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( \newcommand{\kernel}{\mathrm{null}\,}\)

Answers to Selected Exercises

Section 2.1

  1. Only (a), (c), and (e) are statements.

  2. (a) false (b) false (c) false (d) true

  3. (a) πZ (b) 13+23+333242/4 (c) u is not a vowel

    (d) This statement is either true or false.

  4. (a) true (b) true (c) true (d) false (e) false (f) true

  5. By definition, a rational number can be written as a ratio of two integers. After multiplying the numerator by 7, we still have a ratio of two integers. Conversely, given any rational number x, we can multiply the denominator by 7, we obtain another rational number y such that 7y=x. Hence, the two sets 7Q and Q contain the same collection of rational numbers. In contrast, 0Q contains only one number, namely, 0. Therefore, 0QQ.

Section 2.2

  1. (a) pq (b) ¯qr (c) ¯p¯q (d) (pq)¯pq

  2. (a) pq; always false regardless of the value of r.

    (b) pq; always true regardless of the value of r.

    (c) (pq)r; true if r is true, and false if r is false.

    (d) ¯qr; true if r is true, and false if r is false.

  3. (a) false (b) true

  4. (a) (4x)(x7) (b) (4<x)(x7) (c) (4x)(x<7)

Section 2.3

  1. (a) pq (b) rp (c) ¯pq (d) ¯pr (e) (¯pq)r

  2. (a) pq, which is false.

    (b) pr, which is true if r is true, and is false if r is false.

    (c) (pq)r, which is true if r is true, and is false if r is false.

  3. (a) x33x2+x3=0x=3

    (b) x33x2+x3=0x=3

    (c) x=3x33x2+x3=0

  4.  pqrpq(pq)rTTTTTTTFTTTFTFTTFFFFFTTFTFTFFFFFTFTFFFFFpqrpqpr(pq)(pr)TTTTTTTTFTFFTFTTTTTFFTFFFTTTFFFTFTFFFFTFFTFFFFFT

  5. (a) Using a truth table, we find that the implication (pq)(qr) is always true. Hence, no truth value of p would make (pq)(qr) false.

    (b) From a truth table, we find that, (qr)(pq) is false only when p is false. We can draw the same conclusion without using any truth table. An implication is false only when its hypothesis (in this case, qr) is true and its conclusion (in this case, pq) is false. For qr to be true, we need both q and r to be true. Now q is true and pq is false require p to be false.

Section 2.4

  1. (a) pq (b) r¯p (c) r(q¯p) (d) r(pq)

  2. (a) pq, which is false.

    (b) pr, which is true if r is true, and is false if r is false.

    (c) (pq)r, which is true if r is true, and is false if r is false.

  3. (a) true (b) false (c) false (d) false

  4. We say n is odd if and only if n=2q+1 for some integer q.

Section 2.5  

  1. pqpq¯pq¯p¯q¯p¯qTTTFFFFTFTFFTFTTTFTFFTFFTTTT

  2. Only (b) is a tautology, as indicated in the truth tables below.

    (a) pq¯p¯pq(¯pq)pTTFTTTFFFTFTTTFFFTTF

    (b) pqpq¯qp¯q(pq)(p¯q)TTTFFTTFFTTTFTTFTTFFFTTT

    (c) pqrpq(pq)rTTTTTTTFTFTFTFTTFFFTFTTTTFTFTFFFTTTFFFTF

  3. The proofs are displayed below without explanations. Be sure to fill them in.

    (b) (pq)r¯pqr(¯p¯q)r¯p(¯qr)p(¯qr)

    (c) (p¯q)(p¯r)(¯p¯q)(¯p¯r)¯p(¯q¯r)¯p¯qr¯p(qr)

  4. (a)

    Converse: If triangle ABC is a right triangle, then ABC is isosceles
      and contains an angle of 45 degrees.
    Inverse: If triangle ABC is not isosceles or does not contain an angle
      of 45 degrees, then ABC is not a right triangle.
    Contrapositive: If triangle ABC is not a right triangle, then ABC is not isosceles
      or does not contain an angle of 45 degrees.

    (b)

    Converse: If quadrilateral ABCD is both a rectangle and a rhombus,
      then ABCD is a square.
    Inverse: If quadrilateral ABCD is not a square,
      then it is not a rectangle or not a rhombus.
    Contrapositive: If quadrilateral ABCD is not a rectangle or not a rhombus,
      then ABCD is not a square.
  5. (a) true (b) true (c) false

  6. Only (b).

  7. (a) pq (b) p¯q (c) pq

Section 2.6

  1. (a) There exists an integer n such that n is prime and n is even.

    (b) For all integers n, if n>2, then n is prime or n is even.

    (c) There exists an integer n such that n is prime, and either n is even or n>2.

    (d) For all integers n, if n is prime and n is even, then n2.

  2. (a) true (b) true (c) false (d) false (e) true

  3. (a) x<0y,zR(y<zxyxz)

    (b) xZ[¯p(x)¯q(x)]

    (c) x,yR[p(x,y)¯q(x,y)]

  4. (a)

    x,yR(x+y=y+x)
    x,yR(x+yy+x)
    There exist real numbers x and y such that x+yy+x.

    (b)

    xR+yR(y2=x)
    xR+yR(y2x)
    There exists a positive real number x such that for all real numbers y, y2x.

    (c)

    yRxZ(2x2+1>x2y)
    yRxZ(2x2+1x2y)
    For every real number y, there exists an integer x such that 2x2+1x2y.
  5. The statement “a square must be a parallelogram” means, symbolically, PQRS(PQRS is a squarePQRS is a parallelogram), but the statement “a square must not be a parallelogram” means PQRS(PQRS is a squarePQRS is not a parallelogram). The second statement is not the negation of the first. The correct negation, in symbol, is PQRS(PQRS is a squarePQRS is a parallelogram). In words, it means “there exists a square that is not a parallelogram.”

