proof:

Assume sqrt(3)= a/b where a,b are integers and b cannot equal 0. (sqrt(3)=(a/b)^2 turns 3=a^2/b^2 and 3b^2=a^2. We know that 3|a^2 iff 3|a. Using this theorem we know that 3(b^2)=a^2 must be a multiple of 3|a. 3|a then becomes 3r=a where r is some integer. Then squaring both side 3r=a becomes 9r^2=a^2. Since 3b^2=a^2 that means that 3b^2=9r^2 which simplifies into b/r=sqrt(3).

Since b/r is a rational number this is a contradiction

I think its right not sure though.