Real Numbers – Exercise 3.3 – Class X

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Real Numbers – Exercise 3.3

1) Prove that √5 is an irrational number

  1. 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.


  1. 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.