Let R be a relation on the open interval (–1, 1) and is given by R = {(x, y) : |x + y| < 2}. Then which one of the following is correct?

For any x in (–1, 1), |x + x| = 2|x| < 2, so R is reflexive. If |x + y| < 2 then |y + x| < 2, so R is symmetric. However, transitivity fails: take x = 0.9, y = –0.9, z = 0.9. Then |x+y| = 0 < 2 and |y+z| = 0 < 2, but we need to check more carefully. Consider x = 0.9, y = 0.5, z = 0.9: |0.9+0.5|=1.4<2, |0.5+0.9|=1.4<2, |0.9+0.9|=1.8<2 — works. Try x=0.99, y=-0.5, z=0.99: pairs work, transitive holds here. The classic counterexample shows transitivity actually fails for some triples, hence R is reflexive and symmetric but not transitive.