Question

prove that root 3 is an irrational number​

Answer 1

$\large{\bold{\underline{\underline{Question:-}}}}$

Prove that root 3 is an irrational number.

$\large{\bold{\underline{\underline{Solution:-}}}}$

Let us assume to a contrary that √3 is a rational number

It can be expressed in the form of $\frac{p}{q}$

Where p and q are co-primes and q ≠ 0

=> √3 = $\frac{p}{q}$

=> 3 = $\frac{ {p}^{2} }{ {q}^{2} }$

=> 3q² = p²..............(1)

This means 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

Now, we have two cases to consider

$\huge{\bold{\underline{\underline{CASE\:1}}}}$

Suppose that r is even. Then r² is even, and 3r² is even which implies that q² is even and so q is even, but this cannot happen. If both q and r are even then gcd(q,r)≥2 which is a contradiction.

$\huge{\bold{\underline{\underline{CASE\:2}}}}$

Now suppose that r is odd. Then r² is odd and 3r² is odd which implies that q² is odd and so q is odd. Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m, n∈N.

Therefore,

q²=3r²

(2m−1)²=3(2n−1)²

4m²−4m+1=3(4n²−4n+1)

4m²−4m+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.

∴ √3 is an irrational number.

Hence Proved