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Centralizer of a subset, 112 Chain, 194 Circle, 2 Class, equation, 135 equivalence, 9 R-,9 Closed with respect to multiplication, i.e. product of elements in the set belong to the set. Codomain, 12, 13 Common part, 3 Commutative groupoid, 29 Commutator, 112 Commutator subgroup, 112 Commute: a and b commute if ab = ba. Complement, 234 Component, p-, 191 Composition factors, 164 Composition of maps, 17 Composition series, 158, 163 Congruence, 269 Conjugacy classes, 171 Conjugate, subgroups, 131 Conjugates, H-, 134 274 Correspondence theorem, 120 Coset, 108 double-ended, 265 representation, 219 representative, 219 Countable, 14 Countably infinite, 14 Cycles, 167 Cyclic group, 101 Degree of permutational representation, 217 Dependent, 192 Derived subgroup, 112 Difference of sets, 4 Different products, 245 Dihedral group, 75 infinite, 257 of degree 4, Table 5.2, 152 Dimension of vector space, 89 Direct product, 143 external, 143 internal, 146 Direct summand, 178 Direct sums, finite, 178 infinite, 182 Divisible group, 205, 209 Element, 1 order of, 103 Epimorphism, 42 Equivalence relation, 8 Extension of groups, 232-234 splitting, 234-238 Extension of mapping to homomorphism, 185, 250 Extension property, free abelian groups, 185 free groups, 250 Factor groups, 114 Factor of a factor theorem (third isomorphism theorem), 121 Factors of a subnormal series, 158 Faithful representation, 218 Family of indexed sets, 182 Fields of complex numbers, 86 Finitely generated abelian groups, 197 subgroups of, 202 factor groups of, 204 Finitely generated group, 98, 197, 227 Finitely presented, 253 Finite presentation, 253 Four group, 144, 148 INDEX 275 Free abelian group, 186 rank of, 193 Free group, 246 intersections of finitely generated subgroups, 265 length of an element, 248 rank of, 255 subgroups, 260 subgroups of finite index, 264 Free on a set, 246 Free set of generators, 246 Freely generated, 186, 246 Frobenius' representation, 225 Frobenius' theorem, 224 Full linear group, 90 Fundamental theorem of arithmetic, 269 Fundamental theorem of finitely generated abelian groups, 197 Galois
group, 159 Generators, free set of, 246 Greatest common divisor, 269 Group, abelian, 177-212 alternating, 62 automorphism, 83 commutative, 29 commutator, 112 cyclic, 101 definition of, 50 derived, 112 extension, 234, 238 finitely generated, 98, 197, 227 free, 246 isometry, 64, 67 linear, 90 matrices, 81 mixed,188 Mobius transformations, 77 nilpotent, 142 number of groups of given order, 156 primary, = p-group, 190 Priifer groups, 191 quaternion, 151 simple, 158, 163 symmetric, 56 table, 22 torsion, 188 torsion-free, 188 Groupoid, 26 abelian, 29 associative, 29 commutative, 29 equality of, 28 finite, 29 identity, 30 infinite, 29 order, 29 table of, 22 Groups of order p, 148 p2,147 2p, 149 8, 150 Groups of order (cont.) 12, 152-156 15, 131 Groups of small order, 148-156 Hamiltonian group = quaternion group, 151 Higher central series = upper central series, 142 Homomorphism, natural, 115 Homomorphism, of a groupoid, 40 of a group, 94 Homomorphism theorem (first isomorphism theorem), 117 Identical products, 245 Identity, of a groupoid, 30 of a group, 50 Iff = if and only if Image, 12, 13 Independence, independen~ 192 Index of a subgroup, 110 Index 2, 116 Indexed family, 182 Indexing set, 182 Initial segments, 261 Inner automorphism (see Automorphism.) Intersection, 3 Intersection, of normal subgroups, 114 of subgroups, 55 Invariant subgroup = normal subgroup, 111 Invariants of finitely generated abelian group type of abelian group, 200 Inverse, 32 Inverse pairs, 246 Inversion, 60 Isometries, of line, 64 of plane, 67 Isomorphism, of group, 94 of groupoid, 42 Isomorphism theorems, first, 117 second, 125 subgroup, 125 third, 121 Jordan-Holder theorem, 164 Kernel,117 Klein four group, 144, 148 Lagrange's theorem, 109 Largest set, 3 Least common multiple multiple, 269 Length of, lowest common composition series, 164 cycle, 167 element in free group, 248 Linear
fractional transformation Mobius transformation, 77 Linear group, 90 Mal'cev's theorem, 230 276 INDEX Map or mapping, bijection, 14 codomain, 12, 13 composition, 17 definition, 11, 13 domain, 12, 13 equality, 13 matching, 14 one-to-one, 14 onto, 14 restriction of, 14 Matching, 14 Matrices, 81 Matrices, groups of, 81 Maximal independent set, 193 Maximal set, 194 Metacyclic, 243 Mixed group, 188 Mobius transformations, 77 Monomorphism, 42 Multiplication in a set, 2 Multiplication table, 22 Natural homomorphism, 115 Negative, 178 Nielsen-Schreier theorem, 260 Nilpotent group, 142 Normal closure, 253 Normal subgroup, 111 Normal subgroup generated by a set, 253 Normalizer, 112, 133 Odd permutation, 60 One-to-one mapping, 14 Onto mapping, 14 Operation, binary, 19 Order of a group, 50 of a groupoid, 29 of an element, 103 Ordered pair, 2 Partition, 9 P"', group of type = Priifer group, 191 p-component, 191 Periodic = torsion group, 188 Permutation, 56 even, 60 group = symmetric group, 56 odd, 60 Permutational representation, 216 degree of, 217 p-group, 139, 190 p-primary group = p-group, 190 p-Priifer group, 191 main theorem, 206 Pre image, 12, 120 Presentation, 253 finite, 253 finitely presented, 253 group of a, 253 of a group, 253 Primary component p-component, 191 Product, cartesian, 6 direct, 143 reduced, 245 Products, different, 245 identical, 245 of subsets of a group, 109 Proper subgroup, 106 Priifer group, 191 Quaternions = Hamiltonian groups, 151 Quotient group = factor group, 114 Range, 12 Rank, free abelian group, 193 free group, 255 torsion-free abelian group, 193 R-block,9 R-class,9 Reduced product, 245 Reflection, 68, 69 Reflexive property, 9 Representation, coset, 219 degree of, 217 Frobenius, 225 permutational, 216 right-regular, 216 Representative of a coset, 219 Representatives of the equivalence classes, 134 Right-regular representation, 216 Rotation, 68, 69 Schreier's
subgroup theorem, 260 Schreier transversal, 261 Schur's theorem, 242 Semigroup, 29 Series, upper central series, 142 ascending central, composition, 158 solvable, 158 subnormal, 158 Set, 1 Simple group, 158, 163, 172 Smallest set, 3 Solvable, group, 158, 161 by radicals, 159 series, 158 word problem, 246 Split, 234 Splitting extension, 234, 238 Steinitz exchange theorem, 192 Subgroup, commutator, 112 conjugate, 131 definition, 54 derived, 112 generated by a set, 98 invariant, = normal subgroup, 111 isomorphism theorem (second isomorphism theorem), 125 normal, 111 Subgroup (cont.) of index 2, 116 proper, 106 Sylow, 130 theorem for cyclic groups, 105, 126 theorem for free groups, 260 torsion, 189 Subnormal series, 158 factors of, 158 Subset, 2 subgroup generated by, 98 product of, 109 Sum, direct, 178 finite, 178 infinite, 182 Sum of vectors, 89 Sylow subgroup, 130 Sylow's theorems, first, 130 second and third, 131 Symmetric group, definition, 56 of degree 4, 59 of degree 5 or more, 172 Symmetric property, 9 Symmetry groups, 73 INDEX 277 Symmetry groups of an algebraic structure, 83 Table, multiplication, 22 Three-cycle, 172 Torsion group, 188 subgroup, 189 Torsion-free group, 188 Transfer, 240 Transitive property, 9 Translation, 68, 69 Transposition, 167 Transversal, 219 Schreier, 261 Type of a finitely generated abelian group, 200 Union, 3 Unit element = identity element, 30 Upper central series, 142 Vector space of dimension n, 89 Vector sum, 89 Word problem, 246 Zero element, 178 Zorn's lemma, 194 Symbols and Notations Symbols Automorphism group of G, 84 xg(xg)-l, 224 Alternating group of degree n, 61 Complex numbers, 1 Nonzero complex numbers, 51 Centralizer of A, 112 Cyclic group of order n, 148 Dihedral group of degree n, 76 Subgroup generated by X, 98 Conjugation by Xk' 233 Isometries of plane, 67 Group of isometries of R, 64 Symmetry group of S, 73 Group of isometries of R that move integers to integers,
(X;R) Presentation, 253 now going going to apply the machinery of bounded cohomology to understand actions on S1. Recall that the central extension 0 Z Homeo+ Z (R) Homeo+(S1) 0 deﬁnes the Euler class e ∈ H 2(Homeo+(S1), Z). We have also shown that there is a representative cocycle c(f, g) taking the values in {0, 1}, deﬁned by f ◦ g ◦ Tc(f,g) = ¯f ◦ ¯g, where for any f, the map ¯f is the unique lift to R such that ¯f (0) ∈ [0, 1). Since c takes values in {0, 1}, in particular, it is a bounded cocycle. So we can use it to deﬁne Deﬁnition (Bounded Euler class). The bounded Euler class eb ∈ H 2 b (Homeo+(S1), Z) is the bounded cohomology class represented by the cocycle c. By construction, eb is sent to e via the comparison map c2 : H 2 b (Homeo+(S1), Z) H 2(Homeo+(S1), Z). In fact, the comparison map is injective. So this eb is the unique element that is sent to e, and doesn’t depend on us arbitrarily choosing c as the representative. Deﬁnition (Bounded Euler class of action). The bounded Euler class of an action h : Γ → Homeo+(S1) is h∗(eb) ∈ H 2 b (Γ, Z). By naturality (proof as exercise), h∗(eb) maps to h∗(e) under the comparison map. The bounded Euler class is actually a rather concrete and computable object. Note that if we have an element ϕ ∈ Homeo+(S1), then we obtain a group homomorphism Z → Homeo+(S1) that sends 1 to ϕ, and vice versa. In other words, we can identify elements of Homeo+(S1) with homomorphisms h : Z → Homeo+(S1). Any such homomorphism will give a bounded Euler class h∗(eb) ∈ H 2 b (Z, Z) ∼= R/Z.
Exercise. If h : Z → Homeo+(S1) and ϕ = h(1), then under the isomorphism b (Z, Z) ∼= R/Z, we have h∗(eb) = Rot(ϕ), the Poincar´e rotation number of ϕ. H 2 Thus, one way to think about the bounded Euler class is as a generalization of the Poincar´e rotation number. Exercise. Assume h : Γ → Homeo+(S1) takes values in the rotations Rot. Let χ : Γ → R/Z the corresponding homomorphism. Then under the connecting homomorphism Hom(Γ, R/Z) δ H 2 b (Γ, Z), we have δ(χ) = h∗(eb). 26 3 Actions on S1 IV Bounded Cohomology Exercise. If h1 and h2 are conjugate in Homeo+(S1), i.e. there exists a ϕ ∈ Homeo+(S1) such that h1(γ) = ϕh2(γ)ϕ−1, then 1(e) = h∗ h∗ 2(e), 1(eb) = h∗ h∗ 2(eb). The proof involves writing out a lot of terms explicitly. How powerful is this bounded Euler class in distinguishing actions? We just saw that conjugate actions have the same bounded Euler class. The converse does not hold. For example, one can show that any action with a global ﬁxed point has trivial bounded Euler class, and there are certainly non-conjugate actions that both have global ﬁxed points (e.g. take one of them to be the trivial action). It turns out there is a way to extend the notion of conjugacy so that the bounded Euler class becomes a complete invariant. Deﬁnition (Increasing map of degree 1). A map ϕ : S1 → S1 is increasing of degree 1 if there is some ˜ϕ : R → R lifting ϕ such that ˜ϕ is is monotonically increasing and ˜ϕ(x + 1) = ˜ϕ(x) + 1 for all x ∈ R. Note that there is no continuity assumption made on ϕ. On the other hand, it is
an easy exercise to see that any monotonic map R → R has a countable set of discontinuities. This is also not necessarily injective. Example. The constant map S1 → S1 sending x → 0 is increasing of degree 1, as it has a lift ˜ϕ(x) = [x]. Equivalently, such a map is one that sends a positive 4-tuple to a weakly positive 4-tuple (exercise!). Deﬁnition (Semiconjugate action). Two actions h1, h2 : Γ → Homeo+(S1) are semi-conjugate if there are increasing maps of degree 1 ϕ1, ϕ2 : S1 → S1 such that (i) h1(γ)ϕ1 = ϕ1h2(γ) for all γ ∈ Γ; (ii) h2(γ)ϕ2 = ϕ2h1(γ) for all γ ∈ Γ. One can check that the identity action is semiconjugate to any action with a global ﬁxed point. Recall the following deﬁnition: Deﬁnition (Minimal action). An action on S1 is minimal if every orbit is dense. Lemma. If h1 and h2 are minimal actions that are semiconjugate via ϕ1 and ϕ2, then ϕ1 and ϕ2 are homeomorphisms and are inverses of each other. Proof. The condition (i) tells us that h1(γ)(ϕ1(x)) = ϕ1(h2(γ)(x)). 27 3 Actions on S1 IV Bounded Cohomology for all x ∈ S1 and γ ∈ Γ. This means im ϕ1 is h1(Γ)-invariant, hence dense in S1. Thus, we know that im ˜ϕ1 is dense in R. But ˜ϕ is increasing. So ˜ϕ1 must be continuous. Indeed, we can look at the two limits lim xy ˜ϕ1(x) ≤ lim xy ˜ϕ1(x). But since ˜ϕ1 is increasing, if ˜ϕ1 were discontinuous at y ∈ R, then the inequality would be strict, and hence the image misses a non-
trivial interval. So ˜ϕ1 is continuous. We next claim that ˜ϕ1 is injective. Suppose not. Say ϕ(x1) = ϕ(x2). Then by looking at the lift, we deduce that ϕ((x1, x2)) = {x} for some x. Then by minimality, it follows that ϕ is locally constant, hence constant, which is absurd. We can continue on and then decide that ϕ1, ϕ2 are homeomorphisms. Theorem (F. Ghys, 1984). Two actions h1 and h2 are semiconjugate iﬀ h∗ h∗ 2(eb). 1(eb) = Thus, in the case of minimal actions, the bounded Euler class is a complete invariant of actions up to conjugacy. Proof. We shall only prove one direction, that if the bounded Euler classes agree, then the actions are semi-conjugate. Let h1, h2 : Γ → Homeo+(S1). Recall that c(f, g) ∈ {0, 1} refers to the (normalized) cocycle deﬁning the bounded Euler class. Therefore c1(γ, η) = c(h1(γ), h1(η)) c2(γ, η) = c(h2(γ), h2(η)). are representative cocycles of h∗ 1(eb), h∗ 2(eb) ∈ H 2 b (Γ, Z). By the hypothesis, there exists u : Γ → Z bounded such that c2(γ, η) = c1(γ, η) + u(γ) − u(γη) + u(η) for all γ, η ∈ Γ. Let ¯Γ = Γ ×c1 Z be constructed with c1, with group law (γ, n)(η, m) = (γη, c1(γ, η) + n + m) We have a section s1 : Γ → ¯Γ γ → (γ, 0). We also write δ = (e, 1) ∈ ¯Γ, which generates the copy of Z in ¯Γ. Then we have s1(γη)δc1(
γ,η) = s1(γ)s2(η). Likewise, we can deﬁne a section by s2(γ) = s1(γ)δu(γ). 28 3 Actions on S1 IV Bounded Cohomology Then we have s2(γη) = s1(γη)δu(γη) = δ−c1(γ,η)s1(γ)s1(η)δu(γη) = δ−c1(γ,η)δ−u(γ)s2(γ)δ−u(η)s2(η)δu(γη) = δ−c1(γ,η)−u(γ)+u(γη)−u(η)s2(γ)s2(η) = δ−c2(γ,η)s2(γ)s2(η). Now every element in ¯Γ can be uniuely written as a product s1(γ)δn, and the same holds for s2(γ)δm. Recall that for f ∈ Homeo+(S1), we write ¯f for the unique lift with ¯f (0) ∈ [0, 1). We deﬁne Φi(si(γ)δn) = hi(γ) · Tn. We claim that this is a homomorphism! We simply compute Φi(si(γ)δnsi(η)δm) = Φi(si(γ)si(η)δn+m) = Φi(si(γη)δci(γ,η)+n+m) = hi(γη)Tci(γ,η)Tn+m = hi(γ)hi(η)Tn+m = hi(γ)Tnhi(η)Tm = Φi(si(γ)δn)Φi(si(η)δm). So we get group homomorphisms Φi : ¯Γ → Homeo+ Claim. For any x ∈ R, the map Z (R). ¯Γ →
R g → Φ1(g)−1Φ2(g)(x) v(g, x) = Φ1(g)−1Φ(g)x. is bounded. Proof. We deﬁne We notice that v(gδm, x) = Φ1(gδm)−1Φ2(gδm)(x) = Φ1(g)−1T−mTmΦ2(g) = v(g, x). Also, for all g, the map x → v(g, x) is in Homeo+ Z (R). Hence it is suﬃcient to show that γ → v(s2(γ), 0) 29 3 Actions on S1 IV Bounded Cohomology is bounded. Indeed, we just have v(s2(γ), 0) = Φ1(s2(γ)−1Φ2(s2(γ))(0) = Φ1(s1(γ)δu(γ))−1Φ2(s2(γ))(0) −1 = δ−u(γ)h1(γ) h2(γ)(0) −1 = −u(γ) + h1(γ) (h2(γ)(0)). But u is bounded, and also −1 h1(γ) (h2(γ)(0)) ∈ (−1, 1). So we are done. Finally, we can write down our two quasi-conjugations. We deﬁne ˜ϕ(x) = sup g∈¯Γ v(g, x). Then we verify that Reducing everything modulo Z, we ﬁnd that ˜ϕ(Φ2(h)x) = Φ1(h)(ϕ(x)). ϕh2(γ) = h1(γ)ϕ. The other direction is symmetric. 3.2 The real bounded Euler class The next thing we do might be a bit unexpected. We are going to forget that the cocycle c takes values in Z, and view it as an element in the real bounded cohomology group. Deﬁnition (Real bounded Euler class). The real bounded Euler class is the b (Homeo+(S1), R) obtained
