Must the difference between 2 rational numbers be a rational number?

a/b - c/d with a,b,c,d as integers and b,d ≠ 0. a/b - c/d -> ad/bd - bc/bd --> (ad - bc)/bd
The difference will always be rational. This is because the denominator will never be 0. Everything is closed under some kind of operation, and the answers will always be able to be written as a ratio.
Hope that helped.

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AL2006 AL2006 · Answer 2 · 2015-10-10T00:45:43+02:00

Yes. The difference between 2 rational numbers must be a rational number.

The first rational number can always be written as a fraction of integers.
Let's call it   A/B .

The second rational number can always be written as a fraction of integers.
Let's call that one C/D .

   Their difference is A/B - C/D .

   'BD' is a common denominator.

   A/B = AD/BD

   C/D = CB/BD

   A/B - C/D = AD/BD - CB/BD =

      (AD - CB) / BD

Each term in that ugly thing /\ is a product of integers,
so it must be must be another integer.

Therefore the whole ugly fraction is another rational number.