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26. Mar 2023•0 gefällt mir•2 views

26. Mar 2023•0 gefällt mir•2 views

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S Calculator Problem 1.2 Graph the inverse sine function, y= sin-1 x No Calculator Problem 2.3 Multiple Choice: circle one. B. sin sin equals sin( 43 sin[sin | does not exist. B, equals 3 A. sin sin -1 equals C. Choices A and B are incorrect. B. sin-(sin 2) equals l A. sin (sin2T) does not exist. sin\"(sin 2r) D. sin\"\' (sin2x) equals 0 C. equals 2 B. sin (sin2) equals 2 i. A. sin (sin2) does not exist. sin\"(sin 2) equals D. Choices A, B and C are incorre C. -2 . Solution i) B sin(pi/3) = sqrt(3)/2 ii) D sin^-1 (sin(2*pi)) = sin^-1 (0) = 0 iii) C sin^-1(sin(2)) = pi - 2 because 2>pi/2.

- 1. Show that if A and B are sets, A is uncountable, and A B, then B is uncountable. Solution Suppose that A and B are sets, A is uncountable, and A B. Assume further that B is countable. Then there is a bijective function f from , the set of natural numbers, onto B. Let X = f-1(A) = {n : f(n) A}. Clearly A is nonempty since it is uncountable, and so X is nonempty. Further note that X is not a finite set, because if it were, then f(X) = f(f-1(A)) = A would also be finite since the restriction of f to the set X is a bijection onto its image f(X) = A. We now define a sequence of sets X1, X2, X3, … recursively as follows: 1) Let X1 = X. 2) Given Xj, let nj be the smallest element of Xj. 3) Define Xj + 1 := Xj{nj}. We claim that this process is well-defined. This can be demonstrated with mathematical induction. Since X1 = X is a nonempty subset of , it follows by the well-ordering principle that X has a smallest element n1. Now suppose that Xk is an infinite subset of with a smallest element nk where k is a positive integer. Then Xk + 1 = Xk{nk} is clearly also an infinite subset of , and so it has a smallest element nk + 1 because of the well-ordering principle. This shows that the recursion is well-defined. Now define a function g: A such that g(x) = f(nx). Our goal now is to show that g is one-to-one and onto. First, observe that if a and b are positive integers such that a b, then Xb Xa. This can be proved inductively. Indeed, for if b - a = 0, then clearly Xb = Xa. Now if Xa + k Xa for some positive integer k, then if b - a = k + 1, then Xb = Xa + k + 1 = Xa + k{na + k} Xa + k Xa. To see that g is one-to-one, let p and q be two distinct positive integers. Without loss of generality, suppose p < q. Then Xq Xp + 1 = Xp{np} Xp, and so nq, the smallest of Xq, is also an element of Xp. Thus nq np, but since Xq does not contain np, we must have that nq and np are different. Hence, bearing in mind that f is one-to-one, g(q) = f(nq) f(np) = g(p), and so g is one-to-one. As for g being onto, let y be an arbitrary element of A. Since f is onto, there is a positive integer m X such that f(m) = y. Now let Y = {j X : j < m}, and let r be the cardinality of Y (which is necessarily smaller than m). It is evident that Xr + 1 contains none of the elements of Y but still contains m, and so nr + 1 = m. Therefore g(r + 1) = f(nr + 1) = f(m) = y, and thus g is onto. Consequently, there is a bijection g from onto A, and so A is countable. Yet, A is uncountable, so we arrive at a contradiction. Therefore, B must be countable.