Question

show that
$\sqrt{is \: an \: irrational}$
​

Answer 1

Answer:

irrational

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∣b2[∵2∣2c2]

⇒2∣b...(ii)

From (i) and (ii), we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This means that our supposition is wrong.

Hence, 2

​ is an irrational number.