Answers
( \newcommand{\kernel}{\mathrm{null}\,}\)
Answers to Selected Exercises
Section 2.1
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Only (a), (c), and (e) are statements.
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(a) false (b) false (c) false (d) true
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(a) π∉Z (b) 13+23+33≠32⋅42/4 (c) u is not a vowel
(d) This statement is either true or false.
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(a) true (b) true (c) true (d) false (e) false (f) true
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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, 0Q≠Q.
Section 2.2
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(a) p∧q (b) ¯q∧r (c) ¯p∨¯q (d) (p∨q)∧¯p∧q
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(a) p∧q; always false regardless of the value of r.
(b) p∨q; always true regardless of the value of r.
(c) (p∧q)∨r; true if r is true, and false if r is false.
(d) ¯q∧r; true if r is true, and false if r is false.
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(a) false (b) true
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(a) (4≤x)∧(x≤7) (b) (4<x)∧(x≤7) (c) (4≤x)∧(x<7)
Section 2.3
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(a) p⇒q (b) r⇒p (c) ¯p⇒q (d) ¯p⇒r (e) (¯p∧q)⇒r
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(a) p⇒q, which is false.
(b) p⇒r, which is true if r is true, and is false if r is false.
(c) (p∨q)⇒r, which is true if r is true, and is false if r is false.
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(a) x3−3x2+x−3=0⇒x=3
(b) x3−3x2+x−3=0⇒x=3
(c) x=3⇒x3−3x2+x−3=0
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pqrp∧q(p∧q)∨rTTTTTTTFTTTFTFTTFFFFFTTFTFTFFFFFTFTFFFFFpqrp∨qp∧r(p∨q)⇒(p∧r)TTTTTTTTFTFFTFTTTTTFFTFFFTTTFFFTFTFFFFTFFTFFFFFT
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(a) Using a truth table, we find that the implication (p∧q)⇒(q∨r) is always true. Hence, no truth value of p would make (p∧q)⇒(q∨r) false.
(b) From a truth table, we find that, (q∧r)⇒(p∧q) 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, q∧r) is true and its conclusion (in this case, p∧q) is false. For q∧r to be true, we need both q and r to be true. Now q is true and p∧q is false require p to be false.
Section 2.4
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(a) p⇔q (b) r⇔¯p (c) r⇔(q∧¯p) (d) r⇔(p∧q)
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(a) p⇔q, which is false.
(b) p⇔r, which is true if r is true, and is false if r is false.
(c) (p∨q)⇔r, which is true if r is true, and is false if r is false.
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(a) true (b) false (c) false (d) false
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We say n is odd if and only if n=2q+1 for some integer q.
Section 2.5
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pqp∨q¯p∨q¯p¯q¯p∧¯qTTTFFFFTFTFFTFTTTFTFFTFFTTTT
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Only (b) is a tautology, as indicated in the truth tables below.
(a) pq¯p¯p∨q(¯p∨q)⇒pTTFTTTFFFTFTTTFFFTTF
(b) pqp⇒q¯qp⇒¯q(p⇒q)∨(p⇒¯q)TTTFFTTFFTTTFTTFTTFFFTTT
(c) pqrp⇒q(p⇒q)⇒rTTTTTTTFTFTFTFTTFFFTFTTTTFTFTFFFTTTFFFTF
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The proofs are displayed below without explanations. Be sure to fill them in.
(b) (p∧q)⇒r≡¯p∧q∨r≡(¯p∨¯q)∨r≡¯p∨(¯q∨r)≡p⇒(¯q∨r)
(c) (p⇒¯q)∧(p⇒¯r)≡(¯p∨¯q)∧(¯p∨¯r)≡¯p∨(¯q∧¯r)≡¯p∨¯q∨r≡¯p∧(q∨r)
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(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. -
(a) true (b) true (c) false
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Only (b).
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(a) p∧q (b) p∧¯q (c) p∧q
Section 2.6
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(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 n≤2.
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(a) true (b) true (c) false (d) false (e) true
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(a) ∃x<0∃y,z∈R(y<z∧xy≤xz)
(b) ∃x∈Z[¯p(x)∧¯q(x)]
(c) ∃x,y∈R[p(x,y)∧¯q(x,y)]
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(a)
∀x,y∈R(x+y=y+x) ∃x,y∈R(x+y≠y+x) There exist real numbers x and y such that x+y≠y+x. (b)
∀x∈R+∃y∈R(y2=x) ∃x∈R+∀y∈R(y2≠x) There exists a positive real number x such that for all real numbers y, y2≠x. (c)
∃y∈R∀x∈Z(2x2+1>x2y) ∀y∈R∃x∈Z(2x2+1≤x2y) For every real number y, there exists an integer x such that 2x2+1≤x2y. -
The statement “a square must be a parallelogram” means, symbolically, ∀PQRS(PQRS is a square⇒PQRS is a parallelogram), but the statement “a square must not be a parallelogram” means ∀PQRS(PQRS is a square⇒PQRS is not a parallelogram). The second statement is not the negation of the first. The correct negation, in symbol, is ∃PQRS(PQRS is a square∧PQRS is a parallelogram). In words, it means “there exists a square that is not a parallelogram.”
