Chapter 1: Real Numbers — Class 10 Mathematics NCERT Solutions | CBSE

Chapter 1: Real Numbers – NCERT Class 10 Mathematics Solutions | CBSE

Chapter 1 : Real Numbers

Important notes

(i) Euclid’s Division Lemma : Given positive integers and , there exist unique integers and satisfying ,

(ii) Fundamental Theorem of Arithmetic : Every composite number can be expressed ( factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

(iii) The prime factorisation of a natural number is unique, except for the order of its factors.

(iv) HCF = Product of the smallest power of each common prime factor in the numbers.
(v) LCM = Product of the greatest power of each prime factor, involved in the numbers .

(vi) If two positive integers and ,then .

i.e ., HCF of two numbers × LCM of two numbers = One number × Other number

(vii) The product of three numbers is not equal to the product of their HCF and LCM.

(viii) A number is called rational if it can be written in the form , where p and q are integers . Example : 0 , – 5 , , …… etc .

(ix) A number is called irrational if it cannot be written in the form , where p and q are integers . Example : ……… etc.

(x) Let be a prime number. If divides , then divides , where is a positive integer.

(xi) The sum or difference of a rational and an irrational number is irrational .
(xii) The product and quotient of a non-zero rational and irrational number is irrational.

EXERCISE 1.1

1. Use Euclid’s division algorithm to find the HCF of :

(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 225

Solution : (i) Since 225 > 135

Using Euclid’s division algorithm , we have

Therefore, the HCF of 135 and 225 is 45 .

(ii) Since 38220 > 196

Using Euclid’s division algorithm , we have

Therefore, the HCF of 196 and 38220 is 195 .

(iii) Since 867 > 225

Using Euclid’s division algorithm , we have

Therefore , HCF(867 , 255) is 51 .

2. Show that any positive odd integers is of the form or , where is some integer.

Solution : let , be any positive odd integers and .

Using Euclid’s algorithm , we have

So, 0 , 1 , 2 , 3 , 4 or 5 .

If then is an even numbers .

If then is an odd numbers .

If then is an even numbers .

If then is an odd numbers .

If then is an even numbers .

If then is an odd numbers .

Therefore, any positive odd integer is of the form , or .

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade . The two groups are to march in the same number of columns . What is the maximum number of columns in which they can march ?

Solution : .

We find the maximum number of column of (616 , 32) .

Using Euclid’s division algorithm , we have

Therefore , the HCF (616 , 32) is 8 .

Thus , the maximum number of column is 8 .

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form orfor some integer . [ Let be any positive integer then it is of the form or . Now square each of these and show that they can be rewritten in the form or .]

Solution : let, be any positive integers and .

We apply the Euclid’s division algorithm ,

So , , or .

If then

, where

If then

, where

If then

,where

Thus, the square of any positive integer is either of the form or for some integer .

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form or .

Solution : let, be any positive integer and.

Using Euclid’s algorithm , we have

Therefore , 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 or 8 .

If then

,where for some integer .

If then

,where for some integer q .

If then

,where for some integer q.

Therefore ,the cube of any positive integer is of the form , or , for some integer .

EXERCISE 1.2

1. Express each number as a product of its prime factors :

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solution : (i) We have,

(ii) We have ,

(iii) We have ,

(iv) We have ,

(v) We have ,

2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers .

(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

Solution : (i) 26 and 91

We have , and

HCF (26 , 91) 13 and LCM (26 , 91)

Now ,

Verified .

(ii) 510 and 92

We have ,

and

HCF(510 , 92)

and LCM(510 , 92)

Now ,

Verified .

(iii) 336 and 54

We have ,

and

HCF(336 , 54)

and LCM(336 , 54)

Now ,

Verified .

3. Find the LCM and HCF of the following integers by applying the prime factorization method .

(i) 12 , 15 and 21 (ii) 17 , 23 and 29 (iii) 8 , 9 and 25

Solution : (i) 12 , 15 and 21

We have ,

HCF (12 , 15 , 21) 3

and LCM (12 , 15 , 21)

(ii) 17 , 23 and 29

We have ,

HCF (17 , 23 , 29)

and LCM (17 , 23 , 29)

(iii) 8 , 9 and 25

We have ,

HCF(8 , 9 , 25)

and LCM (8 , 9 , 25)

4. Given that HCF (306 , 657) = 9 , find LCM(306 , 657) .

Solution : We have ,

5. Check whether can end with the digit 0 for any natural number .

Solution : We have ,

The prime factors of does not contain in factor , where are positive integers .Therefore, does not end with the digit 0 .

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

Solution: We have ,

is a composite number.

and

is a composite number .

7. There is a circular path around a sports field . Sonia takes 18 minutes to drive one round of the field , while Ravi takes 12 minutes for the same . Suppose they both start at the same point and at the same time, and go in the same direction . After how many minutes will they meet again at the starting point ?

Solution : The required number of minutes is LCM (18 , 12) . We find the LCM by prime factorization method ,

and

Therefore, the LCM (18 , 12)

Hence, Sonia and Ravi will meet again at the starting point after 36 minutes .

EXERCISE 1.3

1. Prove that is irrational .

Solution : Let, us assume to the contrary that is rational .There exists co-prime integers and () such that

Therefore , is divisible by 5 . So, is also divisible by 5 .

Let , for some integer c .

From and , we get

So, is divisible by 5 . So, is also divisible by 5 .

Therefore, and have at least as a common factor . But this contradicts the fact that and are co-prime. This contradiction has arisen because of our incorrect assumption that is rational . So, we conclude that is irrational .

2. Prove that is irrational .

Solution : Let us assume , to the contrary that is rational .

We can find co-prime and ( ) such that

Since, 2 , and are integers , is rational and so, is rational .

But this contradicts the fact that is irrational . So , we conclude that is irrational .

3. Prove that the following are irrationals :

(i) (ii) (iii)

Solution : (i) Let us assume , to the contrary , that is rational . We can find co-prime and ( ) such that

Since 2 , and are integers, is rational and so, is rational . But this contradicts the fact that is irrational . So, is an irrational .

(ii) Let us assume , to the contrary , that is rational . We can find co-prime and ( ) such that

Since 7 , and are integers , is rational and so, is rational . But this contradicts the fact that is irrational .So , is an irrational .

(iii) Let us assume , to the contrary , that is rational . We can find co-prime and ( ) such that

Since and are integers , is rational and so, is rational . But this contradicts the fact that is irrational . So , is an irrational .

EXERCISE 1.4

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion :

(i) (ii) (iii) (iv) (v)

(vi) (vii) (viii (ix) (x)

Solution : (i) We have ,