Question

Prove that √2 is irrational.​

Answer 1

Step-by-step explanation:

Let us assume on the contrary that 2 is a rational number. Then, there exist positive integers a and b such that

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

⇒(2)2=(ba)2 

⇒2=b2a2 

⇒2b2=a2 

⇒2∣a2[∵2∣2b2 and 2b2=a2] 

⇒2∣a...(i) 

⇒a=2c for some integer c

⇒a2=4c2 

⇒2b2=4c2[∵2b2=a2] 

⇒b2=2c2 

⇒2∣b

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Answer 2

Answer:

To prove that √2 is irrational by the contradiction method, we first assume that √2is a rational number. Now, if it is a rational number, there exist two co-prime integers x and y, such that √2 = x/y, where x and y have no other common factors except 1 and y ≠ 0. So, our equation is √2 = x/y