A First Course in Mathematical Analysis by DAVID ALEXANDER BRANNAN


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An Þ n ða1 þ a2 þ Á Á Á þan Þ=n 1: (3) Now, replacing each term ai by lai for any non-zero number l does not alter the left-hand side of the inequality (3). It follows that it is sufficient to prove the inequality (2) in the special case when the product of the terms ai is 1. Hence it is sufficient to prove the following statement P(n) for each natural number n: P(n): For any positive real numbers ai with a1a2 . . an ¼ 1, then a1 þ a2 þ Á Á Á þ an ! n. First, the statement P(1) is obviously true.

N To check part 2 of the strategy, we need to show that, if M0 < 1, then there is some natural number n such that 1 1 À 2 > M0: (1) n However 1 1 1 À 2 > M0 , 1 À M0 > 2 n n 1 , < n2 ðsince 1 À M 0 > 0Þ 1rÀffiffiffiffiffiffiffiffiffiffiffiffiffiffi M0 1 1 , >0 0Þ: We can certainly choose n so that this final inequality holds, by the Archimedean Property of R, and so we can choose n so that inequality (1) holds. & Hence 1 is the least upper bound of E. That is, 1 ¼ sup E. Remark Although we used double-headed arrows in this solution, the actual proof required only the implications going from right to left.

The existence of roots Just as we usually take for granted the basic arithmetical operations with real numbers, so we usually assume that,pgiven any positive real number a, there is ffiffiffi a unique positive real number b ¼ a such that b2 ¼ a. We now discuss the justification for this assumption. First, here is a geometrical justification. Given line segments of lengths 1 and a, we can construct a semi-circle with diameter a þ 1 as shown. For each positive integer pffiffiffin, we can also construct n as follows: 1 1 1 1 b 5 6 a 1 4 3 1 2 7 Using similar triangles, we see that a b ¼ ; b 1 and so 1 1 b2 ¼ a: This shows that there should be a positive real number b such that b2 ¼ a, so that the length of the vertical line segment pffiffiffi pffiffiffi in the figure can be described exactly by the expression a.

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