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hw3sol - Math 319/320 Homework 3 solutions Problem 1 Show...

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Math 319/320 Homework 3 solutions Problem 1. Show that for all real numbers x and y , | | x | - | y | | ≤ | x - y | . By the triangle inequality, we have | x | = | x - y + y | ≤ | x - y | + | y | = ⇒ | x | - | y | ≤ | x - y | and | y | = | y - x + x | ≤ | y - x | + | x | = | x - y | + | x | = ⇒ -| x - y | ≤ | x | - | y | . Combining the two inequalities, we get -| x - y | ≤ | x | - | y | ≤ | x - y | = ⇒ | | x | - | y | | ≤ | x - y | . Problem 2. Let ( F, + , · ) be an ordered field. Carefully prove that x 2 + 1 > 0 for all x F. Let us first check that 0 < 1. If not, since 0 6 = 1 we must have 1 < 0 (by O1). Then 1 - 1 < 0 - 1 or 0 < - 1 (by O3). This implies 1 · ( - 1) < 0 · ( - 1) or - 1 < 0 (by O4). So we have 0 < - 1 and - 1 < 0, which is a contradiction. Thus 0 < 1. Now let us prove that 0 < x 2 + 1 for every x F . We distinguish three cases: (i) If x = 0, then 0 < 1 = x 2 + 1. (ii) If 0 < x , then 0 · x < x · x or 0 < x 2 (by O4). So 1 < x 2 + 1 (by O3) so 0 < x 2 + 1 (by O2). (iii) If x < 0, then x - x < - x or 0 < - x (by O3). This implies x · ( - x ) < 0 · ( - x ) or - x 2 < 0 (by O4). So - x 2 + x 2 < 0 + x 2 or 0 < x 2 (by O3). So, again 0 < x 2 + 1 as in case (ii). Problem 3. In each case, find all the upper bounds (if any) and the least upper bound of S R : S = { a, b, c } , where a > b > c
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