Discrete Mathematics

Introductory Discrete Mathematics Week 3 Homework
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Determine whether each of the following relations is a function with domain {1, 2, 3}. For any relation that is not a function, explain why it isn’t.
f = {(1, 1), (2, 1), (3, 1), (4, 1), (3,3)}
Solution: This is not a function there are two different pairs of the form (3,-) in f
f = {(1, 2), (2, 3), (4, 2)}
Solution: With the given domain, this is not a function since a pair of the form (3,-) does not exist.
f = {(1, 1), (1,2), (1,3), (1,4)}
Solution: This is not a function since more than one pair of the form (1,-) exists.
e. f = {(1, 4), (2, 3), (3,2), (4,1)}
Solution: This is function since the pairs of the (1,-) and (3,-) exists.
Suppose A is the set of students currently registered at the University of Calgary, B is the set of professors at the University of Calgary, and C is the set of courses currently being offered at the University of Calgary. Under what conditions is each of the following function?
{(a, c) | a is taking a course from b}
Solution: Not a function. Each student is taking courses handled by b1 and b2, then the sets have (a, b1) and (a, b2) where a represents each student.
{(a, c) | a’s first class each week is in c}
Solution: This is a function. We assume that every student takes at least one course
{(a, c) | a has a class in c on Friday afternoon}
Solution: Not a function
12. Let S = {1, 2, 3, 4, 5} and define f : S ⟶ Z by
f x=x2+1 If x is even 2x-5 if x is odd Express f as a subset of S x Z. Is f one-to-one?
Solution
If f(x) for each value is computed, and if f(x) ≠ f(y) for all x ≠ y, then it is 1-1.

The subset is given by {(1,f(1)), (2,f(2)),…,(5, f(5))}. Since 1 is odd, f(1) = (2×1)-5=-3Since 2 is even, f(2) =22+1=5 and so on…13. The addition and multiplication of real numbers are functions add, mult: R x R ⟶ R, where
add (x,y) = x+ y; mult ((x,y) = xy.
Is add one-to-one? Is it onto?
Solution: Add is not a 1-1 since for instance, add(1,1)= add(0,2), but (1,1) is not equal to (0,2). Furthermore, it is onto ∀ yϵR, the equation y=addx, is the same as x=(y,1) Is mult one-to-one? Is it onto?
Solution: It is not a 1-1, because for the set (1,4) for instance, mult(1,4)= mult(2,2) but (1,4)≠(2,2). It is onto because ∀ yϵR, the equation y=multx, is the same as x=(y,1) 16. Define h : A ⟶ A by h(x) = x2 + 2. Determine (with reasons) whether f is one-to-one and whether it is
onto in each of the following cases.
A = Z
Not one-to-one, h(1)=3h(-1) but 1≠-1. Not onto because 1∉mgh:1ϵZ, when hxis solved, x2=-1, and a solution for this does not exist in Ζ
A = N
This is a one-to-one because if h(x1) = h(x2), then x2 1+2=x2 2+2, so x2 1=x2 2 and x1=±x2
19. Find the domain and range of each of the given functions of a real variable. In each case, determine
Whether the function one-to-one and whether it is onto.
g : R ⟶ R defined by g (x) = x|x|
gx=x2 if x≥0, -x2 if x<0 the domain of g is therefore R. If y≥0, then y = g(x) while if y <0, y = g(-). Because mg g = R, g is onto. Also, further investigations shows that g is one-to-one.
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1. Let A = {1, 2, 3, 4, 5}. Find the inverse of each of the following functions f : A ⟶ A.
b. f = { (1, 2), (2, 4) (3, 3), (4, 1), (5, 5) }
f-1=1,4,2,1,3,3,4,2,5,55. Let f, g : R ⟶ Z be the floor and ceiling functions, respectively. Compute f ᴏ g (x) and g ᴏ f (x).
Solution:
Because g(x) is an integer, f o g(x) = g(x) and, g o f(x) =f(x)6. Graph each of the following functions and find each inverse. Specify the domain and range of each inverse.
Specify the domain and range of each inverse.
f : R ⟶ R given by f (x) = 3x + 5
y=3x+5x=3y+53y=x-5y=x-53Hence; f-1x=x-53Domain: All real numbers
Range = [0,+∞)
Figure 1 – Graphs of f (x) = 3x + 5 and f (x) = x3 – 2
f : R ⟶ R given by f (x) = x3 – 2
y=x3-2x=y3-2y3=x+2y=3x+2Hence; f-1x=3x+2Domain: All real numbers from -2
Range = [-2,+∞)
9. Suppose A is the set of all married people, mother: A ⟶ A is the function that assigns to each married
Person his or her mother, and father and spouse have similar meanings. Give sensible interpretations of
Each of the following:
(c) Father ᴏ mother
Maternal grandfather
(e) Spouse ᴏ mother
Father
(g) Spouse ᴏ spouse
You
11. Let S = {1, 2, 3, 4} and define functions f, g : S ⟶ S by f = { (1, 3), (2, 2), (3, 4), (4, 1) } and g = { (1, 4), (2, 3),
(3, 1), (4, 2) }. Find
b.f ᴏ g-1ᴏ g =f ={ (1, 3), (2, 2), (3, 4), (4, 1) }c. g ᴏ fᴏ g-1={1,2,2,4,3,3,4,1}19. Let S = {1, 2, 3, 4, 5} and let f, g, h: S ⟶ S be the functions defined by
f = {(1, 2), (2, 1), (3, 4), (4, 5), (5, 3)}
g = {(1, 3), (2, 5), (3, 1), (4, 2), (5, 4)}
h = {(1, 2), (2, 2), (3, 4), (4, 3), (5, 1)}.
Find f ᴏ g and g ᴏ f. Are these functions equal?
fog=1,4,2,3,3,2,4,1,5,5; g o f=1,5,2,3,3,2,4,4,5,1 . From this, fog≠gofExplain why f and g have inverse but h does not.
f-1=1,2,2,1,3,5,4,3,5,4; g-1=1,3,2,4,3,1,4,5,5,2.
The inverses exist for g and f because they are one-to-one. h is not a one-to-one, hence no inverse.