Solution: (a) We show: (∗) If 0 < x < y, then x2 < y2. Indeed, we may multiply the inequality x < y with the positive quantity x and get x2 < xy. We also may multiply x < y with the positive quantity y and get xy < y2. By transitivity, we obtain x2 < y2. From (∗), uniqueness follows: Assuming x2 = y2 = a and x, y > 0, we can rule out x < y (which would imply x2 < y2) and y < x (which would impl...