Section 3.1
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Placing six dominoes horizontally in each row covers the entire chessboard.
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Let f(x)=x3−12x+2. From the following chart x−4−3−2−101234f(x)−141218132−9−14−718 we conclude there x3−12x+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.
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n=3.
Section 3.2
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No, 23+1=9 is composite.
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According to (i), the number √2 is irrational. It follows from (ii) that 4√2=√√2 is also irrational. Applying (ii) one more time, we conclude that 8√2=√4√2 is irrational.
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(a) The statement is false, because (−3)2>(−2)2, but −3≯−2.
(b) The statement is false, because when n=41, n2+n+41=412+41+41=41(41+1+1)=41⋅43 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 a≠b such that a2+b2=2ab. Then 0=a2−2ab+b2=(a−b)2 would have implied that a=b. This contradicts the assumption that a≠b. Therefore, a2+b2≠2ab.
Suppose (p⇒q)∨(p⇒¯q) is false for some logical statements p and q. For a disjunction to be false, we need
p⇒q 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
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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 12⋅224=1. Hence, the identity holds when n=1. Assume the identity holds when n=k for some integer k≥1; that is, assume 13+23+33+⋯+k3=k2(k+1)24 for some integer k≥1. 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
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We proceed by induction on n. When n=1, the product n(n+1)(n+2) becomes 1⋅2⋅3=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 k≥1; that is, assume that k(k+1)(k+2) is a multiple of 3 for some integer k≥1. 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.
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(b) Sn=1−1(n+1)! for all integers n≥1.
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(b) Tn=n+12n+3 for all integers n≥0.
Section 3.6
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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=1⋅1=1. Hence, the identity holds when n=1. Assume the identity holds when n=k for some integer k≥1; that is, assume F21+F22+F23+⋯+F2k=FkFk+1 for some integer k≥1. 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
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(a) {−5,−4,−3,−2,−1,0,1,2,3} (b) {1,2,3} (c) {0,−2,3} (d) {−3,3}
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(a) {n∈Z∣n<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}\}}
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(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.
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(a) (−4,7) (b) (−4,7] (c) (0,7]
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(a) 10 (b) 11 (c) 7
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(a) true (b) true (c) true (d) false
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(a) It is incorrect to write (3,7]=3<x≤7 because (3,7] is a set, but 3<x≤7 is a logical statement.
(b) No, because both {x∈R∣x2<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 “{x∈R∣x2<0}=∅.”
Section 4.2
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(a) true (b) true (c) true (d) true (e) true (f) false
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We have Z⊆N because every integer n is also a rational number, as we can write it as the rational number n1.
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Yes, this is the transitive property.
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(e) {∅,{a},{{b}},{a,{b}}}
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(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
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(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,…}
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(a) false (b) false
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(a) E∩D (b) ¯E∪B
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For example, take A={x}, and B={{x},x}.
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Assume A⊆C and B⊆C, we want to show that A∪B⊆C. In this regard, let x∈A∪B, we want to show that x∈C as well. Since x∈A∪B, the definition of set union asserts that either x∈A or x∈B.
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Case 1: If x∈A, then A⊆C implies that x∈C.
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Case 2: If x∈B, then B⊆C implies that x∈C.
In both cases, we find x∈C. This proves that A∪B⊆C.
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(a) The notation ∩ is used to connect two sets, but “x∈A” and “x∈B” are both logical statements. We should also use ⇔ instead of ≡. The statement should have been written as “x∈A∧x∈B⇔x∈A∩B.”
(b) If we read it aloud, it sounds perfect: If x belongs to A and B, then x belongs to A∩B. 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 “x∈A,” which is a logical statement. But, after ∧, we have “B,” which is a set, and not a logical statement. It should be written as “x∈A∧x∈B⇒x∈A∩B.”
Section 4.4
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(a) {(−2,0),(−2,4),(2,0),(2,4)}
(b) {(−2,−3),(−2,0),(−2,3),(−2,−3),(−2,0),(−2,3)}
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2⋅2⋅2⋅3=24.
