1. Decide for each of the following expressions: Is it a function? If so, f is a function. (i) Domain: R. Codomain: R. Range: R. (iii) Yes surjective.

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1 Homework 2 2/14/2018 SOLUTIONS Exercise Decide for each of the following expressions: Is it a function? If so, (i) what is its domain, codomain, and image? (iii) is it surjective? (ii) is it injective? (iv) is it invertible? (a) f : R R defined by x x 3 f is a function. (i) Domain: R. Codomain: R. Range: R. (ii) Yes injective. (iii) Yes surjective. (iv) Yes invertible. (b) f : R R defined by x x f is not a function, since, for example, 1 1 and 1. (c) f : Z Z Q defined by (a, b) a/b f is not a function since it s not defined on Z {0}. (d) f : R Z Z defined by (r, z) r z f is a function. (i) Domain: R Z. Codomain: Z. Range: Z. (ii) f is not injective. For example, (2, 1) and (1.5, 1) both map to 2. (iii) f is surjective, since for every integer z, (1, z) z. (iv) f is not invertible since its not injective. 1

2 2 f (e) k j i c b a f is a function. (i) Domain: {i, j, k}. Codomain: {a, b, c}. Range: {a, b, c}. (ii) f is injective. (iii) f is surjective. (iv) f is invertible. A B f k c (f) j b f is not a function since j has no image and i has two elements in its image. i a A B 2. For those examples which were not functions in the previous problems, can you restrict the domain and/or codomain to make them functions? i.e. pick a reasonable subset of the given domain/codomain such that the expression is a function on that domain. (b) Some examples: Restrict the domain to R 0, or to R 0, or to {1}, or.... (c) Some examples: Restrict the domain to R Z >0, or to R (Z >0 Z <0, or to { 3} {2}, or.... (f) Restrict the domain to {k} (only possibility). 3. Choose a domain and codomain for the following expressions to make them into functions satisfying (i) f is an injective but not surjective function; (ii) f is a surjective but not injective function; (iii) f is an invertible function; (iv) f is not a function; where f is given by: (a) f(x) = x (i) injective but not surjective: Domain R >0, Codomain R. (ii) surjective but not injective: Domain R, Codomain R 0. (iii) invertible function: Domain R >0, Codomain R >0. (iv) not a function: Domain R >0, Codomain {0}. (b) f(x) = 1/x (i) injective but not surjective: Domain R >0, Codomain R. (ii) surjective but not injective: Not possible within C. (iii) invertible function: Domain R >0, Codomain R >0.

3 3 (iv) not a function: Domain R, Codomain anything. (c) f(x) = 6 (i) injective but not surjective: Domain {1}, Codomain R. (ii) surjective but not injective: Domain {1, 2}, Codomain {6}. (iii) invertible function: Domain {1}, Codomain {6}. (iv) not a function: Domain {1, 2}, Codomain {0}. (d) f(x, y) = xy (i) injective but not surjective: Domain Z >0 {1}, Codomain R. (ii) surjective but not injective: Domain Z >0 {0}, Codomain {0}. (iii) invertible function: Domain {0} {0}, Codomain {0}. (iv) not a function: Domain {1} {1}, Codomain {±1}. (e) f is determined by the set {(1, 5), (2, 3), (1, 10), (3, 3)}. (i) injective but not surjective: Domain {1}, Codomain {5, 6}. (ii) surjective but not injective: Domain {2, 3}, Codomain { 3}. (iii) invertible function: Domain {1}, Codomain {5}. (iv) not a function: Domain {5}, Codomain anything. 4. Let f be a function from the set A to the set B. Let S and T be subsets of A. Show that (a) f(s T ) = f(s) f(t ): Proof. First we show that f(s T ) f(s) f(t ): If x f(s T ), then x = f(a) for a S T. So either x = f(a) for a S, or x = f(a) for a T. Thus x f(s) f(t ), implying that f(s T ) f(s) f(t ). Next, we show that f(s T ) f(s) f(t ): If x f(s) f(t ), then x f(s) f(s T ) or x f(t ) f(s T ). So x f(s T ). Thus f(s T ) f(s) f(t ). So f(s T ) = f(s) f(t ). (b) f(s T ) f(s) f(t ). Proof. If x f(s T ) then x = f(a) for a S and a T. So x f(s) and x f(t ), implying x f(s) f(t ). Thus f(s T ) f(s) f(t ). Can you think of an example where the inclusion in part (b) could be proper (not equal)? and Example: Let f : {a, b} {x} be the function f(a) = f(b) = x. With S = {a} and T = {b}, f(s T ) = f( ) = f(s) f(t ) = {x} {x} = {x}. Exercise 7. Let f : A B and g : B C be functions. Draw some pictures (like our previous example) and make some conjectures about the following questions.

