Question

Prove the following are irrationals (i) \( \sqrt{5} \) (ii) \( 5-\sqrt{3} \)  (iii) \( 7 \sqrt{3} \)

Answer 1

(i) Let us demonstrate that √5 is an irrational number by proving the theorem.

Let’s take it for granted that the number √5 is a rational number. If √5 is rational, then it may be expressed as an equation of the type a/b, here integers are a and b. 

√5/1 = a/b

√5b = a

Bringing both sides into balance,

5b² = a²

b² = a²/5 —- (1)

This indicates that a² is divided by 5.

That indicates that it can also divide a.

a/5 = c

a = 5c

On squaring, we get

a² = 25c²

Replace a² in the equation with its value (1).

5b² = 25c²

b² = 5c²

b²/5 = c²

This indicates that b² may be divided by 5, and hence, b can likewise be divided by 5. Since this is the case, a and b share the factor 5 in common. However, this goes against the fact that a and b are in the coprime position. The fact that we made the assumption that √5 is a rational number led to the formation of this contradiction. As a result, we are forced to the conclusion that √5 is irrational.

Therefore √5 is an irrational number.

(ii)  Let us assume the given number be rational and we will write the given number in p/q form

⇒ 5 − √3 = p/q

⇒ √3 = 5q − p/q

We observe that LHS is irrational and RHS is rational, which is not possible.

This is contradiction.

Hence our assumption that given number is rational is false

⇒ 5 − √3 is irrational

(iii) Let us assume that 3√7 is rational.

3√7 = a/b

Rearranging, we get √7 = a/3b

Since 3, a and b are integers, a/3b can be written in the form of p/q, so a/3b is rational, and so √7 is rational.

But this contradicts that √7 is irrational. So, we conclude that 3√7 is irrational.