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(a) {(−2,∅),(−2,{−2}),(−2,{2}),(−2,{−2,2}),(2,∅),(2,{−2}),(2,{2}),(2,{−2,2})}
Section 4.5
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⋂∞n=1An=[0,2), ⋃∞n=1An=(−1,∞).
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⋂∞n=0Cn=∅, ⋃∞n=0Cn=N∪{0}.
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⋂n∈NEn=E0={0}, ⋃n∈NEn=Z.
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⋃i∈IAi=[1,∞), ⋂i∈IAi={1}.
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⋂x∈(1,2)(1−2x,x2)=[−1,1], ⋃x∈(1,2)(1−2x,x2)=(−3.4).
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⋂r∈(0,∞)Ar={(0,0)}, ⋃r∈(0,∞)Ar=R∗×R+∪{(0,0)}.
Section 5.1
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(a) 3 (b) 3 (c) 3 (d) 1
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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.
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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
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(a) 23, 1 (b) −11, 1 (c) −6, 13
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This is an immediate consequence of Corollary [cor:divalgo].
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(a) Let n be any integer. Then nmod.
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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.
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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.
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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.
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(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
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(a) 1\cdot 27+ 0\cdot 81= 27 (b) -3\cdot 24+ 1\cdot 84= 12 (c) -35\cdot1380+16\cdot3020= 20
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1, 2, 17, and 34.
Section 5.5
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Since -3\cdot(2n+1)+2\cdot(3n+2) = 1, we deduce that \gcd(2n+1,3n+2)=1.
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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
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(a) 3^2\cdot5^2\cdot7 (b) 2\cdot3^2\cdot7^2\cdot11
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(a) 81 (b) 168
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Every 50 days.
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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}.
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(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
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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.
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The sum is 9, and the product is 7.
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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.
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Both methods give 4^{45}=1 in \mathbb{Z}_{11}.
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(a) 9
Section 6.1
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-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}
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[0,\infty).
Section 6.2
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\big[\frac{7}{3},\infty\big).
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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.
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(a) Yes, because no division by zero will ever occur.
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-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}
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(a) 7 (b) 7 (c) 3
Section 6.3
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(a) No. For example, f(0)=f(2)=1.
(b) Yes, since g'(x)=3x^2-4x=x(3x-4)>0 for x>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}.
-
(a) One-to-one (b) Not one-to-one
-
(a) Not one-to-one (b) One-to-one
-
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}
-
(a) One-to-one (b) Not one-to-one (c) Not one-to-one
Section 6.4
-
(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.
-
(b) Not onto (c) Onto
-
(b) Not onto (c) Onto
-
No, because we have at most two distinct images, but the codomain has four elements.
-
(a) Onto (b) Not onto (c) Not onto
Section 6.5
-
(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\}
-
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\}
-
(a) [20,26); \{20,23,26\} (b) [-3,-\frac{4}{3}\big); \{-2\}
-
(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\}
-
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).
-
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).
-
\{0,1,4,9\}; \{0,\pm1,\pm2,\pm3\}.
Section 6.6
-
Only (e) is bijective.
-
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).
-
{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\(. \cr}
-
{s^{-1}}:{(-\infty,-3)}\to{\mathbb{R}}, where s^{-1}(x) = \frac{1}{2}\, \ln\left(\frac{4-x}{7}\right).
-
(a) {u^{-1}}:{\mathbb{Q}}\to{\mathbb{Q}}, u^{-1}(x)=(x+2)/3
-
The images under {\alpha^{-1}}:{\{a,b,c,d,e,f,g,h\}}\to {\{1,2,3,4,5,6,7,8\}} are given below. \begin{array}{|c||*{8}{c|}} \hline x & a & b & c & d & e & f & g & h \\ \hline \alpha^{-1}(x)& 2 & 5 & 8 & 3 & 6 & 7 & 1 & 4 \\ \hline \end{array}
Section 6.7
-
Both f\circ g and g\circ f are from \mathbb{R} to \mathbb{R}, where (f\circ g)(x)=15x^2+19, and (g\circ f)(x)=75x^2-30x+7.
-
We do not need to find the formula of the composite function, as we can evaluate the result directly: f(g(f(0))) = f(g(1)) = f(2) = -5.
