Question

how to prove
$\sqrt{3}$
is irrational number
​

Answer 1

ANSWER:

• √3 is an irrational number.

GIVEN:

• Number = √3

TO PROVE:

• √3 is an Irrational number.

SOLUTION:

Let √3 be a rational number which can be expressed in the form of p/q where p and q have no common factor other than 1.

=> √3 = p/q

=> √3q = p

Squaring both sides , We get;

=> (√3q)² = (p)²

=> 3q² = p². ...(i)

=> 3 divides p²

=> 3 divides p ......(ii)

Let p = 3m in eq(i) we get;

=> 3q² = (3m)²

=> 3q² = 9m²

=> q² = 3m²

=> 3 divides q²

=> 3 divides q. ....(iii)

From (ii) and (iii)

=> 3 is the common factor of both p and q.

=> Thus our contradiction is wrong.

=> √3 is an irrational number.

Answer 2

To Prove :

• √3 is irrational

Theorem to be used :

• If ‘p’ is a prime number and ‘p’ divides a² , then ‘p’ divides ‘a’ , where ‘a’ is a positive integer.

Proof :

Let us assume , to the contrary , that √3 is rational.

Therefore , we can define √3 as :

$\sf \implies \sqrt{3} = \dfrac{a}{b} \: \: (where \: a \: and \: b \: are \: coprime ) \\\\ \sf \implies b\sqrt{3} = a$

Squaring both sides we have :

$\sf \implies 3b^{2} = a^{2}$

From above we get 3 divides a² , so 3 also divides ‘a’.

$\sf \implies a = 3m \: \: ( Let \: m \: be \: any \: positive \: integer)$

Again squaring both sides we have :

$\sf \implies a^{2} = (3m)^{2} \\\\ \sf \implies 3b^{2} = 9m^{2} \\\\ \sf \implies b^{2} = 3m^{2}$

From above we get 3 divides b² , so 3 also divides ‘b’

Thus ‘a’ is divisible by 3 and also ‘b’ is divisible by 3. It contradicts the assumption that ‘a’ and ‘b’ are coprime .

This contradiction has raised due to the incorrect assumption that √3 is rational

So , it can be concluded that √3 is irrational.