Question

prove that
$\sqrt{3}$
is irrational ​

Answer 1

✴ola!!✴

⤵⤵Solution⤵⤵

Let Root 3 be rational , Therefore root 3 = a/b , where (a,b) = 1 , b is not equal to 0.

Squaring,

3b^2 = a^2...............(1)

3 divides L.H.S. of (1) therefore, 3 divides R.H.S of (1)

therefore, 3 divides a^2 ,i.e. 3 divides a therefore, a = 3k

therefore, From (1) , 3b^2 = 9k^2 => b^2 = 3k^2

therefore, a and b both are multiple of 3 , which is a contradiction.

[because, a and b has no common factor]

therefore , Root 3 is irrational number.

Hence Proved !!

tysm❤❤

Answer 2

√3 is irrational number

__________ [TO PROVE]

Let us assume that, √3 is a rational number

√3 = $\dfrac{a}{b}$

Here, a and b are co-prime numbers.

• Squaring on both sides, we get

=> (√3)² = $( { \dfrac{a}{b}) }^{2}$

=> 3 = $\dfrac{ {a}^{2} }{ {b}^{2} }$

=> 3b² = a² ________( eq 1)

Clearly;

a² is divisible by 3

So, a is also divisible by 3

Now, let integer be c.

=> a = 3c

• Squaring on both sides.

=> a² = 9c²

=> 3b² = 9c² [From (eq 1)]

=> b² = 3c² _______(eq 2)

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

3 divides b² and 3 divide a² also.

So, our assumption is wrong.

√3 is irrational number.

________________________________