NCERT10th MathematicsExercise 1.3Topic: 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) √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 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/q2√5 = (p/q) - 32√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 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 fact that √2 is irrational.
4) So, 1/√2 is an irrational number.
(ii) 7√5
1) Let 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 fact that √5 is irrational.
4) So, 7√5 is an irrational number.
(iii) 6 + √2
1) Let 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 fact that √2 is irrational.
4) So, 6 + √2 is irrational number.
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