83
(viii) If X is an innite set, then the family of all nite subsets of X
forms a subring of the Boolean ring B( X ).
Solution. True.
3.2 Prove that a commutative ring R has a unique one 1; that is, i
33
1.82
(i)
Prove that 10q + r is divisible by 7 if and only if q 2r is divisible
by 7.
Solution. If 10q + r 0 mod 7, then 15q + 5r 0 mod 7,
and so q 2r 0 mod 7. Conversely, if q 2r 0 mod 7,
then 3q 6
86
(iii) Prove that the ring of Gaussian integers is a domain.
Solution. The set Gaussian integers form a subring of C, and
hence it is a domain.
3.12 Prove that the intersection of any family of subr
29
A similar argument shows that if pe | pr i for i 1, then pe | pr m i .
By the fundamental theorem of arithmetic, the total number of factors p
occurring on each side must be the same. Therefore, th
42
2.12 Let X = cfw_x1 , . . . , xm and Y = cfw_ y1 , . . . , yn be nite sets, where the xi are
distinct and the y j are distinct. Show that there is a bijection f : X Y if
and only if | X | = |Y |;
142
4.76
(i)
Write x 15 1 as a product of irreducible polynomials in F2 [x ].
Solution. Absent.
(ii) Find an irreducible quartic polynomial g (x ) F2 [x ], and use it to
dene a primitive 15th root of
41
(ii) If f : R R is a bijection whose graph consists of certain points
(a , b) [of course, b = f (a )], prove that the graph of f 1 is
cfw_(b, a ) : (a , b) f .
Solution. By denition, f 1 (b) = a if
141
Solution. Absent.
(ii) Prove that
r
| Br (u )| =
i =0
n
(q 1)i .
i
Solution. Absent.
If C An is an (n , M , d )-code,
(iii) (Gilbert-Varshamov bound.)
where |A| = q , prove that
qn
d 1 n
i =0 i
(q
40
2.5 Let A and B be sets, and let a A and b B . Dene their ordered pair
as follows:
(a , b) = cfw_a , cfw_a , b.
If a A and b B , prove that (a , b ) = (a , b) if and only if a = a and
b = b.
Soluti
140
Solution. If Av = cv , then Am v = cm v for all m 1. Hence, if Am = 0,
then cm = 0 (because eigenvectors are nonzero). Thus, c = 0.
Conversely, if all the eigenvalues of an n n matrix A are 0, the
143
5.5
There is a circular castle whose diameter is unknown; it is provided with four gates, and two lengths out of the north gate there
is a large tree, which is visible from a point six lengths eas
43
Solution. Let z Z . Since g is surjective, there is y Y with
g ( y ) = z ; since f is surjective, there is x X with f (x ) = y . It
follows that (g f )(x ) = g ( f (x ) = g ( y ) = z , and so g f i
144
(ii) Show that one root of f ( X ) = X 3 + X 2 36 is an integer and
nd the other two roots. Compare your method with Cardanos
formula and with Vi` tes trigonometric solution.
e
Solution. Corollary
44
(ii) If B Y , prove that f 1 ( B ) = f 1 ( B ) , where B denotes the
complement of B .
Solution. Absent.
2.18 Let f : X Y be a function. Dene a relation on X by x x if f (x ) =
f (x ). Prove that i
145
Solution. We know that cos is a root of f (x ) = 4x 3 3x r , where
r = cos 3 . In particular, if = + 120 , then cos is a root of 4x 3
3x cos 3( + 120 ). But the addition formula for cosine gives
45
(vii) Every transposition is an even permutation.
Solution. False.
(viii) If a permutation is a product of 3 transpositions, then it cannot
be a product of 4 transpositions.
Solution. True.
(ix) If
146
5.13 Find the roots of x 3 6x + 4.
Solution. We have q = 6, r = 4, D = 16, and 3 = 2 + 2i . Use De
Moivres theorem to nd cube roots of 2 + 2i :
(1 + i )3 = 2 + 2i .
Hence, = 1 + i , = 1 i , and a
46
(ii) If kr n , where 1 < r n , prove that the number of permutations Sn , where is a product of k disjoint r -cycles, is
11
k ! r k [n (n
1) (n kr + 1).]
Solution. Absent.
2.25
(i)
If is an r -cyc
147
Thus, m = 2, = 3, and we have the factorization
x 4 15x 2 20x 6 = (x 2 + 4x + 3)(x 2 4x 2).
The quadratic formula applied to each of the two factors gives the desired
roots:
3, 1, 2 + 6, 2 6.
In p
39
2.2 If A and B are subsets of a set X , prove that A B = A B , where
B = X B is the complement of B .
Solution. This is one of the beginning set theory exercises that is so easy
it is difcult; the
139
(iv) Prove that every hermitian matrix A over C is diagonalizable.
Solution. Absent.
4.55 If A is an m n matrix over a eld k , prove that rank( A) d if and
only if A has a nonsingular d d submatri
38
1.102 My Uncle Ben was born in Pogrebishte, a village near Kiev, and he claimed
that his birthday was February 29, 1900. I told him that this could not be,
for 1900 was not a leap year. Why was I w
130
(iii) Gaussian equivalent matrices have the same column space.
Solution. False.
(iv) The matrix A =
10 0
0 1 1
00 1
is nonsingular.
Solution. True.
(v) Every nonsingular matrix over a eld is a pro
30
(iv) For all rationals a , b, c, prove p (a , b) p (a , c) + p (c, b).
Solution. p (a , b) p (a , c) + p (c, b) because
p (a , b) = a b p = (a c) + (c b)
maxcfw_ a c p , c b p
ac p + cb p
= p (a
131
(ii) Prove that lies in the column space of A if and only if rank([ A| ])
= rank( A).
Solution. Let A be Gaussian equivalent to an echelon matrix U ,
so that there is a nonsingular matrix P with P
31
c = b = db , so that a m = b . Thus, a divides b ; as
(a , b ) = 1, we have a divides , by Corollary 1.40. Write =
a k , and observe that
c = db
= db a k = (db )(da )k /d = [ab/d ]k .
Therefore, ab
132
(ii) Every matrix over R is similar to innitely many different matrices.
Solution. False.
(iii) If S and T are linear transformations on the plane R2 that agree on
two nonzero points, then S = T .
133
x
(ii) Prove that integration S : V3 V4 , dened by S ( f ) = 0 f (t ) dt ,
is a linear transformation, and nd the matrix A = X 4 [ S ] X of in3
tegration.
Solution. Absent.
4.34 If Sn and P = P is
134
prove that there exists a unique solution if and only if det( A) = 0.
Solution. Absent.
(ii) If V is a vector space with basis X = v1 , v2 , dene T : V V
by T (v1 ) = a v1 + bv2 and T (v2 ) = cv1