Question

prove that v3 is an inrrtional number​

Answer 1

Answer:

Proof

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

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 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 left hand 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