Question

√3 is irrational prove that

Answer 1

we will prove this by contradiction

Let √3 be rational where √3 = p/q ( where p and q are co prime)

√3 = p/q

on squaring both sides

$3 = {p}^{2} \div {q} ^{2}$
$3 {q}^{2} = p {}^{2}$
as we know 3 is divisible by 3q^2

it is divisible by p^2

p is prime

3 is divisible by p

let p be 3k ( where k is real)

$3 {q}^{2} = ({3k})^{2}$
$3 {q }^{2} = {9k}^{2}$
${q}^{2} = 3 {k}^{2}$
as we know 3 is divisible by 3k^2

3 is divisible by q^2

q is prime

3 is divisible by q

so

both p and q have common factors

There is contradiction because we have let that p and q are co prime

our assumption is wrong

√3 is irrational

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HOPE IT HELPS

Answer 2

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.