by change of coeﬃcients from Z → R. class eb The real bounded Euler class of an action h : Γ → Homeo+(S1) is the pullback R ∈ H 2 h∗(eb R) ∈ H 2 b (Γ, R). A priori, this class contains less information that the original Euler class. However, it turns out the real bounded Euler class can distinguish between very diﬀerent dynamical properties. Recall that we had the Gersten long exact sequence 0 Hom(Γ, R/Z) δ H 2 b (Γ, Z) → H 2 b (Γ, R). By exactness, the real bounded Euler class vanishes if and only if eb is in the image of δ. But we can characterize the image of δ rather easily. Each homomorphism χ : Γ → R/Z gives an action by rotation, and in a previous exercise, we saw the bounded Euler class of this action is δ(χ). So the image of δ is exactly the bounded Euler classes of actions by rotations. On the other hand, we know that the bounded Euler class classiﬁes the action up to semi-conjugacy. So we know that 30 3 Actions on S1 IV Bounded Cohomology Corollary. An action h is semi-conjugate to an action by rotations iﬀ h∗(eb R) = 0. We want to use the real bounded Euler class to classify diﬀerent kinds of actions. Before we do that, we ﬁrst classify actions without using the real bounded Euler class, and then later see how this classiﬁcation is related to the real bounded Euler class. Theorem. Let h : Γ → Homeo+(S1) be an action. Then one of the following holds: (i) There is a ﬁnite orbit, and all ﬁnite orbits have the same cardinality. (ii) The action is minimal. (iii) There is a closed, minimal, invariant, inﬁnite, proper subset K S1 such that any x ∈ S1, the closure of the orbit h(Γ)x contains K. We will provide a proof sketch. More details can be found in Hector–H
irsch’s Introduction to the geometry of foliations. Proof sketch. By compactness and Zorn’s lemma, we can ﬁnd a minimal, nonempty, closed, invariant subset K ⊆ S1. Let ∂K = K \ ˚K, and let K be the set of all accumulation points of K (i.e. the set of all points x such that every neighbourhood of x contains inﬁnitely many points of K). Clearly K and ∂K are closed and invariant as well, and are contained in K. By minimality, they must be K or empty. (i) If K = ∅, then K is ﬁnite. It is an exercise to show that all orbits have the same size. (ii) If K = K, and ∂K = ∅, then K = ˚K, and hence is open. Since S1 is connected, K = S1, and the action is minimal. (iii) If K = K = ∂K, then K is perfect, i.e. every point is an accumulation point, and K is totally disconnected. We deﬁnitely have K = S1 and K is inﬁnite. It is also minimal and invariant. Let x ∈ S1. We want to show that the closure of its orbit contains K. Since K is minimal, it suﬃces to show that h(Γ)x contains a point in K. If x ∈ K, then we are done. Otherwise, the complement of K is open, hence a disjoint union of open intervals. For the sake of explicitness, we deﬁne an interval of a circle as follows — if a, b ∈ S1 and a = b, then (a, b) = {z ∈ S1 : (a, z, b) is positively oriented}. Now let (a, b) be the connected component of S1 \ K containing x. Then we know a ∈ K. We observe that S1 \ K has to be the union of countably many intervals, and moreover h(Γ)a consists of end points of these open intervals. So h(Γ)a is a countable set. On the other hand, since K is perfect, we know K is uncountable. The point is that
this allows us to pick some element y ∈ K \ h(Γ)a. 31 3 Actions on S1 IV Bounded Cohomology Since a ∈ K, minimality tells us there exists a sequence (γn)n≥1 such that h(γn)a → y. But since y ∈ h(Γ)a, we may wlog assume that all the points {h(γn)a : n ≥ 1} are distinct. Hence {h(γn)(a, b)}n≥1 is a collection of disjoint intervals in S1. This forces their lengths tend to 0. We are now done, because then h(γn)x gets arbitrarily close to h(γn)a as well. We shall try to rephrase this result in terms of the real bounded Euler class. It will take some work, but we shall state the result as follows: Corollary. Let h : Γ → S1 be an action. Then one of the following is true: (i) h∗(eb R) = 0 and h is semi-conjugate to an action by rotations. (ii) h∗(eb R) = 0, and then h is semi-conjugate to a minimal unbounded action, i.e. {h(γ) : γ ∈ Γ} is not equicontinuous. Observe that if Λ ⊆ Homeo+(S1) is equicontinuous, then by Arzela–Ascoli, its closure ¯Λ is compact. To prove this, we ﬁrst need the following lemma: Lemma. A minimal compact subgroup U ⊆ Homeo+(S1) is conjugate to a subgroup of Rot. Proof. By Kakutani ﬁxed point theorem, we can pick an U -invariant probability measure on S1, say µ, such that µ(S1) = 2π. We parametrize the circle by p : [0, 2π) → S1. We deﬁne ϕ ∈ Homeo+(S1) by where s ∈ [0, 2π) is unique with the property that ϕ(p(t)) = p(s), µ(p([0, s)) = t. One then veriﬁes
that ϕ is a homeomorphism, and ϕU ϕ−1 ⊆ Rot. Proof of corollary. Suppose h∗(eb previous trichotomy. R) = 0. Thus we are in case (ii) or (iii) of the We ﬁrst show how to reduce (iii) to (ii). Let K S1 be the minimal h(Γ)invariant closed set given by the trichotomy theorem. The idea is that this K misses a lot of open intervals, and we want to collapse those intervals. We deﬁne the equivalence relation on S1 by x ∼ y if {x, y} ⊆ ¯I for some connected component I of S1 \ K. Then ∼ is an equivalence relation that is h(Γ)-invariant, and the quotient map is homeomorphic to S1 (exercise!). Write i : S1/ ∼→ S1 for the isomorphism. In this way, we obtain an action of ρ : Γ → Homeo+(S1) which is minimal, and the map ϕ : S1 pr S1/ ∼ i S1 intertwines the two actions, i.e. ϕh(γ) = ρ(γ)ϕ. Then one shows that ϕ is increasing of degree 1. Then we would need to ﬁnd ψ : S1 → S1 which is increasing of degree 1 with ψρ(γ) = h(γ)ψ. 32 3 Actions on S1 IV Bounded Cohomology But ϕ is surjective, and picking an appropriate section of this would give the ψ desired. So h is semi-conjugate to ρ, and 0 = h∗(eb Thus we are left with ρ minimal, with ρ∗(eb R) = ρ∗(eb R) = 0. We have to show that ρ is not equicontinuous. But if it were, then ρ(Γ) would be contained in a compact subgroup of Homeo+(S1), and hence by the previous lemma, would be conjugate to an action by rotation. R). The following theorem gives us a glimpse of what unbounded actions look like: Theorem (Ghys, Margulis). If ρ : Γ → Homeo+(S1) is
an action which is minimal and unbounded. Then the centralizer CHomeo+(S1)(ρ(Γ)) is ﬁnite cyclic, say ϕ, and the factor action ρ0 on S1/ϕ ∼= S1 is minimal and strongly proximal. We call this action the strongly proximal quotient of ρ. Deﬁnition (Strongly proximal action). A Γ-action by homeomorphisms on a compact metrizable space X is strongly proximal if for all probability measures µ on X, the weak-∗ closure Γ∗µ contains a Dirac mass. For a minimal action on X = S1, the property is equivalent to the following: – Every proper closed interval can be contracted. In other words, for every interval J ⊆ S1, there exists a sequence (γn)n≥1 such that diam(ρ(γn)J) → 0 as n → ∞. Proof of theorem. Let ψ commute with all ρ(γ) for all γ ∈ Γ, and assume ψ = id. Claim. ψ has no ﬁxed points. Proof. Otherwise, if ψ(p) = p, then ψ(ρ(γ)p) = ρ(γ)ψ(p) = ρ(γ)(p). Then by density of {ρ(γ)p : γ ∈ Γ}, we have ψ = id. Hence we can ﬁnd ε > 0 such that length([x, ψ(x)]) ≥ ε for all x by compact- ness. Observe that ρ(γ)[x, ψ(x)] = [ρ(γ)x, ρ(γ)ψ(x)] = [ρ(γ)x, ψ(ρ(γ)x)]. This is just an element of the above kind. So length(ρ(γ)[x, ψ(x)]) ≥ ε. Now assume ρ(Γ) is minimal and not equicontinuous. Claim. Every point x ∈ S1 has a neighbourhood that can be contracted. Proof. Indeed, since ρ(Γ) is not equicontinuous, there exists ε > 0, a sequence (γn)n≥1 and intervals Ik such that length
(Ik) 0 and length(ρ(γn)In) ≥ ε. Since we are on a compact space, after passing to a subsequence, we may assume that for n large enough, we can ﬁnd some interval J such that length(J) ≥ ε 2 and J ⊆ ρ(γn)In. But this means ρ(γn)−1J ⊆ In. 33 3 Actions on S1 IV Bounded Cohomology So J can be contracted. Since the action is minimal, ρ(γ)J = S1. γ∈Γ So every point in S1 is contained in some interval that can be contracted. We shall now write down what the homeomorphism that generates the centralizer. Fix x ∈ S1. Then the set Cx = {[x, y) ∈ S1 : [x, y) can be contracted} is totally ordered ordered by inclusion. Deﬁne Then ϕ(x) sup Cx. [x, ϕ(x)) = Cx. This gives a well-deﬁned map ϕ that commutes with the action of γ. It is then an interesting exercise to verify all the desired properties. – To show ϕ is homeomorphism, we show ϕ is increasing of degree 1, and since it commutes with a minimal action, it is a homeomorphism. – If ϕ is not periodic, then there is some n such that ϕn(x) is between x and ϕ(x). But since ϕ commutes with the action of Γ, this implies [x, ϕn(x)] cannot be contracted, which is a contradiction. Exercise. We have ρ∗(eb) = kρ∗ 0(eb), where k is the cardinality of the centralizer. Example. We can decompose PSL(2, R) = PSO(2)AN, where A = λ 0 0 λ−. More precisely, SO(2) × A × N → SL(2, R) is a diﬀeomorphism and induces on PSO(2) × A × N → PSL(2, R). In particular, the inclusion i : PSO(2) → PSL(2, R) induces an isomorphism on the level of π1 ∼=
Z. We can consider the subgroup kZ ⊆ Z. which gives us a covering of PSO(2) and PSL(2, R) that ﬁts in the diagram PSO(2)k p PSO(2) ik i PSL(2, R) p. PSL(2, R) On the other hand, if we put B = A · N, which is a contractible subgroup, we obtain a homomorphism s : B → PSL(2, R)k, and we ﬁnd that PSL(2, R)k ∼= PSO(2)k · s(B). 34 3 Actions on S1 IV Bounded Cohomology So we have PSL(2, R)k s(B) ∼= PSO(2)k. So PSL(2, R)k/s(B) is homeomorphic to a circle. So we obtain an action of PSL(2, R)k on the circle. Now we can think of Γ ∼= Fr as a lattice in PSL(2, R). Take any section σ : Γ → PSL(2, R)k. This way, we obtain an unbounded minimal action with centralizer isomorphic to Z/kZ. Deﬁnition (Lattice). A lattice in a locally compact group G is a discrete subgroup Γ such that on Γ\G, there is a G-invariant probability measure. Example. Let O be the ring of integers of a ﬁnite extension k/Q. Then SL(n, O) is a lattice in an appropriate Lie group. To construct this, we write [k : Q] = r + 2s, where r and 2s are the number of real and complex ﬁeld embeddings of k. Using these ﬁeld embeddings, we obtain an injection SL(n, O) → SL(n, R)r × SL(n, C)s, and the image is a lattice. Example. If X is a complete proper CAT(0) space, then Isom(X) is locally compact, and in many cases conatins lattices. Theorem (Burger, 2007). Let G be a second-countable locally compact group, and Γ < G be a lattice
, and ρ : Γ → Homeo+(S1) a minimal unbounded action. Then the following are equivalent: – ρ∗(eb R) is in the image of the restriction map H 2 bc(G, R) → H 2 b (Γ, R) – The strongly proximal quotient ρss : Γ → Homeo+(S1) extends continu- ously to G. Theorem (Burger–Monod, 2002). The restriction map H 2 an isomorphism in the following cases: bc(G) → H 2 b (Γ, R) is (i) G = G1 × · · · × Gn is a cartesian product of locally compact groups and Γ has dense projections on each individual factor. (ii) G is a connected semisimple Lie group with ﬁnite center and rank G ≥ 2, and Γ is irreducible. Example. Let k/Q be a ﬁnite extension that is not an imaginary quadratic extension. Then we have an inclusion SL(2, O) → SL(2, R)r × SL(2, C)s and is a product of more than one thing. One can actually explicitly compute the continuous bounded cohomology group of the right hand side. Exercise. Let Γ < SL(3, R) be any lattice. Are there any actions by oriented homeomorphisms on S1? Let’s discuss according to ρ∗(eb R). – If ρ∗(eb R) = 0, then there is a ﬁnite orbit. Then we are stuck, and don’t know what to say. 35 3 Actions on S1 IV Bounded Cohomology – If ρ∗(eb R) = 0, then we have an unbounded minimal action. This leads to a strongly proximal action ρss : Γ → Homeo+(S1). But by the above results, this implies the action extends continuously to an action of SL(3, R) on S1. But SL(3, R) contains SO(3), which is a compact group. But we know what compact subgroups of Homeo+(S1) look like, and it eventually follows that the action is trivial. So this case is not possible. 36 4 The relative homological approach IV Bounded Cohomology 4 The relative hom
ological approach 4.1 Injective modules When we deﬁned ordinary group cohomology, we essentially deﬁned it as the right-derived functor of taking invariants. While we do not need the machinery of homological algebra and derived functors to deﬁne group cohomology, having that available means we can pick diﬀerent injective resolutions to compute group cohomology depending on the scenario, and often this can be helpful. It also allows us to extend group cohomology to allow non-trivial coeﬃcients. Thus, we would like to develop a similar theory for bounded cohomology. We begin by specifying the category we are working over. Deﬁnition (Banach Γ module). A Banach Γ-module is a Banach space V together with an action Γ × V → V by linear isometries. Given a Banach Γ-module V, we can take the submodule of Γ-invariants V Γ. The relative homological approach tells us we can compute the bounded cohomology H· b (Γ, R) by ﬁrst taking an appropriate exact sequences of Banach Γ-modules 0 R d(0) E0 d(1) E1 d(2) · · ·, and then take the cohomology of the complex of Γ-invariants 0 EΓ 0 d(1) d(2) EΓ 1 EΓ 2 · · ·. Of course, this works if we take Ek = C(Γk+1, A) and d(k) to be the diﬀerentials we have previously constructed, since this is how we deﬁned bounded cohomology. The point is that there exists a large class of “appropriate” exact sequences such that this procedure gives us the bounded cohomology. We ﬁrst need the following deﬁnition: Deﬁnition (Admissible morphism). An injective morphism i : A → B of Banach spaces is admissible if there exists σ : B → A with – σi = idA; and – σ ≤ 1. This is a somewhat mysterious deﬁnition, but when we have such a situation, this in particular implies im A is closed and B
