TEST BANK FOR A First Course in Abstract Algebra 7th Edition By John B. Fraleigh

1. {
p
3,−
p
3} 2. The set is empty.
3. {1,−1, 2,−2, 3,−3, 4,−4, 5,−5, 6,−6, 10,−10, 12,−12, 15,−15, 20,−20, 30,−30,
60,−60}
4. {−10,−9,−8,−7,−6,−5,−4,−3,−2,−1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
5. It is not a well-defined set. (Some may argue that no element of Z+ is large, because every element
exceeds only a finite number of other elements but is exceeded by an infinite number of other elements.
Such people might claim the answer should be ?.)
6. ? 7. The set is ? because 33 = 27 and 43 = 64.
8. It is not a well-defined set. 9. Q
10. The set containing all numbers that are (positive, negative, or zero) integer multiples of 1, 1/2, or
1/3.
11. {(a, 1), (a, 2), (a, c), (b, 1), (b, 2), (b, c), (c, 1), (c, 2), (c, c)}
12. a. It is a function. It is not one-to-one since there are two pairs with second member 4. It is not onto
B because there is no pair with second member 2.
b. (Same answer as Part(a).)
c. It is not a function because there are two pairs with first member 1.
d. It is a function. It is one-to-one. It is onto B because every element of B appears as second
member of some pair.
e. It is a function. It is not one-to-one because there are two pairs with second member 6. It is not
onto B because there is no pair with second member 2.
f. It is not a function because there are two pairs with first member 2.
13. Draw the line through P and x, and let y be its point of intersection with the line segment CD.
14. a. : [0, 1] ! [0, 2] where (x) = 2x b. : [1, 3] ! [5, 25] where (x) = 5 + 10(x − 1)
c. : [a, b] ! [c, d] where (x) = c + d−c
b−a (x − a)
15. Let : S ! R be defined by (x) = tan((x − 1
2 )).
16. a. ?; cardinality 1 b. ?, {a}; cardinality 2 c. ?, {a}, {b}, {a, b}; cardinality 4
d. ?, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}; cardinality 8
17. Conjecture: |P(A)| = 2s = 2|A|.
Proof The number of subsets of a set A depends only on the cardinality of A, not on what the
elements of A actually are. Suppose B = {1, 2, 3, · · · , s − 1} and A = {1, 2, 3, · · · , s}. Then A has all
the elements of B plus the one additional element s. All subsets of B are also subsets of A; these
are precisely the subsets of A that do not contain s, so the number of subsets of A not containing
s is |P(B)|. Any other subset of A must contain s, and removal of the s would produce a subset of
B. Thus the number of subsets of A containing s is also |P(B)|. Because every subset of A either
contains s or does not contain s (but not both), we see that the number of subsets of A is 2|P(B)|.
2 0. Sets and Relations
We have shown that if A has one more element that B, then |P(A)| = 2|P(B)|. Now |P(?)| = 1, so
if |A| = s, then |P(A)| = 2s.
18. We define a one-to-one map of BA onto P(A). Let f 2 BA, and let (f) = {x 2 A | f(x) = 1}.
Suppose (f) = (g). Then f(x) = 1 if and only if g(x) = 1. Because the only possible values for
f(x) and g(x) are 0 and 1, we see that f(x) = 0 if and only if g(x) = 0. Consequently f(x) = g(x) for
all x 2 A so f = g and is one to one. To show that is onto P(A), let S A, and let h : A ! {0, 1}
be defined by h(x) = 1 if x 2 S and h(x) = 0 otherwise. Clearly (h) = S, showing that is indeed
onto P(A).
19. Picking up from the hint, let Z = {x 2 A | x /2 (x)}. We claim that for any a 2 A, (a) 6= Z. Either
a 2 (a), in which case a /2 Z, or a /2 (a), in which case a 2 Z. Thus Z and (a) are certainly
different subsets of A; one of them contains a and the other one does not.
Based on what we just showed, we feel that the power set of A has cardinality greater than |A|.
Proceeding naively, we can start with the infinite set Z, form its power set, then form the power set
of that, and continue this process indefinitely. If there were only a finite number of infinite cardinal
numbers, this process would have to terminate after a fixed finite number of steps. Since it doesn’t,
it appears that there must be an infinite number of different infinite cardinal numbers.
The set of everything is not logically acceptable, because the set of all subsets of the set of
everything would be larger than the set of everything, which is a fallacy.
20. a. The set containing precisely the two elements of A and the three (different) elements of B is
C = {1, 2, 3, 4, 5} which has 5 elements.
i) Let A = {−2,−1, 0} and B = {1, 2, 3, · · ·} = Z+. Then |A| = 3 and |B| = @0, and A
and B have no elements in common. The set C containing all elements in either A or B is C =
{−2,−1, 0, 1, 2, 3, · · ·}. The map : C ! B defined by (x) = x + 3 is one to one and onto B, so
|C| = |B| = @0. Thus we consider 3 + @0 = @0.
ii) Let A = {1, 2, 3, · · ·} and B = {1/2, 3/2, 5/2, · · ·}. Then |A| = |B| = @0 and A and
B have no elements in common. The set C containing all elements in either A of B is C =
{1/2, 1, 3/2, 2, 5/2, 3, · · ·}. The map : C ! A defined by (x) = 2x is one to one and onto A,
so |C| = |A| = @0. Thus we consider @0 + @0 = @0.
b. We leave the plotting of the points in A × B to you. Figure 0.14 in the text, where there are @0
rows each having @0 entries, illustrates that we would consider that @0 · @0 = @0.