Section 3.1

  1. Placing six dominoes horizontally in each row covers the entire chessboard.

  2. Let f(x)=x312x+2. From the following chart x432101234f(x)141218132914718 we conclude there x312x+2=0 has a solution between 4 and 3, another one between 0 and 1, and a third one between 3 and 4. So it has at least three real solutions.

    Remark. The Fundamental Theorem of Algebra asserts that a real polynomial of degree n has at most n real roots. Hence, the given equation has exactly three real solutions.

  3. n=3.

Section 3.2

  1. No, 23+1=9 is composite.

  2. According to (i), the number 2 is irrational. It follows from (ii) that 42=2 is also irrational. Applying (ii) one more time, we conclude that 82=42 is irrational.

  3. (a) The statement is false, because (3)2>(2)2, but 32.

    (b) The statement is false, because when n=41, n2+n+41=412+41+41=41(41+1+1)=4143 is composite.

Section 3.3

(a) We will prove the contrapositive of the given statement. That is, we will prove that if n is odd, then n2 is odd. If n is odd, we can write n=2q+1 for some integer q. Then n2=(2q+1)2=4q2+4q+1=2(2q2+2q)+1, where 2q2+2q is an integer. This shows that n2 is odd.

(b) Suppose the given statement is false. That is, suppose n2 is even, but n is odd. Since n is odd, n=2q+1 for some integer q. Then n2=(2q+1)2=4q2+4q+1=2(2q2+2q)+1, where 2q2+2q is an integer. This shows that n2 is odd, which contradicts the assumption that n2 is even. Therefore, the given statement must be true.

Suppose there exist some numbers ab such that a2+b2=2ab. Then 0=a22ab+b2=(ab)2 would have implied that a=b. This contradicts the assumption that ab. Therefore, a2+b22ab.

Suppose (pq)(p¯q) is false for some logical statements p and q. For a disjunction to be false, we need

pq to be false, and

p¯q to be false.

They in turn require

p to be true and q to be false, and

p to be true and ¯q to be false.

Having ¯q false would imply q is true, which contradicts what we found. Therefore, the given logical formula is always true, hence, a tautology.

Section 3.4

  1. We proceed by induction on n. When n=1, the left-hand side of the identity reduces to 13=1, and the right-hand side becomes 12224=1. Hence, the identity holds when n=1. Assume the identity holds when n=k for some integer k1; that is, assume 13+23+33++k3=k2(k+1)24 for some integer k1. We want to show that it also holds when n=k+1; that is, we want to show that 13+23+33++(k+1)3=(k+1)2(k+2)24. Using the inductive hypothesis, we find 13+23+33++(k+1)3=13+23+33++k3+(k+1)3=k2(k+1)24+(k+1)3=(k+1)2[k2+4(k+1)]4=(k+1)2(k2+4k+4)4=(k+1)2(k+2)24. Therefore, the identity also holds when n=k+1. This completes the induction.

Section 3.5

  1. We proceed by induction on n. When n=1, the product n(n+1)(n+2) becomes 123=6, which is obviously a multiple of 3. Hence, the claim holds when n=1. Assume the claim holds when n=k for some integer k1; that is, assume that k(k+1)(k+2) is a multiple of 3 for some integer k1. Then we can write k(k+1)(k+2)=3q for some integer q. We want to show that the claim is still valid when n=k+1. That is, we want to show that (k+1)(k+2)(k+3) is also a multiple of 3. So we want to find an integer Q such that (k+1)(k+2)(k+3)=3Q. We note that, using the inductive hypothesis, (k+1)(k+2)(k+3)=k(k+1)(k+2)+3(k+1)(k+2)=3q+3(k+1)(k+2)=3[q+(k+1)(k+2)], where q+(k+1)(k+2) is an integer. Hence, (k+1)(k+2)(k+3) is a multiple of 3. This completes the induction.

  2. (b) Sn=11(n+1)! for all integers n1.

  3. (b) Tn=n+12n+3 for all integers n0.

Section 3.6

  1. We proceed by induction on n. When n=1, the left-hand side of the identity reduces to F21=12=1, and the right-hand side becomes F1F2=11=1. Hence, the identity holds when n=1. Assume the identity holds when n=k for some integer k1; that is, assume F21+F22+F23++F2k=FkFk+1 for some integer k1. We want to show that it also holds when n=k+1; that is, we want to show that F21+F22+F23++F2k+1=Fk+1Fk+2. Using the inductive hypothesis, we find F21+F22+F23++F2k+1=F21+F22+F23++F2k+F2k+1=FkFk+1+F2k+1=Fk+1(Fk+Fk+1)=Fk+1Fk+2. Therefore, the identity also holds when n=k+1. This completes the induction.

Section 4.1

  1. (a) {5,4,3,2,1,0,1,2,3} (b) {1,2,3} (c) {0,2,3} (d) {3,3}

  2. (a) {nZn<0}

    (b) \boldsymbol{\{n\in\mathbb{Z} \mid \mbox{\)n\(is a perfect cube}\}}

    (c) \boldsymbol{\{n\in\mathbb{Z} \mid \mbox{\)n\(is a perfect square}\}}

  3. (a) Z (d) 5Z (f) 4+6Z

    Remark. We cannot write (b) as Z3 and (c) as Z2, because Z3 and Z2 mean something else. If we drop 0 from (e), then {4,8,12,}=4N. However, the inclusion of 0 makes it harder to describe (d) in the form of 4S.

