NCERT Solutions for Class 9 Maths Exercise 1.3 Question 6
Understanding the Question 🧐
This question asks us to be detectives! We need to examine several examples of rational numbers (fractions) that result in terminating decimal expansions. By observing these examples, our goal is to guess the special property that the denominator, &&q&&, must have for this to happen.
There are two important conditions for our examples:
- The fraction &&\frac{p}{q}&& must be in its simplest form (&&p&& and &&q&& have no common factors other than 1).
- It must have a terminating decimal expansion.
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. Can you guess what property q must satisfy?
Step 1: Gathering Evidence (Examples) 📝
Let’s collect some examples that fit the conditions:
- &&\frac{1}{2} = 0.5&& (Denominator &&q = 2&&)
- &&\frac{3}{4} = 0.75&& (Denominator &&q = 4&&)
- &&\frac{7}{5} = 1.4&& (Denominator &&q = 5&&)
- &&\frac{13}{8} = 1.625&& (Denominator &&q = 8&&)
- &&\frac{9}{10} = 0.9&& (Denominator &&q = 10&&)
- &&\frac{17}{20} = 0.85&& (Denominator &&q = 20&&)
- &&\frac{3}{25} = 0.12&& (Denominator &&q = 25&&)
Step 2: Analyzing the Denominators (Finding the Pattern) 🕵️♀️
Now, let’s break down each denominator into its prime factors to see if there’s a hidden pattern.
- &&q = 2 \implies 2^1&&
- &&q = 4 \implies 2 \times 2 = 2^2&&
- &&q = 5 \implies 5^1&&
- &&q = 8 \implies 2 \times 2 \times 2 = 2^3&&
- &&q = 10 \implies 2 \times 5&&
- &&q = 20 \implies 4 \times 5 = 2^2 \times 5^1&&
- &&q = 25 \implies 5 \times 5 = 5^2&&
The pattern is clear! The prime factors of all these denominators are only 2s or 5s, or a combination of both. There are no 3s, 7s, 11s, or any other prime factors.
Step 3: The Guess (Formulating the Property) ✅
Based on our analysis, we can make an educated guess.
The property that &&q&& must satisfy is: The prime factorization of the denominator &&q&& must only have powers of 2 or powers of 5 or both.
Mathematically, we can say that &&q&& must be of the form:
&&q = 2^n 5^m&&
where &&n&& and &&m&& are whole numbers (i.e., &&n, m \ge 0&&).
Conclusion
These ncert solutions lead us to a fundamental rule in number theory. For a simplified rational number &&\frac{p}{q}&& to be expressed as a terminating decimal, its denominator &&q&& must have only 2s and/or 5s as its prime factors.
&&\frac{13}{8} = \frac{13}{2^3} = \frac{13 \times 5^3}{2^3 \times 5^3} = \frac{13 \times 125}{10^3} = \frac{1625}{1000} = 1.625&&
- The fraction &&\frac{p}{q}&& must first be in its simplest form.
- The prime factors of the denominator &&q&& can only be 2 or 5.
- If any other prime factor (like 3, 7, 11, etc.) exists in the denominator, the decimal will be non-terminating and repeating.
FAQ
Q: What property must the denominator ‘q’ satisfy for a fraction ‘p/q’ to have a terminating decimal?
A: For a rational number &&\frac{p}{q}&& in its simplest form, the prime factorization of the denominator &&q&& must only contain powers of 2 or powers of 5, or both. In other words, &&q&& must be of the form &&2^n 5^m&&, where &&n&& and &&m&& are whole numbers (0, 1, 2, …).
Q: Why are the prime factors 2 and 5 so important for terminating decimals?
A: Our number system is base-10, and the prime factors of 10 are 2 and 5. A terminating decimal is a fraction that can be written with a denominator that is a power of 10 (like 10, 100, 1000, etc.). This is only possible if the original denominator’s prime factors are exclusively 2s and 5s.
Q: Will the fraction &&\frac{1}{6}&& terminate? Why or why not?
A: No, &&\frac{1}{6}&& will not terminate. It is in its simplest form, and its denominator is &&6&&, which has a prime factorization of &&2 \times 3&&. The presence of the prime factor 3 (a number other than 2 or 5) means the decimal expansion will be non-terminating and repeating (&&0.1666…&&).
Q: Will the fraction &&\frac{7}{80}&& terminate? Why or why not?
A: Yes, &&\frac{7}{80}&& will terminate. It is in its simplest form, and the prime factorization of its denominator is &&80 = 16 \times 5 = (2^4) \times 5&&. Since the only prime factors are 2 and 5, the decimal expansion will terminate.
Q: What are whole numbers?
A: Whole numbers are the set of non-negative integers, which includes 0. The set is &&\{0, 1, 2, 3, …\}&&. When we say &&n&& and &&m&& are whole numbers in the rule &&2^n 5^m&&, it means that &&n&& or &&m&& can be 0.
Further Reading
This property is a cornerstone of understanding rational numbers. To learn more, you can refer to the official NCERT textbooks and explore the relationship between fractions and decimal expansions.
