Question

Proved that root 3 is an irrational number
please answer fast​

Answer 1

Step-by-step explanation:

Given:

• Number is √3

To Prove:

• √3 is an irrational number.

Proof: Let us assume, to the contrary, that √3 is rational.and let its simplest form be a/b

Then, a and b are integers having no common factors other than 1, and b ≠ 0

Now, √3 = a/b { Squaring on both the sides }

→ √3² = (a/b)²

→ 3 = a²/b²

→ 3b² = a² .....(1) { By cross multiplication }

→ 3 divides a² [ Since, 3 divides 3b² ]

→ 3 divides a

• [ Since, 3 is prime and 3 divides a² => 3 divides a ]

Let a = 3c fir some integer c

• Putting a = 3c in Equation (1) , we get •

$\small\implies{\sf }$ 3b² = 9c²

$\small\implies{\sf }$ b² = 9c²/3

$\small\implies{\sf }$ b² = 3c²

$\small\implies{\sf }$ 3 divides b² [ Since, 3 dives 3c² ]

$\small\implies{\sf }$ 3 divides b [ 3 is prime and 3 divides b² => 3 divides b ]

This, 3 is a common factor of a and b

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction arises by assuming that √3 is rational.

Hence, We conclude that √3 is irrational.