4 4 1. Is g f always a function? Yes. 2. What are the conditions on f, g, A, B, and/or C for g f to be surjective? It must be the case that g is surjective. Then for each c C, f(a) g 1 (c) must be nonempty. 3. What are the conditions on f, g, A, B, and/or C for g f to be injective? It must be the case that f is injective, and that g restricted to f(a) must also be injective. 4. What are the conditions on f, g, A, B, and/or C for g f to be bijective? In this case, f must be injective, and g restricted to f(a) must be bijective. 5. What are the conditions on f, g, A, B, and/or C for f g to be a function? In order for f to be defined on g(b), we must have g(b) A 6. If g and g f are injective, is it necessarily true that f is injective? Yes, as we guessed before, for g f to be injective at all, f must be injective. 7. If g and g f are surjective, is it necessarily true that f is surjective? No. Exercise 8. Show that if both f : A B and g : B C are surjective functions, then g f is also surjective. Proof. Pick any c C. Since g is a surjective function, there is some b B such that g(b) = c. And since f is a surjective function, there is some a A such that f(a) = b. So g(f(a)) = g(b) = c, so that c (g f)(a). Since this was true for any c C, we have C (g f)(a). Therefore g f is surjective. Exercise Decide if the following sequences are arithmetic, geometric, or neither. If it s arithmetic or geometric, find a formula expressing the nth term of the sequence. (a) 5, 1, 3, 7, 11,... Check: 4 = 1 5 = 3 1 = 7 ( 3) = 11 ( 7) =, so the sequence is arithmetic, given by a n = 4n + 5 (starting at a 0 ). (b) 1, 4, 9, 16, 25,... Check: the differences and ratios are not constant ( and 4/1 9/4), so this sequence is neither arithmetic nor geometric. (c) 3, 9, 27, 81, 243,... Check: 3 = 9/3 = 27/9 = 81/27 = 243/81 =, so this sequence is geometric, given by a n = 3 3 n (starting at a 0 ). (d) The sequence satisfying a 0 = 2 and a n = 1 2 a n 1. This sequence starts out like 2, 1, 1/2, 1/4, 1/8,.... Check: a n /a n 1 = 1 2 a n 1/a n 1 = 1 2. So this sequence is geometric with a n = 2(1/2) n (starting at a 0 ).

5 (e) The sequence satisfying a 0 = 1 and a n = a n This sequence starts out like 1, 4, 9, 14,.... Check: a n a n 1 = a n a n 1 = 5, so this sequence is arithmetic with a n = 1 + 5n (starting at a 0 ). (f) The sequence satisfying a 0 = 1, a 1 = 1 and a n = a n 2 a n 1. This sequence starts out like 1, 1, 1, 1, 1, 1, 1, 1,.... Check: a n /a n 1 = 1/ 1 or 1/1, and is either way the constant 1. So this sequence is geometric, given by a n = ( 1) n (starting from a 0 ). 2. For each of the following, try to find a closed formula. (a) The sequence satisfying a 0 = 2 and a n = 1 2 a n 1. a n = 1 2 a n 1 for n = 0, 1, 2,.... (see part (1)(d)) (b) The sequence satisfying a 0 = 1 and a n = a n a n = a n for n = 0, 1, 2,.... (see part (1)(e)) (c) The sequence satisfying a 0 = 2 and a n = a n 1. This sequence starts out like 2, 2, 2, 2, 2,..., so that it is a geometric sequence with a n = 2( 1) n for n = 0, 1, 2, Exercise 10. (1) Calculate the following: 8 (a) (1 + ( 1) j ) = = 10 (b) j=0 2 j= 1 i=2 j= 1 i=2 3 (2i + 3j) (2i + 3j) = (( j) + ( j)) j= 1 2 = (10 + 6j) j= 1 = (10 + 6( 1)) + (10 + 6(0)) + (10 + 6(1)) + (10 + 6(2)) = ( ) = 52 (c) (d) 2 = = 6 j { 1,4,15} j j {z Z z 2} Since {z Z z 2} = { 2, 1, 0, 1, 2}, we have j = = 0 j {z Z z 2}

