ALTERNATIVE SECTION 12.2 SUPPLEMENT TO BECK AND GEOGHEGAN S ART OF PROOF

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1 ALTERNATIVE SECTION 12.2 SUPPLEMENT TO BECK AND GEOGHEGAN S ART OF PROOF MICHAEL P. COHEN Remark. The purpose of these notes is to serve as an alternative Section 12.2 for Beck and Geoghegan s Art of Proof. I am using these notes for my own course (Spring 2018 Intro to Abstract Mathematics) because I would rather establish decimal expansions using supremums as the main tool, instead of limits. We will establish the same theorems with the same numbering as in 12.2 of Art of Proof, and many definitions are taken word-for-word from the book, but the content in these notes may be taught without covering Chapter 10 of Art of Proof and without defining a limit. I don t mind if other instructors want to use these notes for similar purposes. Definition. A nonnegative decimal is a sequence (m, d 1, d 2, d 3,...) where m 0 is an integer and each d n is a digit, that is, an integer between 0 and. Given such a nonnegative decimal, for any natural number n, we may define the finite sum m.d 1 d 2...d n = m + d j 10 j. j=0 If the set {m.d 1 d 2...d n : n N} is bounded above, then we set x = sup{m.d 1 d 2...d n : n N} and we say that m.d 1 d 2 d 3... represents the real number x, or else that m.d 1 d 2 d 3... is the decimal expansion of x. Traditionally, we write x = m.d 1 d 2 d 3... A decimal expansion of a negative real number x is defined by placing a minus sign in front of a decimal expansion of the positive number x. Since the decimal expansion is defined as the supremum of some set, the first question we need to consider is, is a decimal expansion always a well-defined concept? In other words, do we know that the set {m.d 1 d 2...d n : n N} is necessarily bounded above? The next few theorems will address this question, and perform some important and useful computations along the way. Lemma. The real number 1 is represented by 0.d 1d 2 d 3... where d n = 1 for every n N. In other words, = 1. Proof. Denote A = {0.d 1 d 2...d n : n N} = 1 10 j : n N. Our job is to consider the set A. We must first show A is bounded, whence sup A exists and the expression makes sense; secondly, we need to show that sup A = 1. By Proposition 4.13, for any n N we have 1

2 2 MICHAEL P. COHEN 0.d 1 d 2...d n = 1 10 j ( ) j = j=0 = 1 (1/10)n (1/10) = 10 ( 1 1 ) 10 n+1 1 = 1 ( 10 1 ) 10 n 1 = 1 ( 1 1 ) From the above calculation, it is plain to see that the set A is bounded above by the number 1. Therefore by the Completeness Axiom 8.52, the decimal expansion is defined, and moreover we have: Suppose for the sake of a contradiction that < 1. Set ɛ = ( ), so ɛ is a strictly positive real number (perhaps a very small one). By Proposition 10.4, there exists N N so large that 1 N < ɛ. Since 10N 1 > N, we also have < 1 10 N N < ɛ, and therefore 0.d 1 d 2...d N = 1 ( 1 1 ) 10 N > 1 (1 ɛ) = 1 ( ) = which is a contradiction because 0.d 1 d 2...d N A, and is defined to be an upper bound for A. This contradiction ensures that = 1, as claimed. Corollary. Let D be any fixed digit. Then the real number D d n = D for every n N. In other words, is represented by 0.d 1d 2 d 3... where 0.DDDDD... = D. Proof. Set A 1 = { : there are n many 1 s} and A D = {0.DDD...D : there are n many D s}. The previous lemma ensures that 1 = sup A 1. Our goal in this proof is to show that D = sup A D. Consider x A D. It is clear that 1 D x A 1, and therefore 1 D x 1 = sup A 1. But then x D. Since x was taken arbitrarily from A D, we have shown that D is an upper bound for

