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Unformatted text preview: ECE 3250 THE POWER OF THE CONTINUUM Fall 2008 My main goal is to demonstrate that R , the set of real numbers, is uncountably infinite. Ill employ a classic proof technique known as Cantors diagonal argument. Invoking Cantors argument requires some other results about the real numbers that are interesting and useful in their own right, and Ill elaborate on those as they arise. Before discussing decimal expansions of real numbers, which will play a crucial role in what follows, Id like to remind you about an extraordinarily useful piece of machinery that youve learned about before. Its the geometric series. Well be talking more formally later on about series (particularly power series), but for now consider this expression: X n =0 n . Here, is some given real or complex number. The official meaning of the series expression is lim N N 1 X n =0 n . Of course, the limits existence is not guaranteed. Observe, however, that (1 ) N 1 X n =0 n = 1 N , so that N 1 X n =0 n = 1 N 1 . If   < 1, the righthand side converges as N to 1 / (1 ), and hence so does the lefthand side. The bottom line is that X n =0 n = 1 1 if   < 1 . Id like now to address decimal expansions of real numbers. Ill focus on the open unit interval (0 , 1) = { x R : 0 < x < 1 } . It turns out that any x (0 , 1) has at least one decimal expansion x = .a 1 a 2 a 3 a 4 a 5 ... , where each a n is a natural number between 0 and 9, inclusive. The expansion means x = X n =1 a n 10 n = lim N N X n =1 a n 10 n . A decimal expansion of x is one way of representing x as the limit of a sequence of rational numbers. If you define q N = N X n =1 a n 10 n , then each q N is rational and the sequence { q N } is a Cauchy sequence (check this for yourself) that converges to x as n . Some decimal expansions terminate; for those expansions, theres a smallest M > such that a n = 0 for n > M . It happens that if x has a terminating decimal expansion, then x has at least one other decimal expansion. To wit, suppose x = .a 1 a 2 ...a M is a terminating expansion for x with a M 6 = 0 and a n = 0 for all n > M . Then a non terminating expansion for x is x = .a 1 a 2 a 3 ...a M 1 ( a M 1)999999 ... , 1 2 where the 9s go on forever. The M th decimal place has a M 1 in it; in other words, you decrement the last nonzero decimal place in x s terminating expansion by 1, and you replace all the trailing zeroes in x s terminating expansion by 9s. As an example, . 183746285647 = . 1837462856469999999999999999999999 ... The geometric series makes this work. Rewrite the second decimal expansion above as a 1 10 1 + a 2 10 2 + + a M 1 10 ( M 1) + a M 10 M 10 M + X n = M +1 9 10 n ....
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 '04
 LIPSON

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