**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} =p^{2}

5q^{2} = p^{2} ,

q^{2} = ^{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 p^{2} = 25 r^{2}

But p^{2} = 5q^{2}

5q^{2} = 25r^{2}

q^{2} = 5r^{2}

5 divides q^{2}

√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 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.