#1**+1 **

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

CPhill
Apr 5, 2017