  4. (a) (4,7) (b) (4,7] (c) (0,7]

  5. (a) 10 (b) 11 (c) 7

  6. (a) true (b) true (c) true (d) false

  7. (a) It is incorrect to write (3,7]=3<x7 because (3,7] is a set, but 3<x7 is a logical statement.

    (b) No, because both {xRx2<0} and are sets, so we should use an equal sign to compare them. The notation only applies to logical statements. The correct way to say it is “{xRx2<0}=.”

Section 4.2

  1. (a) true (b) true (c) true (d) true (e) true (f) false

  2. We have ZN because every integer n is also a rational number, as we can write it as the rational number n1.

  3. Yes, this is the transitive property.

  4. (e) {,{a},{{b}},{a,{b}}}

  5. (a) False, because the set {a} cannot be found in {a,b,c} as an element.

    (b) False, because a, the sole element in {a}, cannot be found in {{a},b,c} as an element.

    (c) False. For {a}({{a},b,c}), the set {a} must be a subset of {{a},b,c}}. This means a must belong to {{a},b,c}, which is not true.

Section 4.3

  1. (a) {4,3,2,1,0,1,2,3,4}

    (b) {3,2,1,0,1,2,3,4}

    (c) {3,2,1,0,1,2,3,}

  2. (a) false (b) false

  3. (a) ED (b) ¯EB

  4. For example, take A={x}, and B={{x},x}.

  5. Assume AC and BC, we want to show that ABC. In this regard, let xAB, we want to show that xC as well. Since xAB, the definition of set union asserts that either xA or xB.

    • Case 1: If xA, then AC implies that xC.

    • Case 2: If xB, then BC implies that xC.

    In both cases, we find xC. This proves that ABC.

  6. (a) The notation is used to connect two sets, but “xA” and “xB” are both logical statements. We should also use instead of . The statement should have been written as “xAxBxAB.”

    (b) If we read it aloud, it sounds perfect: If x belongs to A and B, then x belongs to AB. The trouble is, every notation has its own meaning and specific usage. In this case, is not exactly a replacement for the English word “and.” Instead, it is the notation for joining two logical statements to form a conjunction. Before , we have “xA,” which is a logical statement. But, after , we have “B,” which is a set, and not a logical statement. It should be written as “xAxBxAB.”

Section 4.4

  1. (a) {(2,0),(2,4),(2,0),(2,4)}

    (b) {(2,3),(2,0),(2,3),(2,3),(2,0),(2,3)}

  2. 2223=24.

  3. (a) {(2,),(2,{2}),(2,{2}),(2,{2,2}),(2,),(2,{2}),(2,{2}),(2,{2,2})}

Section 4.5

  1. n=1An=[0,2), n=1An=(1,).

  2. n=0Cn=, n=0Cn=N{0}.

  3. nNEn=E0={0}, nNEn=Z.

  4. iIAi=[1,), iIAi={1}.

  5. x(1,2)(12x,x2)=[1,1], x(1,2)(12x,x2)=(3.4).

  6. r(0,)Ar={(0,0)}, r(0,)Ar=R×R+{(0,0)}.

Section 5.1

  1. (a) 3 (b) 3 (c) 3 (d) 1

  2. We claim that the subset (3,5) does not have a smallest element. To see why, suppose it has a smallest element x. The midpoint between 3 and x is the number 3+x2, and 3<3+x2<x<5. This means 3+x2 is also inside the interval (3,5), and is smaller than x. This contradicts the minimality of x. Thus, the interval (3,5) does not have a smallest element. Consequently, the interval (3,5] is not well-ordered.

  3. We know that N is well-ordered. Since 2N is a subset of N, and 2N is clearly nonempty, we conclude from Problem [ex:PWO-04] that 2N is also well-ordered.

Section 5.2

  1. (a) 23, 1 (b) 11, 1 (c) 6, 13

  2. This is an immediate consequence of Corollary [cor:divalgo].

  3. (a) Let n be any integer. Then nmod.

    • Case 1: if n\bmod3=0, then n=3q for some integer q, and n^3-n = (3q)^3-3q = 27q^3-3q = 3(9q^2-q), where 9q^2-q is an integer.

    • Case 2: if n\bmod3=1, then n=3q+1 for some integer q, and n^3-n = (3q+1)^3-(3q+1) = 27q^3+27q^2+6q = 3(9q^3+9q^2+2q), where 9q^3+9q^2+2q is an integer.

    • Case 2: if n\bmod3=2, then n=3q+2 for some integer q, and n^3-n = (3q+2)^3-(3q+2) = 27q^3+54q^2+33q+6 = 3(9q^3+18q^2+11q+2), where 9q^3+18q^2+11q+2 is an integer.

    In all three cases, we have shown that n^3-n is a multiple of 3.

    (b) We note that n^3-n = n(n^2-1) = n(n-1)(n+1) = (n-1)n(n+1) is a product of three consecutive integers. As we have seen in Problem [ex:divalgo-04], any three consecutive integers must contain a multiple of 3. It follows that their product is also a multiple of 3.

  4. (a) s+t (b) 4

Section 5.3

Assume a\mid b and c\mid (-a). There exist integers x and y such that b=ax and -a=cy. Then b = ax = (-a)(-x) = cy\cdot(-x) = (-c)\cdot xy, where xy is an integer. Thus, (-c)\mid b.

There are three cases, depending on the remainder when an integer is divided by 3.

(3q)^2 = 9q^2 = 3\cdot3q^2.

(3q+1)^2 = 9q^2+6q+1 = 3(3q^2+2q)+1.

(3q+2)^2 = 9q^2+12q+4 = 9q^2+12q+3+1 = 3(3q^2+4q+1)+1.