= i(A)⊕ker σ. In usual homological algebra, we don’t meet these kinds of things, because our vector spaces always have complements. However, here we need them. Deﬁnition (Injective Banach Γ-module). A Banach Γ-module is injective if for any diagram i B A α E 37 4 The relative homological approach IV Bounded Cohomology where i and α are morphisms of Γ-modules, and i is injective admissible, then there exists β : B → E a morphism of Γ-modules such that i B β A α E commutes and β ≤ α. In other words, we can extend any map from a closed complemented subspace of B to E. Deﬁnition (Injective resolution). Let V be a Banach Γ-module. An injective resolution of V is an exact sequence V E0 E1 E2 · · · where each Ek is injective. Then standard techniques from homological algebra imply the following theorem: Theorem. Let E· be an injective resolution of R Then H·(E·Γ) ∼= H· b (Γ, R) as topological vector spaces. In case E· admits contracting homotopies, this isomorphism is semi-norm decreasing. Unsurprisingly, the deﬁning complex for bounded cohomology were composed of injective Γ-modules. Lemma. – ∞(Γn) for n ≥ 1 are all injective Banach Γ-modules. – ∞ alt(Γn) for n ≥ 1 are injective Banach Γ-modules as well. This is a veriﬁcation. More interestingly, we have the following Proposition. The trivial Γ-module R is injective iﬀ Γ is amenable. As an immediate corollary, we know that if Γ is amenable, then all the higher bounded cohomology groups vanish, as 0 → R → 0 → 0 → · · · is an injective resolution. Proof. (⇒) Suppose A is injective. Consider the diagram R i ∞(Γ), R 38 4 The relative homological approach IV Bounded Cohomology where i(t) is the constant function t. We need to verify that i is an admissible injection
. Then we see that σ(f ) = f (e) is a left inverse to i and σ ≤ 1. Then there exists a morphism β : ∞(Γ) → R ﬁlling in the diagram with β ≤ idR = 1 and in particular β(1Γ) = 1 Since the action of Γ on R is trivial, this β is an invariant linear form on Γ, and we see that this is an invariant mean. (⇐) Assume Γ is amenable, and let m : ∞(Γ) → R be an invariant mean. Consider a diagram i B A α R as in the deﬁnition of injectivity. Since i is an admissible, it has a left inverse σ : B → A. Then we can deﬁne β(v) = m{γ → α(σ(γ∗v))}. Then this is an injective map B → R and one can verify this works. This theory allows us to study bounded cohomology with more general coeﬃcients. This can also be extended to G a locally-compact second-countable groups with coeﬃcients a G-Banach module E which is the dual of a continuous separable Banach module Eb. This is more technical and subtle, but it works. 4.2 Amenable actions In Riemannian geometry, we have the Hodge decomposition theorem. It allows us to understand the de Rham cohomology of a Riemannian manifold in terms of the complex of harmonic forms, whose cohomology is the complex itself. In bounded cohomology, we don’t have something like this, but we can produce a complex such that the complex is equal to the cohomology in the second degree. The setting is that we have a locally-compact second-countable group G with a non-singular action on a standard measure space (S, M, µ). We require that the action map G × S → S which is measurable. Moreover, for any g ∈ G, the measure g∗µ is equivalent to µ. In other words, the G-action preserves the null sets. Example. Let M be a smooth manifold. Then the action of Diﬀ(M ) on M is non-singular. We want to come up with a notion
similar to amenability. This is what we call conditional expectation. Deﬁnition (Conditional expectation). A conditional expectation on G × S is a linear map M : L∞(G × S) → L∞(S) such that (i) M (1) = 1; (ii) If M ≥ 0, then M (f ) ≥ 0; and 39 4 The relative homological approach IV Bounded Cohomology (iii) M is L∞(S)-linear. We have a left G-action on L∞(G × S) given by the diagonal action, and also a natural on L∞(S). We say M is G-equivariant if it is G-equivariant. Deﬁnition (Amenable action). A G-action on S is amenable if there exists a G-equivariant conditional expectation. Note that a point (with the trivial action) is a conditional G-space if G is amenable itself. Example. Let H be a closed subgroup of G, then the G action on G/H is amenable iﬀ H is amenable. Theorem (Burger–Monod, 2002). Let G × S → S be a non-singular action. Then the following are equivalent: (i) The G action is amenable. (ii) L∞(S) is an injective G-module. (iii) L∞(Sn) for all n ≥ 1 is injective. So any amenable G-space can be used to compute the bounded cohomology of G. Corollary. If (S, µ) is an amenable G-space, then we have an isometric isomorphism H·(L∞(Sn, µ)G, dn) ∼= H·(L∞ Example. Let Γ < G be a lattice in SL(n, R), say. Let P < G be a parabolic subgroup, e.g. the subgroup of upper-triangular matrices. We use L∞ alt((G/P )n)Γ to compute bounded cohomology of Γ, since the restriction of amenable actions to closed subgroups is amenable. We have alt(Sn, µ)G, dn) ∼= Hb(G, R). 0 L∞(G/p)�
� Lalt((G/P )2)Γ Lalt((G/P )3)Γ · · · R 0 0 So we know that H 2 a Banach space. b (Γ, R) is isometric to Z(L∞ alt((G/P )3)Γ). In particular, it is 40 Index Index C(Γk+1, A), 14 C(Γk, A)Γ, 15 Cb(Γk+1, A), 20 G-equivariant, 40 H k(Γ, A), 15 R(ϕ), 13 Γ-module Banach, 37 injective Banach, 37 cl(a), 9 ∞(G, A), 4 QH(G, A), 4 QHh(G, R), 5 Homeo+(S1), 12 Homeo+ Z (R), 12 Rot, 12 d(k), 14 eb, 26 k-coboundary, 15 inhomogeneous, 16 k-cochain, 14 k-cocycle, 15 inhomogeneous, 16 k-th bounded cohomology, 20 action amenable, 40 minimal, 27 semiconjugate, 27 strongly proximal, 33 unbounded, 32 admissible morphism, 37 amenable action, 40 amenable group, 23 Banach Γ-module, 37 injective, 37 Bockstein homomorphism, 22 bounded cohomology, 20 bounded Euler class, 26 real, 30 canonical semi-norm, 20 central extension, 12, 17 chain complex, 14 coboundary, 15 inhomogeneous, 16 IV Bounded Cohomology cochain, 14 cocycle, 15 inhomogeneous, 16 commutator length, 9 comparison map, 20 complex, 14 conditional expectation, 39 conjugate semi-, 27 defect, 4 diﬀerential, 14 Euler class, 19 bounded, 26 real bounded, 30 function homogeneous, 4 Gersten long exact sequence, 22 group cohomology, 15 homegeneous function, 4 homogeneous k-cochain, 14 increasing map, 27 degree 1, 27 inhomogeneous k-coboundaries, 16 inhomogeneous k-cocycles, 16 inhomogeneous cochains, 16 injective Banach Γ-module, 37 injective resolution, 38 lattice, 35 long exact sequence Gersten, 22 mean, 23 minimal action, 27 normalized cocycle
, 17 open interval of circle, 31 orientation-preserving map, 12 Poincare translation quasimorphism, 13 positively-oriented triple, 11 quasi-homomorphism, 4 41 Index IV Bounded Cohomology defect, 4 real bounded Euler class, 30 rotation number, 13 stable commutator length, 9 strongly proximal action, 33 strongly proximal quotient, 33 semiconjugate action, 27 unbounded action, 32 42D MODEL Using the model The examples we study in this section are simple, but the methods are far reaching. This will become more apparent as we explore the applications of algebra in the Focus on Modeling sections that follow each chapter starting with Chapter 1. ■ Using Algebra Models We begin our study of modeling by using models that are given to us. In the next subsection we learn how to make our own models. E X AM P L E 1 \| Using a Model for Pay Use the model P NW to answer the following question: Aaron makes $10 an hour and worked 35 hours last week. How much did he get paid? S E C T I O N P.1 \| Modeling the Real World with Algebra 3 ▼ SO LUTI O N We know that N 35 h and W $10. We substitute these values into the formula. P NW Model 35 10 350 Substitute N = 35, W = 10 Calculator So Aaron got paid $350. ✎ Practice what you’ve learned: Do Exercise 3. ▲ E X AM P L E 2 \| Using a Model for Pay Use the model P NW to solve the following problem: Neil makes $9.00 an hour tutoring mathematics in the Learning Center. He wants to earn enough money to buy a calculus text that costs $126 (including tax). How many hours does he need to work to earn this amount? ▼ SO LUTI O N We know that Neil’s hourly wage is W $9.00 and the amount of pay he needs to buy the book is P $126. To ﬁnd N, we substitute these values into the formula. P NW 126 9N N 126 9 N 14 Model Substitute P = 126, W = 9.00 Divide by 9 Calculator So Neil must work 14 hours to buy this book. ✎ Practice what you’ve learned: Do Exercise 7. ▲ E X AM P L E 3 \| Using an Elevation-Temperature Model A mountain climber uses the model T 20
10h to estimate the temperature T (in C) at elevation h (in kilometers, km). (a) Make a table that gives the temperature for each 1-km change in elevation, from elevation 0 km to elevation 5 km. How does temperature change as elevation increases? (b) If the temperature is 5C, what is the elevation? ▼ SO LUTI O N (a) Let’s use the model to ﬁnd the temperature at elevation h 3 km. T 20 10h 20 10 10 1 3 Model Substitute h = 3 2 Calculator So at an elevation of 3 km the temperature is 10C. The other entries in the following table are calculated similarly. 4 CHAPTER P \| Prerequisites Elevation (km) Temperature (ºC) 0 1 2 3 4 5 20 10 0 10 20 30 We see that temperature decreases as elevation increases. (b) We substitute T 5C in the model and solve for h. T 20 10h 5 20 10h 15 10h 15 10 h 1.5 h Model Substitute T = 5 Subtract 20 Divide by –10 Calculator The elevation is 1.5 km. ✎ Practice what you’ve learned: Do Exercise 11. ▲ ■ Making Algebra Models In the next example we explore the process of making an algebra model for a real-life situation. E X AM P L E 4 \| Making a Model for Gas Mileage The gas mileage of a car is the number of miles it can travel on one gallon of gas. (a) Find a formula that models gas mileage in terms of the number of miles driven and the number of gallons of gasoline used. (b) Henry’s car used 10.5 gallons to drive 230 miles. Find its gas mileage. Thinking About the Problem Let’s try a simple case. If a car uses 2 gallons to drive 100 miles, we easily see that gas mileage 100 2 50 mi/gal So gas mileage is the number of miles driven divided by the number of gallons used. ▼ SO LUTI O N (a) To ﬁnd the formula we want, we need to assign symbols to the quantities involved: In Words In Algebra Number of miles driven Number of gallons used Gas mileage (mi/gal) N G M 12 mi/gal 40 mi/gal S E C T I O N P.1 \| Modeling the Real World with Algebra 5 We can express the model as follows: gas mileage number of
miles driven number of gallons used M N G Model (b) To get the gas mileage, we substitute N = 230 and G = 10.5 in the formula. M N G 230 10.5 21.9 Model Substitute N = 230, G = 10.5 Calculator The gas mileage for Henry’s car is 21.9 mi/gal. ✎ Practice what you’ve learned: Do Exercise 21. ▲ P. ▼ CONCE PTS 1. The model L 4S gives the total number of legs that S sheep have. Using this model, we ﬁnd that 12 sheep have L legs. 2. Suppose gas costs $3 a gallon. We make a model for the cost C of buying x gallons of gas by writing the formula C. ▼ SKI LLS 3–12 ■ Use the model given to answer the questions about the object or process being modeled. ✎ 3. The sales tax T in a certain county is modeled by the formula T = 0.08x. Find the sales tax on an item whose price is $120. ✎ 4. A company ﬁnds that the cost C (in dollars) of manufacturing x compact discs is modeled by C 500 0.35x Find the cost of manufacturing 1000 compact discs. 5. A company models the proﬁt P (in dollars) on the sale of x CDs by P 0.8x 500 Find the proﬁt on the sale of 1000 CDs. 6. The volume V of a cylindrical can is modeled by the formula where r is the radius and h is the height of the can. Find the volume of a can with radius 3 in. and height 5 in. 3 in. 5 in. 7. The gas mileage M (in mi/gal) of a car is modeled by M N/G where N is the number of miles driven and G is the number of gallons of gas used. (a) Find the gas mileage M for a car that drove 240 miles on 8 gallons of gas. (b) A car with a gas mileage M 25 mi/gal is driven 175 miles. How many gallons of gas are used? 8. A mountain climber models the temperature T (in F) at eleva- tion h (in ft) by T 70 0.003h (a) Find the temperature T at an elevation of 1500 ft. (b) If the temperature is 64F, what is the elevation? 9
. The portion of a ﬂoating iceberg that is below the water surface is much larger than the portion above the surface. The total volume V of an iceberg is modeled by V 9.5S V pr 2h where S is the volume showing above the surface. 6 CHAPTER P \| Prerequisites (a) Find the total volume of an iceberg if the volume showing 15. The average A of two numbers a and b above the surface is 4 km3. (b) Find the volume showing above the surface for an iceberg with total volume 19 km3. 10. The power P measured in horsepower (hp) needed to drive a certain ship at a speed of s knots is modeled by P 0.06s3 (a) Find the power needed to drive the ship at 12 knots. (b) At what speed will a 7.5-hp engine drive the ship? ✎ 11. An ocean diver models the pressure P (in lb/in2) at depth d (in ft) by P 14.7 0.45d (a) Make a table that gives the pressure for each 10-ft change in depth, from a depth of 0 ft to 60 ft. (b) If the pressure is 30 lb/in2, what is the depth? 12. Arizonans use an average of 40 gallons of water per person each day. (a) Find a model for the number of gallons W of water used by x Arizona residents each day. (b) Make a table that gives the number of gallons of water used for each 1000-person increase in population, from 0 to 5000. (c) Estimate the population of an Arizona town whose water usage is 140,000 gallons per day. 13–20 ■ Write an algebraic formula that models the given quantity. 13. The number N of days in „ weeks 14. The number N of cents in q quarters 16. The average A of three numbers a, b, and c 17. The cost C of purchasing x gallons of gas at $3.50 a gallon 18. The amount T of a 15% tip on a restaurant bill of x dollars 19. The distance d in miles that a car travels in t hours at 60 mi/h 20. The speed r of a boat that travels d miles in 3 hours ▼ APPLICATIONS 21. Cost of a Pizza A pizza parlor charges $12 for a cheese ✎ pizza and $1 for each topping. (a) How much does a 3-
topping pizza cost? (b) Find a formula that models the cost C of a pizza with n toppings. (c) If a pizza costs $16, how many toppings does it have? n=1 n=4 22. Renting a Car At a certain car rental agency a compact car rents for $30 a day and 10¢ a mile. (a) How much does it cost to rent a car for 3 days if the car is driven 280 miles? (b) Find a formula that models the cost C of renting this car for n days if it is driven m miles. (c) If the cost for a 3-day rental was $140, how many miles was the car driven? 23. Energy Cost for a Car The cost of the electricity needed to drive an all-electric car is about 4 cents per mile. The cost of the gasoline needed to drive the average gasoline-powered car is about 12 cents per mile. (a) Find a formula that models the energy cost C of driving x miles for (i) the all-electric car and (ii) the average gasoline-powered car. (b) Find the cost of driving 10,000 miles with each type of car. 24. Volume of Fruit Crate A fruit crate has square ends and is twice as long as it is wide (see the ﬁgure below). (a) Find the volume of the crate if its width is 20 inches. (b) Find a formula for the volume V of the crate in terms of its width x. 2x x x 25. Cost of a Phone Call A phone card company charges a $1 connection fee for each call and 10¢ per minute. (a) How much does a 10-minute call cost? (b) Find a formula that models the cost C of a phone call that lasts t minutes. (c) If a particular call cost $2.20, how many minutes did the call last? (d) Find a formula that models the cost C (in cents) of a phone call that lasts t minutes if the connection fee is F cents and the rate is r cents per minute. 26. Grade Point Average In many universities students are given grade points for each credit unit according to the following scale: A B 4 points 3 points S E C T I O N P.2 \| Real Numbers and Their Properties 7 C D F 2 points 1 point 0 point For example, a grade of A in a 3-unit course earns 4 3 12
