Question

Prove that square root 3 is irrational.

Answer 1

Solution:

Let us assume that, √3 is a rational number of simplest form $\frac{a}{b}$, having no common factor other than 1.

√3 = $\frac{a}{b}$

On squaring both sides, we get ;

3 = $\frac{a^{2}}{b^{2}}$

⇒ a² = 3b²

Clearly, a² is divisible by 3.

So, a is also divisible by 3.

Now, let some integer be c.

⇒ a = 3c

Substituting for a, we get ;

⇒ 3b² = 3c

Squaring both sides,

⇒ 3b² = 9c²

⇒ b² = 3c²

This means that, 3 divides b², and so 2 divides b.

Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b have no common factor other than 1.

This contradiction has arises because of our assumption that √3 is rational.

So, we conclude that √3 is irrational.

Answer 2

Answer:

Let √3 be a rational number  

i.e. √3 = a/b where a,b ∈ integers having no common factor other then 1 and b≠0

= √3 = a/b

square both the sides

= 3= a²/b2

= a² = 3b²

= b² = a²/3

= 3 divides a²

= 3 divides a²

let a²= 3c

= b²=9c²/3

= b²= 3c²

= c²= b²/3

= 3 divides b²

= 3 divides b

thus 3 is a common factor of a and b

this contradicts the fact that a and b are coprime numbers i.e. having no common factor other then 1.

therefore √3 is not a rational number  

hence it is irrational

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Step-by-step explanation: