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Math 101. Rumbos Fall 2012 1 Solutions to Assignment #7 1. Let π, π β β. Prove that π < π if and only if π < π+π < π. 2 Proof: Assume that π < π. Then, adding π to both sides of the inequality yields 2π < π + π, from which we get that π+π . 2 On the other hand, adding π to both sides of π < π, we obtain π< π + π < 2π, which implies that π+π < π. 2 Hence, π < π implies that π< π+π < π. 2 Conversely, assume that π+π < π. 2 Multiplying by the positive number 2 then yields π< 2π < π + π < 2π. Adding βπ to both sides of the ο¬rst inequality gives π < π. 2. Prove that between any two rational numbers there is at least one rational number. Math 101. Rumbos Fall 2012 2 Proof: Let π and π denote rational numbers and suppose that π < π. Then, by the result of Problem 1, π+π π< < π, 2 π+π where is a rational number since β is a ο¬eld. 2 3. Prove that between any two rational numbers there are inο¬nitely many rational numbers. Proof: Let π and π denote rational numbers with π < π. Assume, by way of contradiction, that there are only a ο¬nite number, π, of rational numbers π1 , π2 , . . . , ππ ; in other words, π1 , π2 , . . . , ππ are the only rational numbers between π and π. By trichotomy, since these rational numbers are distinct, we may assume that they are ordered as follows π < π1 < π2 < β β β < ππ < π. Applying the result of Problem 2 to ππ and π, we obtain a rational number, π, such that ππ < π < π. Thus, there are π + 1 rational numbers between π and π. This is a contradiction. Thus, there are inο¬nitely many rational numbers between π and π. 4. Given two subsets, π΄ and π΅, of real numbers, the union of π΄ and π΅ is the set π΄ βͺ π΅ deο¬ned by π΄ βͺ π΅ = {π₯ β β β£ π₯ β π΄ or π₯ β π΅} Assume that π΄ and π΅ are nonβempty and bounded above. Prove that sup(π΄βͺπ΅) exists and sup(π΄ βͺ π΅) = max{sup(π΄), sup(π΅)}, where max{sup(π΄), sup(π΅)} denotes the largest of sup(π΄) and sup(π΅). Proof: Assume that π΄ and π΅ are nonβempty and bounded above. Then, their suprema, sup(π΄) and sup(π΅), respectively, exist, by the completeness axiom. The assumption that π΄ and π΅ are nonβempty also implies that π΄ βͺ π΅ is nonβ empty. Math 101. Rumbos Fall 2012 3 Let π₯ be an arbitrary element in π΄ βͺ π΅. Then, either π₯ β π΄ or π₯ β π΅. If π₯ β π΄, the π₯ β©½ sup(π΄). (1) On the other hand, if π₯ β π΅, then π₯ β©½ sup(π΅). (2) Thus, if π₯ β π΄ βͺ π΅, it follows from (1) and (2) that π₯ β©½ max{sup(π΄), sup(π΅)}, since sup(π΄) β©½ max{sup(π΄), sup(π΅)} and sup(π΅) β©½ max{sup(π΄), sup(π΅)}. Thus, max{sup(π΄), sup(π΅)} is an upper bound for π΄ βͺ π΅. Thus, by the completeness axiom sup(π΄ βͺ π΅) exists and sup(π΄ βͺ π΅) β©½ max{sup(π΄), sup(π΅)}. (3) Next, observe that π΄βπ΄βͺπ΅ and π΅ β π΄ βͺ π΅, from which we get that sup(π΄) β©½ sup(π΄ βͺ π΅) and sup(π΅) β©½ sup(π΄ βͺ π΅). Consequently, max{sup(π΄), sup(π΅)} β©½ sup(π΄ βͺ π΅). (4) Combining the inequalities in (3) and (4) yields the result. 5. Given two subsets, π΄ and π΅, of real numbers, the intersection of π΄ and π΅ is the set π΄ β© π΅ deο¬ned by π΄ β© π΅ = {π₯ β β β£ π₯ β π΄ and π₯ β π΅} Is it true that sup(π΄ β© π΅) = min{sup(π΄), sup(π΅)}? Math 101. Rumbos Fall 2012 4 Here, min{sup(π΄), sup(π΅)} denotes the smallest of sup(π΄) and sup(π΅). Solution: The problem here is that the intersection of π΄ and π΅ might be empty. If this is the case sup(π΄ β© π΅) will not be deο¬ned. Thus, if π΄ β© π΅ = β , the answer to the question is no. On the other hand, if π΄ β© π΅ β= β , then, since π΄β©π΅ βπ΄ and π΄ β© π΅ β π΅, sup(π΄ β© π΅) β©½ sup(π΄) and sup(π΄ β© π΅) β©½ sup(π΅), It then follows that sup(π΄ β© π΅) β©½ min{sup(π΄), sup(π΅)}. However, this inequality can be strict. For instance, let π΄ = {0, 1} and π΅ = {0, 2}. Then, π΄ β© π΅ = {0}; so sup(π΄ β© π΅) = 0, sup(π΄) = 1 and sup(π΅) = 2. Thus, min{sup(π΄), sup(π΅)} = 1 > 0 = sup(π΄ β© π΅). Thus, in general, the answer to the question is no. β‘