Homework-3-3

CSC-208 Home Work 3 1. What is a function from X to Y? A function from set X to set Y is a rule that associates each element in X with exactly one element in Y. 2. Define a one-to-one function. Give an example of a one-to-one function. Explain how to use an arrow diagram to determine whether a function is one-to-one. A one-to-one function assigns unique outputs to distinct inputs. For example, f(x) = 3x is a one-to-one function. In an arrow diagram, if no two arrows point to the same element in the codomain, it's one-to-one. 3. Define an onto function. Give an example of a onto function. Explain how to use an arrow diagram to determine whether a function is onto. An onto function covers the entire codomain with its range. For example, the function h(x) = 2x, with the domain and codomain being real numbers, is onto. In an arrow diagram, if every element in the codomain has at least one arrow pointing to it, it's onto.

4. Determine if function f x ( ) = 3 x 2 - 3 x + 1 is one-to-one, onto, or both. Give justification for your answer. The function f(x) = 3x^2 - 3x + 1 is not one-to-one because it's a quadratic function, and quadratic functions are generally not one-to-one. For example, when you expand f(x), you'll get two solutions for x for most values in the codomain, indicating that multiple inputs can map to the same output. 5. Prove that if n is an odd integer then n 2 4 ê ë ê ú û ú = n - 1 2 æ è ç ö ø ÷ n + 1 2 æ è ç ö ø ÷ . Right side of equation {n - 1/2}{n + 1/2} = (2k + 1 - 1/2)(2k + 1 + 1/2) = (2k + 1/2)(2k + 2/2) = (2k + 1/2)(2k + 1) = 4k^2 + 2k + k + 1/2 = 4k^2 + 3k + ½ Left side of equation [n^2/4] = [(2k + 1)^2/4] = (4k^2 + 4k + 1)/4

11. Given t n = 3 n - 2 a. Find t 7 t 7=3 7−2=21−2=19 ⋅ t7=19 b. Find t 100 t 100=3 100−2=300−2=298 ⋅ t 100=298 c. Find t i i = 1 3 å t 1 ∣∣ t =1=3 1−2=3−2=1 ⋅ t 1 ∣ t =2=3 2−2=6−2=4 ⋅ t 1 ∣∣ t =2=3 2−2=6−2=4 ⋅ t 1 ∣ t =3=3 3−2=9−2=7 ⋅ t 1 ∣∣ t =3=3 3−2=9−2=7 ⋅ Σ t =13 t 1=1+4+7=12 Σ t =13 t 1 is equal to 12. d. t i i = 3 7 Õ t 1 ∣∣ i =3=3 ⋅ 3−2=9−2=7 t 1 ∣ t =4=3 ⋅ 4−2=12−2=10 t 1 ∣∣ i =4=3 ⋅ 4−2=12−2=10 t 1 ∣ t =5=3 ⋅ 5−2=15−2=13 t 1 ∣∣ i =5=3 ⋅ 5−2=15−2=13 t 1 ∣ t =6=3 ⋅ 6−2=18−2=16 t 1 ∣∣ i =6=3 ⋅ 6−2=18−2=16 t 1 ∣ t =7=3 ⋅ 7−2=21−2=19 t 1 ∣∣ i =7=3 ⋅ 7−2=21−2=19 ∏ i =37 t 1=7 ⋅ 10 ⋅ 13 ⋅ 16 ⋅ 19

12. A relation R on a set X={1, 2, 3, 4, 5} defined by the rule ( x, y ) ÎR if 3 divides x-y. List the elements of R and the elements of R − 1 . 1. (1, 1): 3 divides (1 - 1) = 0, so (1, 1) is in R. 2. (2, 2): 3 divides (2 - 2) = 0, so (2, 2) is in R. 3. (3, 3): 3 divides (3 - 3) = 0, so (3, 3) is in R. 4. (4, 4): 3 divides (4 - 4) = 0, so (4, 4) is in R. 5. (5, 5): 3 divides (5 - 5) = 0, so (5, 5) is in R. The elements of R and its inverse, R ¹, are the same in this case because all pairs ⁻ in R satisfy the rule, and the inverse of a relation contains the same pairs as the original relation. 13. Given R={(1,1), (2,2), (3,3), (4,4), (5,5), (1,3) , (3,1) } is a relation on the set X={1,2,3,4,5}. Determine if R is and equivalence relation. Show your work. If you decide R is an equivalence relation then list the equivalence classes. R is an equivalence relation because it satisfies the properties of reflexivity and symmetry. The equivalence classes are [1] = {1}, [2] = {2}, [3] = {3}, [4] = {4}, [5] = {5}, and [1, 3] = {1, 3}.

R ={(1,2),(1,4),(3,2),(3,4),(4,4)} 17. How can we quickly determine whether a relation is R is antisymetric by examining the matrix of R relative to some ordering? Look at the main diagonal (where x and y are the same). If any diagonal entry is 1, the relation is not antisymmetric.

18. Given two relations R 1 = { ( 1 ,x ) , ( 1 , y ) , ( 2 , x ) , ( 3 , x ) } and R 2 = { ( x ,b ) , ( y,b ) , ( y ,a ) , ( y ,c ) } with orderings 1,2,3;x,y;abc Answering the following questions. a. Find the matrix A 1 of relation R 1 relative to the orderings. The matrix A for relation R relative to the orderings is: ₁ ₁ A = | 0 1 | ₁ | 1 1 | b. Find the matrix A 2 of relation R 2 relative to the orderings. The matrix A for relation R relative to the orderings is: ₂ ₂ A = | 0 1 0 | ₂ | 1 1 1 | | 0 0 0 | c. Find the matrix product A 1 A 2 . The matrix product A A is: ₁ ₂ A A = | 1 1 0 | ₁ ₂ | 2 2 1 | d. Use the result of part c to find the matrix of the relation R 2 ∨ 1 ¿ ¿ . The matrix of the relation R R is the result from part c: ₂◦ ₁ R R = | 1 1 0 | ₂◦ ₁ | 2 2 1 |