Question

Prove that

√2

is an irrational number? ​

Answer 1

this is your answer dear have a good day dear

$ANSWER</p><p></p><p>Let us assume on the contrary that 2 is a rational number. Then, there exist positive integers a and b such that</p><p>2=ba where, a and b, are co-prime i.e. their HCF is 1</p><p>⇒(2)2=(ba)2 </p><p>⇒2=b2a2 </p><p>⇒2b2=a2 </p><p>⇒2∣a2[∵2∣2b2 and 2b2=a2] </p><p>⇒2∣a...(i) </p><p>⇒a=2c for some integer c</p><p>⇒a2=4c2 </p><p>⇒2b2=4c2[∵2b2=a2] </p><p>⇒b2=2c2 </p><p>⇒2∣b2[∵2∣2c2] </p><p>⇒2∣b...(ii)</p><p>From (i) and (ii), we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This means that our supposition is wrong.</p><p>Hence,  \sqrt[?]{?} 2 is an irrational number.</p><p></p><p>$