ML Aggarwal Solution Class 9 Chapter 1 Rational and Irrational Numbers Exercise 1.2

 Exercise 1.2

Question 1

Sol :

Let √5 be a rational number then

5=pq where p,q are integers q≠0 and p, q have no common factors (except 1)

5=p2q2

p2=5q2..(i)

As "5" divides 5q2.So, '5' divides pand '5' is prime 

5 divides p

Let p=5m , where m is an integer

Substituting the value of 'p' in (i)

(5m)2=5q2

25m2=5q2

q2=5m2

As 5 divides 5mso '5' divides qbut 5 is prime 

5 divides q

Thus p and q have a common factor 5. This contradicts that 'p' and 'q' have no common factor (except 1)

Hence √5 is not rational number

So. we conclude √5 is a rational number


Question 2

Sol :

Let √7 be a rational number

7=pq , where p, q are integers , q≠0 and p, q have no common factors (except 1)

7=p2q2

p2=7q2..(i)

As 7 divides 7q2, so 7 divides p2 and 7 is a prime

7 divides p

Let p=7 m where 'm' is an integer

Substitute this value of 'p' in (i) , we have

(7m)2=7q2

4m2=7q2

q2=7m2

As 7 divides 7m2, so 7 divides qbut 7 is a prime number

7 divides q

Thus , 'p' and 'q' have a common factor 7. Thus contradicts that p and q have no common factor (except 1)

Hence √7  is not a rational number

So, we conclude √7 is rational number


Question 3

Sol :

Let √6 be a rational number

6=pq, where p and q ave integers q≠0 and p, q have no common factors (except 1)

6=p2q2

p2=6q2...(i)

As '2' divides 6q2. So, 2 divides pbut 2 is prime 

2 divides p

Let p=2m , where 'm' is an integer

Substitute this value of 'p' in (i)

(2m)2=6q2

4m2=6q2

2m2=3q2

'2' divides '2m2', So 2 divides '3q2'

2 should either divide '3' or divide q2

But 2 should does not divide 3

Therefore, 2 divides q2 and 2 is a prime

2 divides q

Thus, p and q have a common factor 2 . This contradicts that 'p' and 'q' have no common factors (except 1)

Hence, 6 is not rational number.

So, we conclude 6 is irrational number.


Question 4

Sol :

Let 111 be a ratitnal number.

111=pq, where 'p' , 'q' are integers, q≠0 and p, q have no common factors (except 1)

111=p2q2

q2=11p2...(i)

As 11 divides 11p2,So 11 divides qbut 11 is a prime

 11 divides q

q=11m where 'm' is an integer

(11m)2=11p2

121m2=11p2

p2=11m2

As 11 divides 11m2.So 11 divides pbut 11 is a prime

11 divides p

Thus ,p and q have a common factor 11. This contradicts the fact that p and q has no common factors (except 1)

Hence , 111 is not rational number

So, we conclude 111 is irrational number


Question 5

Sol :

Let √2 be a rational number

2=pq where p and q are integers q≠0 and p and q have no common factors (except 1)

2=p2q2

p2=2q2..(i)

As '2' divides 2q,so 2 divides pbut 2 is prime

2 divides p

Let p=2m, where 'm' is a integer

Substitute this value of p in (i)

(2m)2=2q2

4m2=2q2

q2=2m2

As 2 divides 2m2, so 2 divides qbut 2 is prime 2 divides q

Thus , p and q have a common factor 2. This contradicts the fact that p and q has no common factor (Except 1)

Hence, √2 is not rational number

So, we conclude √2 is irrational number

Let us assume 3-√2 is rational number, say r

Thus, 32=r3r=2

As 'r' is rational , 3-r is rational 2 is rational

This contradicts the fact thas 2 is irrational

Hence, our assumption is wrong. Therefore, 32 is an irrational number


Question 6

Sol :

Let √3 be a rational number

3=pq where p and q are integers q≠0 and p and q have no common factors (except 1)
3=p2q2

p2=3q2..(i)

As '3' divides 3q2, So 3 divides p2 but 3 is a prime.

3 divides p

Let p=3 m, where 'm' is an integer

Substituting this value of p in (i)

(3m)2=3q2

9m2=3q2

q2=3m2

As '3' divides 3 m2,so '3' divides q2 but 3 is not prime.

3 divides q

Thus , p and q have a common factor 3. This contradicts the fact that p and q has no common factor (except 1)

Hence √3 is not rational number

So, we conclude 3 is irrational number.

Let us assume 253 is a rational number, say r

Thus 253=r3=52r

As r is rational 52r is rational 3 is rational

This contradicts the fact that 3 is irrational

Hence our assumption is wrong. Therefore, 253 is irrational number


Question 7

Sol :

Let √5 be a rational number

5=pq, where p and q are integers , q0 and p, q have no common factors (except 1)
5=p2q2
p2=5q2...(i)

As 5 divides 5q2, So, 5 divides p2 but 5 is a prime 
5 divides p
Let p=5m where m is an integer
Substitute this value of m in (i)
(5m)2=5q2
25m2=5q2
q2=5m2

As 5 divides 5 m2, 5 divides q2 but 5 is a prime 5 divides q

Thus, p and q have a common factor 5 . This contradicts the fact that p and q has no common factors (except 1)

Hence √5 is not rational number 

So, we conclude √5 is irrational number

Let us assume -3+2√5 is a rational number , say r
Thus -3+2√5
-3-r=2√5
5=(3+r)2

As 'r' is rational (3+r2) is rational 

√5 is rational . This contradict the fact that √5 is irrational 

Hence , our assumption is wrong. Therefore, -3+2√5 is irrational number

Question 8

Sol :

(i) Let 5+√2 is rational number, Say r

5+√2=r
√2=r-5

As 'r' is rational, r-5 is rational 
√2 is rational. This contradicts the fact that √2 is irrational. Hence , our assumption is wrong. Therefore , 5+√2 is an irrational number

(ii) Let 3-5√3 is rational number, Say r
353=r53=3r
3=(3r5)

As r is rational (3r5) is rational 3 is rational
This contradicts the fact that 3 is irrational.
Hence, our assumption is wrong. Therefore, 353 is an irrational number

(iii) Let 2√3-7 is rational number, Say r
237=r23=r+7
3=r+72

As 'r' is a rational (r+72) is rational 3 is rational

This contradicts the fact that 3 is irrational.

Hence, our assumption is wrong. Therefore, 237 is an irrational number.


(iv) Let √2+√5 is a rational number. Say r
2+5=r
5=r2
(5)2=(r2)2 (on squaring both sides)
5=r2+(2)22×r×2  [(ab)2=a2+b22ab]
5=r2+222r
22r=r23
2=r232r

As r is rational r23 is rational (r232r) is rational
√2 is rational
This contradicts the fact that 2 is irrational

Hence, our assumption is wrong. Therefore (2+5) is an irrational number.

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