In each case, we have shown that the square of an integer is of the form 3k or 3k+1.

Section 5.4

  1. (a) 1\cdot 27+ 0\cdot 81= 27 (b) -3\cdot 24+ 1\cdot 84= 12 (c) -35\cdot1380+16\cdot3020= 20

  2. 1, 2, 17, and 34.

Section 5.5

  1. Since -3\cdot(2n+1)+2\cdot(3n+2) = 1, we deduce that \gcd(2n+1,3n+2)=1.

  2. Let a, b, and c be positive integers such that a\mid c, b\mid c, and \gcd(a,b)=1. Then there exist integers x and y such that c=ax and c=by; and there exist integers s and t such that sa+tb=1. It follows that c = c\cdot 1 = c(sa+tb) = csa+ctb. Using c=ax and c=by, we find c = csa+ctb = by\cdot sa+ax\cdot tb = ab(ys+xt), where ys+xt is an integer. Thus, ab\mid c.

Section 5.6

  1. (a) 3^2\cdot5^2\cdot7 (b) 2\cdot3^2\cdot7^2\cdot11

  2. (a) 81 (b) 168

  3. Every 50 days.

  4. Assume x\in 10\mathbb{Z}\cap15\mathbb{Z}, then x\in10\mathbb{Z} and x\in15\mathbb{Z}. This means x is a multiple of both 10 and 15. Consequently, x is a multiple of \lcm(10,15)=30, which means x\in30\mathbb{Z}. Thus, 10\mathbb{Z}\cap15\mathbb{Z} \subseteq 30\mathbb{Z}.

    Next, assume x\in 30\mathbb{Z}, then x is a multiple of 30. Consequently, x is a multiple of 10, as well as a multiple of 15. This means x\in10\mathbb{Z}, and x\in15\mathbb{Z}. As a result, x\in 10\mathbb{Z}\cap15\mathbb{Z}. Thus, 30\mathbb{Z} \subseteq 10\mathbb{Z}\cap15\mathbb{Z}. Together with 10\mathbb{Z}\cap15\mathbb{Z} \subseteq 30\mathbb{Z}, we conclude that 10\mathbb{Z}\cap15\mathbb{Z} = 30\mathbb{Z}.

  5. (a) When p is divided by 4, its remainder is 0, 1, 2, or 3. But p is odd, hence, p is of the form 4k+1 or 4k+3 for some integer k. Since p\geq3, we also need k to be a nonnegative integer.

    (b) When p is divided by 6, its remainder is 0, 1, 2, 3, 4, or 5. But p is odd, hence, p is of the form 6k+1, 6k+3, or 6k+5. We rule out the form 6k+3 because this would make p a multiple of 3. Hence, p is of the form 6k+1 or 6k+5 for some nonnegative integer k.

Section 5.7

  1. The addition and multiplication tables for \mathbb{Z}_8 are listed below. \begin{array}{|c||*{8}{c|}}\hline + & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline\hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 0 \\ \hline 2 & 2 & 3 & 4 & 5 & 6 & 7 & 0 & 1 \\ \hline 3 & 3 & 4 & 5 & 6 & 7 & 0 & 1 & 2 \\ \hline 4 & 4 & 5 & 6 & 7 & 0 & 1 & 2 & 3 \\ \hline 5 & 5 & 6 & 7 & 0 & 1 & 2 & 3 & 4 \\ \hline 6 & 6 & 7 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 7 & 7 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \end{array} \qquad\qquad \begin{array}{|c||*{8}{c|}}\hline \cdot & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline\hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 2 & 0 & 2 & 4 & 6 & 0 & 2 & 4 & 6 \\ \hline 3 & 0 & 3 & 6 & 1 & 4 & 7 & 2 & 5 \\ \hline 4 & 0 & 4 & 0 & 4 & 0 & 4 & 0 & 4 \\ \hline 5 & 0 & 5 & 2 & 7 & 4 & 1 & 6 & 3 \\ \hline 6 & 0 & 6 & 4 & 2 & 0 & 6 & 4 & 2 \\ \hline 7 & 0 & 7 & 2 & 5 & 4 & 3 & 2 & 1 \\ \hline \end{array} Only 1, 3, 5, and 7 have multiplicative inverses. In fact,1^{-1}=1, 3^{-1}=3, 5^{-1}=5, and 7^{-1}=7.

  2. The sum is 9, and the product is 7.

  3. From the following computation \begin{array}{|c|l|} \hline \mbox{$m$ (mod~7)} & \hfil\mbox{$m^2+1$ (mod~7)} \\ \hline 0 & 0^2+1 = 1 \\ \pm1 & 1^2+1 = 2 \\ \pm2 & 2^2+1 = 5 \\ \pm3 & 3^2+1 = 10 \equiv 3 \\ \hline \end{array} we determine that m^2+1\not\equiv0 (mod 7). Hence, m^2+1 is not a multiple of 7 for all integers m.

  4. Both methods give 4^{45}=1 in \mathbb{Z}_{11}.

  5. (a) 9

Section 6.1

  1. -24pt \begin{array}[t]{|c||*{6}{c|}} \hline x & 5.7 & \pi & e & -7.2 & -0.8 & 9 \\ \hline \lfloor x \rfloor & 5 & 3 & 2 & -8 & -1 & 9 \\ \lceil x \rceil & 6 & 4 & 3 & -7 &\phantom{-}0 & 9 \\ {[x]} & 6 & 3 & 3 & -7 & -1 & 9 \\ \hline \end{array}

  2. [0,\infty).

Section 6.2

  1. \big[\frac{7}{3},\infty\big).

  2. Only g is a well-defined function. The image f(4) is undefined, and there are two values for h(3). Hence, both f and h are not well-defined functions.

