Symmetry with Sums of Proper Fractions

If a is relatively prime to b, then b-a is also relatively prime to b. (ref)

Consider the sum a/b+c/d, where both fractions are reduced, that is, a is relatively prime to b and c is relatively prime to d.

Then the sum (b-a)/b+(d-c)/d is also the sum of two reduced fractions.

(b-a)/b+(d-c)/d=1-a/b+1-c/d=2-(a/b+c/d)

So if a/b+c/d is a reducible sum then so is (b-a)/b+(d-c)/d. Better yet, their sum is always 2.