142-NCERT-10-1-Real Numbers - Ex-1.3

NCERT

10th Mathematics

Exercise 1.3

Topic: 1 Real Numbers

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EXERCISE 1.3

1. Prove that √5 is irrational.  

Explanation:

1) We can prove this using indirect proof. It is also known as proof by contradiction.

Solution:

1) Let if possible, consider that √5 be a rational number.

2) So, √5 = p/q where p and q are coprime and q ≠ 0. That is, there is no common factor other than 1.

(√5)² = (p/q)²

5 = (p)²/(q)²

q² = (p)²/5  ---------- equation 1

3) This shows that 5 divides (p)²  i.e. 5 also divides p.

4) So we have, p/5 = r (say).

(p)²/25 = (r)²

(p)² = 25 (r)²  ---------- equation 2

5) From Equation 1 and Equation 2, we have

(q)² = 25 (r)²/5

(q)² = 5 (r)²

(q)²/5 = (r)²

6) This shows that 5 divides (q)²  i.e. 5 also divides q.

7) This shows that 5 is a common factor of p and q, which contradicts our assumption that √5 is a rational number.

8) So, √5 is an irrational number.

Q2. Prove that 3 + 2√5 is irrational.

Explanation:

1) We can prove this using indirect proof. It is also known as proof by contradiction.

Solution:

1) Let if possible, consider that 3 + 2√5 be a rational number.

2) So, 3 + 2√5 = p/q where p and q are co-primes and q ≠ 0. That is, there is no common factor other than 1.

3 + 2√5 = p/q 

2√5 = (p/q) - 3

2√5 = [(p-3q)/q]

√5 = (p-3q)/2q

3) As p and q are co-primes, (p-3q)/2q is rational, so √5 is also rational which contradicts the actual fact that √5 is irrational.

4) So, 3 + 2√5 is an irrational number.

Q3. Prove that the following are irrationals :

(i) 1/√2    (ii) 7√5    (iii) 6 + √2

Explanation:

1) We can prove this using indirect proof. It is also known as proof by contradiction.

Solution:

(i) 1/√2

1) Let if possible, consider that 1/√2 be a rational number.

2) So, 1/√2 = p/q where p and q are co-primes and q ≠ 0. That is, there is no common factor other than 1.

1/√2 = p/q

√2 = q/p

3) As p and q are co-primes, q/p is rational, so √2 is also rational which contradicts the actual fact that √2 is irrational.

4) So, 1/√2 is an irrational number.

(ii) 7√5

1) Let if possible, consider that 7√5 be a rational number.

2) So, 7√5 = p/q where p and q are co-primes and q ≠ 0. That is, there is no common factor other than 1.

7√5 = p/q

√5 = p/7q

3) As p and q are co-primes, p/q is rational, so √5 is also rational which contradicts the actual fact that √5 is irrational.

4) So, 7√5 is an irrational number.

(iii) 6 + √2

1) Let if possible, consider that 6 + √2 be a rational number.

2) So, 6 + √2 = p/q where p and q are co-primes and q ≠ 0. That is, there is no common factor other than 1.

6 + √2 = p/q 

√2 = (p/q) - 6

√2 = (p-6q)/q

3) As p and q are co-primes, (p-6q)/q is rational, so √2 is also rational which contradicts the actual fact that √2 is irrational.

4) So, 6 + √2 is irrational number.

Click here for ⇨ NCERT-10-1-Real Numbers - Ex-1.4

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