Question

prove that there is no rational number whose square is 3.

Answer 1

Let sqrt(3) = a/b where a and b are integers and they are in lowest terms, meaning they have no common factors, WE know if they had common factors we can always find another expression that has no common factors by canceling out the common ones. So a/b does exist.squaring both sides of sqrt(3) = a/b gives us 3 = a2/b2 so a2 = 3b2Now a2 must have its factors as even powers of primes.(because of the exponent 2 on the left and both sides must have the same total power since they are equal). So it musthave 32 as one of its factors. (can't be just 3 since that is 31 which is an odd exponent and when added to the 2 in b2 gives us an odd exponents also) This means a must have 3 as a factor. Now 3b2 has a factor 32 so b2 has afactor 3. But if b2 has a factor 3, then since powers must be even,it has 32 as a factor. This means that b must have 3 as a factor also.So we have shown that both a and b have 3 as a factor. But we statedthat a/b has no common

factors and we just found one, namely 3. So we have a contradiction

if 32 is a factor then 3 must be also)