  3. (a) Yes, because no division by zero will ever occur.

  4. -24pt \begin{array}[t]{|c||c|c|c|c|} \hline x & 1 & 2 & 3 & 4 \\ \hline p(x) & 3 & 1 & 2 & 2 \\ \hline \end{array} 0.4in \begin{array}[t]{|c||c|c|c|c|} \hline x & 1 & 2 & 3 & 4 \\ \hline q(x) & 2 & 3 & 1 & 3 \\ \hline \end{array}

  5. (a) 7 (b) 7 (c) 3

Section 6.3

  1. (a) No. For example, f(0)=f(2)=1.

    (b) Yes, since g'(x)=3x^2-4x=x(3x-4)>0 for x>2.

  2. Because the domain and the codomain are half-open intervals, we need to be careful with the inclusion and exclusion of the endpoints. We can use the graph displayed below on the left.

    We find f(x) = \frac{3}{2}\,x+\frac{1}{2}.

  3. (a) One-to-one (b) Not one-to-one

  4. (a) Not one-to-one (b) One-to-one

  5. There are twelve one-to-one functions from \{1,2\} to \{a,b,c,d\}. The images of 1 and 2 under them are listed below. \begin{array}{|c||*{12}{c|}} \hline & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 & f_7 & f_8 & f_9 & f_{10} & f_{11} & f_{12} \\ \hline\hline 1 & a & a & a & b & b & b & c & c & c & d & d & d \\ \hline 2 & b & c & d & a & c & d & a & b & d & a & b & c \\ \hline \end{array}

  6. (a) One-to-one (b) Not one-to-one (c) Not one-to-one

Section 6.4

  1. (a) Yes! It is not easy to express x in terms of y from the equation y=x^3-2x^2+1. However, from its graph, we can tell that the y-values cover all the possible real values in the codomain.

    (b) No, because g(x)\geq1.

  2. (b) Not onto (c) Onto

  3. (b) Not onto (c) Onto

  4. No, because we have at most two distinct images, but the codomain has four elements.

  5. (a) Onto (b) Not onto (c) Not onto

Section 6.5

  1. (a) f_1(A)=\{a,b\}, f_1^{-1}(B)=\{2,3,4,5\}

    (b) f_2(A)=\{a,c\}, f_2^{-1}(B)=\{2,4\}

    (c) f_3(A)=\{b,d\}, f_3^{-1}(B)=\emptyset

    (d) f_4(A)=\{e\}, f_4^{-1}(B)=\{5\}

  2. The images of s are tabulated below. \begin{array}{|c||*{12}{c|}} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hline s(x) & 7 & 11 & 3 & 7 & 11 & 3 & 7 & 11 & 3 & 7 & 11 & 3 \\ \hline \end{array} (a) \{3,11\} (b) \{0,3,6,9\} (c) \{3,7,11\}

  3. (a) [20,26); \{20,23,26\} (b) [-3,-\frac{4}{3}\big); \{-2\}

  4. (a) \big\{\frac{3}{5},\frac{9}{5},\frac{27}{5}, 3,9,27,15,45,135\big\} (b) \{(-3,2)\} (c) \mathbb{N}\times\{0\}

  5. For a function to be well-defined, each row sum must be 1. For the function to be one-to-one, each column sum must be at most 1. For the function to be onto, each column sum must be at least 1 (hence, no column sum is zero).

  6. Let y\in f(C_1)-f(C_2), we want to show that y\in f(C_1-C_2) as well. Since y\in f(C_1)-f(C_2), we know there exists x\in A such that f(x)=y. Having y\in f(C_1)-f(C-2) means y\in f(C_1) but y\notin f(C_2). Hence, x\in C_1 but x\notin C_2. In other words, x\in C_1-C_2. This leads to y=f(x)\in f(C_1-C_2). This completes the proof that f(C_1)-f(C_2) \subseteq f(C_1-C_2).

  7. \{0,1,4,9\}; \{0,\pm1,\pm2,\pm3\}.

Section 6.6

  1. Only (e) is bijective.

  2. Their inverse functions {f^{-1},g^{-1}}:{(4,7)}\to{(1,3)} are defined by f^{-1}(x) = \frac{2}{3}\left(x-\frac{ 5}{2}\right), \qquad\mbox{and}\qquad g^{-1}(x) = -\frac{2}{3}\left(x-\frac{17}{2}\right).

  3. {g^{-1}}:\to{[4,7]}{[1,3]}, where g^{-1}(x) = \cases{ x-3 & if\)4x < 5\(, \cr \textstyle \frac{1}{2}(11-x) & if\)5x7\(. \cr}

  4. {s^{-1}}:{(-\infty,-3)}\to{\mathbb{R}}, where s^{-1}(x) = \frac{1}{2}\, \ln\left(\frac{4-x}{7}\right).

  5. (a) {u^{-1}}:{\mathbb{Q}}\to{\mathbb{Q}}, u^{-1}(x)=(x+2)/3

  6. The images under {\alpha^{-1}}:{\{a,b,c,d,e,f,g,h\}}\to {\{1,2,3,4,5,6,7,8\}} are given below. \begin{array}{|c||*{8}{c|}} \hline x & a & b & c & d & e & f & g & h \\ \hline \alpha^{-1}(x)& 2 & 5 & 8 & 3 & 6 & 7 & 1 & 4 \\ \hline \end{array}

Section 6.7

  1. Both f\circ g and g\circ f are from \mathbb{R} to \mathbb{R}, where (f\circ g)(x)=15x^2+19, and (g\circ f)(x)=75x^2-30x+7.

  2. We do not need to find the formula of the composite function, as we can evaluate the result directly: f(g(f(0))) = f(g(1)) = f(2) = -5.

