3333_HW1_solutions.pdf - Math 3333 Homework 1 Solutions Name Peoplesoft ID Show your work If a problem requires a proof explain and justify your steps

# 3333_HW1_solutions.pdf - Math 3333 Homework 1 Solutions...

This preview shows page 1 - 4 out of 12 pages.

Math 3333 Homework 1 Solutions Name: Peoplesoft ID: Show your work. If a problem requires a proof, explain and justify your steps carefully. Homework papers should be legible and neat, and the pages should be stapled together in the correct order. Illegible work may not be graded. Homework should be submitted in class on the indicated due date. Submissions by email or to the math office will not be accepted.
1. Recall that the absolute value of a real number x is defined as follows: | x | = x if x 0 , - x if x < 0 . Give a mathematical proof for each of the statements below. In all cases, you may use basic facts about inequalities and the algebra of real numbers. In particular, you may assume that the product of nonnegative integers is nonneg- ative, and that the product of a nonnegative number and a nonpositive number is nonpositive. (a) | x | 2 = x 2 for all real numbers x . (Hint: Use a proof by cases.) Assume x R . We will divide the proof into two cases. Case 1: Assume x 0. Then, | x | = x . Therefore, | x | 2 = x 2 . Case 2: Assume x < 0. Then, | x | = - x . Therefore, | x | 2 = ( - x ) 2 = x 2 . In both cases, we have | x | 2 = x 2 . Therefore, | x | 2 = x 2 for all x R . (b) x ≤ | x | for all real numbers x . (Hint: Use a proof by cases.) Assume x R . We will divide the proof into two cases. Case 1: Assume x 0. Then, | x | = x . Therefore, x ≤ | x | . Case 2: Assume x < 0. Then, | x | = - x . Therefore, x < 0 < - x = | x | . Therefore, x ≤ | x | . In both cases, we have x ≤ | x | . Therefore, x ≤ | x | for all x R .
(c) | xy | = | x || y | for all real numbers x and y . Assume x, y R . By part (a) we have | xy | 2 = ( xy ) 2 = x 2 y 2 = | x | 2 | y | 2 = ( | x || y | ) 2 That is, | xy | 2 = ( | x || y | ) 2 . Therefore, | xy | = | x || y | or | xy | = -| x || y | . Since | xy | ≥ 0 and | x || y | ≥ 0, the equation | xy | = -| x || y | is satisfied if and only if x = 0 or y = 0. In either case, the former equation | xy | = | x || y | is satisfied as well. We conclude that | xy | = | x || y | for all real numbers x, y R . Alternate proof (by cases): Assume x, y R . We will divide the proof into four cases. Case 1: Assume x 0 and y 0. Then, xy 0. Therefore, | x | = x , | y | = y , and | xy | = xy . It follows that | xy | = xy = | x || y | . Case 2: Assume x 0 and y < 0. Then, xy 0. Therefore, | x | = x , | y | = - y , and | xy | = - xy . It follows that | xy | = - xy = x ( - y ) = | x || y | . Case 3: Assume x < 0 and y 0. Then, xy < 0. Therefore, | x | = - x , | y | = y , and | xy | = - xy . It follows that | xy | = - xy = ( - x ) y = | x || y | .

#### You've reached the end of your free preview.

Want to read all 12 pages?

• Fall '08
• Staff
• Order theory, max S