grade points and a grade of B in a 5-unit course earns 3 5 15 grade points. A student’s grade point average (GPA) for these two courses is the total number of grade points earned divided by the number of units; in this case the GPA is 12 15 1 (a) Find a formula for the GPA of a student who earns a grade of A in a units of course work, B in b units, C in c units, D in d units, and F in f units. /8 3.375. 2 (b) Find the GPA of a student who has earned a grade of A in two 3-unit courses, B in one 4-unit course, and C in three 3-unit courses. P.2 Real Numbers and Their Properties LEARNING OBJECTIVES After completing this section, you will be able to: ■ Classify real numbers ■ Use properties of real numbers ■ Use properties of negatives ■ Add, subtract, multiply, and divide fractions ■ Types of Real Numbers Let’s review the types of numbers that make up the real number system. We start with the natural numbers: 1, 2, 3, 4,... The integers consist of the natural numbers together with their negatives and 0:..., 3, 2, 1, 0, 1, 2, 3, 4,... We construct the rational numbers by taking ratios of integers. Thus, any rational number r can be expressed as r m n where m and n are integers and n 0. Examples are 1 2 3 7 46 46 1 0.17 17 100 The different types of real numbers were invented to meet speciﬁc needs. For example, natural numbers are needed for counting, negative numbers for describing debt or below-zero temperatures, rational numbers for concepts such as “half a gallon of milk,” and irrational numbers for measuring certain distances, such as the diagonal of a square. (Recall that division by 0 is always ruled out, so expressions such as and are undeﬁned.) There are also real numbers, such as, that cannot be expressed as a ratio of integers and are therefore called irrational numbers. It can be shown, with varying degrees of difﬁculty, that these numbers are also irrational: 12 3 0 0 0 13 15 23 2 p 3 p 2 8 CHAPTER P \| Prerequisites The set of all real numbers is usually denoted by the symbol. When we use
the word number without qualiﬁcation, we will mean “real number.” Figure 1 is a diagram of the types of real numbers that we work with in this book. Rational numbers, 1 2 3 -, 7 46, 0.17, 0.6, 0.317 Irrational numbers,œ3,œ5 3 œ2 π,, 3 π2 Integers Natural numbers..., −3, −2, −1, 0, 1, 2, 3,... Every real number has a decimal representation. If the number is rational, then its cor- responding decimal is repeating. For example, 0.5000... 0.50 2 3 0.3171717... 0.317 9 7 157 495 1 2 0.66666... 0.6 1.285714285714... 1.285714 (The bar indicates that the sequence of digits repeats forever.) If the number is irrational, the decimal representation is nonrepeating: 12 1.414213562373095... p 3.141592653589793... If we stop the decimal expansion of any number at a certain place, we get an approximation to the number. For instance, we can write p 3.14159265 where the symbol is read “is approximately equal to.” The more decimal places we retain, the better our approximation. FIGURE 1 The real number system A repeating decimal such as x 3.5474747... is a rational number. To convert it to a ratio of two integers, we write 1000x 3547.47474747... 10x 35.47474747... 990x 3512.0 x 3512 990 Thus,. (The idea is to multiply x by appropriate powers of 10 and then subtract to eliminate the repeating part.) E X AM P L E 1 \| Classifying Real Numbers Determine whether each given real number is a natural number, an integer, a rational number, or an irrational number. (a) 999 (b) 6 5 (c) 6 3 (d) 125 (e) 13 SO LUTI O N (a) 999 is a positive whole number, so it is a natural number. (b) is a ratio of two integers, so it is a rational number. equals 2, so it is an integer. (
c) (d) (e) 6 5 6 3 125 13 tional number. equals 5, so it is a natural number. is a nonrepeating decimal (approximately 1.7320508075689), so it is an irra- ✎ Practice what you’ve learned: Do Exercise 5. Operations on Real Numbers Real numbers can be combined using the familiar operations of addition, subtraction, multiplication, and division. When evaluating arithmetic expressions that contain several of these operations, we use the following conventions to determine the order in which the operations are performed: S E C T I O N P.2 \| Real Numbers and Their Properties 9 1. Perform operations inside parentheses ﬁrst, beginning with the innermost pair. In dividing two expressions, the numerator and denominator of the quotient are treated as if they are within parentheses. 2. Perform all multiplications and divisions, working from left to right. 3. Perform all additions and subtractions, working from left to right. E X AM P L E 2 \| Evaluating an Arithmetic Expression Find the value of the expression 8 10 ▼ SO LUTI O N these are treated as if they are inside parentheses: First we evaluate the numerator and denominator of the quotient, since 8 10 Evaluate numerator and denominator 2 1 5 9 1 1 3 4 18 # 14 21 28 7 2 2 Evaluate quotient Evaluate parentheses Evaluate products Evaluate difference ▲ ✎ Practice what you’ve learned: Do Exercise 7. ■ Properties of Real Numbers We all know that 2 3 3 2. We also know that 5 7 7 5, 513 87 87 513, and so on. In algebra we express all these (inﬁnitely many) facts by writing a b b a where a and b stand for any two numbers. In other words, “a b b a” is a concise way of saying that “when we add two numbers, the order of addition doesn’t matter.” This fact is called the Commutative Property of Addition. From our experience with numbers we know that the properties in the following box are also valid. PROPERTIES OF REAL NUMBERS Property Commutative Properties a b b a ab ba Associative Properties a b c a b c 1 2 2 ab 1 2 c a bc 1 2 Distributive Property b c a 1 b c ab ac 2 a ab ac 1 1 Example Description
When we add two numbers, order doesn’t matter. When we multiply two numbers, order doesn’t matter When we add three numbers, it doesn’t matter which two we add ﬁrst. When we multiply three numbers, it doesn’t matter which two we multiply ﬁrst When we multiply a number by a sum of two numbers, we get the same result as multiplying the number by each of the terms and then adding the results. 10 CHAPTER P \| Prerequisites The Distributive Property is crucial because it describes the way addition and multiplication interact with each other. The Distributive Property applies whenever we multiply a number by a sum. Figure 2 explains why this property works for the case in which all the numbers are positive integers, but the property is true for any real numbers a, b, and c. 2(3+5) FIGURE 2 The Distributive Property 2%3 2%5 Don’t assume that a is a negative number. Whether a is negative or positive depends on the value of a. For example, if a 5, then a 5, a negative number, but if 5 5 a 5, then a 1 (Property 2), a positive number. 2 E X AM P L E 3 \| Using the Properties of Real Numbers 7 3 2 3 2 3 7 2 (a) 1 Associative Property of Addition Commutative Property of Addition 2x 6 (b) 2 1 x 3 2 Distributive Property Simplify c (c ax bx 2 ax bx ay by a b y 2 ay by 1 1 Distributive Property Distributive Property 2 Associative Property of Addition In the last step we removed the parentheses because, according to the Associative Property, the order of addition doesn’t matter. ✎ Practice what you’ve learned: Do Exercise 15. ▲ ■ Addition and Subtraction The number 0 is special for addition; it is called the additive identity because a 0 a for any real number a. Every real number a has a negative, a, that satisﬁes a 0. Subtraction is the operation that undoes addition; to subtract a number from another, we simply add the negative of that number. By deﬁnition a 2 1 To combine real numbers involving negatives, we use the following properties. a b a b 1 2 PROPERTIES OF NEGATIVES 3. Property 1 a a 1. 2 1 a
2. 21 1 a b 5. 1 a b 6 ab a b b a ab 2 Example 21 Property 6 states the intuitive fact that a b and b a are negatives of each other. Prop- erty 5 is often used with more than two terms.2 \| Real Numbers and Their Properties 11 E X AM P L E 4 \| Using Properties of Negatives Let x, y, and z be real numbers. 3 2 (a) 1 x 2 (b) 2 1 x y z (c Property 5: –(a + b) = –a – b Property 5: –(a + b) = –a – b Property 5: –(a + b) = –a – b Property 2: –(–a) = a 2 ▼ ■ Multiplication and Division The number 1 is special for multiplication; it is called the multiplicative identity because a 1 a for any real number a. Every nonzero real number a has an inverse, 1/a, that 1. Division is the operation that undoes multiplication; to divide by a satisﬁes a number, we multiply by the inverse of that number. If b 0, then by deﬁnition 1//b 1 We write a as simply a/b. We refer to a/b as the quotient of a and b or as the fraction a over b; a is the numerator, and b is the denominator (or divisor). To combine real numbers using the operation of division, we use the following properties. 2 2 # 5 10 3 # 7 21 # 7 14 2 15 35 Example #, # so 2 9 3 6 # Description When multiplying fractions, multiply numerators and denominators. When dividing fractions, invert the divisor and multiply. When adding fractions with the same denominator, add the numerators. When adding fractions with different denominators, ﬁnd a common denominator. Then add the numerators. Cancel numbers that are common factors in the numerator and denominator. Cross multiply. 29 35 The word algebra comes from the ninth-century Arabic book Hisâb al-Jabr w’al-Muqabala, written by al-Khowarizmi. The title refers to transposing and combining terms, two processes that are used in solving equations. In Latin translations the title was shortened to Aljabr, from which we get the word algebra. The author’s name itself made its
way into the English language in the form of our word algorithm. PROPERTIES OF FRACTIONS 2. 1. Property #. 4. ac bd # d a c b a b c ad bc bd 5. ac bc a b a b c d 6. If, then ad bc When adding fractions with different denominators, we don’t usually use Property 4. Instead, we rewrite the fractions so that they have the smallest possible common denominator (often smaller than the product of the denominators), and then we use Property 3. This denominator is the Least Common Denominator (LCD) described in the next example. E X AM P L E 5 \| Using the LCD to Add Fractions Evaluate: 5 36 7 120 ▼ SO LUTI O N Factoring each denominator into prime factors gives 36 22 32 # and 120 23 3 5# # 12 CHAPTER P \| Prerequisites We ﬁnd the least common denominator (LCD) by forming the product of all the factors that occur in these factorizations, using the highest power of each factor. Thus, the LCD is 23 32 5 360. So # # 5 36 7 120 5 # 10 36 # 10 7 # 3 120 # 3 71 360 50 360 21 360 ✎ Practice what you’ve learned: Do Exercise 29. Property 3: Adding fractions with the same denominator ▲ Use common denominator Property ; ; Property Property 19–22 ■ Rewrite the expression using the given property of real numbers. 19. Commutative Property of Addition, x 3 P. ▼ CONCE PTS 1. Give an example of each of the following: (a) A natural number (b) An integer that is not a natural number (c) A rational number that is not an integer (d) An irrational number 2. Complete each statement and name the property of real numbers you have used. (a) ab (bc) a 3. To add two fractions, you must ﬁrst express them so that they have the same. 4. To divide two fractions, you the divisor and then multiply. ▼ SKI LLS 5–6 ■ List the elements of the given set that are (a) natural numbers (b) integers (c) rational numbers (d) irrational numbers ✎ 5. 6. 5 5 0, 10, 50, 22 7, 0.538, 17, 1.23, 1 1.
001, 0.333..., p, 11, 11, 13 3, 23 2 6 15, 116, 3.14, 15 3 6 7–10 ■ Evaluate the arithmetic expression. ✎ 7. 8 10 12 1 5 6 # 7 9 8 A 7 2B 4 # 15 8 2 1 17 2 # 3 2 2 4 9. 10. 3 1 2 4 11–18 ■ State the property of real numbers being used. 11. 7 10 10 7 3 5 3 5 12. 2 2 1 2 1 2 2y 3z 1 2 x 2y 2 1 A B 2 1 5x 1 x a 1 ✎ 13. 14. 15. 16. 17. 1 2x 2 1 3 y 1 a b c 2 18. 7 1 3z x 2A 2B 2 3 15x 2x 2 a b 7c 2 20. Associative Property of Multiplication, 7 21. Distributive Property, 4 22. Distributive Property, 5x 5y A B 2 1 3x 2 1 23. 3 23–28 ■ Use properties of real numbers to write the expression without parentheses. x y 1 2 2m 2 1 2x 4y 5 21 a b 1 6y 4 31 3a 1 b c 2d 25. 4 28. 24. 26. 27. 21 8 2 2 2 2 29–40 ■ Perform the indicated operations. ✎ 29. 31. 33. 35. 37. 39. 3 10 4 15 2 3 3 5 6 3 2 2B 3A 3 1 4B 5B 30. 32. 34. 36. 38. 40. 1 6 1 2B 1 3B 1 2 3BA.25 1 2 A 1 8 2 5 1 10 1 12 1 9 1 2 3 15 41–42 ■ Express each repeating decimal as a fraction. (See the margin note on page 8.) 41. (a) 0.7 42. (a) 5.23 (b) 0.28 (b) 1.37 (c) 0.57 (c) 2.135 S E C T I O N P.3 \| The Real Number Line and Order 13 tional? In general, what can you say about the sum of a rational and an irrational number? What about the product? 46. Commutative and Noncommutative Operations We have seen that addition and multiplication are both commutative operations. (a) Is subtraction commutative? (b) Is
division of nonzero real numbers commutative? (c) Are the actions of putting on your socks and putting on your shoes commutative? (d) Are the actions of putting on your hat and putting on your coat commutative? (e) Are the actions of washing laundry and drying it commutative? (f) Give an example of a pair of actions that is commutative. (g) Give an example of a pair of actions that is not commutative. ▼ APPLICATIONS 43. Area of a Garden Mary’s backyard vegetable garden measures 20 ft by 30 ft, so its area is 20 30 600 ft2. She decides to make it longer, as shown in the ﬁgure, so that the area increases to A 20 30 x. Which property of real numbers 2 1 tells us that the new area can also be written A 600 20x? 30 ft x 20 ft ▼ DISCOVE RY • DISCUSSION • WRITI NG 44. Sums and Products of Rational and Irrational Numbers Explain why the sum, the difference, and the product of two rational numbers are rational numbers. Is the product of two irrational numbers necessarily irrational? What about the sum? 45. Combining Rational Numbers with Irrational Numbers 1 Is 2 22 1 rational or irrational? Is 2 rational or irra- # 22 P.3 The Real Number Line and Order LEARNING OBJECTIVES After completing this section, you will be able to: ■ Graph numbers on the real line ■ Use the order symbols,,, ■ Work with set and interval notation ■ Find and use absolute values of real numbers ■ Find distances on the real line ■ The Real Line The real numbers can be represented by points on a line, as shown in Figure 1. The positive direction (toward the right) is indicated by an arrow. We choose an arbitrary reference point O, called the origin, which corresponds to the real number 0. Given any convenient unit of measurement, each positive number x is represented by the point on the line a distance of x units to the right of the origin, and each negative number x is represented by the point x units to the left of the origin. Thus, every real number is represented by a point on the line, and every point P on the line corresponds to exactly one real number. The number associated with the point P is called the coordinate of P, and the line is then called a coordinate line, or a real number line, or
