Question

show that √3is irrational​

Answer 1

Hey there!

The answer to your question is easy.

So, here is the answer to your question.

SOLUTION :

let us assume √3 as Rational.

Then, it will be of the form a/b , [where a, b are integers and b≠ 0.

Again, let a and b have no common factor other than 1.

: √3 = a/b , where a and b are coprime integers.

On squaring both sides, we get:

3 = a²/b² => 3b² = a² ...................(i)

=> 3 divides a² => 3 divides a [ Theorem ]

Then, a can be written as 3m, where m is an integer.

On putting a = 3m in Eq.(i) ,we get

3b² = (3m) ² => 3b² = 9 m² =>b² = 3m²

So, 3 divisibles b² => 3 divisibles b [Theorum]

Thus 3 is a common factor of a and b.

But this contradicts the fact that a and b have no common factor other than 1. The contradiction arise due to our wrong assumption.

Hence√3 is an irrational number.

Hope it helped!!!

Answer 2

SOLUTION :

Let us assume that √3 is an rational number.

So,

$\bf \sqrt{3}=\frac{p}{q}$

[∴ p and q are integers and q ≠ 0, and HCF(p,q) = 1 ]

$\bf \rightarrow {(\sqrt{3})}^{2} ={(\frac{p}{q})}^{2}$

[∴ by squaring both the sides]

$\bf \rightarrow 3= \frac{p^{2}}{q^{2}}$

$\bf \rightarrow 3q^{2}=p^{2} \:\:\:......(i)$

Here,

3 is a factor of p²

so, 3 is also a factor of p

Let p = 3k

(p)² = (3k)²    [squaring both sides]

p² = 9k²   ........(ii)

by comparing equation(i) and equation(ii), we get

3q² = 9k²

⇒ q² = (9k²) ÷ 3

⇒ q² = 3k²

Here,

3 is a factor of q²

so, 3 is also a factor of q

From the above discussion we found that 3 is a common factor of both (p) and (q)

i.e. HCF(p,q) = 3

This contradicts our fact that if √3 is a rational number than HCF(p,q) = 1

So,

√3 cannot be a rational number

Hence,

√3 is an irrational number. [Proved]