  3. (a) {g\circ f}:{\mathbb{Z}}\to{\mathbb{Q}}, (g\circ f)(n)=1/(n^2+1)

    (b) {g\circ f}:{\mathbb{R}}\to{(0,1)}, (g\circ f)(x)=x^2/(x^2+1)

  4. (a) {g\circ f}:{\{1,2,3,4,5\}}\to{\{1,2,3,4,5\}},

    \phantom{(a)} (g\circ f)(1)=2, (g\circ f)(2)=5, (g\circ f)(3)=1, (g\circ f)(4)=3, (g\circ f)(5)=4

  5. {g\circ f}:{\mathbb{Z}}\to{\mathbb{Z}}, (g\circ f)(n) = \cases{ 3(2n-1) & if\)n0\(, \cr 2n+1 & if\)n < 0\(. \cr}

  6. (a) {f\circ g}:{\mathbb{Z}}\to{\mathbb{Z}}, (f\circ g)(n) = 3-n

    \phantom{(a)} {(f\circ g)^{-1}}:{\mathbb{Z}}\to{\mathbb{Z}}\), (f\circ g)^{-1}(n) = 3-n

    \phantom{(a)} {f^{-1}:}{\mathbb{Z}}\to{\mathbb{Z}}, f^{-1}(n) =2-n

    \phantom{(a)} {g^{-1}}:{\mathbb{Z}}\to{\mathbb{Z}}, g^{-1}(n) = n-1

    \phantom{(a)} {g^{-1}\circ: f^{-1}}{\mathbb{Z}}\to{\mathbb{Z}}, (g^{-1}\circ f^{-1})(n) = 3-n

Section 7.1

  1. (a)

    \begin{array}{cc} & \begin{array}{cccc} 1 & 2 & 3 & 6 \end{array} \\ \noalign{\medskip} \begin{array}{c} 1 \\ 2 \\ 3 \\ 6 \end{array} & \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \end{array}

    (b)

    0pt

    (140,60)(-10,-10) (0, 0)(40,0)4 (0,40)(40,0)4 (-10,-20)(20,20)1 (-10,40)(20,20)1 ( 30,-20)(20,20)2 ( 30,40)(20,20)2 ( 70,-20)(20,20)3 ( 70,40)(20,20)3 (110,-20)(20,20)6 (110,40)(20,20)6 ( 0,0)( 1,1) 40 ( 40,0)(-1,1) 40 ( 0,0)( 2,1) 80 ( 40,0)( 1,1) 40 ( 0,0)( 3,1)120 ( 40,0)( 2,1) 80 ( 80,0)(-2,1) 80 (120,0)(-3,1)120 ( 80,0)(-1,1) 40 (120,0)(-2,1) 80 ( 80,0)( 1,1) 40 (120,0)(-1,1) 40

    0.6in

    0pt \begin{array}{cc} & \begin{array}{cccc} 1 & 2 & 3 & 6 \end{array} \\ \noalign{\medskip} \begin{array}{c} 1 \\ 2 \\ 3 \\ 6 \end{array} & \left(\begin{array}{cccc} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right) \end{array}

    (c)

    0pt

    (140,60)(-10,-10) (0, 0)(40,0)4 (0,40)(40,0)4 (-10,-20)(20,20)1 (-10,40)(20,20)1 ( 30,-20)(20,20)2 ( 30,40)(20,20)2 ( 70,-20)(20,20)3 ( 70,40)(20,20)3 (110,-20)(20,20)6 (110,40)(20,20)6 ( 0,0)( 1,1) 40 ( 0,0)( 2,1) 80 ( 40,0)( 1,1) 40 ( 0,0)( 3,1)120 ( 40,0)( 2,1) 80 ( 80,0)( 1,1) 40

    0.6in

    0pt \begin{array}{cc} & \begin{array}{cccc} 1 & 2 & 3 & 6 \end{array} \\ \noalign{\medskip} \begin{array}{c} 1 \\ 2 \\ 3 \\ 6 \end{array} & \left(\begin{array}{cccc} 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right) \end{array}

  2. (a) \mbox{domain}=\mbox{image}=\{1,2,3,6\}.

    (b) \mbox{domain}=\mbox{image}=\{1,2,3,6\}.

    (c) \mbox{domain}=\{1,2,3\}, \mbox{image}=\{2,3,6\}.

  3. 0pt

    (220,60)(-10,-10) (0, 0)(40,0)6 (0,40)(40,0)6 (-10,-20)(20,20) 1 (-10,40)(20,20) 1 ( 30,-20)(20,20) 2 ( 30,40)(20,20) 2 ( 70,-20)(20,20) 4 ( 70,40)(20,20) 4 (110,-20)(20,20) 5 (110,40)(20,20) 5 (150,-20)(20,20)10 (150,40)(20,20)10 (190,-20)(20,20)20 (190,40)(20,20)20 ( 0,0)( 1,1) 40 ( 40,0)( 1,1) 40 ( 0,0)( 2,1) 80 ( 0,0)( 3,1)120 ( 40,0)( 3,1)120 ( 0,0)( 4,1)160 ( 40,0)( 4,1)160 ( 0,0)( 5,1)200 ( 80,0)( 3,1)120 (120,0)( 1,1) 40 (120,0)( 2,1) 80 (160,0)( 1,1) 40

