Real Numbers – Exercise 3.3 – Class X

Previous exercise Real Numbers – Exercise 3.2

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Real Numbers – Exercise 3.3

1) Prove that √5 is an irrational number

2. Prove that the following are irrational numbers.

(i) 2√3

(ii) ^√7/₄

(iii) 3 + √5

(iv) √2 + √5

(v) 2√3 – 4

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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.

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2. 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 2^p/_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.

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