Question

prove that √3 is an irrational number ​

Answer 1

Answer:

Let us assume on the contrary that 3 is a rational number. 

Then, there exist positive integers a and b such that

3=ba where, a and b, are co-prime i.e. their HCF is 1

Now,

3=ba

⇒3=b2a2 

⇒3b2=a2 

⇒3 divides a2[∵3 divides 3b2] 

⇒3 divides a...(i) 

⇒a=3c for some integer c

⇒a2=9c2 

⇒3b2=9c2[∵a2=3b2] 

⇒b2=3c2 

⇒3 divides b2[∵3 divides 3c2] 

⇒3 divides b.

Answer 2

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 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.

Hope I helped you..