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This also means that E3 cannot have a maximum element. Problem 1 Sketch the following sets, and determine which are bounded above, and which have a maximum element: (a) E1 ¼ (À1, 1]; (b) E2 ¼ f1 À 1n : n ¼ 1; 2; . g; 2 (c) E3 ¼ {n : n ¼ 1, 2, . }. 99. . 9 or y ¼ 12(x þ 2). 2 is not a maximum element, since 2 2 = E1. 1: Numbers 24 Similarly, we define lower bounds. For example, the interval (0, 2) is bounded below by 0, since 0 for all x 2 ð0; 2Þ: x; However, 0 does not belong to (0, 2), and so 0 is not a minimum element of (0, 2).

Ever since learning to count you have been familiar with the sequence of natural numbers 1; 2; 3; 4; 5; 6; . : You have also encountered many other sequences of numbers, such as 2; 4; 6; 8; 10; 12; . ; 1 1 1 1 1 1 ; ; ; ; ; ; . : 2 4 8 16 32 64 We begin our study of sequences with a definition and some notation. Definition A sequence is an unending list of real numbers a1 ; a2 ; a3 ; . : The real number an is called the nth term of the sequence, and the sequence is denoted by fan g: In each of the sequences above, we wrote down the first few terms and left you to assume that subsequent terms were obtained by continuing the pattern in an obvious way.

You saw in part (b) of 1 Example 6 that 2n 1 þ 1n : Our second inequality is of considerable use in various branches of pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffi Analysis. In Problem 3 you proved that aþb a2 þ b2 , for a, b 2 R. We 2 can rewrite this inequality in the equivalent form ða þ bÞ2 2ða2 þ b2 Þ or ða þ bÞ2 ða2 þ b2 Þð12 þ 12 Þ. The Cauchy–Schwarz Inequality is a generalisation of this result to 2n real numbers. Theorem 2 Cauchy–Schwarz Inequality For any real numbers a1 ; a2 ; . ; an and b1 ; b2 ; . ; bn ; we have ða1 b1 þ a2 b2 þ Á Á Á þ an bn Þ2 À 2 ÁÀ Á a1 þ a22 þ Á Á Á þ a2n b21 þ b22 þ Á Á Á þ b2n : We give the proof of Theorem 2 at the end of the sub-section.

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