By George D. Birkhoff, Ralph Beatley

A hugely advised high-school textual content via eminent students

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Describing cornerstones of arithmetic, this uncomplicated textbook provides a unified method of algebra and geometry. It covers the tips of complicated numbers, scalar and vector items, determinants, linear algebra, workforce concept, permutation teams, symmetry teams and points of geometry together with teams of isometries, rotations, and round geometry.

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**Example text**

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 !