Class 10 RD Sharma Solutions- Chapter 1 Real Numbers - Exercise 1.3

In Chapter 1 of Class 10 Mathematics, RD Sharma explores the topic of "Real Numbers". This chapter delves into various properties of the real numbers including their classification and how they can be expressed as rational and irrational numbers. Exercise 1.3 focuses on understanding and solving the problems related to the Euclidean division algorithm prime factorization and highest common factor (HCF) of the numbers using these methods.

Real Numbers

The Real numbers include both rational and irrational numbers. Rational numbers can be expressed as fractions or decimals that either terminate or repeat while irrational numbers have non-repeating, non-terminating decimals. Real numbers form the backbone of many mathematical operations and are used to describe quantities along a continuous number line. The properties of real numbers like closure, commutativity, associativity and distributivity play an essential role in simplifying complex mathematical problems.

Question 1. Express each of the following integers as a product of its prime.

i) 420

ii) 468

iii) 945

iv) 7325

Solution:

Let us express each of the numbers as a product of prime factors. 

i) 420

Performing prime factorisation of the number, we get,

420 = 2 × 2 × 3 × 5 × 7

ii) 468

Performing prime factorisation of the number, we get,

468 = 2 × 2 × 3 × 3 × 13

iii) 945

Performing prime factorisation of the number, we get,

945 = 3 × 3 × 3 × 5 × 7

iv) 7325

Performing prime factorisation of the number, we get,

7325 = 5 × 5 × 293

Question 2. Determine the prime factorisation of each of the following positive integer:

i) 20570

ii) 58500

iii) 45470971

Solution:

Let us express each of the numbers as a product of prime factors. 

i) 20570

Performing prime factorisation of the number, we get,

20570 = 2 × 5 × 11 × 11 × 17

ii) 58500

Performing prime factorisation of the number, we get,

58500 = 2 × 2 × 3 × 3 × 5 × 5 × 5 × 13

iii) 45470971

Performing prime factorisation of the number, we get,

45470971 = 7 × 7 × 13 × 13 × 17 × 17 × 19

Question 3. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution:

Both of these numbers have a common factor of 7. Also, every number is divisible by 1.

7 × 11 × 13 + 13 = (77 + 1) × 13 = 78 × 13

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = (7 × 6 × 4 × 3 × 2 + 1) × 5 = 1008 × 5

Composite numbers are those numbers which have at least one more factor other than 1.

Now,

Both of these numbers are even. Therefore, the given two numbers are composite numbers

Question 4. Check whether 6n can end with the digit 0 for any natural number n.

Solution:

Since, 6n = (2 × 3)n

6n = 2n × 3n

Any number can end with 0 if it divisible by 10 or 5 and 2 together. The, prime factorisation of 6n does not contain 5 and 2 as a pair of factors. 

Therefore, 6n can never end with the digit 0 for any natural number n.

Read more :

• Real Numbers
• Real-Life Applications of Real Numbers

Summary

Exercise 1.3 | Set 2 of RD Sharma's Class 10 Mathematics textbook focuses on the properties of rational and irrational numbers. This section covers topics such as identifying rational and irrational numbers, operations on rational and irrational numbers, and proving the irrationality of certain numbers. Students learn to classify numbers as rational or irrational, understand the closure properties of rational and irrational numbers under various operations, and solve problems involving the representation of numbers in decimal form. The exercise also includes questions on finding rational numbers between given numbers and proving statements related to rational and irrational numbers.