In Class Problem Set #15

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In Class Problem Set #15 CSE 1400 and MTH 2051 Fall 2012 Relations Definitions A relation is a set of ordered pairs (x, y) where x is related to y. Let denote a relational symbol. Write x y to express the predicate x is related to y. Reflexive: A relation is reflexive if for all x, x x. A relation is if for all x and y, x y implies y x. A relation is anti if for all x and y, x y and y x implies x = y. A relation is if for all x, y and z, x y and y z implies x z. ( x)(x x) Symmetric: ( x, y)((x y) (y x)) Anti: ( x, y)(((x y) (y x)) (x = y)) Transitive: ( x, y, z)(((x y) (y z)) (x z)) Problems 1. Use Boolean algebra to show that (x y) (y x) is equivalent to (y x) (x y). (x y) (y x) (x y) (y x) (y x) (x y) 0 0 1 1 1 1 0 1 1 0 1 1 1 0 0 1 0 0 1 1 1 0 1 0 Note that (x y) (y x) and (y x) (x y) are two equivalent ways to state is a relation. 2. Use Boolean algebra to show that ((x y) (y x)) (x = y) is equivalent to (x = y) ((x y) (y x)).

in class problem set #15 2 (x y) (y x) x = y (x y) (y x) x = y x = y (x y) (y x) 0 0 0 0 1 0 1 1 1 0 0 1 0 1 1 0 1 1 0 1 0 0 1 0 1 1 1 0 1 1 0 1 1 0 1 1 1 0 0 0 1 0 1 1 1 1 0 1 0 1 1 0 1 1 1 1 0 1 0 0 1 0 0 1 1 1 1 1 1 0 1 0 Note that ((x y) (y x)) (x = y) and (x = y) ((x y) (y x)) are two equivalent ways to state is an anti relation. 3. Use Boolean algebra to show that ((x y) (y z)) (x z) equivalent to (x z) ((x y) (y z)) (x y) (y z) (x z) (x y) (y z) (x z) (x z) (x y) (y z) 0 0 0 0 1 0 1 1 1 0 0 1 0 1 1 0 1 1 0 1 0 0 1 0 1 1 1 0 1 1 0 1 1 0 1 1 1 0 0 0 1 0 1 1 1 1 0 1 0 1 1 0 1 1 1 1 0 1 0 0 1 0 0 1 1 1 1 1 1 0 1 0 Note that ((x y) (y z)) (x z) and (x z) ((x y) (y z)) are two equivalent ways to state is a relation. 4. For each relation (on the set R of real numbers), identify if it is reflexive,, anti, or. (a) Equality: x = y x = x is True, so equality is reflexive If x = y, then y = x is True, so equality is If x = y and y = x, then x = y is True, so equality is anti If x = y and y = z, then x = z is True, so equality is (b) Not equal: x = y x = x is False, so not equal is not reflexive

in class problem set #15 3 If x = y, then y = x is True, so not equal is If x = y and y = x, then x = y is False, so equal is not anti If x = y and y = z, then x = z is False, so not equality is not (c) Less than: x < y x < x is False, so less than is not reflexive If x < y, then y < x is False, so less than is not If (x < y and y < x), then x = y is True, so less than is anti (The conditional is True because the premise (x < y and y < x) is False) If x < y and y < z, then x < z is True, so less than is (d) Greater than or equal: x y x x is True, so greater than or equal is reflexive If x y, then y x is False, so greater than or equal is not If (x y and y x), then x = y is True, so greater than or equal is anti If x y and y z, then x z is True, so greater than or equal is (e) Equal magnitude: x = y x = x is True, so equal magnitude is reflexive If x = y, then y = x is True, so equal magnitude is If ( x = y and y = x ), then x = y is False, so equal magnitude is not anti If x = y and y = z, then x = z is True, so equal magnitude is (f) Approximately equal (within ɛ > 0): x y ɛ x = x is True, so equal magnitude is reflexive If x = y, then y = x is True, so equal magnitude is If ( x = y and y = x ), then x = y is False, so equal magnitude is not anti

in class problem set #15 4 If x = y and y = z, then x = z is True, so equal magnitude is 5. For each relation (on the set of geometric figures), identify if it is reflexive,, anti, or. (a) Parallel lines: L 0 L 1 L 0 L 0 is True, so parallel lines is reflexive If L 0 L 1, then L 1 L 0 is True, so parallel lines is If (L 0 L 1 and L 1 L 0 ), then L 0 = L 1 is False, so parallel lines is not anti If L 0 L 1 and L 1 L 2, then L 0 L 2 is True, so parallel lines is (b) Perpendicular lines: L 0 L 1 L 0 L 0 is False, so perpendicular lines is not reflexive If L 0 L 1, then L 1 L 0 is True, so perpendicular lines is If (L 0 L 1 and L 1 L 0 ), then L 0 = L 1 is False, so perpendicular lines is not anti If L 0 L 1 and L 1 L 2, then L 0 L 2 is False, so perpendicular lines is not (c) Similar triangles: T 0 T 1 (every angle in T 0 is equal to a corresponding angle in T 1 ) T 0 T 0 is True, so similar triangles is reflexive If T 0 T 1, then T 1 T 0 is True, so similar triangles is If (T 0 T 1 and T 1 T 0 ), then T 0 = T 1 is False, so similar triangles is not anti If T 0 T 1 and T 1 T 2, then T 0 T 2 is True, so similar triangles is (d) Congruent triangles: T 0 = T1 (every angle and side in T 0 is equal to a corresponding angle and side in T 1 ) T 0 = T0 is True, so congruent triangles is reflexive If T 0 = T1, then T 1 = T0 is True, so congruent triangles is If (T 0 = T1 and T 1 = T0 ), then T 0 = T 1 is False, so congruent triangles is not anti

If T 0 = T1 and T 1 = T2, then T 0 = T2 is True, so congruent triangles is in class problem set #15 5