Math 523H #1 - Math 523H Homework#1 Section 1.1 1.1#1 Prove...

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Math 523H Homework #1 Section 1.1 1.1 #1. Prove that if x and y are real numbers, then | xy | = | x || y | . Proof. Suppose that x , y R . Case 1: x < 0 , y < 0 . Then xy > 0, so | xy | = xy . But | x | = - x and | y | = - y , so | x || y | = ( - x )( - y ) = xy by Prop 1.1.1(c). Thus, | xy | = xy = | x || y | . Case 2: x < 0 , y > 0 . Then xy < 0, so | xy | = - ( xy ). But | x | = - x and | y | = y , so | x || y | = ( - x ) y = - ( xy ) by Prop 1.1.1(c). Thus, | xy | = - ( xy ) = | x || y | . Case 3: x > 0 , y < 0 . Then xy < 0, so | xy | = - ( xy ). But | x | = x and | y | = - y , so | x || y | = x ( - y ) = - ( xy ) by Prop 1.1.1(c). Thus, | xy | = - ( xy ) = | x || y | . Case 4: x > 0 , y > 0 . Then xy > 0, so | xy | = xy . But | x | = x and | y | = y , so | x || y | = xy . Thus, | xy | = xy = | x || y | . Case 5: x = 0. Then | x | = 0. | xy | = | 0 · y | = | 0 | = 0 = 0 · | y | = | x || y | . Similarly if y = 0, since we can simply reverse the roles of x and y. Therefore, x, y R , | xy | = | x || y | . 1.1 #3. Prove that - ( - x ) = x for all real numbers x . Proof. Let x R . By (P4), we know that there exists - x such that x + ( - x ) = 0. By (P1), x + ( - x ) = ( - x ) + x , so ( - x ) + x = 0. By (P4), we also know that there exists - ( - x ) such that ( - x ) + ( - ( - x )) = 0. By Prop 1.1.1(b), additive inverses are unique, so we have - ( - x ) = x . 1.1 #4. Prove that all real numbers z satisfy z · 0 = 0 = 0 · z . Proof. Let z R . By (P9), z · (0 + 0) = z · 0 + z · 0. But 0 + 0 = 0, so z · (0 + 0) = z · 0. We thus have z · 0 + z · 0 = z · 0. By (P4), - ( z · 0) exists such that z · 0 + ( - ( z · 0)) = 0. By Prop 1.1.1(a), we can say that z · 0 + z · 0 + ( - ( z · 0)) = z · 0 + ( - ( z · 0)). By (P2) and (P4), z · 0 = 0. By (P5), multiplication is commutative, so z · 0 = 0 · z . Thus,
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