Claim: The difference between two rational numbers always is a rational number
Proof: You have a/b - c/d with a,b,c,d being integers and b,d not equal to 0.
Then:
a/b - c/d ----> ad/bd - bc/bd -----> (ad - bc)/bd
Since ad, bc, and bd are integers since integers are closed under the operation of multiplication and ad-bc is an integer since integers are closed under the operation of subtraction, then (ad-bc)/bd is a rational number since it is in the form of 1 integer divided by another and the denominator is not eqaul to 0 since b and d were not equal to 0. Thus a/b - c/d is a rational number.
Another definition of rational number is a real number $x$ such that there exists a non-zero integer $n$ for which $nx$ is an integer.
Since $bd(a/b-c/d) =ad-bc $ is an integer, $(a/b-c/d) $ is rational.