Question

prove that root 3 is irrational number​

Answer 1

Let, on the contrary, √3 be a rational number of the form of p/q, where p and q both are co-prime integers and q ≠ 0.

Then,

√3 = p/q

Squaring both sides,

3 = p²/q²

⇒ p² = 3q²

This means that 3 is a factor of p², this further means that 3 is a factor of p too.

So, p can be written as 3a for some integer a.

p = 3a

⇒ p² = 9a²

But, p² = 3q²

⇒ 9a² = 3q²

⇒ q² = 3a²

This means that q² is divisible by 3, this further means that 3 is a factor of q too.

Initially, we had taken that p and q are co primes, but now we got that p and q both are divisible by 3.

Hence, our assumption is wrong. This contradiction occurred because we had taken √3 as a rational number. Hence, √3 is not a rational number. Thus, √3 is an irrational number.

Answer 2

Statement:

$\sf{prove \: that \: \sqrt{3} \: is \: an \: irrational \: number}$

Proof :

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number