ML Aggarwal Solutions Class 9 Math First Chapter Rational and Irrational Numbers Exercise 1.2

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.