ML Aggarwal Solutions Class 9 Math 1st Chapter Rational and Irrational Numbers Exercise 1.2
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions First Chapter Rational and Irrational Numbers Exercise 1.2. APC Solution Class 9 Exercise 1.2.
(1) Prove that √5 is an irrational number.
Solution:
Let, √5 be a rational number then.
√5 = P/q where q ≠ o
Or, 5 = p2/q2
Or, p2 = 5q2 —- (i)
∴ There is no common factor between Poq except 1.
∴ As 5 divides 5q2, So 5 divides p2 but 5 is prime number.
∴ 5 divides P.
Let, P = 5m, where m is an integer.
Substituting this value of P in equation we get,
(5m)2 = 5q2
Or, 25m2 = 5q2
Or, 5m = q
As 5 divides 5q2 so 5 divides q2 but 5 is prime
∴ 5 divides q
Thus p & q has a common factor 5. This contradicts p & q has no common factors.
Hence, √5 is an irrational number.
(2) Prove that √7 is an irrational number.
Solution:
Let, √7 be a rational number then,
√7 = P/q where q ≠ o.
Or, 7 = p2/q2
∴ Or, p2 = 7q2 —– (i)
∴ p2 & q2 has no common factors except,
As 7 divides 7q2, so 7 divides p2 but 7 is prime.
∴ 7 divides p
Now, Let, P = 7cm
Substituting, the value of P in equation (i) we get,
(7m)2 = 7q2
Or, 49m2 = 7q2
Or, 7m2 = q2
As 7 divides 7m2, so 7 divides q2 but 7 is point
∴ 7 divides q
Thus, p & q has one common factor 7 which, contradicts that p & q has no common factor.
Hence, √7 is an irrational numbers.
(3) Prove that √6 is an irrational number.
Solution:
Let, √6 be a rational number than,
√6 = p/q, then q≠0
P & q has no common factors,
∴ 6 = p2/q2
Or, p2 = 6q2 —- (i)
As 2 divides 6q2, 2 divides p2 & 2 is prime
∴ 2 divides p.
Let, P = 2m, where m is an integer substituting the value of p in equation (i) we get.
(2k)2 = 2q2
Or, 4k2 = 2q
Or, 2k2 = q2
As 2 divides 2k2 So 2 divides q2& 2 is prime.
∴ 2 divides q
Thus, P & q has at least one common factor 2 which contradicts that p & q has no common factors.
Hence, √6 is in irrational number.
(4) Prove that 1/√11 is an irrational number.
Solution:
Let, 1/√11 be a rational numbers then 10.
1/√11 = p/q, where q ≠ o = p & q have no common factors.
Or, 1/11 = p2/q2
Or, q2 = 11p2 —– (i)
As, 11 divides p2, so 11 divides q2 but 11 is prime.
∴ 11 divides q
Let, q = 11/m where m is an integer.
Substituting the value of q in equation (i) we get,
(11m)2= 11p2
Or, 121 m2 = 11p2
Or, 11m2 = p2
As 11 divided m2 so 11 divides p2 but 11 is prime.
∴ 11 divides p.
Now, p & q have a common factor 11 which contradicts that p & q does not have any common factor.
Hence, 1/√11 is an irrational number.
(5) Prove that √2 is an irrational number. Hence, show that 3 – √2 is an irrational number.
Solution:
Let, √2 be a rational number then.
√2 = p/q, where q ≠ o; p & q has no common factor.
Or, 2 = p2/q2
Or, p2 = 2q2 —- (i)
As 2 divides q2 so 2 divides p2 but 2 is prime.
∴ 2 divides p.
Let, P = 2m where m is an integer,
Substituting the value of p in equation (i) we get,
(2m)2 = 2q2
Or, 4m2 = 2q2
Or, 2m2 = q2
As 2 divides m2 so 2 divides q2 but 2 is prime.
∴ 2 divides q
∴ p & q has one common factor 2 which contradicts. P & q have no common factor.
Hence √2 is an irrational number.
Now, let as assume 3 – √2 as a rational number.
Then, 3 – √2 = r
Or, 3 – r = √2
As r is rational number,
∴ 3 – r is a rational number.