simply a real line. Often we identify the point with its coordinate and think of a number as being a point on the real line. _4.9 _4.7 _3.1725 _2.63 _ 2 œ∑ 1 16_ 1 8 1 4 1 2 œ∑3 œ∑5 œ∑ 2 _5 _4 _3 _2 _1 0 1 2 FIGURE 1 The real line _4.85 0.3∑ π 3 4.2 4.4 4.9999 4 4.3 5 4.5 14 CHAPTER P \| Prerequisites ■ Order on the Real Line The real numbers are ordered. We say that a is less than b and write a b if b a is a positive number. Geometrically, this means that a lies to the left of b on the number line. (Equivalently, we can say that b is greater than a and write b a.) The symbol a b (or b a) means that either a b or a b and is read “a is less than or equal to b.” For instance, the following are true inequalities (see Figure 2): p 3 12 2 2 2 7 7.4 7.5 _π œ∑2 7.4 7.5 FIGURE 2 _4 _3 _2 _1 AM P L E 1 \| Graphing Inequalities (a) On the real line, graph all the numbers x that satisfy the inequality x 3. (b) On the real line, graph all the numbers x that satisfy the inequality x 2. ▼ SO LUTI O N (a) We must graph the real numbers that are smaller than 3—those that lie to the left of 3 on the real line. The graph is shown in Figure 3. Note that the number 3 is indicated with an open dot on the real line, since it does not satisfy the inequality. (b) We must graph the real numbers that are greater than or equal to 2: those that lie to the right of 2 on the real line, including the number 2 itself. The graph is shown in Figure 4. Note that the number 2 is indicated with a solid dot on the real line, since it satisﬁes the inequality. ✎ Practice what you’ve learned: Do Exercises 17 and 19. ▲ ■ Sets and Intervals A set is a collection of objects, and these objects
are called the elements of the set. If S is a set, the notation a S means that a is an element of S, and b S means that b is not an element of S. For example, if Z represents the set of integers, then 3 Z but p Z. Some sets can be described by listing their elements within braces. For instance, the set A that consists of all positive integers less than 7 can be written as We could also write A in set-builder notation as A 5 1, 2, 3, 4, 5, 6 6 0 3 FIGURE 3 _2 0 FIGURE 4 A x 5 0 x is an integer and 0 x 7 6 which is read “A is the set of all x such that x is an integer and 0 x 7.” If S and T are sets, then their union S T is the set that consists of all elements that are in S or T (or in both). The intersection of S and T is the set S T consisting of all elements that are in both S and T. In other words, S T is the common part of S and T. The empty set, denoted by, is the set that contains no element. E X AM P L E 2 \| Union and Intersection of Sets If S and S V., T 1, 2, 3, 4, 5 6 4, 5, 6, 7 6, and V 5 5 6, 7, 8 6 5, ﬁnd the sets S T, S T, T64748 1, 2, 3, 4, 5, 6, 7, 8 14243 123 S V a b FIGURE 5 The open interval a, b 1 2 a b FIGURE 6 The closed interval a.3 \| The Real Number Line and Order 15 ▼ SO LUTI O N 1, 2, 3, 4, 5, 6, 5 6 All elements in S or T 6 Elements common to both S and T S and V have no element in common ✎ Practice what you’ve learned: Do Exercise 27. ▲ Certain sets of real numbers, called intervals, occur frequently in calculus and correspond geometrically to line segments. For example, if a b, then the open interval from a to b consists of all numbers between a and b and is denoted by the symbol (a, b). Using set-builder notation, we can write a, b x a x b 0 Note that the endpoints, a and
b, are excluded from this interval. This fact is indicated by the parentheses in the interval notation and the open circles on the graph of the interval in Figure 5. 5 6 2 1 1 2 The closed interval from a to b is the set a, b x a x b 3 0 Here the endpoints of the interval are included. This is indicated by the square brackets in the interval notation and the solid circles on the graph of the interval in Figure 6. It 3 is also possible to include only one endpoint in an interval, as shown in the table of intervals below. 5 6 4 4 We also need to consider inﬁnite intervals, such as a x a, q x 1 2 5 0 6 This does not mean that q (“inﬁnity”) is a number. The notation stands for the set of all numbers that are greater than a, so the symbol q simply indicates that the interval extends indeﬁnitely far in the positive direction. a, q 1 2 The following table lists the nine possible types of intervals. When these intervals are discussed, we will always assume that a b. Notation Set description Graph a, b 1 a, b 3 4 a, b 3 2 a, b 1 4 a, q 1 a, q 3 2 2 2 q, b 1 q, b 1 4 q} 5 x x b 5 (set of all real numbers AM P L E 3 \| Graphing Intervals Express each interval in terms of inequalities, and then graph the interval. (a) (b) 3 3 1, 2 x 5 2 1.5, 4 4 3, q x 5 1 x 2 6 1.5 x 4 6 0 0 _1 1 x (c) 6 ✎ Practice what you’ve learned: Do Exercise 33. _3 5 2 0 3 x 2 1.5 4 0 0 0 ▲ 16 CHAPTER P \| Prerequisites E X AM P L E 4 \| Finding Unions and Intersections of Intervals No Smallest or Largest Number in an Open Interval Any interval contains inﬁnitely many numbers—every point on the graph of an interval corresponds to a real number. In the closed interval ”0, 1’, the smallest number is 0 and the largest is 1, but the open interval Ó0, 1Ô contains no smallest or largest number. To see this, note that 0.01 is close to zero
but 0.001 is closer, 0.0001 closer yet, and so on. So we can always ﬁnd a number in the interval Ó0, 1Ô closer to zero than any given number. Since 0 itself is not in the interval, the interval contains no smallest number. Similarly, 0.99 is close to 1, but 0.999 is closer, 0.9999 closer yet, and so on. Since 1 itself is not in the interval, the interval has no largest number. 0 0.01 0 0.001 0.1 0.01 0 0.0001 0.001 \| _3 \|=3 \| 5 \| =5 _3 0 5 Graph each set. (a) 2, 7 3 1, 3 2 1 ▼ SO LUTI O N (a) The intersection of two intervals consists of the numbers that are in both intervals. 1, 3 1 2, 7 (b) 2 4 3 4 Therefore, 1, 3 1 2 2 This set is illustrated in Figure 7. 1 x 3 and 2 x 7 2 x 3 2, 3 6 6 3 2 (b) The union of two intervals consists of the numbers that are in either one interval or the other (or both). Therefore, 1, 3 1 2 2 This set is illustrated in Figure 8. 1 x 3 or 2 x 7 1 x 7 1, 7 6 6 1 4 (1, 3) 0 1 3 0 0 2 [2, 3) 2 3 (1, 3) 0 1 3 [2, 7] [2, 7] 7 0 2 (1, 7] 0 1 7 7 FIGURE 7 1, 3 1 2 2, 7 3 4 FIGURE 8 1, 3 1 2 2, 7 3 4 ✎ Practice what you’ve learned: Do Exercise 47. ▲ ■ Absolute Value and Distance The absolute value of a number a, denoted by a, is the distance from a to 0 on the real number line (see Figure 9). Distance is always positive or zero, so we have a 0 for every number a. Remembering that a is positive when a is negative, we have the following deﬁnition. FIGURE 9 DEFINITION OF ABSOLUTE VALUE If a is a real number, then the absolute value of a is a if a 0 a if a 0 a e 0 0 E X AM P L E 5 \| Evaluating Absolute Values of Numbers (a) (b) (c
.3 \| The Real Number Line and Order 17 (d) 0 12 1 3 p 12 1 3 p 0 1 p 3 0 (e) 2 ✎ Practice what you’ve learned: Do Exercise 53. 1 1 0 since 12 1 1 12 1 0 since p 3 1 3 p 0 2 2 ▲ When working with absolute values, we use the following properties. PROPERTIES OF ABSOLUTE VALUE Property a 1. 0 0 0 Example 3 3 0 0 0 2. 3. 4. a 0 0 a ab 12 3 ` 12 3 0 0 0 0 Description The absolute value of a number is always positive or zero. A number and its negative have the same absolute value. The absolute value of a product is the product of the absolute values. The absolute value of a quotient is the quotient of the absolute values. What is the distance on the real line between the numbers 2 and 11? From Figure 10 13 From this observation we make the following deﬁnition (see we see that the distance is 13. We arrive at this by ﬁnding either 2 or 1 Figure 11). 11 13. 11 2 2 2 1 DISTANCE BETWEEN POINTS ON THE REAL LINE If a and b are real numbers, then the distance between the points a and b on the real line is a From Property 6 of negatives it follows that b a a b. This conﬁrms that, as we would expect, the distance from a to b is the same as the distance from b to a. 0 0 E X AM P L E 6 \| Distance Between Points on the Real Line The distance between the numbers 8 and 2 is 8 2 10 10 a, b d 1 2 0 0 0 We can check this calculation geometrically, as shown in Figure 12. ✎ Practice what you’ve learned: Do Exercise 61. ▲ 13 _2 0 FIGURE 10 \| b-a \| a 11 b FIGURE 11 Length of a line segment is b a 10 _8 0 2 FIGURE 12 P. ▼ CONCE PTS 1. Explain how to graph numbers on a real number line. 2. If a b, how are the points on a real line that correspond to in set-builder notation and in interval notation. 4. Explain the difference between the following two sets of the numbers a and b related to each other? numbers: 3. The set of numbers between but not including 2 and 7 can
be written as follows: A 2, 5 4 3 B 2, 5 1 2 5. The symbol x stands for the 0 is not 0, then the sign of 0 is always x 0 0. of the number x. If x 18 CHAPTER P \| Prerequisites 6. The absolute value of the difference between a and b is (geo- metrically) the between them on the real number line. ✎ ▼ SKI LLS,, or 7–8 ■ Place the correct symbol 2 1 7 (c) 3.5 2 0.67 (b) 3 0.67 (b) 7. (a) 3 8. (a) (c) 7 2 2 3 2 3 in the space. 7 2 0 0.67 0 0 0.67 0 9–16 ■ State whether each inequality is true or false. 9. 6 10 12 1.41 10. 12 13 10 11 p 3 1.1 1.1 11. 13. 15. 12. 1 2 1 14. 8 9 16. 8 8 17–20 ■ On a real number line, graph the numbers that satisfy the inequality. x 1 2 18. x 4 17. 19. x 3 20. x 0 21–24 ■ Find the inequality whose graph is given. 21. 22. 23. 24. _2 3 _ 2 1 0 0 0 0 5 25–26 ■ Write each statement in terms of inequalities. 25. (a) x is positive. (b) t is less than 4. (c) a is greater than or equal to p. (d) x is less than and is greater than 5. 1 3 (e) The distance from p to 3 is at most 5. 26. (a) y is negative. (b) z is greater than 1. (c) b is at most 8. (d) „ is positive and is less than or equal to 17. (e) y is at least 2 units from p. 6 1, 2, 3, 4, 5, 6, 7 27–30 ■ Find the indicated set if A B 5 27. (a) A B 28. (a) B C 29. (a) A C 30. (a) A B C 6 2, 4, 6, 8 5 (b) A B (b) B C (b) A C (b 31–32 ■ Find the indicated set if A 31. (a) B C 32. (ab
) B C (b 33–38 ■ Express the interval in terms of inequalities, and then graph the interval. 3, 0 34. 33. 35. 37. 2 1 2, 8 2 2, q 3 3 2 36. 38. 2, 8 4 1 6, 1 2 q, 1 1 2 39–44 ■ Express the inequality in interval notation, and then graph the corresponding interval. 39. 41. 43. x 1 2 x 1 x 1 40. 42. 44. 1 x 2 x 5 5 x 2 45–46 ■ Express each set in interval notation. 45. (a) (b) 46. (a) −3 −3 0 0 (b) −2 0 0 5 5 2 47–52 ■ Graph the set. 1, 1 47. ✎ 1 2 2, 0 4, 6 q, 4 4 1 2 0, 8 3 2 4, q 1 2 48. 50. 52. 2 2, 0 4, 6 2 q, 6 1 3 1 2 4 1, 1 1 2 3 0, 8 2 2, 10 1 2 49. 51. 3 1 53–58 ■ Evaluate each expression. ✎ 53. (a) ƒ 100 ƒ 54. (a) ƒ 25 5 ƒ 55. (a) ƒ 6 ƒ ƒ 4 ƒ 56. (a) ƒ 57. (a) 58. (a) ` 2 ƒ 12 ƒ # 6 ƒ 2 1 6 24 ` 2 (b) (b) (b) (b) (b) (b) ƒ 73 ƒ ƒ 10 p ƒ 1 ƒ 1 ƒ 1 1 ƒ 3B1 A 7 12 12 7 ` ` 1 ƒ 1 ƒ ƒ 15 2 59–62 ■ Find the distance between the given numbers. 59. 60. −3 −2 −1 −3 −2 −, 8, 9, 10 6 ✎ 61. (a) 2 and 17 (b) 3 and 21 8 and 3 (c) 10 11 62. (a) 7 21 15 and 1 (b) 38 and 57 (c) 2.6 and 1.8 ▼ APPLICATIONS 63. Temperature Variation The bar graph on the next page shows the daily high temperatures for Omak, Washington, and Geneseo, New York
, during a certain week in June. Let TO represent the temperature in Omak and TG the temperature in Geneseo. Calculate for each day shown. Which of these two values gives more information? TG and ƒ TO TG TO ƒ S E C T I O N P.4 \| Integer Exponents 19 Omak, WA Geneseo, NY ▼ DISCOVE RY • DISCUSSION • WRITI NG 66. Signs of Numbers Let a, b, and c be real numbers such that a b 0, and c 0 (a) a (d) a b (g) c 2. Find the sign of each expression. (b) c (e) c a (h) abc (c) bc (f) a bc 67. Limiting Behavior of Reciprocals Complete the tables. What happens to the size of the fraction 1/x as x gets large? As x gets small? x 1/x x 1/x 1 2 10 100 1000 1.0 0.5 0.1 0.01 0.001 68. Irrational Numbers and Geometry Using the following on a number line. ﬁgure, explain how to locate the point 26? Can you locate List some other irrational numbers that can be located this way. by a similar method? What about 25 22 œ∑2 1 1 2 _1 80 75 70 65 Sun Mon Tue Wed Day Thu Fri Sat 64. Fuel Consumption Suppose an automobile’s fuel consumption is 28 mi/gal in city driving and 34 mi/gal in highway driving. If x denotes the number of city miles and y the number of highway miles, then the total miles this car can travel on a 15-gallon tank of fuel must satisfy the inequality 1 28 x 1 34 y 15 Use this inequality to answer the following questions. (Assume the car has a full tank of fuel.) (a) Can the car travel 165 city miles and 230 highway miles before running out of gas? (b) If the car has been driven 280 miles in the city, how many highway miles can it be driven before running out of fuel? 65. Mailing a Package The post ofﬁce will accept only pack- ages for which the length plus the “girth” (distance around) is no more than 108 inches. Thus, for the package in the ﬁgure, we must have L 2 x y 108 1 (a)
Will the post ofﬁce accept a package that is 6 in. wide, 8 in. deep, and 5 ft long? What about a package that measures 2 ft by 2 ft by 4 ft? 2 (b) What is the greatest acceptable length for a package that has a square base measuring 9 in. by 9 in? L x y 6 in. 8 in. 5 ft=60 in. P.4 Integer Exponents LEARNING OBJECTIVES After completing this section, you will be able to: ■ Use exponential notation ■ Simplify expressions using the Laws of Exponents ■ Write numbers in scientiﬁc notation In this section we review the rules for working with exponent notation. We also see how exponents can be used to represent very large and very small numbers. ■ Exponential Notation A product of identical numbers is usually written in exponential notation. For example, # 5 5 5 is written as 53. In general, we have the following deﬁnition. # 20 CHAPTER P \| Prerequisites EXPONENTIAL NOTATION If a is any real number and n is a positive integer, then the nth power of a is an a a # #p# a 14243 n factors The number a is called the base, and n is called the exponent. E X AM P L E 1 \| Exponential Notation Note the distinction between 3 4 2 1 and 34. In 4 the exponent applies 2 to 3, but in 34 the exponent applies only to 3. 3 1 (a) (b) A 1 2BA 5 A 4 1 2B 3 2 1 34 1 1 1 1 1 2B 2BA 2BA 2BA # 32 # 3 2 1 81 # 2 3 1 2 81 (c) ✎ Practice what you’ve learned: Do Exercises 13 and 15. 2 1 ▲ We can state several useful rules for working with exponential notation. To discover the rule for multiplication, we multiply 54 by 52: 54 52 Ó5 5 5 5ÔÓ5 5Ô 5 5 5 5 5 5 56 542 1442443 123 144424443 4 factors 2 factors 6 factors It appears that to multiply two powers of the same base, we add their exponents. In general, for any real number a and any positive integers m and n we have aman Óa a # #p# # aÔÓa a #p# 1442443 1442443 14424
