Question

prove that there is no rational number such that s^2=3

Answer 1

lets say, there exists such s, that s^2=3. Then there must exist 2 integers a and b that s=a/b. Therefore \[a^2/b^2=3\] and \[b^2=a^2/3\]\[=>b=a/\sqrt3\]. a must be integer and \[\sqrt3\] is not integer, so b cannot be an integer. We get a contradiction which means that s is not a rational number.