Question

proce √3 is irrational​

Answer 1

Answer:

Sol:

Let us assume that √3 is a rational number.

That is, we can find integers a and b (≠ 0) such that √3 = (a/b)

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

√3b = a

⇒ 3b2=a2 (Squaring on both sides) → (1)

Therefore, a2 is divisible by 3

Hence ‘a’ is also divisible by 3.

So, we can write a = 3c for some integer c.

Equation (1) becomes,

3b2 =(3c)2

⇒ 3b2 = 9c2

∴ b2 = 3c2

This means that b2 is divisible by 3, and so b is also divisible by 3.

Therefore, a and b have at least 3 as a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that √3 is rational.

So, we conclude that √3 is irrational.

Step-by-step explanation:

Answer 2

Answer:

we have to prove

$\sqrt{3}$

is irrational.

let as assume the opposite,

i.e.,

$\sqrt{3}$

is rational

hence

$\sqrt{3}$

can be written in the form a/b

where a and b (b is not equal to 0) are co prime (no common factor other than 1)

hence,

$\sqrt{3} = a \div b$

$\sqrt{3b} = a$

squaring both side(

$\sqrt{3b {}^{2} } = a {}^{2}$

$3b {}^{2} = a {}^{2}$