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Math 104: Homework 2 Solutions Math 104: Homework 2 Solutions The following problems are taken from the textbook, and listed according to the same numbering: 6.2 Show that if α and β are Dedekind cuts, then so is α + β = fr1 + r2 : r1 2 α and r2 2 βg. Solution: Let α; β ⊂ Q be Dedekind cuts. To show α + β is a Dedekind cut, we must verify three properties. First, we claim α + β is a nonempty proper subset of Q. To see this, note that each of α and β is nonempty, so we can find r1 2 α, r2 2 β, meaning r1 + r2 2 α + β, so α + β is nonempty. On the other hand, since α and β are Dedekind cuts we must have α 6= Q and β 6= Q, meaning we can find M1 2 Q with M1 62 α and M2 2 Q with M2 62 β. We claim that in fact r1 < M1 for all r1 2 α and r2 < M2 for all r2 2 β. This is because clearly we cannot have r1 = M1 or r2 = M2 by assumption that M1 62 α and M2 62 β. On the other hand, we also can't have r1 > M1 because then property (ii) of Dedekind cuts would require M1 2 α. Similarly we cannot have r2 > M2. So since r1 < M1 for all r1 2 α and r2 < M2 for all r2 2 β we must have r1 + r2 < M1 + M2, and since r1 and r2 were arbitrary this shows that s < M1 + M2 for all s 2 α + β. In particular, this means M1 + M2 62 α + β, so α + β is indeed a proper subset of Q. The second property we must verify is that α + β is \closed downward," i.e. if s1 2 α + β and s2 2 Q with s2 < s1 then s2 2 α + β as well. To see this, let s1 = r1 + r2 2 α + β (i.e. with r1 2 α and r2 2 β) and let s2 2 Q with s2 < s1. Then this means s2 < r1 + r2 so s2 − r2 < r1. Then since α is a Dedekind cut and therefore closed downward, we must have s2 − r2 2 α. Then we must have (s2 − r2) + r2 2 α + β, so indeed s2 2 α + β as desired. The final property we must verify is that α + β has no maximal element. To see this, let s1 2 α + β and we will show there exists some s2 2 α + β with s2 > s1. This is because, if s1 = r1 + r2 for some r1 2 α and r2 2 β then since α has no maximal element we can find 0 0 0 r1 2 α with r1 > r1. Then if we take s2 = r1 + r2 we indeed have s2 2 α + β with s2 > s1, so α + β has no maximal element. 6.3 (a) Show α + 0∗ = α for all Dedekind cuts α. Solution: Let α ⊂ Q be an arbitary Dedekind cut. By definition, the Dedekind cut ∗ ∗ 0 ⊂ Q is the set consisting of all negative rational numbers. Now let s 2 α + 0 , say ∗ s = r1 + r2 with r1 2 α and r2 2 0 . Then since r2 < 0 we have r1 + r2 < r1 so since α ∗ is closed downward we have s = r1 + r2 2 α, showing that α + 0 ⊂ α. On the other hand, let r1 2 α. Then since α has no maximal element we can find ∗ r2 2 α with r2 > r1. This means r1 − r2 < 0 and therefore r1 − r2 2 0 . So then ∗ ∗ ∗ r2 + (r1 − r2) 2 α + 0 , so r1 2 α + 0 , meaning α ⊂ α + 0 as well. Since containment holds in both directions, we have the equality of sets α = α + 0∗. (b) We claimed, without proof, that addition of Dedekind cuts satisfies property A4. Thus if α is a Dedekind cut, there is a Dedekind cut −α such that α + (−α) = 0∗. How would you define −α? Solution: The \naive" approach would be to attempt the definition −α = {−r : r 2 αg. However, a quick sketch of what this would look like on the number line should be enough to convince one that this will not work, since Dedekind cuts are supposed to essentially resemble intervals of the form (−∞; x) and this definition would produce something of the form (−x; 1). However, there is a simple solution to this issue, since (again, as we can see intuitively from a sketch) the naive definition actually happens to produce what 1 looks like the complement of the set we are aiming for { i.e. it produces the set of all rational numbers greater than the place where