HW6Solutions - Homework 6 Solutions Section 7.1 2. Let X =...

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Homework 6 Solutions Section 7.1 2. Let X = {1, 3, 5} and Y = {a, b, c, d}. Define g: X —> Y by the arrow diagram in problem 2 on page 399. a. Write the domain of g and the co-domain of g. Domain of g = {1, 3, 5}, co-domain of g = {a, b, c, d}. b. Find g(1), g(3), and g(5). g(1) = b, g(3) = b, g(5) = b. c. What is the range g? Range of g = {b}. d. Is 3 and inverse image of a? Is 1 an inverse image of b? No, 3 is not an inverse image of a. Yes, 1 is an inverse image of b. e. What is the inverse image of b? of c? Inverse image of b = {1, 3, 5}. Inverse image of c = Ø. f. Represent g as a set of ordered pairs. g = { (1, b), (3, b), (5, b)}. 13. Let A = {1, 2, 3, 4, 5} and define F as described in problem 13 on page 400. Find the following: b. F(Ø). (Size of null set is zero; zero is even). F(Ø) = 0. d. F({2, 3, 4, 5}). F({2, 3, 4, 5}) = 0. 32. Student C tries to define a function h: Q –> Q by the rule: h(m/n) = m 2 /n with for all integers m and n with n 0. Student D claims that h is not well defined. Justify Student D’s claim. Proof Suppose on the contrary that g is well defined. Then g(1/2) = g(3/6), since 1/2 = 3/6. But g(1/2) = 1 2 /2 = 1/2 and g(3/6) = 3 2 /6 = 9/6 = 3/2, which contradicts g(1/2) = g(3/6). The contradiction was based on the supposition that g is well defined. Therefore the supposition is false; that is, g is not well defined. Section 7.2 8. Let X = {a, b, c} and Y = {w, x, y, z}. Define functions H and K by the arrow diagrams in problem 8 on page 417. a. Is H one-to-one? Why or why not? Is it onto? Why or why not? H is not one-to-one because F(b) = F(c) = y. H is not onto because F(e) x for any element e in X. Also, H is not onto because F(e) z for any element e in X. b. Is K one-to-one? Why or why not? Is it onto? Why or why not?
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K is one-to-one because no two different elements of X are sent by K to the same element of Y. K is not onto because K(e) z for any element e in X. 9. Let X = {1, 2, 3}, Y = {1, 2, 3, 4}, and Z = {1, 2}.
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This note was uploaded on 06/12/2011 for the course MATH 103 taught by Professor Wouters during the Spring '08 term at Wisc Oshkosh.

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HW6Solutions - Homework 6 Solutions Section 7.1 2. Let X =...

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