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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.


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.


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

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  • $\begingroup$ Yep that's right. $\endgroup$ – Cameron Williams Sep 29 '15 at 0:47
  • $\begingroup$ I don't downvote, and I'm not sure exactly why someone did, but you technically didn't ask a question. And I'm not sure why this is prefaced with "best answer" $\endgroup$ – pjs36 Sep 29 '15 at 1:42

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.


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