3 ALTERNATIVE SECTION 12.2 SUPPLEMENT TO BECK AND GEOGHEGAN S ART OF PROOF 3 A D. On the other hand, let M be an arbitrary upper bound for A D. If y A 1, then Dy A D ; therefore Dy M; therefore y M D. This means M D is an upper bound for A 1. It follows then that 1 M D, since 1 = sup A 1. Thus, D M. Since M was arbitrary, D is the least upper bound for A D. Corollary = 1. Proposition (Proposition 12.4). Let m Z 0 and let (d k ) k=1 be a sequence of digits. Then the set {m.d 1 d 2...d n : n N} is bounded above. In fact, it is bounded above by m + 1. Proof. Again, denote A = {m.d 1 d 2...d n : n N} = {m + d j 10 j : n N}. By the previous corollary, for any n N, we have m + d j 10 j m + 10 j m + 1. Therefore A is bounded above by m + 1. Corollary. For every nonnegative decimal (m, d 1, d 2, d 3,...), there exists x R 0 such that x = m.d 1 d 2 d 3... In other words, every decimal expansion defines a real number. Proof. Because the set A = {m.d 1 d 2...d n : n N} is bounded above, the Completeness Axiom 8.52 implies that x = sup A is well-defined. Proposition (Proposition 12.5). Let n N. Suppose (d k ) k=1 d k = 0 whenever 1 k < n. Then is a sequence of digits such that 0.d 1 d 2...d n 1 d n d n+1... = d n d n n 1. Theorem (Theorem 12.6). Every real number has a decimal expansion. Proof. We will prove this theorem for nonnegative real numbers. The general case follows easily: if x < 0, just get a decimal expansion of x and precede it with a minus sign. So let x 0. We will recursively define a decimal expansion m.d 1 d 2 d 3... of x. Let m be the smallest integer for which x < m + 1. Then m x (otherwise m was not chosen minimally). Next, let d 1 be the smallest element of { k Z 0 : x < m + k + 1 }. 10 Then 0 d 1 (otherwise m was not chosen minimally) and m + d1 10 x (otherwise d 1 was not chosen minimally). In summary, m + d 1 10 x < m + d Now we define the remaining digits recursively. Assume that d 1, d 2,..., d n have been defined and satisfy the inequalities

4 4 MICHAEL P. COHEN m + Then let d n be the smallest element of Then n 1 d j 10 j x < m + d j 10 j + d n + 1 k Z 0 : x < m + d j 10 j + k n+1 n+1 m + d j 10 j x < m + d j 10 j + d n n+1 which allows the recursive definition to continue. Thus we obtain a sequence (m, d 1, d 2,...), and we set y = m.d 1 d 2 d 3... It remains for us to verify that x = y. To see this, first observe that our recursive definition ensures that for every n N, we have 0 x m + d j 10 j < 1 Since x m.d 1 d 2...d n 0 for every n N, we have that x is an upper bound for A = {m.d 1 d 2...d n : n N}. Therefore x y = sup A. Suppose for a contradiction that y < x, and set ɛ = x y > 0. By Proposition 10.4, there exists N N so large that 1 N < ɛ. But then, for this N we have N x m + d j 10 j < 1 10 N < 1 N < ɛ = y x, which implies y < m.d 1 d 2 d 3...d N, contradicting the fact that y is an upper bound for A. Thus we must have x = y, which completes the proof. Proposition (Proposition 12.7). Let m.d 1 d 2 d 3... represent 1 R. Then either m = 1 and every d k equals 0, or m = 0 and every d k =. In other words, 1 can be represented by or 0..., and by no other decimal. Proof. It is easy to check that represents 1, and we know 0... represents 1 by a previous corollary. We must show that there are no other decimal expansions of 1. So let m.d 1 d 2 d 3... be some other decimal expansion. We will show it never equals 1. If m 2 or m 1 then this expansion differs from 1 be at least 1, so we just have to consider the cases m = 0 and m = 1. Case 1: Suppose m = 0. Let N be the smallest subscript l 1 for which d l <. (There has to be at least one such subscript, or else the expansion would just be 0...) Our choice of N implies that d N 8, and d j = for all j < N. Therefore, for every n > N we have (using Proposition 12.5):

5 ALTERNATIVE SECTION 12.2 SUPPLEMENT TO BECK AND GEOGHEGAN S ART OF PROOF 5 m.d 1 d 2 d 3...d N...d n = N 1 N 1 10 j + d N 10 N + j=n+1 10 j N + 10 N N 10 j. d j 10 j Since this is true for all sufficiently large n, we have shown that m.d 1 d 2 d (where N many s appear in the latter expansion). So m.d 1 d 2 d Case 2: Suppose m = 1. Let N now be the smallest subscript l 1 for which d l > 0. (It exists, otherwise the expansion is ) Then m.d 1 d 2 d 3... m.d 1 d 2...d N 1 d N = d N > 1. The above proof contains all the necessary ingredients for the following more general theorem. Theorem (Theorem 12.8). Let m.d 1 d 2 d 3... and n.e 1 e 2 e 3... be two different decimal expansions of the same positive real number. (1) If m < n, then n = m + 1, every e k is 0 and every d k is. (2) If m = n, let N denote the smallest subscript where d N e N. If d N < e N then e N = d N + 1, e j = 0 for all j > N, and d j = for all j > N. Corollary (Corollary 12.). If r R has two different decimal expansions then r is a rational number. Remark (Major Takeaways). The major takeaways from this section are the following: (1) Every decimal expansion defines a real number. (2) Every real number has some decimal expansion. (3) Every real number has at most two distinct decimal expansions.