Question

prove that√3 is irrational​

Answer 1

here is ur previous question's answer chidu.....hope it helps...

Answer 2

To prove:-

• √3 is an irrational number.

Solution:-

Let us assume that √3 is a rational number. i.e. it can be written in form of $\sf{ \dfrac{p}{q}}$ where, q ≠ 0 and p and q are co prime ( no common factor other than 1)

Therefore,

$\bullet \; \sf{ \sqrt{3} = \dfrac{p}{q}}$

$\implies \sf{ \sqrt{3} q = p}$

Squaring both sides:-

$\implies \sf{( \sqrt{3} q)^2 = (p)^2}$

$\implies$ 3q² = p²

♦ 3 divides p²

♦ so, 3 also divides p ----(1)

Lets p = 3c

Now we know that,

$\implies$ 3q² = p²

$\bullet \;$ Putting p = 3c

$\implies$ 3q² = (3c)²

$\implies$ 3q² = 9c²

$\sf{ q^2 = \dfrac{9c^2}{3}}$

$\implies$ q² = 3c²

Hence,

♦ 3 divides q²

♦ so, 3 also divides q ----(2)

From eq (1) and (2) :-

★ 3 divides both p and q.

♦ Hence, 3 is a common factor or p and q.

♦ Therefore, p and q are not coprime.

★ Hence, our supposition is wrong.

By contradiction,

$\bold{\underline{\underline{\sf{\red{\dag \; \sqrt{3} \; is \; an \; irrational \; number.}}}}}$

$\rule{200}{2}$