Let g(P, Q) = max([math]x_P,x_Q[/math]) + max([math]y_P,y_Q[/math]).[br][br]A) Prove or disprove that g is a metric:[br][br][b][u]Counter Example:[/u][/b][br][br][b][u]Metric Axiom 1:[/u][/b] Let P and Q be points where P = (1, 2) and Q = (1, 2). Thus, P = Q.[br][br]Notice that: g(P, Q) = max(1, 1) + max(2, 2) = 1 + 2 = 3. Obviously, we can see this fails Metric Axiom 1 because g(P, Q) [math]\ne[/math] 0.[br][br]Therefore, g does not define a metric.[br][br][b][u][color=#ff0000]The rest of this is for communication only and not part of the proof: [/color][/u][/b][br][br][b][u]Metric Axiom 2:[/u][/b] Let P = (a, b) and Q = (x, y). We need to show that g(P, Q) = g(Q, P).[br][br]Notice that: g(P, Q) = max(a, x) +max(b, y) and g(Q, P) = max(x, a) + max(y, b). In both cases, 'max' will choose the greater value of each pair which will be the same number regardless of the individual coordinates. Therefore, g(P, Q) = g(Q, P).[br][br][br][b]Metric Axiom 3:[/b] Let P, Q, and Z be points where P = (a, b), Q = (c, d), and Z = (e, f).[br][br]Notice that g(P, Q) + g(Q, Z) = max(a, c) + max(b, d) + max(c, e) + max(d, f) [math]\ge[/math] g(P, Z) = max(a, e) + max(b, f). This is the case because the 'max' function always chooses the largest of two values. Since g(P, Q) + g(Q, Z) is the sum of four 'max' functions, we know that its sum must always be greater than the sum of two 'max' functions containing on the same set of values: {a, b, c, d, e, f}. [br][br][br]