# The difference between two rational numbers always is a rational number [duplicate]

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:

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.

## marked as duplicate by Zev Chonoles, user147263, ncmathsadist, Ross Millikan, Eric WofseySep 29 '15 at 2:55

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.