43 m n factors # aÔ a a a m factors n factors #p# # a amn Thus, aman amn. We would like this rule to be true even when m and n are 0 or negative integers. For instance, we must have 20 23 203 23 But this can happen only if 20 1. Likewise, we want to have 2 544 50 1 4 1 and this will be true if 54 1/54. These observations lead to the following deﬁnition. 54 54 54 ZERO AND NEGATIVE EXPONENTS If a 0 is any real number and n is a positive integer, then a0 1 and an 1 an E X AM P L E 2 \| Zero and Negative Exponents (a) (b) 0 1 4 7B A x1 1 1 x1 x 3 1 2 1 2 1 (c) 1 8 ✎ Practice what you’ve learned: Do Exercise 19.4 \| Integer Exponents 21 ■ Rules for Working with Exponents Familiarity with the following rules is essential for our work with exponents and bases. In the table the bases a and b are real numbers, and the exponents m and n are integers. LAWS OF EXPONENTS Law 1. 2. 3. 4. 5. aman amn am an am amn n amn 1 2 ab 1 2 a b b a n anbn an bn n Example 32 # 35 325 37 35 32 352 33 5 32 5 310 32 2 3 # 4 2 32 # 42 32 3 42 4 b 2 2 a 1 1 # Description To multiply two powers of the same number, add the exponents. To divide two powers of the same number, subtract the exponents. To raise a power to a new power, multiply the exponents. To raise a product to a power, raise each factor to the power. To raise a quotient to a power, raise both numerator and denominator to the power. ▼ P RO O F O F LAW 3 am If m and n are positive integers, we have 1442443 m factors a # a # p p 1442443 1442443 1442443 factors m factors 144444444424444444443 n groups of factors m factors a # a # p # a amn 1442443 mn factors The cases for which m 0 or n 0 can be proved by using the deﬁnition of negative ▲ exponents. ▼
P RO O F O F LAW 4 If n is a positive integer, we have n ab 1 2 ab 2 1 144424443 1442443 1442443 ab ab anbn 2 2 1 n factors n factors n factors Here we have used the Commutative and Associative Properties repeatedly. If n 0, ▲ Law 4 can be proved by using the deﬁnition of negative exponents. You are asked to prove Laws 2 and 5 in Exercise 97. E X AM P L E 3 \| Using Laws of Exponents (a) (b) (c) (d) (e) (f) x4x7 x47 x11 y4y7 y47 y3 1 y3 c9 c5 b4 2 3x 1 1 c95 c4 # 5 b20 5 b4 3 33x3 27x3 5 2 x5 25 x5 32 x 2 b a Law 1: aman = am+n Law 1: aman = am+n Law 2: am/an = amn Law 3: (am)n = amn Law 4: (ab)n = anbn Law 5: (a/b)n = an/bn ✎ Practice what you’ve learned: Do Exercises 37, 41, and 45. ▲ 22 CHAPTER P \| Prerequisites E X AM P L E 4 \| Simplifying Expressions with Exponents Simplify: (a) 2a3b2 3ab4 3 2 2 1 1 (b) 3 x y b a 4 y2x z b a ▼ SO LUTI O N 2a3b2 (a) 3ab4 1 2 1 3 2 4 2 3 33a3 b4 1 27a3b12 2 2 1 a3a3b2b12 3 2a3b2 2 1 1 2a3b2 27 2 2 2 1 54 a6b14 1 (b) 3 x y b a 4 y 2x z b a 4x z4 y 8x 4 z4 Law 4: (ab)n = anbn Law 3: (am)n = amn Group factors with the same base Law 1: aman = amn Laws 5 and 4 Law 3 x 3x 7y 5 z4 Group factors with the same base Laws 1 and 2 ✎ Practice what you’ve learned: Do Exercises 47 and 61. ▲ When simplifying an expression
, you will ﬁnd that many different methods will lead to the same result; you should feel free to use any of the rules of exponents to arrive at your own method. We now give two additional laws that are useful in simplifying expressions with negative exponents. LAWS OF EXPONENTS Law 6. 7 an bm bm an Example 32 45 45 32 Description To raise a fraction to a negative power, invert the fraction and change the sign of the exponent. To move a number raised to a power from numerator to denominator or from denominator to numerator, change the sign of the exponent AW 7 Using the deﬁnition of negative exponents and then Property 2 of fractions (page 11), we have an bm 1/an 1/bm 1 an # bm 1 bm an ▲ You are asked to prove Law 6 in Exercise 98. E X AM P L E 5 \| Simplifying Expressions with Negative Exponents Eliminate negative exponents and simplify each expression. y 3z3 b 6st 4 2s2t 2 (b) (a) 2 a MATHEMATICS IN THE MODERN WORLD Although we are often unaware of its presence, mathematics permeates nearly every aspect of life in the modern world. With the advent of modern technology, mathematics plays an ever greater role in our lives. Today you were probably awakened by a digital alarm clock, made a phone call that used digital transmission, sent an e-mail message over the Internet, drove a car with digitally controlled fuel injection, listened to music on a CD or MP3 player, then fell asleep in a room whose temperature is controlled by a digital thermostat. In each of these activities mathematics is crucially involved. In general, a property such as the intensity or frequency of sound, the oxygen level in a car’s exhaust emission, the colors in an image, or the temperature in your bedroom is transformed into sequences of numbers by sophisticated mathematical algorithms. These numerical data, which usually consist of many millions of bits (the digits 0 and 1), are then transmitted and reinterpreted. Dealing with such huge amounts of data was not feasible until the invention of computers, machines whose logical processes were invented by mathematicians. The contributions of mathematics in the modern world are not limited to technological advances. The logical processes of mathematics are now used to analyze complex problems in the social, political, and life sciences in new and surprising ways. Advances in mathematics continue to be made
, some of the most exciting of these just within the past decade. In other Mathematics in the Modern World vignettes, we will describe in more detail how mathematics affects us in our everyday activities. S E C T I O N P.4 \| Integer Exponents 23 ▼ SO LUTI O N (a) We use Law 7, which allows us to move a number raised to a power from the numer- ator to the denominator (or vice versa) by changing the sign of the exponent. 4 moves to denominator t and becomes t4. 2 moves to numerator s and becomes s2. 6st 4 2s2t 2 6s s2 t 4 2t 2 3s3 t6 Law 7 Law 1 (b) We use Law 6, which allows us to change the sign of the exponent of a fraction by inverting the fraction. y 3z3 b a 2 3z3 y b a 9z6 y2 2 Law 6 Laws 5 and 4 ✎ Practice what you’ve learned: Do Exercises 63 and 67. ▲ ■ Scientiﬁc Notation Exponential notation is used by scientists as a compact way of writing very large numbers and very small numbers. For example, the nearest star beyond the sun, Proxima Centauri, is approximately 40,000,000,000,000 km away. The mass of a hydrogen atom is about 0.00000000000000000000000166 g. Such numbers are difﬁcult to read and to write, so scientists usually express them in scientiﬁc notation. SCIENTIFIC NOTATION A positive number x is said to be written in scientiﬁc notation if it is expressed as follows: x a 10n where 1 a 10 and n is an integer For instance, when we state that the distance to the star Proxima Centauri is 4 1013 km, the positive exponent 13 indicates that the decimal point should be moved 13 places to the right: 4 1013 40,000,000,000,000 Move decimal point 13 places to the right. When we state that the mass of a hydrogen atom is 1.66 1024 g, the exponent 24 indicates that the decimal point should be moved 24 places to the left: 1.66 1024 0.00000000000000000000000166 Move decimal point 24 places to the left. E X AM P L E 6 \| Changing from Decimal to Scientiﬁc Notation Write each number in scient
iﬁc notation. (a) 56,920 (b) 0.000093 24 CHAPTER P \| Prerequisites ▼ SO LUTI O N (a) 56,920 5.692 104 123 (b) 0.000093 9.3 105 123 4 places 5 places ✎ Practice what you’ve learned: Do Exercises 73 and 75. E X AM P L E 7 \| Changing from Decimal to Scientiﬁc Notation Write each number in decimal notation. (a) 6.97 109 (b) 4.6271 106 ▼ SO LUTI O N (a) 6.97 109 6,970,000,000 1442443 9 places (b) 4.6271 106 0.0000046271 14243 6 places Move decimal 9 places to the right Move decimal 6 places to the left ✎ Practice what you’ve learned: Do Exercises 81 and 83. ▲ ▲ Scientiﬁc notation is often used on a calculator to display a very large or very small number. For instance, if we use a calculator to square the number 1,111,111, the display panel may show (depending on the calculator model) the approximation 1.234568 12 or 1.234568 E12 Here the ﬁnal digits indicate the power of 10, and we interpret the result as 1.234568 1012 or EE to enter the exponent. To use scientiﬁc notation on a calculator, use a key labeled or EXP EEX For example, to enter the number 3.629 1015 on a TI-83 calculator, we enter 3.629 2ND EE 15 and the display reads 3.629E15 E X AM P L E 8 \| Calculating with Scientiﬁc Notation If a 0.00046, b 1.697 1022, and c 2.91 1018, use a calculator to approximate the quotient ab/c. ▼ SO LUTI O N We could enter the data using scientiﬁc notation, or we could use laws of exponents as follows: ab c 1 4.6 104 1.697 1022 2 1 2.91 1018 2 1 4.6 1.697 2 1 2.91 2.7 1036 2 1042218 We state the answer correct to two signiﬁcant ﬁgures because
the least accurate of the given numbers is stated to two signiﬁcant ﬁgures. ✎ Practice what you’ve learned: Do Exercise 91. ▲ P. ▼ CONCE PTS 1. Using exponential notation, we can write the product # # # 5 5 5 5 5 5 as # #. 4. When we divide two powers with the same base, we 2. In the expression 34, the number 3 is called the, and the number 4 is called the. the exponents. So 35 32. 3. When we multiply two powers with the same base, we the exponents. So 34 35. # 5. When we raise a power to a new power, we the exponents. So 2 34 1 2. 6. Express the following numbers without using exponents. (a) 21 1 (c) 1 2B A (b) 23 # 7. 52 5 ▼ SKI LLS 7–28 ■ Evaluate each expression. # 8. 23 22 6 3 23 10. 1 2 13. 32 14. 11. 2 1 0 0 2 ✎ 2 1 107 104 23 30 3 2B A 2 # 9 16 54 # 52 16. ✎ 19. 22. 0 21 3 5 3B A 2 3B A 0 6 A 2 3B 25. 4 1 9B 13B A 27. 22 23 A 17. 20. 23. 3 29–46 ■ Simplify each expression. 26. 32 # 42 # 5 24 # 33 # 25 28. 31 33 29. 32. 35. 38. ✎ 41. x8x2 x5x3 y10y0 y7 4 2z z 3z1 z a2a4 1 44. 2z2 1 2 3 2 5z10 30. 33. 4y5 3y2 2 1 1 „2„4„ 2 6 36. 39. 42. x6 x10 3 2y2 2 2a3a2 1 1 ✎ 45. a2 4 b a 31. 34. ✎ 37. x2x6 5z3z4 z a9a2 a 4 2 3 40. 43. 46. 2 2 2 8x 3z 1 1 6z2 2 1 2 3x4 4x2.4 \| Integer Exponents 25 65. 2 u1√ u3√2 1 1 2 3 2 2 1 3a b3 b ✎
67. 69. 71. a a a 3 y 5x2 b q1r1s2 r5sq8 b 1 66. 68. 70. 72. rs2 1 2 r3s2 3 2 2 1 a a a 3 2 x2y 2y3 b 2a1b a2b3 b xy2z3 x2y3z4 b 3 ✎ 2 23 9. 2 1 12. 60 1 3B 3 32 15. 18. A 1 4 3 2 2 21. 24. 2 1 4B A 2 1 2B A 4 5 2B A ✎ ✎ ✎ ✎ 73–80 ■ Write each number in scientiﬁc notation. 73. 69,300,000 75. 0.000028536 77. 129,540,000 74. 7,200,000,000,000 76. 0.0001213 78. 7,259,000,000 79. 0.0000000014 80. 0.0007029 81–88 ■ Write each number in decimal notation. 81. 3.19 105 83. 2.670 108 85. 7.1 1014 87. 8.55 103 82. 2.721 108 84. 9.999 109 86. 6 1012 88. 6.257 1010 89–90 ■ Write the number indicated in each statement in scientiﬁc notation. 89. (a) A light-year, the distance that light travels in one year, is about 5,900,000,000,000 mi. (b) The diameter of an electron is about 0.0000000000004 cm. (c) A drop of water contains more than 33 billion billion molecules. 90. (a) The distance from the earth to the sun is about 93 million miles. (b) The mass of an oxygen molecule is about 0.000000000000000000000053 g. (c) The mass of the earth is about 5,970,000,000,000,000,000,000,000 kg. 3 2 47–72 ■ Simplify the expression and eliminate any negative exponent(s). ✎ ✎ 47. 49. 51. 53. 55. 57. 59. ✎ 61. ✎ 63. 4x2y4 1 b4 2 1 3ab3 2 x5y 2 2a2b5 4 2 1 5x2y3 s2t2 2 2 rs 1 1 2 1 3x
2y5 2 1 2 s2t 1 2rs2 3 3 2 2 1 2 1 6y3z 2yz2 2 48. 50. 52. 54. 56. 58. B 2 A 2 1 2 a3z4 1 4 s7t B 1 5a2b5 1 3u3√ 2 2 1 3 2 2 2 x2y5 1 1 1 8 a2z 2 A 2 s3t2 2a3b2 2 3 2 1 2u2√3 1 3x1y2 1 2x3y4 x5y3 16 t4 2 1 2 xy2z3 x2y2z 1 1 a2 b b a 4 3 2 2 5 a 8a3b4 2a5b5 3 a3b2 c3 b 60. 1 2 2√ 3„ 2 2 3„ √ 2x3y2 2 z3 b 62. 64. x4z2 4y5 b a a 5xy2 x1y3 91–96 ■ Use scientiﬁc notation, the Laws of Exponents, and a calculator to perform the indicated operations. State your answer correct to the number of signiﬁcant digits indicated by the given data. 7.2 109 1 21 1.062 1024 1.806 1012 8.61 1019 2 21 1.295643 109 2 3.610 1017 2.511 106 2 2 1 73.1 1.6341 1028 2 1 0.0000000019 2 1 1 1 91. 92. 93. 94. 95. 0.0000162 1 2 1 594,621,000 0.01582 2 0.0058 2 1 2 1 3.542 106 2 5.05 104 12 9 2 96. 1 1 97. Prove the given Laws of Exponents for the case in which m and n are positive integers and m n. (a) Law 2 (b) Law 5 98. Prove Law 6 of Exponents. 26 CHAPTER P \| Prerequisites ▼ APPLICATIONS 99. Distance to the Nearest Star Proxima Centauri, the star nearest to our solar system, is 4.3 light-years away. Use the information in Exercise 89(a) to express this distance in miles. 100. Speed of Light The speed of light is about 186,000 mi/s. Use the information in Exercise 90(a)
to ﬁnd how long it takes for a light ray from the sun to reach the earth. 101. Volume of the Oceans The average ocean depth is 3.7 103 m, and the area of the oceans is 3.6 1014 m2. What is the total volume of the ocean in liters? (One cubic meter contains 1000 liters.) 102. National Debt As of January 2001, the population of the United States was 2.83 108, and the national debt was 5.736 1012 dollars. How much was each person’s share of the debt? 103. Number of Molecules A sealed room in a hospital, measuring 5 m wide, 10 m long, and 3 m high, is ﬁlled with pure oxygen. One cubic meter contains 1000 L, and 22.4 L of any gas contains 6.02 1023 molecules (Avogadro’s number). How many molecules of oxygen are there in the room? 104. Body-Mass Index The body-mass index is a measure that medical researchers use to determine whether a person is overweight, underweight, or of normal weight. For a person who weighs W pounds and who is H inches tall, the bodymass index B is given by B 703 W H2 A body-mass index is considered “normal” if it satisﬁes 18.5 B 24.9, while a person with body-mass index B 30 is considered obese. (a) Calculate the body-mass index for each person listed in the table, then determine whether he or she is of normal weight, underweight, overweight, or obese. Person Weight Height Brian Linda Larry Helen 295 lb 105 lb 220 lb 110 lb 5 ft 10 in. 5 ft 6 in. 6 ft 4 in. 5 ft 2 in. (b) Determine your own body-mass index. 105. Interest on a CD A sum of $5000 is invested in a 5-year certiﬁcate of deposit paying 3% interest per year, compounded monthly. After n years, the amount of interest I that has accumulated is given by I 5000 1.0025 12n 1 3 1 2 4 Complete the following table, which gives the amount of interest accumulated after the given number of years. Year Total interest $152.08 308.79 1 2 3 4 5 ▼ DISCOVE RY • DISCUSSION • WRITI NG 106. How Big Is a Billion
