+129847 Here's one way :
Let us suppose that we have positive integers a and b such that
√3 = a / b where a/b is in lowest terms and a > b ...... square both sides
3 = a^2 / b^2
3b^2 = a^2
2b^2 + b^2 = a^2
2b^2 = a^2 - b^2 (1)
We have three cases to consider
Case A ....a and b are both even......but this is impossible because a/b would not be reduced to lowest terms if this were true since both would be divisible by 2
Case B .......a is even and b is odd....but, in (1), 2b^2 is even but a^2 - b^2 is odd....and this is impossible [Note the same result is produced if a is even and b is odd]
Case C ......the hardest to prove......let a, b be odd....then we have that
2b^2 = a^2 - b^2
2b^2 = (a + b) (a - b)
However.....if a and b are both odd......then (a + b) and (a - b) are both even....so....let (a + b) = 2m and (a - b) = 2n ......and we have that
2b^2 = 2m * 2n
2b^2 = 4mn
b^2 = 2mn
But if b is odd, then so is b^2.......but 2mn is even......so this is a contradiction
Hence.......there are no positive integers a, b of the same or different parity such that
√3 = a / b
