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`.

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