    0pt \begin{array}[t]{cc} & \begin{array}{cccccc} 1 & 2 & 4 & 5 & 10 & 20 \end{array} \\ \noalign{\medskip} \begin{array}{c} 1 \\ 2 \\ 4 \\ 5 \\ 10 \\ 20 \end{array} & \left(\begin{array}{cccccc} 0 & 1 & 1 & 1 &\; 1 \;&\; 1 \\ 0 & 0 & 1 & 0 &\; 1 \;&\; 1 \\ 0 & 0 & 0 & 0 &\; 0 \;&\; 1 \\ 0 & 0 & 0 & 0 &\; 1 \;&\; 1 \\ 0 & 0 & 0 & 0 &\; 0 \;&\; 1 \\ 0 & 0 & 0 & 0 &\; 0 \;&\; 0 \end{array}\right) \end{array}

  4. \begin{array}[t]{cc} & \begin{array}{cccc} \;\;\;\;\emptyset & \{1\} & \{2\} & \{1,2\} \end{array} \\ \noalign{\medskip} \begin{array}{c} \emptyset \\ \{1\} \\ \{2\} \\ \{1,2\} \end{array} & \left(\begin{array}{cccc} 0 &\;\;0\;&\;\;0&\;\;\;0 \\ 0 &\;\;1\;&\;\;0&\;\;\;1 \\ 0 &\;\;0\;&\;\;1&\;\;\;1 \\ 0 &\;\;1\;&\;\;1&\;\;\;1 \end{array}\right) \end{array}

Section 7.2

  1. (a) Reflexive, symmetric, antisymmetric, and transitive.

    (b) Irreflexive, and symmetric.

    (c) Irreflexive, and transitive.

  2. (a) Antisymmetric.

    (b) Reflexive, symmetric, and transitive.

    (c) Irreflexive, symmetric, and transitive.

  3. Reflexive, symmetric, and transitive.

  4. Antisymmetric, and transitive.

  5. Irreflexive, and antisymmetric.

  6. Symmetric.

  7. (a) A is not reflexive because (X,X)\notin A if X\neq\emptyset.

    (b) A is not irreflexive because (\emptyset,\emptyset)\in A.

    (c) No. For example, consider S=\{a,b,c\}, X=\{a\}, Y=\{b\}, and Z=\{a,c\}. Then (X,Y)\in A, (Y,Z)\in A, but (X,Z)\notin A.

    (d)

    0pt

    (300,60)(-10,-10) (0, 0)(40,0)8 (0,40)(40,0)8 (-10,-20)(20,20)\emptyset (-10, 40)(20,20)\emptyset ( 30,-20)(20,20)\{a\} ( 30, 40)(20,20)\{a\} ( 70,-20)(20,20)\{b\} ( 70, 40)(20,20)\{b\} (110,-20)(20,20)\{c\} (110, 40)(20,20)\{c\} (140,-20)(40,20)\{a,b\} (140, 40)(40,20)\{a,b\} (180,-20)(40,20)\{a,c\} (180, 40)(40,20)\{a,c\} (220,-20)(40,20)\{b,c\} (220, 40)(40,20)\{b,c\} (260,-20)(40,20)\{a,b,c\} (260, 40)(40,20)\{a,b,c\} ( 0, 0)( 0,1) 40 ( 0, 0)( 1,1) 40 ( 0, 0)(5,1)200 ( 0, 0)( 2,1) 80 ( 0, 0)(6,1)240 ( 0, 0)( 3,1)120 (0,0, 280,40) ( 0, 0)( 4,1)160 ( 40, 0)(-1,1) 40 ( 40, 0)( 1,1) 40 ( 40, 0)( 2,1) 80 ( 40, 0)( 5,1)200 ( 80, 0)(-2,1) 80 ( 80, 0)(-1,1) 40 ( 80, 0)( 2,1) 80 ( 80, 0)( 1,1) 40 (120, 0)(-3,1)120 (120, 0)(-1,1) 40 (120, 0)(-2,1) 80 (120, 0)( 1,1) 40 (160, 0)(-1,1) 40 (160, 0)(-4,1)160 (200, 0)(-3,1)120 (200, 0)(-5,1)200 (240, 0)(-6,1)240 (240, 0)(-5,1)200 (280,0, 0,40)

    18pt

    0pt \quad \begin{array}[t]{cc} & \begin{array}{*{8}{c}} \rotatebox{90}{\)\(} & \rotatebox{90}{\){a}\(} & \rotatebox{90}{\){b}\(} & \rotatebox{90}{\){c}\(} & \rotatebox{90}{\){a,b}\(} & \rotatebox{90}{\){a,c}\(} & \rotatebox{90}{\){b,c}\(} & \rotatebox{90}{\){a,b,c}\(} \end{array} \\ \noalign{\medskip} \begin{array}{c} \emptyset \\ \{a\} \\ \{b\} \\ \{c\} \\ \{a,b\} \\ \{a,c\} \\ \{b,c\} \\ \{a,b,c\} \end{array} & \setlength{\arraycolsep}{7.7pt} \left(\begin{array}{cccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right) \end{array}

  8. (a) Symmetric.

    (b) Reflexive, and symmetric.

  9. (a) Reflexive, antisymmetric, and transitive.

    (b) Reflexive, symmetric, and transitive.

    (c) Symmetric.

  10. (a) Reflexive, antisymmetric, and transitive.

    (b) Symmetric.

    (c) Symmetric, and transitive.

  11. (a) Reflexive, and transitive.

    (b) Symmetric,

    (c) Reflexive, symmetric, and transitive.

  12. (a) Symmetric, and transitive.

    (b) Reflexive, symmetric, and transitive.

    (c) Reflexive, and transitive.

Section 7.3

  1. (a) The equivalence classes are of the form \{3-k,3+k\} for some integer k. For instance, [3]=\{3\}, [2]=\{2,4\}, [1]=\{1,5\}, and [-5]=\{-5,11\}.