-
(a) {g\circ f}:{\mathbb{Z}}\to{\mathbb{Q}}, (g\circ f)(n)=1/(n^2+1)
(b) {g\circ f}:{\mathbb{R}}\to{(0,1)}, (g\circ f)(x)=x^2/(x^2+1)
-
(a) {g\circ f}:{\{1,2,3,4,5\}}\to{\{1,2,3,4,5\}},
\phantom{(a)} (g\circ f)(1)=2, (g\circ f)(2)=5, (g\circ f)(3)=1, (g\circ f)(4)=3, (g\circ f)(5)=4
-
{g\circ f}:{\mathbb{Z}}\to{\mathbb{Z}}, (g\circ f)(n) = \cases{ 3(2n-1) & if\)n0\(, \cr 2n+1 & if\)n < 0\(. \cr}
-
(a) {f\circ g}:{\mathbb{Z}}\to{\mathbb{Z}}, (f\circ g)(n) = 3-n
\phantom{(a)} {(f\circ g)^{-1}}:{\mathbb{Z}}\to{\mathbb{Z}}\), (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
-
(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}
-
(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\}.
-
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}
-
\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
-
(a) Reflexive, symmetric, antisymmetric, and transitive.
(b) Irreflexive, and symmetric.
(c) Irreflexive, and transitive.
-
(a) Antisymmetric.
(b) Reflexive, symmetric, and transitive.
(c) Irreflexive, symmetric, and transitive.
-
Reflexive, symmetric, and transitive.
-
Antisymmetric, and transitive.
-
Irreflexive, and antisymmetric.
-
Symmetric.
-
(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}\(} \end{array} \\ \noalign{\medskip} \begin{array}{c} \emptyset \\ \{a\} \\ \{b\} \\ \{c\} \\ \{a,b\} \\ \{a,c\} \\ \{b,c\} \\ \{a,b,c\} \end{array} & \setlength{\arraycolsep}{7.7pt} \left(\begin{array}{cccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right) \end{array}
-
(a) Symmetric.
(b) Reflexive, and symmetric.
-
(a) Reflexive, antisymmetric, and transitive.
(b) Reflexive, symmetric, and transitive.
(c) Symmetric.
-
(a) Reflexive, antisymmetric, and transitive.
(b) Symmetric.
(c) Symmetric, and transitive.
-
(a) Reflexive, and transitive.
(b) Symmetric,
(c) Reflexive, symmetric, and transitive.
-
(a) Symmetric, and transitive.
(b) Reflexive, symmetric, and transitive.
(c) Reflexive, and transitive.
Section 7.3
-
(a) The equivalence classes are of the form \{3-k,3+k\} for some integer k. For instance, [3]=\{3\}, [2]=\{2,4\}, [1]=\{1,5\}, and [-5]=\{-5,11\}.
(b) There are three equivalence classes: [0]=3\mathbb{Z}, [1]=1+3\mathbb{Z}, and [2]=2+3\mathbb{Z}.
-
(a) True
(b) False
(c) [\{1,5\}] = \big\{ \{1\}, \{1,2\}, \{1,4\}, \{1,5\}, \{1,2,4\}, \{1,2,5\}, \{1,4,5\}, \{1,2,4,5\} \big\}
(d) [X] = \{(X\cap T)\cup Y \mid Y\in\wp(\overline{T})\}. In other words, S\sim X if S contains the same element in X\cap T, plus possibly some elements not in T.
-
(a) Yes, with [(a,b)] = \{(x,y) \mid y=x+k \mbox{ for some constant\)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.
-
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
-
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
-
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.
-
(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
-
B=\big\{\emptyset,\{a\},\{a,b\},\{a,b,c\},\{a,b,c,d\}\big\}.
Section 8.2
-
6.
-
70.
-
7\cdot5+7\cdot4+5\cdot4
-
4^5, 4^5-3\cdot4^2
-
(a) 52^4 (b) 39^4 (c) 4\cdot13^4 (d) 4\cdot48\cdot52^3 (e) 52^4-48^4
-
(a) 9\cdot10^3 (b) 8\cdot9^3 (c) 9\cdot10^3-8\cdot9^3 (d) 9\cdot10
-
(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
-
62^8, P(62,8).
-
P(14,5).
-
p(7,3)\cdot P(10,3) + P(7,3)\cdot P(11,3) + P(10,3)\cdot P(11,3).
-
P(11,7)\cdot3!/7.
Section 8.4
-
\binom{6}{3}\binom{8}{3}.
-
(a) at least 5 (b) at least 7
-
10.
-
(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}
-
(a) 8! (b) \binom{8}{2}\,P(8,2)\, \big[\binom{6}{2}\,P(8,2)+2\cdot7\cdot6\cdot7+7\cdot6\big]
-
\binom{16}{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
-
(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
-
(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}
-
\sum_{k=0}^n \binom{n}{k} r^k = (1+r)^n
-
(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)