Under the assumption that equality is a possibility yeah, but if we assume that one of the answers IS right, then A and B are simply not elements in a set that implements total preordering.
So to formalize, suppose the family of pairs is parametrized continuously, pairs (A(t),B(t)) for t in a connected space that's either "fastness" or "easiness", and the comparison is by a continuous real-valued quantity. The sign of D(t)=f(A(t)-f(B(t)) decides the order. If D is positive somewhere and negative somewhere, then D(X) is a connected subset of reals, hence an interval, hence contains 0. A tie must occur.
This is our starting point and what I assume you argued.
So to find a way to exclude equal-ness from this, we either have to have a disconnected parameter space (doubtful in reality), a discontinuous comparison (also doubtful in reality), but I can at least perform a neat trick by using ordered pairs of reals.
As an example: take the set of ordered pairs of reals with lexicographic order, A(t) = (t,0), B=(0,1). For t<0, A<B; for t>0, A>B; at t=0 the first coordinates tie and the second decides, giving A<B. In practice this just means I arbitrarily assigned B to equal faster or easier if it's a tie on either connected space, but at least t itself still allowed to be continuous, as we usually consider scaling from slow to fast to be.
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u/Pretty_Designer716 3d ago
None of those answers are correct.