    (b) There are three equivalence classes: [0]=3\mathbb{Z}, [1]=1+3\mathbb{Z}, and [2]=2+3\mathbb{Z}.

  2. (a) True

    (b) False

    (c) [\{1,5\}] = \big\{ \{1\}, \{1,2\}, \{1,4\}, \{1,5\}, \{1,2,4\}, \{1,2,5\}, \{1,4,5\}, \{1,2,4,5\} \big\}

    (d) [X] = \{(X\cap T)\cup Y \mid Y\in\wp(\overline{T})\}. In other words, S\sim X if S contains the same element in X\cap T, plus possibly some elements not in T.

  3. (a) Yes, with [(a,b)] = \{(x,y) \mid y=x+k \mbox{ for some constant\)k\(}\}. In other words, the equivalence classes are the straight lines of the form y=x+k for some constant k.

    (b) No. For example, (2,5)\sim(3,5) and (3,5)\sim(3,7), but (2,5)\not\sim(3,7). Hence, the relation \sim is not transitive.

  4. We find [0] = \frac{1}{2}\,\mathbb{Z} = \{\frac{n}{2} \mid n\in\mathbb{Z}\}, and [\frac{1}{4}] = \frac{1}{4}+\frac{1}{2}\,\mathbb{Z} = \{\frac{2n+1}{4} \mid n\in\mathbb{Z}\}.

Section 7.4

  1. The Hasse diagram is shown below.

    (180,230) (80,20)(80, 70)2(0,0)( 0,70)2(-6,5)60 (20,90)(80,-70)2(0,0)( 0,70)2( 6,5)60 (10,90)(80,-70)2(0,0)(80,70)2( 0,1)50 ( 80, 0)(20,20) 1 ( 0, 70)(20,20) 2 ( 80, 70)(20,20) 3 (160, 70)(20,20) 5 (-10,140)(40,20) 6 ( 70,140)(40,20)10 (150,140)(40,20)15 ( 70,210)(40,20)30

  2. Let a\in B, since B\subseteq A, we also find a\in A. Since (A,\preceq) is a poset, the relation \preceq on A is reflexive, hence, a\preceq a. This shows that \preceq is still reflexive when restricted to B. Antisymmetry and transitivity are proved with a similar argument.

  3. (b) The Hasse diagram is shown below.

    (180,160) (10,90)(160,0)2(0,1)50 ( 80,20)(-6,5)60 (100,20)( 6,5)60 (160,90, 20,140) ( 20,90, 160,140) ( 80, 0)(20,20)0 ( 0, 70)(20,20)-1 (160, 70)(20,20)1 ( 0,140)(20,20)-2 (160,140)(20,20)2

  4. B=\big\{\emptyset,\{a\},\{a,b\},\{a,b,c\},\{a,b,c,d\}\big\}.

Section 8.2

  1. 6.

  2. 70.

  3. 7\cdot5+7\cdot4+5\cdot4

  4. 4^5, 4^5-3\cdot4^2

  5. (a) 52^4 (b) 39^4 (c) 4\cdot13^4 (d) 4\cdot48\cdot52^3 (e) 52^4-48^4

  6. (a) 9\cdot10^3 (b) 8\cdot9^3 (c) 9\cdot10^3-8\cdot9^3 (d) 9\cdot10

  7. (a) 8^6 (b) 8\cdot7\cdot6\cdot5\cdot4\cdot3 (c) 0 (d) 8^6-4^6 (e) 4\cdot8^4 (f) 7^5

Section 8.3

  1. 62^8, P(62,8).

  2. P(14,5).

  3. p(7,3)\cdot P(10,3) + P(7,3)\cdot P(11,3) + P(10,3)\cdot P(11,3).

  4. P(11,7)\cdot3!/7.

Section 8.4

  1. \binom{6}{3}\binom{8}{3}.

  2. (a) at least 5 (b) at least 7

  3. 10.

  4. (a) \binom{14}{4} (b) \binom{14}{4}-\binom{11}{4} (c) \binom{3}{2}\binom{7}{1}\binom{4}{1} +\binom{3}{1}\binom{7}{2}\binom{4}{1} +\binom{3}{1}\binom{7}{1}\binom{4}{2}

  5. (a) 8! (b) \binom{8}{2}\,P(8,2)\, \big[\binom{6}{2}\,P(8,2)+2\cdot7\cdot6\cdot7+7\cdot6\big]

  6. \binom{16}{7}.

  7. (a) \binom{52}{5} (b) 4 \,\binom{13}{2}\,13^3 (c) 13\,\binom{4}{2}\binom{12}{3}\,4^3 (d) 13\,\binom{4}{3}\binom{12}{2}\,4^2

    (e) 13\,\binom{4}{3}\,12\,\binom{4}{2} (f) 10\cdot(4^5-1) (g) 4\,\big[\binom{13}{5}-10\big] (h) 4\cdot10

Section 8.5

  1. (a) x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

    (b) s^6-6s^5t+15s^4t^2-20s^3t^3+15s^2t^4-6st^5+t^6

    (c) a^4+12a^3b+54a^2b^2+108ab^3+81b^4

  2. (a) \binom{4}{2}=6 (b) -\binom{9}{3}\,3^6\left(\frac{2}{5}\right)^3 =-\frac{489888}{125} (c) 0 (d) -\binom{6}{3}\,3^3\left(\frac{5}{7}\right)^3 =-\frac{67500}{343}

  3. \sum_{k=0}^n \binom{n}{k} r^k = (1+r)^n

  4. (c) k^2 = 2\binom{k}{2} + \binom{k}{1} (d) \sum_{k=1}^n k^2 = \frac{1}{6}\,n(n+1)(2n+1)


This page titled Answers is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Harris Kwong (OpenSUNY) .

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