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 p2 and '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 5m2 so '5' divides q2 but 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=7⋅q2..(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=7⋅q2
4m2=7q2
q2=7m2
As 7 divides 7m2, so 7 divides q2 but 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=6⋅q2...(i)
As '2' divides 6q2. So, 2 divides p2 but 2 is prime
2 divides p
Let p=2m , where 'm' is an integer
Substitute this value of 'p' in (i)
(2m)2=6⋅q2
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 1√11 be a ratitnal number.
1√11=pq, where 'p' , 'q' are integers, q≠0 and p, q have no common factors (except 1)
⇒111=p2q2
⇒q2=11⋅p2...(i)
As 11 divides 11p2,So 11 divides q2 but 11 is a prime
11 divides q
q=11m where 'm' is an integer
(11m)2=11p2
121⋅m2=11⋅p2
p2=11⋅m2
As 11 divides 11m2.So 11 divides p2 but 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 , 1√11 is not rational number
So, we conclude 1√11 is irrational number
Question 5
Sol :
Let √2 be a rational number
2=p2q2
p2=2⋅q2..(i)
As '2' divides 2q2 ,so 2 divides p2 but 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 q2 but 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, 3−√2=r⇒3−r=√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, 3−√2 is an irrational number
Question 6
Sol :
Let √3 be a rational number
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 25√3 is a rational number, say r
Thus 25√3=r⇒√3=52⋅r
As r is rational 52r is rational √3 is rational
This contradicts the fact that √3 is irrational
Hence our assumption is wrong. Therefore, 25√3 is irrational number
Question 7
Sol :
Let √5 be a rational number
Question 8
Sol :
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