Question

plzz Help Karo yarrooo​

Answer 1

Step-by-step explanation:

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

or

Therefore there exists no rational number r such that r2=3. Hence the root of 3 is an irrational number.