challset1sol

# Nn 12n 1 find a claim a 012 whole numbers proof a 012

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, ,..., } n,n +1,...,2n +1}. Find A claim: A {0,1,2,...,} = Whole numbers Proof: A {0,1,2,...} Let x A x A for some n = n This means that x {n n 2n , x is a positive integer and x {0,1,2,... }. {0,1,2,... } A . Let n {0,1,2,... }. By the definition of A , A = n n=0 n n=0 n n=0 n n=0 n 0 0 0 0 n n=0 n n 0 U U U U U 1 1 {n, n +1,..., 2n +1}, n A , and hence n A Claim: A Proof: If not, suppose m A . This means that m A for all n. However, m A +1, m + 2, ..., 2(m +1) +1}. This is a contradiction. n n n=0 n n=0 n n=0 n m+1 = ∅ = U I I . . { m 10.

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a x x n . , ,..., }. ), ),..., )}. ) X and Y are finite sets. Suppose f:X Y is one to one. It follows from this that card (f(X)) = card(X). Since f(X) Y , card (f(X)) card (Y). Putting these facts together, card(X) = card(f(X)) card(Y). b. Suppose f:X Y is onto, i.e. f(X) = Y. claim: card (f(X)) card (X). proof: Let X = {x f(X) = {f(x f(x f(x Even if all of the f(x are distinct, card(f(X)) is at most card (X). Hence card (f(X)) card(X). With these facts, it now follows that card(Y) = card(f(X)) card(X). c. Suppose f:X Y is one to one and onto. From part(a), card (X) 1 1 2 n i 2 card(Y) and from part b , card (Y) card (X). Putting these together, card (X) = card (Y). d. Suppose card (X) = card (Y) and f: X Y is one to one. We want to prove that f is onto, i.e. f(X) = Y. Since f:X Y is one to one, card (X) = card(f(X)). By hypothesis, card (X) = card(Y), and since f(X) Y, with card (f(X)) = card (Y), it follows that f(X) can' t be a proper subset, and we must conclude f(X) = Y. Suppose f:X Y is onto, i.e. f(X) = Y and card(X) = card(Y). We want to prove that f:X Y is one to one. Card (f(X)) = card(Y) = card(X). If card(f(X)) = card(X), then the elements of {f(x f(x f(x 1 2 n ), ),..., )} must be distinct. If not then for some i,and j, f(x f(x and card(f(X)) < n = card (X). It now follows that f:X Y is one to one. i j ) ), = 11. It is not hard to see that n must be at least 10, in order to for n < 2 because n for n = 1,2,...,9. From the graph of y = log log for x > 2, so log n if n > 10. Log n) < n because n < 2 (you can prove this by induction). n < n log n) because log (n) > 1 for n 10. nlog n) < nxn = n n We still need to show that n for n 10. It's not easy to do this by induction, so we will use a different method. Since log is an increasing function, it suffices to prove that 3 n 3 2 2 2 2 n 2 2 2 2 3 3 2 , ( ), ( ) ( ) ( ( ( . ( ) < < 2 1 1 2 n n x x x log n n) < log n. Use calculus. Define f(x) = x - 3log show that f(x) is increasing for x 10. f'(x) = 1- 3 x For x 10 , f'(x) 0. 2 3 2 2 ( ) log ( ( ) ( ), ln( ) . = = 3 2 1 2 2 n x
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• Fall '06
• miller
• Natural number, Prime number, positive integer

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