hw7solutions23b

# hw7solutions23b - Anna Medvedovsky [email protected] Math...

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Anna Medvedovsky Math 23b / Spring 2009 HW #7 solutions 1. Let y > x . (a) Prove that there is a natural number n such that n ( y - x ) > 1 . The numbers y - x and 1 are both positive reals, so the archimedean property tells us that there exists an n N satisfying n ( y - x ) > 1. (b) Prove that there is an integer m such that m > nx m - 1 . We’ll be using the well-ordering property to ﬁnd m . But the classic version of WOP (any nonempty subset of the positive integers has a least element) will be tricky to use here because we don’t know that nx is positive. There are a few ways to deal with this problem, but the one we’ll take is to prove a lemma. Lemma: If S is a nonempty subset of Z and S is bounded below, then S has a least element. Proof: Let r be a lower bound for S . First we need to ﬁnd an integral lower bound b for S . If r is nonnegative, set b = 0 Z . Then b r , so b is also a lower bound for S : for any s S , we have s r b . And if r is negative, use the archimedean property to ﬁnd a natural number z such that z · 1 > - r , and set b = - z . In this case b < r , so that b is again a lower bound for S . Note that b Z . Now deﬁne a set T = { s - b : s S } by translating S by b . I claim that T is a nonempty subset of the positive integers. First oﬀ, since S Z and b Z we know that every element of T is an integer. Moreover, since b is a lower bound for S , we have s - b > 0 for all s S , so that in fact every element of T is positive. Finally, since S is nonempty, T is nonempty also. Since T is a nonempty subset of the positive integers, we know by well-ordering that T has a least element; call it t . Then t + b S , and I claim that t + b is the least element of S . Indeed, for every s S , we have s - b T , so that s - b t . Therefore s t + b for all s S , and t + b is the least element of S , as claimed. Back to the problem at hand: ﬁnding m such that m > nx m - 1. Let S = { s Z : s > nx } . Then S is a subset of Z , and S is clearly bounded below by nx . So all we need to do to WOP this S is to show that S is nonempty. Again we take cases. If nx 0, then 1 > nx , so 1 S . And if nx is positive, then using the archimedean property we can ﬁnd a natural number s such that s · 1 > nx . Then s S . In either case, S is nonempty, so it has a least element, by the lemma above. Let

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hw7solutions23b - Anna Medvedovsky [email protected] Math...

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