RD Sharma Class 10 Solutions Chapter 1 - Real Numbers (Ex 1.6) Exercise 1.6 - Free PDF
FAQs on RD Sharma Class 10 Solutions Chapter 1 - Real Numbers (Ex 1.6) Exercise 1.6
1. What is the key method taught in RD Sharma Class 10 Chapter 1, Exercise 1.6, to check if a rational number has a terminating decimal expansion?
The key method, without performing actual long division, involves two main steps. First, ensure the rational number, represented as p/q, is in its simplest form. Second, find the prime factorization of the denominator, q. If the prime factorization of q is in the form 2ⁿ5ᵐ, where 'n' and 'm' are non-negative integers, then the rational number has a terminating decimal expansion. If any other prime factor exists, it is non-terminating.
2. According to the concepts in Exercise 1.6, after how many decimal places will the expansion of the rational number 43/(2⁴ × 5³) terminate?
To find the number of decimal places, we examine the powers of 2 and 5 in the denominator's prime factorization. The denominator is 2⁴ × 5³. The number of decimal places is determined by the highest power among the factors of 2 and 5. In this case, the highest power is 4 (from 2⁴). Therefore, the decimal expansion of this number will terminate after precisely 4 decimal places.
3. Why does a fraction like 6/15 need to be simplified before checking for a terminating decimal?
This is a crucial step to avoid incorrect conclusions. If you check the denominator of 6/15 directly, you get 15 = 3 × 5. The presence of the factor '3' would suggest a non-terminating decimal. However, the correct method is:
Step 1: Simplify the fraction. The fraction 6/15 simplifies to 2/5 by dividing both numerator and denominator by their common factor, 3.
Step 2: Analyse the new denominator. The denominator of the simplified fraction is now 5.
Step 3: Check its prime factors. The prime factorization is 5¹, which fits the required 2ⁿ5ᵐ form (with n=0, m=1). Thus, 6/15 has a terminating decimal expansion (0.4).
4. What is the mathematical reason that only prime factors of 2 and 5 in the denominator lead to a terminating decimal?
The reason is based on the structure of our decimal number system, which is base-10. A terminating decimal is a fraction whose denominator can be written as a power of 10 (e.g., 10, 100, 1000). The prime factors of 10 are 2 and 5. Therefore, any power of 10 will only have prime factors of 2 and 5 (e.g., 100 = 10² = (2×5)² = 2²×5²). When a denominator has only factors of 2 and 5, we can always multiply the numerator and denominator by an appropriate number of 2s or 5s to convert the denominator into a power of 10, resulting in a terminating decimal.
5. How would you apply the method from Exercise 1.6 to determine if 77/210 has a terminating or non-terminating repeating decimal expansion?
To solve this problem, we follow the correct step-by-step process:
First, simplify the fraction 77/210. The greatest common divisor is 7. So, 77/210 simplifies to 11/30.
Next, find the prime factorization of the new denominator, 30. The factorization is 2 × 3 × 5.
Finally, inspect the factors. Because the prime factorization includes a '3' in addition to '2' and '5', it is not in the required 2ⁿ5ᵐ form. Therefore, the decimal expansion of 77/210 is non-terminating and repeating.
6. How does the Fundamental Theorem of Arithmetic support the solutions for problems in RD Sharma Exercise 1.6?
The Fundamental Theorem of Arithmetic is the foundation of this method. It states that every composite number has a unique prime factorization. When we analyse a denominator, this theorem guarantees that the set of prime factors we find (e.g., 2, 3, 5 for the number 30) is the only possible set. This uniqueness allows us to confidently conclude whether the denominator fits the 2ⁿ5ᵐ criteria. Without this guarantee of a single, unique factorization, our check would not be definitive.
