Basic Geometry, Third Edition by George D. Birkhoff, Ralph Beatley

By George D. Birkhoff, Ralph Beatley

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Thus in Problems 1 and 2 we chose the particular examples 1 3 and 1000 in place of the of 100 ‘general’ positive number ". ε X n –ε Note that here X need not be an integer; any appropriate real number will serve. X does NOT depend on n, but does in general depend on ". 2: Sequences 44 n Þ For example, consider the sequence {an} where an ¼ ðÀ1 n2 ; n ¼ 1; 2; . .. This sequence is null. For   ðÀ1Þn  1 jan j ¼  2  ¼ 2 ; n n so that, if we make the choice X ¼ p1ffiffi", it is certainly true that jan j < "; for all n > X: In terms of the " À X game, this simply means that whatever choice of " is made by player A, player B can always win by making the choice X ¼ p1ffiffin !

In fact, (0, 2) has no minimum element. Definitions A set E  R is bounded below if there is a real number, m say, called a lower bound of E, such that m x; for all x 2 E: If the lower bound m belongs to E, then m is called the minimum element of E, denoted by min E. Geometrically, the set E is bounded below by m if no point of E lies to the left of m on the real line. Problem 2 Determine which of the following sets are bounded below, and which have a minimum element: (a) E1 ¼ (À1, 1]; (b) E2 ¼ f1 À 1n : n ¼ 1; 2; .

N 1 Prove the inequality 1 þ 1n ! 52 À 2n ; for n ! 1. Hint: consider the first three terms in the binomial expansion. Example 7 Solution Prove that 2n ! n2, for n ! 4. Let P(n) be the statement PðnÞ : 2n ! n2 : First we show that P(4) is true: 24 ! 42. STEP 1 Since 24 ¼ 16 and 42 ¼ 16, P(4) is certainly true. STEP 2 We now assume that P(k) holds for some k ! 4, and deduce that P(k þ 1) is then true. So, we are assuming that 2k ! k2. Multiplying this inequality by 2 we get 2kþ1 ! 2k2 ; so it is therefore sufficient for our purposes to prove that 2k2 !

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