6 6 (e) 5 (a n a n 1 ) where a n = n! n=2 5 (a n a n 1 ) = (a 2 a 1 ) + (a 3 a 2 ) + (a 4 a 3 ) + (a 5 a 4 ) = a 5 a 1 = 5! 1! = 5! 1 n=2 (f) 2000 i = /2 = 2, 001, 000 (2) Use partial sums to explain why, for any sequence a 0, a 1,..., a n, that n a i a i 1 = a n a 0. [Let S n = n a i a i 1. Write out S 1, S 2, S 3, and so on until you see the pattern. Then use the fact that S n = S n 1 + (a n a n 1 ).] Let S n = n a i a i 1, so that S 1 = a 1 a 0 S 2 = a 1 a 0 + a 2 a 1 = a 2 a 0 S 3 = a 1 a 0 + a 2 a 1 + a 3 a 2 = a 3 a 0. Note that S n = S n 1 + (a n a n 1 ). So starting with S 1 = a 1, adding the next term to S n 1 to get S n always cancels the highest index term of S n 1 (a n 1 ) and adds a n. What s left if a n a #2 Determine whether f is a function from Z to R if (a) f(n) = ±n Ans: no, since 1 ±1 (b) f(n) = n Ans: Yes, if we always take the positive root, no if not. (c) f(n) = 1/(n 2 4) Ans: No, since f(2) is not defined. 2.3 #6 Find the domain and range of these functions. (a) the function that assigns to each pair of positive integers the first integer of the pair. Ans: Domain Z >0 Z >0, Range Z >0. (b) the function that assigns to each positive integer its largest decimal digit. Ans: Domain Z >0, Range {0}. (d) the function that assigns to each positive integer the largest integer not exceeding the square root of the integer. Ans: Domain Z >0 Z >0, Range Z > #16 Consider these functions from the set of students in a discrete math class. Under what conditions is the function one-to-one if it assigns to a student his of her (a) mobile phone number Ans: If no two students share a phone

7 7 (b) student identification number Ans: If the registrar hasn t messed up (c) final grade in the class Ans: If no two students get the same final grade (small class!) (d) home town Ans: If no two students come from the same town 2.3 #32 Let f(x) = 2x where the domain is the set of real numbers. (a) f(z) = { even integers } (b) f(n) = { positive even integers } (c) f(r) = R 2.3 #42 Let f be the function from R to R defined by f(x) = x 2. (a) f 1 ({1}) = {±1} (b) f 1 ({x 0 < x < 1}) = {x 0 < x < 1} {x 0 > x > 1} (c) f 1 ({x x > 4}) = {x x > 2} {x x < 2} 2.4 #4 What are the terms a 0, a 1, a 2, and a 3 of the sequence {a n }, where a n equals (a) ( 2) n : a 0 = 1, a 1 = 2, a 2 = 4, and a 3 = 8. (b) 3: a 0 = 3, a 1 = 3, a 2 = 3, and a 3 = 3. (c) n : a 0 = 8, a 1 = 11, a 2 = 23, and a 3 = 71. (d) 2 n + (?2) n : a 0 = 2, a 1 = 0, a 2 = 8, and a 3 = #8 Find at least three different sequences beginning with the terms 3, 5, 7 whose terms are generated by a simple formula or rule. a n = 2n + 3 a n = a n Some examples, each beginning with a 0 : n 2, a 0 = 3 a n = a n 1 + 2f n, a 0 = 3, where f n is the Fibbonacci sequence starting with f 0 = 0, f 1 = 1 The day value of dates that MWFs fall on, beginning with August 3rd, #10 (b,c) Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions. (b) a n = a n 1 a n 2, a 0 = 2, a 1 = 1 2, 1, 3, 2, 1, 3,... (c) a n = 3a 2 n 1, a 0 = 1 1, 3, 3 3, 3 7, 3 15, #12 (b,c)show that the sequence {a n } is a solution of the recurrence relation a n = 3a n 1 + 4a n 2 if (b) a n = 1: Check: a n = 1 = 3(1) + 4(1) = 3a n 1 + 4a n 2.

8 8 (c) a n = ( 4) n : Check: 3a n 1 + 4a n 2 = 3 ( 4) n ( 4) n 2 = 3 ( 4) n 1 ( 4) n 1 = ( 4) n = a n. 2.4 #14 (c,d,e) For each of these sequences find a recurrence relation satisfied by this sequence. (The answers are not unique because there are infinitely many different recurrence relations satisfied by any sequence.) (c) a n = 2n + 3 For example, a 0 = 3 and a n = a n (d) a n = 5 n For example, a 0 = 1 and a n = 5a n 1. (e) a n = n 2 For example, a 0 = 0 and a n = a n 1 +2n 1 (since (n 1) 2 = n 2 2n+1). 2.4 #43 (a,b,c) What are the values of the following products? (a) (b) (c) 10 i=0 i = = 0 8 i = = 1680 i=5 100 ( 1) i 100 ( 1) i = ( 1)( 1) 2 ( 1) 3 ( 1) #44 Express n! using product notation. = ( 1) = ( 1) /2 = ( 1) 5050 = 1 n! = n i.