we are \cutting" the rational line. So we should expect the definition −α = Q n {−r : r 2 αg = fr 2 Q : −r 62 αg to work. There is still one technical problem, however. If the original Dedekind cut α happened to represent a rational number, then its complement Q n α will have a minimum element (equal to the rational number α represents). Under negation, this will unfortunately become a maximal element of −α. However, −α is easy to define in this case, because ∗ in this case we have α = s = fr 2 Q : r < sg for some s 2 Q, so we can just directly ∗ define −α = (−s) = fr 2 Q : r < −sg. While the problem does not require you to verify that this definition is satisfactory, I'll do so here anyway for your edification. Our first task is to show that our definition in fact produces a Dedekind cut. Clearly since we defined −α = (−s)∗ in ∗ the case α = s for some s 2 Q, we have that −α is a Dedekind cut because we already ∗ know all sets of the form q for some q 2 Q are Dedekind cuts, so we focus on the case where α is not rational. To see that −α is nonempty, note that since α is not all of Q we can find r 2 Q with r 62 α. Then we see (since α is not rational), −r 2 −α since r 62 α. On the other hand, we cannot have −α = Q because α is nonempty, and for any r 2 α we have −r 62 −α. To see that −α is closed downward, let r 2 −α and s < r. Then that means −r 62 α, and we also have −s > −r. This means we cannot have −s 2 α because then since α is closed downward that would mean −r 2 α as well. So since −s 62 α we in fact have s 2 −α as desired. To see −α has no maximal element, suppose for the sake of contradiction that it does, say s = max(−α). Then we claim −s = min(Q n α). To see this, let r 2 Q n α. Then since r 62 α we have −r 2 −α, which means −r < s, meaning r > −s, so −s is a lower bound of Q n α. On the other hand we also have since s 2 −α that −s 62 α, meaning −s 2 Q n α, so indeed −s = min(Q n α). However, this means s is the least upper bound for α, i.e. s = sup α, contradicting the assumption that α was not rational. Now that we have verified that our definition of −α indeed produces a Dedekind cut, we still must verify that it satisfies the desired property of being the additive inverse of α, i.e. that α + (−α) = 0∗ as we have defined it. First we will handle the case α = s∗. First ∗ ∗ ∗ ∗ let r 2 α + (−α) = s + (−s) . Then we have r = r1 + r2 where r1 2 s and r2 2 (−s) . ∗ So r1 < s and r2 < −s. But then r1 + r2 < s + −s = 0, so r1 + r2 2 0 , showing s∗ + (−s)∗ ⊆ 0∗. On the other hand, let r 2 0∗. Then s + r=2 < s and −s + r=2 < −s so s + r=2 2 s∗ and −s + r=2 2 (−s)∗. So then r = (s + r=2) + (−s + r=2) 2 s∗ + (−s)∗. So we have shown s∗ + (−s)∗ ⊆ 0∗, completing the proof that s∗ + (−s)∗ = 0∗. Now suppose α is irrational, and let r 2 α + (−α), meaning r = r1 + r2 where r1 2 α and r2 2 −α. Then we claim −r2 > r1. This is because r2 2 −α means −r2 2 Q n α, but if −r2 ≤ r1 we would have −r2 2 α since α is closed downward. So since −r2 > r1 ∗ ∗ we have r1 + r2 < 0, meaning r = r1 + r2 2 0 , showing α + −α ⊆ 0 . On the other hand, let r 2 0∗. Now we claim we can find s 2 α with s − r=2 62 α. This is because if we take an arbitrary q 2 α we can consider the set fq − nr=2 : n 2 Ng, which is unbounded above (by the Archimedean property) and therefore contains some element which is not in α since α itself is bounded above. Then by taking the smallest n for which q − nr=2 62 α we see that q − (n − 1)r=2 gives the desired choice of s. Now if we take r1 = s + r=2 we clearly have r1 2 α since s + r=2 < s. Moreover, if we take r2 = −(s − r=2) = −s + r=2, we have r2 2 −α since −r2 62 α by our choice of s.
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