∴ √2 is a rational number ∵ √2 = 3 – r
But, it contradicts the fact that √2 is an irrational number.
Hence, our assumption is wrong, 3 – √2 is an irrational number.
(6) Prove that √3 is an irrational number. Hence, show that 2/5 √3 is an irrational number.
Solution:
Let, √3 be a rational number then, √3 = P/q : where q ≠ o : p & q has no common factors.
Or, 3 = p2/q2
Or, p2 = 3q2 —- (i)
As 3 divides q2 So 3 divides p2 but 3 is prime.
Let, p = ∴ divides p = 3m where m is an integer.
∴ substituting the value of p in equation (i) we get.
(3m)2 = 3q2
Or, 9m2 = 3q2
Or, 3m2 = q2
As 3 divides m2 so 3 divides q2 but 3 is prime.
∴ 3 divides q.
∴ p & q has one common factor 3 which contradicts. P & q having no common factors.
Hence, √3 is in irrational number.
Now, let, 2/5 √3 be a rational number then, 2/5 √3 = r
Or, √3 = 5r/2
As r is a rational number, 5r/2 is a rational number.
∴ √3 is a rational numbers∵ [√3 = 5/2 r 13]
But, it contradicts the fact that √3 is an irrational number.
∴ initial assumption is wrong 2/5 √3 is an irrational number.
(7) Prove that √5 is an irrational number. Hence, show that -3 + 2√5 is an irrational number.
Solution:
Proof of √5 is irrational number is provided in question no 1 of exercise 1.2.
Now, let -3 +2√5 be a rational number then,
∴ -3 + 2√5 = 2
Or, √5 = r+3/2
Now, since r is a rational number,
∴ r +3 is a rational number.
Hence, r+3/2 is also a rational number.
∴ √5 is a rational number [∵√5 = r+3/2]
But, it contradicts the fact that √5 is an irrational number.
Hence, our initial assumption is wrong.
-3+2√5 is an irrational number.
(8) Prove that the following numbers are irrational:
(i) 5 + √2
(ii) 3 – 5 √3
(iii) 2√3 – 7
(iv) √2 + √5
Solution:
(i) Proof of √2 is an irrational number is provided in question no. 5 in exercise 1.2
Now, let 5 + √2 be a rational number then,
5 + √2 = r
Or, √2 = r – 5
∴ r is rational number
∴ r – 5 is a rational number.
∴ √2 is a rational number [∵ √2 = r – 5]
But, it contradicts the fact that √2 is an irrational number.
Hence, our initial assumption is wrong, 5 + √2 is an irrational number.
(ii) Proof of √3 is an irrational number is provided in question no. 6 of exercise 1.2
Now, let, 3 = 5√3 be rational number.
∴ 3 – 5√3 = r
Or, 3-r/5 = √3
∵ r is a rational number, ∴ 3-r is a rational number.
∴ 3-r is a rational number.
∴ √3 is a rational number [∵ √3 = 3-r/5]
But, √3 is irrational number is a contradiction.
Hence, our initial assumption is wrong.
3 – 5√3 is an irrational number.
(iii) Proof of √3 is an irrational number is provided in question no. 6 exercise 1.2
Now let, 2√3 – 7 be a rational number then,
2√3 – 7 = r
Or, r+7/2 = √3
∵ r is a rational number.
∴ r + 7 is rational number
∴ r+7/2 is rational number
∴ √3 is rational number.
But it contradicts the fact that √3 is an irrational number.
Hence, our initial assumption is wrong.
2√3 – 7 is an irrational number.
(iv) Let, √2 + √5 be a rational number then,
√2 + √5 = r where r ≠ o
Or, √2 = r – √5
Or, (√2)2 = (r – √5)2
Or, 2 = r2 + 5 – 2√5 r
Or, 2√5r = r2 + 3
Or, √5 = r2 + 3/2r
Now, As r is rational number.
∴ r2+3/2r is a rational number.
∴ √5 is a rational number [∵ √5 = r2+3/2r]
But, it contradicts the fact that √5 is an irrational number.
Hence our initial assumption is wrong.
∴ √2 + √5 is an irrational number.