? dollars in 106 2 103 a suitcase and you spend a thousand dollars each day, 1 2 how many years would it take you to use all the money? If you spent at the same rate, how many years would it take you to empty a suitcase ﬁlled with a billion If you have a million dollars? 1 109 1 2 107. Easy Powers That Look Hard Calculate these expressions in your head. Use the Laws of Exponents to help you. (a) 185 95 206 # (b) 0.5 1 6 2 108. Distances between Powers Which pair of numbers is closer together? 1010 and 1050 or 10100 and 10101 109. Signs of Numbers Let a, b, and c be real numbers with a 0, b 0, and c 0. Determine the sign of each expression. (a) b5 (b) b10 (d) b a 3 2 1 (e) b a 1 4 2 (c) ab2c3 a3c3 b6c6 (f) P.5 Rational Exponents and Radicals LEARNING OBJECTIVES: After completing this section, you will be able to: ■ Simplify expressions involving radicals ■ Simplify expressions involving rational exponents ■ Express radicals using rational exponents ■ Rationalize a denominator In this section we learn to work with expressions that contain radicals or rational exponents. S E C T I O N P.5 \| Rational Exponents and Radicals 27 ■ Radicals We know what 2n means whenever n is an integer. To give meaning to a power, such as 24/5, whose exponent is a rational number, we need to discuss radicals. The symbol 1 means “the positive square root of.” Thus, 1a b means b2 a and b 0 It is true that the number 9 has two square roots, 3 and 3, but the notation 19 is reserved for the positive square root of 9 (sometimes called the principal square root of 9). If we want the negative root, we must write 19, which is 3. Since a b2 0, the symbol 1a makes sense only when a 0. For instance, 19 3 because 32 9 and 3 0 Square roots are special cases of nth roots. The nth root of x is the number that, when raised to the nth power, gives x. DEFINITION OF nth ROOT If n is any positive integer, then the principal nth root of a is deﬁ
AM P L E 2 \| Combining Radicals 132 1200 116 # 2 1100 # 2 116 12 110012 412 1012 1412 Factor out the largest squares Property 1 Distributive Property ✎ Practice what you’ve learned: Do Exercise 39. ▲ ■ Rational Exponents To deﬁne what is meant by a rational exponent or, equivalently, a fractional exponent such as a1/3, we need to use radicals. To give meaning to the symbol a1/n in a way that is consistent with the Laws of Exponents, we would have to have So by the deﬁnition of nth root, n a1 1/n n a1 a 2 a1/n 1 2 a1/n 2n a In general, we deﬁne rational exponents as follows. DEFINITION OF RATIONAL EXPONENTS For any rational exponent m/n in lowest terms, where m and n are integers and n 0, we deﬁne am/n If n is even, then we require that a 0. 1n a 1 2 m or, equivalently, am/n 2n am DEﬁNITION OF RATIONAL EXPONENTS With this deﬁnition it can be proved that the Laws of Exponents also hold for rational exponents. E X AM P L E 3 \| Using the Deﬁnition of Rational Exponents 41/2 14 2 (a) (b) 82/3 23 8 2 1 2 22 4 Alternative solution: 82/3 23 82 23 64 4 S E C T I O N P.5 \| Rational Exponents and Radicals 29 (c) 125 1 2 1/3 1 1251/3 1 23 125 1 5 (d) 1 23 x4 1 x4/3 x4/3 ✎ Practice what you’ve learned: Do Exercises 17 and 19. ▲ E X AM P L E 4 \| Using the Laws of Exponents with Rational Exponents (a) a1/3a7/3 a8/3 (b) (c) a2/5a7/5 a3/5 2a3b4 1 a2/57/53/5 a6/5 2 3/2 a3 3/2 23/2 1 12 2 212 a9/2b6 2 3a3 1 3/2 1 1 b4 2 2b4 3
/2 3/2 1 2 (d) 3 2x3/4 y1/3 b a y4 x1/2 b a y4x1/2 23 x3/4 1 y1/3 1 8x9/4 y 3 # 2 3 1 2 # y4x1/2 Law 1: aman am+n Law 1, Law 2: am an amn Law 4: (abc)n anbncn Law 3: (am)n amn Laws 5, 4, and 7 2 Law 3 8x11/4y3 Laws 1 and 2 ✎ Practice what you’ve learned: Do Exercises 53, 61 and 65. ▲ E X AM P L E 5 \| Simplifying by Writing Radicals as Rational Exponents (a) 21x 1 2 1 313 x 2 2x1/2 3x1/3 1 2 1 6x1/21/3 6x5/6 2 (b) xx 1/2 1 2x1x x 3/2 1 x 3/4 2 1/2 2 1/2 Deﬁnition of rational exponents Law 1 Deﬁnition of rational exponents Law 1 Law 3 ✎ Practice what you’ve learned: Do Exercises 71 and 81. ▲ ■ Rationalizing the Denominator It is often useful to eliminate the radical in a denominator by multiplying both numerator and denominator by an appropriate expression. This procedure is called rationalizing the 1a denominator. If the denominator is of the form, we multiply numerator and denomi1a. In doing this, we multiply the given quantity by 1, so we do not change its nator by value. For instance, 1 1a 1 1a # 1 1 1a # 1a 1a 1a a Note that the denominator in the last fraction contains no radical. In general, if the dewith m n, then multiplying the numerator and denominanominator is of the form tor by will rationalize the denominator, because (for a 0) 1n anm 1n am 1n am1n anm 1n amnm 1n an a 01A-W4525.qxd 1/8/08 11:04 AM Page 30 30 CHAPTER P \| Prerequisites E X AM P L E 6 \| Rationalizing Denominators Rationalize the denominator in each fraction. (a)
2 13 (b) 1 23 5 (c) 1 25 5x2 ▼ SO LUTI O N This equals 1. (a) 2 13 # 13 13 2 13 213 3 1 23 5 23 25 5 1 25 x2 25 x3 x # 23 52 23 52 # 25 x3 25 x3 (b) (c) 1 23 5 1 25 x2 Multiply by 13 13 13 # 13 3 Multiply by 23 52 23 52 23 5 # 23 52 23 53 5 Multiply by 25 x 3 25 x 3 25 x2 # 25 x3 25 x5 x ✎ Practice what you’ve learned: Do Exercises 83 and 87. ▲ P. ▼ CONCE PTS 1. Using exponential notation, we can write 2. Using radicals, we can write 51/2 as 23 5 as.. 3. Is there a difference between 252 and 15 2 1 2? Explain. 4. Explain what 43/2 means, then calculate 43/2 in two different ways: 41/2 1 2 or 43 1 2 5. Explain how we rationalize a denominator, then complete the following steps to rationalize 1 13 : # ∞ † 1 13 1 13 † ∞ 6. Find the missing power in the following calculation: 5. ✎ # 5 51/3 ▼ SKI LLS 7–14 ■ Write each radical expression using exponents, and each exponential expression using radicals. Radical expression Exponential expression 10. 11. 12. 13. 14. 25 53 1 2x5 15–24 ■ Evaluate each expression. 15. (a) 116 16. (a) 164 ✎ 17. (a) 24 9 (b) 24 16 (b) 23 64 (b) 24 256 18. (a) 17 128 (b) 113/2 21.5 a2/5 (c) (c) (c) 24 1 16 25 32 26 1 64 (c) 14 24 14 54 148 13 32 27 8 B 4/3 1/2 1/3 2 1 1 A 2/5 2/5 3/2 (c) (c) 4 9B (b) (b) 2 2/3 21. (a) 19. (a) A 20. (a) 10240.1 125 25 64B 1 8B A 8000 4/3 (c) 2 1 (c) 0.0012/3 # (c) 91
/3 151/3 251/3 24. (a) 25–28 ■ Evaluate the expression using x 3, y 4, and z 1. (b) 27 2 (b) 10,0003/2 # # (b) 42/3 62/3 92/3 (b) 0.250.5 1 32B A 1000 22. (a) 1 23. (a) 1001.5 0.75 1 16B 2/3 2/3 (c) A A # 1 2 42/3 25. 27. 2x2 y2 2/3 9x 1 2 2/3 z2/3 2y 2 26. 28. 24 x3 14y 2z 2z xy 1 2 1 1 15 23 72 7. 8. 9. S E C T I O N P.5 \| Rational Exponents and Radicals 31 83–88 ■ Rationalize the denominator. ✎ 83. (a) 84. (a) 85. (a) 86. (a) ✎ 87. (a) 88. (a) 1 16 12 13 1 23 4 1 25 23 1 23 x 1 23 x2 (b) 3 12 (b) (b) (b) (b) (b) 5 12 1 24 3 2 24 3 1 25 x2 1 24 x3 (c) (c) (c) (c) (c) (c) 9 13 2 16 8 25 2 3 24 23 1 27 x3 1 23 x4 ▼ APPLICATIONS 89. How Far Can You See? Because of the curvature of the earth, the maximum distance D that you can see from the top of a tall building of height h is estimated by the formula D 22rh h2 where r 3960 mi is the radius of the earth and D and h are also measured in miles. How far can you see from the observation deck of the Toronto CN Tower, 1135 ft above the ground? CN Tower 1/3 r 29–38 ■ Simplify the expression. Assume the letters denote any real numbers. ✎ ✎ 29. 24 x4 31. 24 16x8 33. 23 x3y 35. 25 a6b7 37. 23 164x 6 39–48 ■ Simplify the expression. ✎ 39. 132 118 41. 1125 145 43. 23 108 23 32 45. 1245 1125 47. 25 96 25 3 30. 25 x10 32.
23 x3y6 34. 2x4y4 36. 23 a2b23 a4b 38. 24 x4y2z2 40. 175 148 42. 23 54 23 16 44. 18 150 46. 23 24 23 81 48. 24 48 24 3 49–68 ■ Simplify the expression and eliminate any negative exponent(s). Assume that all letters denote positive numbers. 49. x3/4x5/4 50. y2/3y4/3 51. ✎ 53. 55. 57. 59. ✎ 61. 1 1 63. ✎ 65. 67. 4b 1/2 8b1/4 2 2/3 3/5 1 2/3 2 1 „4/3„ „1/3 8a6b3/2 2 2/3 8y3 2 x5y1/3 1 1 1 2 2/3 2 1/4 8s3t3 s4t8 2 x8y4 16y4/3 b x2/3 y1/2 b a a1/6b3 x1y b a a a 1/4 5a1/2 2 2 2 52. 54. 56. 58. 60. 62. 64. 2 3a3/4 1 s5/2 1 2 2s5/4 1 s1/2 4a6b8 3/2 2 1/3 1 1 1 1 2 u4√6 2 2x3y1/4 2 32y5z10 1 64y6z12 8y3/4 y3z6 b 2 a 8y3/2 1 1/5 1/3 2 2 1/6 1/6 x2 y3 b 3 x2b1 a3/2y1/3 b a 4y3z2/3 2 66. a a x1/2 b 3/2 9st 1 2 27s3t4 68. 1 2/3 a 2 1/3 1 x3y6 8z4 b 3s2 4t1/3 b 69–82 ■ Simplify the expression and express the answer using rational exponents. Assume that all letters denote positive numbers. ✎ 69. 71. 73. 75. 77. 79. 26 y5 A BA 23 y2 B 2 14 x 513 x B A A 24st326 s3t2 24 x7 24 x3 1xy
24 16xy 16u3√ u√5 B 70. 24 b31b B 72. 74. 76. 78. 80. 21a 23 a2 B B A A 10 25 x3y22x4y16 23 8x2 2x 2a3b 24 a3b2 3 54x2y4 B 2x5y ✎ 81. 23 y1y 82. 2s1s3 90. Speed of a Skidding Car Police use the formula s 130 fd to estimate the speed s (in mi/h) at which a car is traveling if it skids d feet after the brakes are applied suddenly. The number f is the coefﬁcient of friction of the road, which is a measure of the “slipperiness” of the road. The table gives some typical estimates for f. Tar Concrete Gravel Dry Wet 1.0 0.5 0.8 0.4 0.2 0.1 (a) If a car skids 65 ft on wet concrete, how fast was it moving when the brakes were applied? 32 CHAPTER P \| Prerequisites (b) If a car is traveling at 50 mi/h, how far will it skid on wet tar? 91. Sailboat Races The speed that a sailboat is capable of sail- ing is determined by three factors: its total length L, the surface area A of its sails, and its displacement V (the volume of water it displaces), as shown in the sketch. In general, a sailboat is capable of greater speed if it is longer, has a larger sail area, or displaces less water. To make sailing races fair, only boats in the same “class” can qualify to race together. For a certain race, a boat is considered to qualify if 0.30L 0.38A1/2 3V 1/3 16 where L is measured in feet, A in square feet, and V in cubic feet. Use this inequality to answer the following questions. (a) A sailboat has length 60 ft, sail area 3400 ft2, and displace- ment 650 ft3. Does this boat qualify for the race? (b) A sailboat has length 65 ft and displaces 600 ft3. What is the largest possible sail area that could be used and still allow the boat to qualify for this race? A V L 92. Flow Speed in a Channel The speed of water �
��owing in a channel, such as a canal or river bed, is governed by the Manning Equation, V 1.486 A2/3S1/2 p2/3n Here V is the velocity of the ﬂow in ft/s; A is the crosssectional area of the channel in square feet; S is the downward slope of the channel; p is the wetted perimeter in feet (the distance from the top of one bank, down the side of the channel, across the bottom, and up to the top of the other bank); and n is the roughness coefﬁcient (a measure of the roughness of the channel bottom). This equation is used to predict the capacity of ﬂood channels to handle runoff from heavy rainfalls. For the canal shown in the ﬁgure, A 75 ft2, S 0.050, p 24.1 ft, and n 0.040. (a) Find the speed at which water ﬂows through the canal. (b) How many cubic feet of water can the canal discharge per [Hint: Multiply V by A to get the volume of the second? ﬂow per second.] 20 ft 10 ft 5 ft ▼ DISCOVE RY • DISCUSSION • WRITI NG 93. Limiting Behavior of Powers Complete the following tables. What happens to the nth root of 2 as n gets large? What about the nth root of? 1 2 21n n 1 2 5 10 100 1/n 1 2B A n 1 2 5 10 100 Construct a similar table for of n as n gets large? n1/n. What happens to the nth root 94. Comparing Roots Without using a calculator, determine which number is larger in each pair. 1 1 (a) 21/2 or 21/3 (b) 2 2 2 2 (d) 13 5 or 13 (c) 71/4 or 41/3 1/2 or 1 1 1/3 P.6 Algebraic Expressions LEARNING OBJECTIVES After completing this section, you will be able to: ■ Add and subtract polynomials ■ Multiply algebraic expressions ■ Use the Special Product Formulas A variable is a letter that can represent any number from a given set of numbers. If we start with variables such as x, y, and z and some real numbers and we combine them using ad- S E
C T I O N P.6 \| Algebraic Expressions 33 dition, subtraction, multiplication, division, powers, and roots, we obtain an algebraic expression. Here are some examples: 2x2 3x 4 1x 10 y 2z y2 4 A monomial is an expression of the form ax k, where a is a real number and k is a nonnegative integer. A binomial is a sum of two monomials, and a trinomial is a sum of three monomials. In general, a sum of monomials is called a polynomial. For example, the ﬁrst expression listed above is a polynomial, but the other two are not. POLYNOMIALS A polynominal in the variable x is an expression of the form n1 p a1 x a 0 n an1 x an x where a0, a1,..., an are real numbers and n is a nonnegative integer. If an then the polynomial has degree n. The monomials akx k that make up the polynomial are called the terms of the polynomial. 0, Note that the degree of a polynomial is the highest power of the variable that appears in the polynomial. Polynomial Type Terms Degree 2x2 3x 4 x8 5x 8 x x2 1 5x 1 9x5 6 2x3 trinomial binomial four terms binomial monomial monomial 2x2, 3x, 4 x8, 5x 1 2x3, x2, x, 8 5x, 1 9x5 6 2 8 3 1 5 0 ■ Adding and Subtracting Polynomials We add and subtract polynomials using the properties of real numbers that were discussed in Section P.2. The idea is to combine like terms (that is, terms with the same variables raised to the same powers) using the Distributive Property. For instance, 5x7 3x7 5 3 x7 8x7 1 In subtracting polynomials, we have to remember that if a minus sign precedes an expression in parentheses, then the sign of every term within the parentheses is changed when we remove the parentheses: 2 Distributive Property ac bc a b 1 2 c [This is simply a case of the Distributive Property, a b c 2 1 ab ac, with a 1.] b c 1 2 b c E
X AM P L E 1 \| Adding and Subtracting Polynomials (a) Find the sum (b) Find the difference 1 x3 6x2 2x 4 x3 5x2 7x 1 x3 6x2 2x 4 2 1 1 2 x3 5x2 7x. 2. 2 ▼ SO LUTI O N (a) 1 x3 6x2 2x 4 x3 x3 2 2x3 x2 5x 4 2 6x2 5x2 1 1 1 x3 5x2 7x 2 2x 7x 1 2 4 2 Group like terms Combine like terms 34 CHAPTER P \| Prerequisites (b) 1 x3 6x2 2x 4 x3 5x2 7x 2 1 x3 6x2 2x 4 x3 5x2 7x 11x2 9x 4 6x 2 5x 2x 7x 4 2 Distributive Property Group like terms Combine like terms ✎ Practice what you’ve learned: Do Exercise 21. ▲ ■ Multiplying Algebraic Expressions To ﬁnd the product of polynomials or other algebraic expressions, we need to use the Distributive Property repeatedly. In particular, using it three times on the product of two binomials, we get ac ad bc bd 1 1 This says that we multiply the two factors by multiplying each term in one factor by each term in the other factor and adding these products. Schematically, we have 2 1 2 2 1 2 The acronym FOIL helps us to remember that the product of two binomials is the sum of the products of the First terms, the Outer terms, the Inner terms, and the Last terms. (a b)(c d) ac ad bc bd L F O I In general, we can multiply two algebraic expressions by using the Distributive Prop- erty and the Laws of Exponents. E X AM P L E 2 \| Multiplying Binomials Using FOIL 2x 1 1 2 1 3x 5 2 6x2 10x 3x 5 L I O F 6x2 7x 5 Distributive Property Combine like terms ✎ Practice what you’ve learned: Do Exercise 41. ▲ When we multiply trinomials or other polynomials with more terms, we use the Distributive Property. It is also helpful to arrange our work in table form. The next example illustrates both methods.
