Prove that root 3 is irrational

1.	Prove that root 3 is irrational
Answer» ${\bold{\blue{\underline{\red{S}\pink{ol}\green{uti}\purple{on:-}}}}}$ Let √3 be rational and let it's simple form be a/B Then, a and b are integers having no common factor other than 1, and b ≠ 0. Now, √3 = a/b Squaring both side (√3)² = (a/b)² $\rm\longrightarrow\:$ 3 = a²/b² $\rm\longrightarrow\:$ 3b² = a²...............(i) $\rm\longrightarrow\:$ 3 divides a² [∴ 3 divides 3b²] $\rm\longrightarrow\:$ 3 divides a (∴ 3 is PRIME and 3 divides a² $\rm\longrightarrow\:$ 3 divides a ) let a = 3C for some integer c putting a = 3c in eq (i) ,we get 3b² = 9c² $\rm\longrightarrow\:$ b² = 3c² $\rm\longrightarrow\:$ 3 divides b² [∴3 divides 3c²] $\rm\longrightarrow\:$ 3 divides b (∴ 3 is prime and 3 divides b² $\rm\longrightarrow\:$ 3 divides b) Thus, 3 is a common factor of a and b but this contradiction the fact that a and b have no factor other than 1 This contradiction ARISES by assuming √3 is rational. Hence, √3 is irrational.

Answer»

${\bold{\blue{\underline{\red{S}\pink{ol}\green{uti}\purple{on:-}}}}}$

Let √3 be rational and let it's simple form be a/B

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

Now,

√3 = a/b

Squaring both side

(√3)² = (a/b)²

$\rm\longrightarrow\:$ 3 = a²/b²

$\rm\longrightarrow\:$ 3b² = a²...............(i)

$\rm\longrightarrow\:$ 3 divides a² [∴ 3 divides 3b²]

$\rm\longrightarrow\:$ 3 divides a

(∴ 3 is PRIME and 3 divides a² $\rm\longrightarrow\:$ 3 divides a )

let a = 3C for some integer c

putting a = 3c in eq (i) ,we get

3b² = 9c² $\rm\longrightarrow\:$ b² = 3c²

$\rm\longrightarrow\:$ 3 divides b² [∴3 divides 3c²]

$\rm\longrightarrow\:$ 3 divides b

(∴ 3 is prime and 3 divides b² $\rm\longrightarrow\:$ 3 divides b)

Thus,

3 is a common factor of a and b

but this contradiction the fact that a and b have no factor other than 1

This contradiction ARISES by assuming √3 is rational.

Hence, √3 is irrational.