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Math 3000 Section 003 Intro to Abstract Math Homework 8 Department of Mathematical and Statistical Sciences University of Colorado Denver, Spring 2012 Solutions (April 26, 2012) Please note that these solutions are only suggestions; different answers and proofs are also always possible. • Section 9.1: The Definition of a Function 1. Exercise 9.6: In each of the following, a function fi : Ai → R (1 ≤ i ≤ 5) is defined, where the domain Ai consists of all real numbers x for which fi (x) is defined. In each case, determine the domain Ai and the range of fi . √ x (a) f1 (x) = 1 + x2 (b) f2 (x) = 1 − x1 (c) f3 (x) = 3x − 1 (d) f4 (x) = x3 − 8 (e) f5 (x) = x−3 Solution: (a) dom f1 = R, ran f1 = {y ∈ R : y ≥ 1}; (b) dom f2 = R − {0} = {x ∈ R : x 6= 0}, ran f2 = R − {1} = {y ∈ R : y 6= 1}; (c) dom f3 = {x ∈ R : x ≥ 1/3}, ran f3 = R+ 0 = {y ∈ R : y ≥ 0}; (d) dom f4 = ran f4 = R; (e) dom f5 = R − {3} = {x ∈ R : x 6= 3}, ran f5 = R − {1} = {y ∈ R : y 6= 1}. • Section 9.2: The Set of All Functions from A to B 2. Exercise 9.10: (a) Give an example of two sets A and B such that |B A | = 8. (b) For the sets A and B given in (a), provide an example of an element in B A . Solution: For (a), all examples are either of the form A = {a} and B = {b1 , b2 , . . . , b8 }, or of the form A = {a1 , a2 , a3 } and B = {b1 , b2 }. For (b), we need to give an example for a function from A to B, and in the first case all possible eight functions are of the form f = {(a, bi )} for some 1 ≤ i ≤ 8, and in the second case of the form f = {(a1 , bi ), (a2 , bj ), (a3 , bk )} where 1 ≤ i ≤ 2, 1 ≤ j ≤ 2, and 1 ≤ k ≤ 2. • Section 9.3: One-to-One and Onto Functions 3. Exercise 9.18: Is there a function f : R → R that is onto but not one-to-one? Explain your answer. Solution: The correct answer is yes. We discussed in class that functions f : R → R can be onto but not one-to-one (f (x) = x3 + x2 ) and similarly, one-to-one but not onto (f (x) = ex ), neither one-to-one nor onto (f (x) = x2 or f (x) = sin x), or both one-to-one and onto (bijective and invertible, f (x) = x). • Section 9.4: Bijective Functions 4. Exercise 9.22(b): Let f : Z5 → Z5 be a function defined by f ([a]) = [2a + 3]. Determine whether f is bijective. [If you know what “well-defined” means, also do part (a): Show that f is well-defined.] Solution: If we recall that Z5 stands for the integers modulo 5 whose only elements are the five residue classes [a] modulo 5, a = 0, 1, 2, 3, 4, we can explicitly list the element pairs of the function f described by f ([a]) = [2a + 3]: f = {([0], [3]), ([1], [0]), ([2], [2]), ([3], [4]), ([4], [1])}. Because each residue class in the codomain is the unique image of exactly one residue class in the domain of f , this function is bijective. [For (a), to show that f is well-defined, we need to show that the image of [a] under f is independent of the chosen element in the equivalence class [a]. Hence, let b and c be two elements in [a] so that a ≡ b (mod 5) and a ≡ c (mod 5). Because congruences are an equivalence relation, this implies that b ≡ c (mod 5) also. By the laws of modular arithmetic, we obtain that 2b ≡ 2c (mod 5) and 2b+3 ≡ 2c+3 (mod 5) and thus [2b+3] = [2c+3] showing that f ([b]) = f ([c]). Mission complete.] • Section 9.5: Composition of Functions 5. Exercise 9.28: For nonempty sets A, B, and C, let f : A → B and g : B → C be functions. (a) Prove: If g ◦ f is one-to-one, then f is one-to-one. using as many of the following proof techniques as possible: direct proof, proof by contrapositive, proof by contradiction. Math 3000-003 Intro to Abstract Math Homework 7, UC Denver, Spring 2012 (Solutions) 2 (b) Disprove: If g ◦ f is one-to-one, then g is one-to-one. Solution: For (a), you should have found that all three techniques are possible, as illustrated below. Direct Proof. Let g ◦ f be one-to-one so that a = b whenever (g ◦ f )(a) = (g ◦ f )(b). To show that f is one-to-one, we need to show that a = b also whenever f (a) = f (b). Hence, let f (a) = f (b), then g(f (a)) = g(f (b)) because g is a function and thus a = b because g ◦ f is one-to-one. Proof by Contrapositive. For a proof by contrapositive, we need to show that if f is not one-to-one, then g ◦ f is not one-to-one. Hence, let f be not one-to-one, then there exist two distinct elements a 6= b ∈ A such that f (a) = f (b) ∈ B and thus g(f (a)) = g(f (b)) because g is a function. It follows that (g ◦ f )(a) = (g ◦ f )(b) but a 6= b, showing that g ◦ f is not one-to-one. Proof by Contradiction. Let g ◦ f be one-to-one, so that a = b whenever (g ◦ f )(a) = (g ◦ f )(b), and assume that f is not one-to-one. Then there exists a 6= b ∈ A such