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?