1. REAL NUMBERS (NCERT)

NCERT

Chapter 1 : Real Numbers

Exercise 1.1 Complete solution

Exercise 1.2 Complete solution

Exercise 1.3 Complete solution

Exercise 1.4 Complete solution

Algebraic Identities

(i)

(ii)

(iii)

(iv)

(v)

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.

(xiii) Let be a rational number whose decimal expansion terminates. Then we can express in the form , where and are coprime, and the prime factorisation of is of the form , where are non-negative integers.
(xiv). Let be a rational number, such that the prime factorisation of is of the form , where are non-negative integers. Then has a decimal expansion which terminates.
(xv). Let be a rational number, such that the prime factorisation of is not of the form , where are non-negative integers. Then has a decimal expansion which is non-terminating repeating (recurring).

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 ,