Question

prove root3 is irrational​

Answer 1

=> Let us assume to the contrary that √3 is a rational number.

=> It can be expressed in the form of p/q

=> where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p²/q² ( Squaring on both the sides )

⇒ 3q² = p² ………………………………..(1)

=> It means that 3 divides p² and also 3 divides p because each factor should appear two times for the square to exist.

=> So we have p = 3r

=> where r is some integer.

⇒ p² = 9r²………………………………..(2)

=> from equation (1) and (2)

⇒ 3q² = 9r²

⇒ q² = 3r²

Therefore :-

=> q² =3r²

=> ( 2m−1 )² = 3 ( 2n−1 )²

=> 4m² −4m+1 = 3 ( 4n² − 4n+1 )

=> 4m¹ −4m+1 = 12n² − 12n+3

=> 4m² −4m = 12n² − 12n+2

=> 2m² − 2m = 6n² − 6n+1

=> 2( m²−m ) = 2(3n² − 3n)+1

=> We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number r

=> such that r² = 3.

=> Hence the root of 3 is an irrational number.

=> Hence Proved

I hope this will help you dear..

Always stay safe and stay healthy..

Answer 2

Answer:

let √3 is rational .

√3= p/q

taking square root on both sides

(√3)²= p²/q²

p = 3q² ( eqñ 1)

3 divide by p²

3 divide by p

putting 3p = q in eqn 1

3p = 9q

p= 3q

so, our assumption is wrong .

√3 is an irrational.