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Math 101. Rumbos Spring 2010 1 Solutions to Exam 1 (Part I) 1. Provide concise answers to the following questions: (a) A subset, ๐ด, of the real numbers is said to be bounded if there exists a positive real number, ๐ , such that โฃ๐โฃ โฉฝ ๐ for all ๐ ∈ ๐ด. Give the negation of the statement “๐ด is bounded.” Answer: The negation of ”๐ด is bounded ” is For every positive number, ๐ , there exists and an element, ๐, in ๐ด such that โฃ๐โฃ > ๐ . โก (b) Let ๐ด denote a subset of the real numbers and ๐ฝ a positive real number. Give the contrapositive for the following implication: ๐ก ∈ ๐ด ⇒ ๐ก ≤ ๐ − ๐ฝ. Answer: The contrapositive of “๐ก ∈ ๐ด ⇒ ๐ก ≤ ๐ − ๐ฝ” is ๐ก > ๐ − ๐ฝ ⇒ ๐ก โ∈ ๐ด. โก 2. Use the ๏ฌeld and order axioms of the real numbers to prove the following. (a) Let ๐, ๐ ∈ ๐ . If ๐๐ = 0, then either ๐ = 0 or ๐ = 0. Proof: Assume that ๐๐ = 0 and ๐ โ= 0. Then, by Field Axiom (๐น9 ), ๐−1 exists. Multiplying ๐๐ = 0 by ๐−1 on both sides yields ๐−1 (๐๐) = ๐−1 ⋅ 0 = 0, from which we get that ๐ = 0, where we have used the Field Axioms (๐น7 ), (๐น9 ) and (๐น10 ). Math 101. Rumbos Spring 2010 2 (b) Let ๐ ∈ โ. If ๐ > 1, then ๐ < ๐2 . Proof: Assume that ๐ > 1. It then follows that ๐ > 0, since 1 > 0. We also have that ๐ − 1 > 0. Consequently, by the Order Axiom (๐3 ), ๐(๐ − 1) > 0. Thus, by the distributive property, ๐2 − ๐ > 0, from which we get that ๐ < ๐2 . 3. Use the completeness axiom of โ to prove that the set of natural numbers is not bounded above. Deduce, therefore, that for any real number, ๐ฅ, there exists a natural number, ๐, such that ๐ฅ < ๐. Proof: Assume by way of contradiction that โ is a bounded above. Then, since โ is not empty, it follows from the completeness axiom that sup(โ) exists. Thus there must be ๐ ∈ โ such that sup(โ) − 1 < ๐. (1) It follows from the inequality in (1) that sup(โ) < ๐ + 1, where ๐ + 1 ∈ โ. This is a contradiction. Therefore, it must be that case that โ not bounded above. Thus, given any real number, ๐ฅ, there must be a natural number, ๐, such that ๐ฅ < ๐. Otherwise, ๐ โฉฝ ๐ฅ for all ๐ ∈ โ, which would say that ๐ฅ is an upper bound for โ. But we just proved that โ is not bounded above.