Question

Show that root 3 Is
is irrational. ​

Answer 1

Step-by-step explanation:

Let's assume 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

We have two cases to consider now.

Case I

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.

Case II

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.

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 left hand side of this equation is even, while the right hand 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