By DAVID ALEXANDER BRANNAN

ISBN-10: 0511348576

ISBN-13: 9780511348570

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Example text

To do this, we must find an element x in [0, 2) which is greater than M0 . But, if M0 < 2, then there is a real number x such that 0 M

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; . g. Solution We guess that the least upper bound of E is M ¼ 1. Certainly, 1 is an upper bound of E, since 1 1 À 2 1; for n ¼ 1; 2; . : 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.

And x, y 2 Q , then ax ay ¼ axþ y. and x, y 2 Q , then ðax Þy ¼ axy . If a > 0 If a > 0 If x and y are integers, these laws actually hold for all non-zero real numbers a and b. However, if x and y are not integers, then we must have a, b > 0. For 1 example, ðÀ1Þ2 is not defined as a real number. m However, if a is a negative real number, then a n can be defined whenever m 2 Z, n 2 N and mn is reduced to its lowest terms with n odd, as follows m an ¼ Àp ffiffiffiÁm n a : Finally, you may have wondered why we did not mention that each positive number has two nth roots when n is even.

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A First Course in Mathematical Analysis by DAVID ALEXANDER BRANNAN


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