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By DAVID ALEXANDER BRANNAN

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9 for a suitably large number of digits, or it can be 12 ðM 0 þ 2Þ since M 0 < 12 ðM 0 þ 2Þ < 2: 1: Numbers 26 Strategy Given a subset E of R, to show that M is the least upper bound, or supremum, of E, check that: 1. x M, for all x 2 E; 0 0 2. if M < M, then there is some x 2 E such that x > M . GUESS CHECK the value of M, then parts 1 and 2. Notice that, if M is an upper bound of E and M 2 E, then part 2 is automatically satisfied, and so M ¼ sup E ¼ max E. Example 4 Determine the least upper bound of E ¼ f1 À n12 : n ¼ 1; 2; .

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 !

This follows from the fact that, for any x 2 E, we have inf E x sup E. 2. For any bounded interval I of R, let a be its left end-point and b its right end-point. Then inf I ¼ a and sup I ¼ b. Least upper bounds and greatest lower bounds of functions Similar terminology applies to bounds for functions. Definitions Let ƒ be a function defined on an interval I  R. Then: A real number M is the least upper bound, or supremum, of ƒ on I if:  1. M is an upper bound of ƒ(I); 2. if M0 < M, then M0 is not an upper bound of ƒ(I).

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