5.
Complete step-by-step answer:
i) Classify the decimal form of the \[\dfrac{{13}}{5}\] rational numbers into termination and non-terminating recurring types.
Now, use the denominator value of the given rational number:
\[ \Rightarrow 5 = {2^0} \times {5^1}\]
\[\therefore \] The above value of the denominator is in the form of ${2^m} \times {5^n}$ which is given in the hint portion, where m and n are non-negative integers.
So, by the definition of the rational number, the decimal form of the \[\dfrac{{13}}{5}\] will be the terminating type.
ii) Classify the decimal form of the \[\dfrac{2}{{11}}\] rational numbers into termination and non-terminating recurring types.
Now, use the denominator value of the given rational number:
\[ \Rightarrow 11 = {2^0} \times {5^0} \times {11^0}\]
\[\therefore \]The above value of the denominator is not in the form of ${2^m} \times {5^n}$ which is given in the hint portion, where m and n are non-negative integers.
So, by the definition of the rational number, the decimal form of the \[\dfrac{2}{{11}}\] will be non-terminating recurring type.
iii) Classify the decimal form of the \[\dfrac{{29}}{{16}}\] rational numbers into termination and non-terminating recurring types.
Now, use the denominator value of the given rational number:
\[ \Rightarrow 16 = {2^4} \times {5^0}\]
\[\therefore \] The above value of the denominator is in the form of ${2^m} \times {5^n}$ which is given in the hint portion, where m and n are non-negative integers.
So, by the definition of the rational number, the decimal form of the \[\dfrac{{29}}{{16}}\] will be the terminating type.
iv) Classify the decimal form of the \[\dfrac{{17}}{{125}}\] rational numbers into termination and non-terminating recurring types.
Now, use the denominator value of the given rational number:
\[ \Rightarrow 125 = {2^0} \times {5^3}\]
\[\therefore \] The above value of the denominator is in the form of ${2^m} \times {5^n}$ which is given in the hint portion, where m and n are non-negative integers.
So, by the definition of the rational number, the decimal form of the \[\dfrac{{17}}{{125}}\] will be the terminating type.
v) Classify the decimal form of the \[\dfrac{{11}}{6}\] rational numbers into termination and non-terminating recurring types.
Now, use the denominator value of the given rational number:
\[ \Rightarrow 6 = {2^1} \times {5^0} \times {3^1}\]
\[\therefore \] The above value of the denominator is not in the form of ${2^m} \times {5^n}$ which is given in the hint portion, where m and n are non-negative integers.
So, by the definition of the rational number, the decimal form of the \[\dfrac{{11}}{6}\] will be non-terminating recurring type.
Option (ii) and (v) are the correct answers.
Note: Simply, it is possible to express rational numbers in the form of decimal fractions. The meaning of terminating decimals are the numbers which come to an end after some decimal points and the non-terminating decimals define a decimal number that persists indefinitely, with no group of digits repeated indefinitely, is a non-terminating or non-repeating decimal. The denominator of the terminating factors, the prime factors of the denominators are either 2 or 5 or both. The denominator of the non-terminating recurring decimals, they have numbers other than 2 and 5.
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