that f (a) = f (b) ∈ B and thus g(f (a)) = g(f (b)) because g is a function: contradiction E. For (b), we give a counterexample and let A = C = {1}, B = {1, 2}, f = {(1, 1)} and g = {(1, 1), (2, 1)}. Then g ◦ f = f = {(1, 1)} is one-to-one but g is not. Those of you who “proved” that g ◦ f is oneto-one have usually done the subtle mistake to ignore the full domain of g; namely, it is true that the restriction of g onto the range of f is one-to-one: If g ◦f is one-to-one, then ĝ : ran f → C is one-to-one. • Section 9.6: Inverse Functions 6. Exercise 9.36: Let A = R − {1} and define f : A → A by f (x) = x x−1 for all x ∈ A. (a) Prove that f is bijective. (b) Determine f −1 . (c) Determine f ◦ f ◦ f . Solution: For (a), we need to show that f is one-to-one and onto. For the former, let (x, y) ∈ A × A so that f (x) = f (y). It follows that x/(x − 1) = y/(y − 1) which can be rewritten as x(y − 1) = y(x − 1) and readily reduces to x = y, showing that f is one-to-one. For the latter, let y ∈ A and choose x = y/(y − 1) which is well defined because 1 ∈ / A (in particular, note that x = f (y)). It follows that f (x) = x y/(y − 1) y = = =y x−1 y/(y − 1) − 1 y − (y − 1) showing that f is onto (this also shows that f (x) = f (f (y)) = (f ◦ f )(y) = y and thus that f ◦ f = idA , which we will use later). Together, this shows that f is bijective and completes the proof for (a). Alternatively, we could have quoted the result proven in class that a function f is bijective if and only if its inverse f −1 is a function, and concluded the same result from our solution to part (b): To find f −1 , it suffices to write y = x/(x − 1) and then solve this equation for x = y/(y − 1), so that f −1 : A → A is determined by f −1 (y) = f (y) = y/(y − 1) (note that the work here is basically identical to the work necessary to prove that f is onto in part (a)). Finally, for (c) we already observed in part (a) that f ◦ f = idA so that f ◦ f ◦ f = f ◦ idA = idA ◦f = f is identical to the original function. Or do the work: (x/(x−1)) x (x/(x−1))−1 x−(x−1) x = f (x). (f ◦ f ◦ f )(x) = = = (x/(x−1)) x x−1 −1 −1 (x/(x−1))−1 x−(x−1) • Section 9.7: Permutations 1 2 Determine α ◦ β and β −1 . 7. Exercise 9.38: Let α = 2 3 3 4 4 5 5 1 and β = 1 3 2 5 3 2 4 4 5 1 be permutations in S5 . Math 3000-003 Intro to Abstract Math Homework 7, UC Denver, Spring 2012 (Solutions) 3 Solution: First, to determine α ◦ β (remember that this means to first permute according to β and then according to α), we can sort the columns of αso that all numbers in its first row align with those 3 5 2 4 1 numbers in the second row of β, yielding that α = . Then we merely need to combine 4 1 3 5 2 1 2 3 4 5 the first row of β with the second row of α to find that α ◦ β = (of course, you can 4 1 3 5 2 also figure this out without the tedious sorting procedure). Second, to determine β −1 , we can swap the 1 2 3 4 5 −1 two rows of β and then sort its columns so that its first row is again sorted: β = . 5 3 1 4 2 • Additional Exercises for Chapter 9 8. Exercise 9.60: Let f : A → B be a function. For a subset C of A, the image of C under f is the set f (C) = {f (c) : c ∈ C}. (Thus f (A) is the range of f .) Let A1 , A2 ⊆ A. Prove the following. (a) f (A1 ∪ A2 ) = f (A1 ) ∪ f (A2 ) (b) f (A1 ∩ A2 ) ⊆ f (A1 ) ∩ f (A2 ) (c) If f is one-to-one, then f (A1 ∩ A2 ) = f (A1 ) ∩ f (A2 ). Solution: For (a), we need to show that f (A1 ∪ A2 ) ⊆ f (A1 ) ∪ f (A2 ) and f (A1 ) ∪ f (A2 ) ⊆ f (A1 ∪ A2 ). For the former, let b ∈ f (A1 ∪ A2 ), then there is a ∈ A1 ∪ A2 such that f (a) = b. It follows that a is in A1 or A2 (or both), and thus f (a) ∈ f (A1 ) or f (a) ∈ f (A2 ) showing that b = f (a) ∈ f (A1 ) ∪ f (A2 ). For the reverse, let b ∈ f (A1 ) ∪ f (A2 ) so that b is in f (A1 ) or f (A2 ) (or both). It follows that there is a ∈ A1 or a ∈ A2 such that f (a) = b, and thus a ∈ A1 ∪ A2 showing that b = f (a) ∈ f (A1 ∪ A2 ). For (b), let b ∈ f (A1 ∩ A2 ), then there is a ∈ A1 ∩ A2 such that f (a) = b. It follows that a is in both A1 and A2 , and thus f (a) ∈ f (A1 ) and f (a) ∈ f (A2 ) showing that b = f (a) ∈ f (A1 ) ∩ f (A2 ) also. Finally for part (c), together with part (b) it suffices to show that f (A1 ) ∩ f (A2 ) ⊆ f (A1 ∩ A2 ) if f is one-to-one. Hence, let f be one-to-one and b ∈ f (A1 ) ∩ f (A2 ), then there exists a1 ∈ A1 and a2 ∈ A2 such that f (a1 ) = f (a2 ) = b, and a1 = a2 because f is one-to-one. Denote a = a1 = a2 . It follows that a is an element in both A1 and A2 , and thus a ∈ A1 ∩ A2 showing that b = f (a) ∈ f (A1 ∩ A2 ) also. There will be no homework from Chapter 10 (Cardinalities), but here is a Dino Cartoon: Please let me know if you have any questions, comments, corrections, or remarks.