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 Feld. Carefully prove that x 2 + 1 > 0 for all x F. Let us Frst 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
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hw3sol - Math 319/320 Homework 3 solutions Problem 1. Show...

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