Previous exercise Real Numbers – Exercise 3.2
Real Numbers – Exercise 3.3
1) Prove that √5 is an irrational number
- Prove that the following are irrational numbers.
(i) 2√3
(ii) √7/4
(iii) 3 + √5
(iv) √2 + √5
(v) 2√3 – 4
Real Numbers – Exercise 3.3 – Solutions
1) Prove that √5 is an irrational number
Solution:
Let us assume √5 as a rational number.
Therefore it is in the form of p/q,
Then, p & q are integers having no common factor other than 1.
Now √5 = p/q
By cross multiplication √5q = p
Squaring on the both sides (√5q)2 =p2
5q2 = p2 ,
q2 = p^2/5
5 divides p^2/5
√5 divides p———-(1)
p/5 = r for some positive integer r
p = 5r
Squaring on both sides p2 = 25 r2
But p2 = 5q2
5q2 = 25r2
q2 = 5r2
5 divides q2
√5 divides q ————(2)
From (1) and (2), we have,
p and q have at least 5 as a common factor, this contradicts the fact that p and q are co-prime.
Hence √5 is an irrational number.
- Prove that the following are irrational numbers.
(i) 2√3
(ii) √7/4
(iii) 3 + √5
(iv) √2 + √5
(v) 2√3 – 4
Solution:
(i) 2√3
Let us assume that 2√3 is a rational number.
⸫ there exists integers p and q such that, 2√3 = p/q where p and q are co-prime to each other.
⸫√3 = p/q x 1/2
√3 = p/2q
Since p and q are integers then p/2q represents a rational number.
But this a contradiction since RHS is a rational number where as LHS (√3) is an irrational number
Hence our assumption that 2√3 = p/q is a rational number is incorrect.
Thus, 2√3 is an irrational number.
(ii) √7/4
Let us assume that √7/4 is a rational number.
⸫ there exists integers p and q such that, √7/4 = p/q where p and q are co-prime to each other.
⸫√7/4 = p/q
√7 = 4p/q
Since p and q are integers then 4p/q represents a rational number.
But this a contradiction since RHS is a rational number where as LHS (√7) is an irrational number
Hence our assumption that √7/4 = p/q is a rational number is incorrect.
Thus, √7/4 is an irrational number.
(iii) 3 + √5
Let us assume that 3 + √5 is a rational number.
3 + √5 = p/q, where p, q ϵ z, q≠0
√5 = p/q – 3 = (p – 3q)/q
⸫√5 is a rational number
⸫ (p – 3q)/q is a rational number
But √3 is not a r√5ational number
This gives the contradiction
⸫ our assumption that 3 + √5 is a rational number is wrong.
⇒ 3 + √5 is an irrational number.
(iv) √2 + √5
Let us assume that √2 + √5 be a rational number
⇒ √2 + √5 = p/q, where p, q ϵ z, q ≠ 0
⇒ √5 = p/q – √2
By squaring on both the sides, we get,
(√5)2 =(p/q – √2)2
5 = p^2/q^2 – 2(√2)( p/q) + 2
– 2√2 p/q = 5 – 2 – p^2/q^2 =3 – p^2/q^2
2√2 p/q = p^2/q^2 – 3
√2 x 2p/q =( p^2 – 3q^2 /q^2 )
√2 = (p^2 – 3q^2 /q^2 )x q/2p
√2 = p^2 – 3q^2 /2pq
√2 is a rational number
Since p^2 – 3q^2 /2pq is rational number.
But √2 is not a rational number. This leads us to a contradiction
Therefore, our assumption that √2 + √5 is a rational number is incorrect.
⇒ √2 + √5 is an irrational number.
(v) 2√3 – 4
Let us assume that 2√3 – 4 is a rational number.
2√3 – 4= p/q, where p, q ϵ z, q≠0
2√3 = p/q + 4 = (p + 4q)/q
√3 = (p + 4q)/2q
⸫√3 is a rational number
⸫ (p + 4q)/2q is a rational number
But √3 is not a rational number
This gives the contradiction
⸫ our assumption that 2√3 – 4 is a rational number is wrong.
⇒ 2√3 – 4 is an irrational number.