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Continuum

# Continuum - ECE 3250 THE POWER OF THE CONTINUUM Fall 2008...

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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. I’ll employ a classic proof technique known as Cantor’s diagonal argument. Invoking Cantor’s argument requires some other results about the real numbers that are interesting and useful in their own right, and I’ll elaborate on those as they arise. Before discussing decimal expansions of real numbers, which will play a crucial role in what follows, I’d like to remind you about an extraordinarily useful piece of machinery that you’ve learned about before. It’s the geometric series. We’ll be talking more formally later on about series (particularly power series), but for now consider this expression: X n =0 γ n . Here, γ is some given real or complex number. The official meaning of the series expression is lim N →∞ N - 1 X n =0 γ n . Of course, the limit’s existence is not guaranteed. Observe, however, that (1 - γ ) N - 1 X n =0 γ n = 1 - γ N , so that N - 1 X n =0 γ n = 1 - γ N 1 - γ . If | γ | < 1, the right-hand side converges as N → ∞ to 1 / (1 - γ ), and hence so does the left-hand side. The bottom line is that X n =0 γ n = 1 1 - γ if | γ | < 1 . I’d like now to address decimal expansions of real numbers. I’ll 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, there’s a smallest M > 0 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 = 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

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2 where the 9’s 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 9’s. As an example, . 183746285647 = . 1837462856469999999999999999999999 . . . The geometric series makes this work. Re-write 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 . Change the index in the last sum to m = n - ( M + 1) and you get 9 × 10 - ( M +1) X m =0 10 - m = 9 × 10 - ( M +1) 1 1 - 1 / 10 = 10 - M . So the sum of all the 9-terms in the expansion adds up to 10 - M , which cancels the - 10 - M in the expansion, which makes the expansion equal to a 1 10 - 1 + a 2 10 - 2 + · · · + a M - 1 10 - ( M - 1) + a M 10 - M , which is just our first (terminating) expansion for x . The bottom line is that you can expand such an x
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