**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 = p^{2}/q^{2}

Or, p^{2} = 5q^{2} —- (i)

∴ There is no common factor between Poq except 1.

∴ As 5 divides 5q^{2}, So 5 divides p^{2} 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} = 5q^{2}

Or, 25m^{2} = 5q^{2}

Or, 5m = q

As 5 divides 5q^{2} so 5 divides q^{2} 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 = p^{2}/q^{2}

∴ Or, p^{2} = 7q^{2} —– (i)

∴ p^{2 }& q^{2} has no common factors except,

As 7 divides 7q^{2}, so 7 divides p^{2} but 7 is prime.

∴ 7 divides p

Now, Let, P = 7cm

Substituting, the value of P in equation (i) we get,

(7m)^{2} = 7q^{2}

Or, 49m^{2} = 7q^{2}

Or, 7m^{2} = q^{2}

As 7 divides 7m^{2}, so 7 divides q^{2} 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 = p^{2}/q^{2}

Or, p^{2} = 6q^{2} —- (i)

As 2 divides 6q^{2}, 2 divides p^{2 }& 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} = 2q^{2}

Or, 4k^{2} = 2q

Or, 2k^{2} = q^{2}

As 2 divides 2k^{2} So 2 divides q^{2}& 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 = p^{2}/q^{2}

Or, q^{2} = 11p^{2} —– (i)

As, 11 divides p^{2}, so 11 divides q^{2} 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}= 11p^{2}

Or, 121 m^{2} = 11p^{2}

Or, 11m^{2} = p^{2}

As 11 divided m^{2} so 11 divides p^{2} 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 = p^{2}/q^{2}

Or, p^{2} = 2q^{2} —- (i)

As 2 divides q^{2} so 2 divides p^{2} 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} = 2q^{2}

Or, 4m^{2} = 2q^{2}

Or, 2m^{2} = q^{2}

As 2 divides m^{2} so 2 divides q^{2} 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 = p^{2}/q^{2}

Or, p^{2} = 3q^{2} —- (i)

As 3 divides q^{2} So 3 divides p^{2} 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} = 3q^{2}

Or, 9m^{2} = 3q^{2}

Or, 3m^{2} = q^{2}

As 3 divides m^{2} so 3 divides q^{2} 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 = r^{2} + 5 – 2√5 r

Or, 2√5r = r^{2} + 3

Or, √5 = r^{2} + 3/2r

Now, As r is rational number.

∴ r^{2}+3/2r is a rational number.

∴ √5 is a rational number [∵ √5 = r^{2}+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.