E X AM P L E 3 \| Multiplying Polynomials 2x 3 Find the product: x2 5x 4 1 2 1 2 ▼ SO LUTI O N 1: Using the Distributive Property 2x 3 1 2 1 x2 5x 4 2 2 1 3 x2 5x 4 2x 1 2x # x2 2x # 5x 2x # 4 2x3 10x2 8x 2x3 7x2 7x 12 1 1 1 2 x2 5x 4 2 2 1 3x2 15x 12 2 3 # x2 3 # 5x 3 # 4 2 Distributive Property Distributive Property Laws of Exponents Combine like terms ▼ SO LUTI O N 2: Using Table Form x2 5x 4 2x 3 3x2 15x 12 2x3 10x2 8x 2x3 7x2 7x 12 ✎ Practice what you’ve learned: Do Exercise 71. Add like terms First factor Second factor Multiply x 2 5x 4 by 3 Multiply x 2 5x 4 by 2x ▲ See the Discovery Project on page 38 for a geometric interpretation of some of these formulas. S E C T I O N P.6 \| Algebraic Expressions 35 ■ Special Product Formulas Certain types of products occur so frequently that you should memorize them. You can verify the following formulas by performing the multiplications. SPECIAL PRODUCT FORMULAS If A and B are any real numbers or algebraic expressions, then 1. 2. 3. 4. 5. A2 B2 A B A B 21 1 2 2 A2 2AB B2 A B 1 2 2 A2 2AB B2 A B 2 1 3 A3 3A2B 3AB2 B3 A B 1 2 3 A3 3A2B 3AB2 B3 A B 2 1 Sum and difference of same terms Square of a sum Square of a difference Cube of a sum Cube of a difference The key idea in using these formulas (or any other formula in algebra) is the Principle of Substitution: We may substitute any algebraic expression for any letter in a formula. 2, we use Product Formula 2, substituting x 2 for A and y3 for For example, to ﬁnd 1 B, to get x 2 y3 2 x2 y3 1 2 2 x2 1 2 2 2 1 x2 21 y3 2 y3 1 2 2 (A
B)2 A2 2AB B2 E X AM P L E 4 \| Using the Special Product Formulas Use a Special Product Formula to ﬁnd each product. (a) 3x 5 2 (b) x2 2 3 1 2 1 2 ▼ SO LUTI O N (a) Substituting A 3x and B 5 in Product Formula 2, we get 3x 2 (b) Substituting A x2 and B 2 in Product Formula 5, we get 3x 2 1 5 1 2 1 2 1 3x 5 2 2 2 52 9x2 30x 25 x2 2 1 2 3 x2 3 3 3 1 x6 6x4 12x2 8 x2 2 1 2 2 2 1 2 x2 1 2 2 2 1 2 23 ✎ Practice what you’ve learned: Do Exercises 51 and 67. ▲ E X AM P L E 5 \| Using the Special Product Formulas Find each product. 2x 1y (a) 2x 1y 1 2 1 ▼ SO LUTI O N (a) Substituting A 2x and B 2 (b 2x 1y 2x 2 1y 2 4x2 y 1 (b) If we group x y together and think of this as one algebraic expression, we can use 2 1 2 1 2 2 1 in Product Formula 1, we get 1y 2x 1y Product Formula 1 with A x y and B 1 x2 2xy y2 1 1 2 2 12 4 3 1 1 1 4 2 Product Formula 1 Product Formula 2 ✎ Practice what you’ve learned: Do Exercises 65 and 89. ▲ 36 CHAPTER P \| Prerequisites P. ▼ CONCE PTS 1. Which of the following expressions are polynomials? (a) 2x2 3x (b) x3 21x (c) x5 2x4 1 2x 3 2. To add polynomials, we add terms. So 3x2 2x 4 1 2 8x2 x 1 2 1. 3. To subtract polynomials, we subtract terms. So 2x3 9x2 x 10 1 2 x3 x2 6x 8 2 1. 4. Explain how we multiply two polynomials, then perform the x 2 1 following multiplication: x 3 21. 2 5. The Special Product Formula for the “square of a sum” is
2 A B 2 1 2x 3. So 1 6. The Special Product Formula for the “sum and difference 2 2. A B of the same terms” is 21 1 5 x 5 x 2 21 1. A B 2. So ▼ SKI LLS 7–12 ■ Complete the following table by stating whether the polynomial is a monomial, binomial, or trinomial, then list its terms and state its degree. Polynomial Type Terms Degree 7. x2 3x 7 8. 2x5 4x2 9. 8 1 2x7 10. 11. x x2 x3 x4 12x 13 12. 13–18 ■ Determine whether the expression is a polynomial. If it is, state its degree. 13. 2x2 3x 12 15. 17. 1 4 3 x3 15x 1 1 2x3 13x 1 14. 16. 18. 2 x2 4x 6 px5 1 7 x 13 1 17 x 1 2 ✎ 2 2 2 2 2 21. 22. 23. 19. 20. 2x2 3x 5 1 2x2 3x 5 1 19–36 ■ Find the sum, difference, or product. 5 3x 5x 12 12x 7 1 2 1 1 3x2 x 1 1 3x2 x 1 1 x3 6x2 4x 7 1 x 2 4 x 1 24. 3 1 2 1 2 3 x2 3x 5 1 x 1 27. 2x 1 x 3 1 x2 2x 1 1 Ó3x2 2x 4 2 2y 5 28. 3y 2 1 y2 2 30. y 2 2x 5 1 29. x2 25. 8 26. 4 2 2 2 2 1 2 2 2x 8 1 2 7 x 9 1 2 2 31. 2 33. 35. 2 t 2 5t 1 1 3r 2 r 2 9 2 2x2 x 1 r 1 x2 1 t 10 2 2r 1 32. 34. 36. 2 5 1 √ 3 1 3x3 2 2t 3t 4 1 2√ 2 √ 9 1 2 x4 4x2 5 t 3 2 2 2√ 1 2 1 2 37–48 ■ Multiply the algebraic expressions using the FOIL method and simplify. ✎ 37. 39. 41. 43. 45. 47. 2 x 4 1 r 3 1 21 3t
2 1 21 3x 5 1 x 3y 1 21 2r 5s 21 1 x 3 21 r 5 2 7t 4 2 2x 1 21 2x y 2 3r 2s 2 2 38. 40. 42. 44. 46. 48. 2 y 1 1 s 8 1 21 4s 1 21 1 7y 3 21 1 4x 5y 1 21 6u 5√ 21 1 y 5 21 s 2 2 2s 5 2 2y 1 2 3x y 2 u 2√ 2 49–70 ■ Multiply the algebraic expressions using a Special Product Formula and simplify 3x 4 1 2u √ 1 2 2x 3y 2 1 x2 1 1 x 5 1 21 3x 4 1 x 3y 21 1 1x 2 y 2 2 1 2r 3 2 1 1 1 3 x 5 2 3x 4 21 x 3y 2 1x 2 2 2 1 50. 52. 54. 56. 58. 60. 62. 64. 66. 68. 70 2y 1 2 x 3y 1 2 r 2s 2 1 2 y3 1 y 3 1 21 2y 5 1 2u √ 1 1y 12 1 x 3 2 1 3 2y 2 1 21 21 3 3 y 3 2 2y 5 2u √ 2 2 1y 12 2 1 71–90 ■ Perform the indicated operations and simplify. 2 2 2 2 72. 74. 76. 78. 80. 2 1 x 1 2 1 1 2x 2x2 x 1 2 x2 3x 1 2 1 1 x3/2 1 1x 1/ 1x 2x3/4 x1/4 2 2 x1/4 1 c 1 c b a 2 2 x1/2 y1/2 2 1 1a b 2 1 x 2 2 1 2x 5 x2 2x 3 x2 x 1 1 1x y1/3 2 1 x 1x 2 1 y2/3 y5/ x2 a2 1 x2 y2 x2 a2 2 1 x1/2 y1/2 1a b 2h2 1 1 2 1 1 x2/3 1 x2/3 2 1 1 b 1 b 2 1 x2 x 1 2 11 2 x2 x 1 1 2x y 3 1 x y z 1 2 11 21 1 1 2 2 2 x2 2 2 x2 x 1 x 2 2 1 2x y 3 1 2 2 2
2 21 x y z 2 2h2 1 1 2 2 2 ✎ ✎ ✎ ✎ ✎ 49. 51. 53. 55. 57. 59. 61. 63. 65. 67. 69. 71. 73. 75. 77. 79. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90. ▼ APPLICATIONS 91. Volume of a Box An open box is constructed from a 6 in. by 10 in. sheet of cardboard by cutting a square piece from each corner and then folding up the sides, as shown in the ﬁgure. The volume of the box is V x 6 2x 21 1 10 2x 2 (a) Explain how the expression for V is obtained. (b) Expand the expression for V. What is the degree of the re- sulting polynomial? (c) Find the volume when x 1 and when x 2. 6 in. x x x x 10 in. 10 _ 2x x 6 _ 2x x x x 92. Building Envelope The building code in a certain town requires that a house be at least 10 ft from the boundaries of the lot. The buildable area (or building envelope) for the rectangular lot shown in the ﬁgure is given by 2 2 1 A y 20 x 20 1 (a) Explain how the expression for A is obtained. (b) Expand to express A as a polynomial in x and y. (c) A contractor has a choice of purchasing one of two rectangular lots, each having the same area. One lot measures 100 ft by 400 ft, the other 200 ft by 200 ft. Which lot has the larger building envelope? y 10 ft x Building envelope 10 ft S E C T I O N P.6 \| Algebraic Expressions 37 93. Interest on an Investment A 3-year certiﬁcate of deposit pays interest at a rate r compounded annually. If $2000 is invested, then the amount at maturity is A 2000 3 1 r 2 1 (a) Expand the expression for A. What is the degree of the resulting polynomial? (b) Find the amounts A for the values of r in the table. Interest rate r 2% 3% 4.5% 6% 10% Amount A 94. Proﬁt A wholesaler sells graphing calculators. For an order of x calculators, his total cost in dollars is C
50 30x 0.1x 2 and his total revenue is R 50x 0.05x2 (a) Find the proﬁt P on an order of x calculators. (b) Find the proﬁt on an order of 10 calculators and on an order of 20 calculators. ▼ DISCOVE RY • DISCUSSION • WRITI NG 95. An Algebra Error Beginning algebra students sometimes make the following error when squaring a binomial: x 5 1 2 2 x 2 25 (a) Substitute a value for x to verify that this is an error. (b) What is the correct expansion for x 5 1 2 2? 96. Degrees of Sums and Products of Polynomials Make up several pairs of polynomials, then calculate the sum and product of each pair. On the basis of your experiments and observations, answer the following questions. (a) How is the degree of the product related to the degrees of the original polynomials? (b) How is the degree of the sum related to the degrees of the original polynomials? (c) Test your conclusions by ﬁnding the sum and product of the following polynomials: 2x 3 x 3 and 2x 3 x 7 DISCOVERY PR OJECT VISUALIZING A FORMULA Many of the Special Product Formulas that we learned in this section can be “seen” as geometrical facts about length, area, and volume. For example, the ﬁgure shows how the formula for the square of a binomial can be interpreted as a fact about areas of squares and rectangles. b a (a+b)™ a b b a ab b™ a™ a ab b (a+b)™=a™+2ab+b™ a b 2 represent areas. The In the ﬁgure, a and b represent lengths, and a2, b2, ab, and 1 ancient Greeks always interpreted algebraic formulas in terms of geometric ﬁgures, as we have done here. 2 1. Explain how the ﬁgure veriﬁes the formula a2 b2 = a b 1 21. Find a ﬁgure that veriﬁes the formula 1 3. Explain how the ﬁgure veriﬁes the formula a b 3 a3
3a2b 3ab2 b3. 1 2 2 a2 2ab b2. Is it possible to draw a geometric ﬁgure that veriﬁes the formula for 1 4? 2 Explain. 5. (a) Expand a b c 2 1 2. (b) Make a geometric ﬁgure that veriﬁes the formula you found in part (a). 38 S E C T I O N P.7 \| Factoring 39 P.7 Factoring LEARNING OBJECTIVES After completing this section, you will be able to: ■ Factor out common factors ■ Factor trinomials by trial and error ■ Use the Special Factoring Formulas ■ Factor algebraic expressions completely ■ Factor by grouping terms We use the Distributive Property to expand algebraic expressions. We sometimes need to reverse this process (again using the Distributive Property) by factoring an expression as a product of simpler ones. For example, we can write x2 4 x 2 1 21 x 2 2 We say that x 2 and x 2 are factors of x2 4. ■ Common Factors The easiest type of factoring occurs when the terms have a common factor. E X AM P L E 1 \| Factoring Out Common Factors Factor each expression. (a) 3x2 6x (b) 8x4y2 6x3y3 2xy4 ▼ SO LUTI O N (a) The greatest common factor of the terms 3x2 and 6x is 3x, so we have (b) We note that 3x2 6x 3x x 2 1 2 8, 6, and 2 have the greatest common factor 2 x4, x3, and x have the greatest common factor x y2, y3, and y4 have the greatest common factor y2 So the greatest common factor of the three terms in the polynomial is 2xy2, and we have 8x4y2 6x3y3 2xy4 2xy2 1 2xy2 4x3 2xy2 2 2 1 2 1 4x3 3x2y y2 1 1 2 3x2y 2xy2 1 2 1 y2 2 2 ✎ Practice what you’ve learned: Do Exercises 7 and 9. ▲ E X AM P L E 2 \| Factoring Out a Common Factor 5 Factor: x 3 2x 4 1 21 x 3 1 2 2 Check Your Answer (a
) Multiplying gives 3x2 6x x 2 3x 1 2 (b) Multiplying gives 4x3 3x2y y2 2xy2 1 2 8x4y2 6x3y3 2xy4 ✔ ✔ 40 CHAPTER P \| Prerequisites ▼ SO LUTI O N 2x 4 The two terms have the common factor x 3 ✎ Practice what you’ve learned: Do Exercise 11. 2x 4 3 1 2x 1 2 1 2 1 2 2 1 2 Distributive Property Simplify ▲ ■ Factoring Trinomials To factor a trinomial of the form x2 bx c, we note that x r x2 x s r s x rs 2 1 so we need to choose numbers r and s so that r s b and rs c. 2 1 2 1 E X AM P L E 3 \| Factoring x2 bx c by Trial and Error Factor: x2 7x 12 Check Your Answer ▼ SO LUTI O N We need to ﬁnd two integers whose product is 12 and whose sum is 7. By trial and error we ﬁnd that the two integers are 3 and 4. Thus, the factorization is ✔ Multiplying gives 7x 12 ax 2 bx c factors of a qx s px r 2 21 1 factors of c Check Your Answer Multiplying gives 2x 1 3x 5 1 2 1 2 6x2 7x 5 ✔ x2 7x 12 x 3 1 2 1 x 4 2 factors of 12 ✎ Practice what you’ve learned: Do Exercise 13. ▲ To factor a trinomial of the form ax2 bx c with a 1, we look for factors of the form px r and qx s: px r ax2 bx c qx s Therefore, we try to ﬁnd numbers p, q, r, and s such that pq a, rs c, ps qr b. If these numbers are all integers, then we will have a limited number of possibilities to try for p, q, r, and s. pqx2 ps qr x rs 2 1 1 2 1 2 E X AM P L E 4 \| Factoring ax2 bx c by Trial and Error Factor: 6x2 7x 5 # ▼ SO LUTI O N We can factor 6 as 6 1 or 3 2, and 5 as 5 1 or 5
these possibilities, we arrive at the factorization # # # 1 1 2. By trying factors of 6 6x2 7x 5 3x 5 1 2 1 2x 1 2 factors of 5 ✎ Practice what you’ve learned: Do Exercise 17. ▲ E X AM P L E 5 \| Recognizing the Form of an Expression Factor each expression. (a) x2 2x 3 (b) 5a 1 1 ▼ SO LUTI O N x 1 (a) x2 2x 3 2 (b) This expression is of the form x 3 21 1 2 2 2 1 5a 1 2 3 Trial and error.7 \| Factoring 41 represents 5a 1. This is the same form as the expression in part (a), so it will where factor as 1 3 2 1 5a 1 1 1. 2 5a 1 2 2 1 2 2 3 1 5a 1 3 1 5a 2 3 4 3 1 2 5a 2 2 1 2 5a 1 1 4 2 ✎ Practice what you’ve learned: Do Exercise 19. ▲ ■ Special Factoring Formulas Some special algebraic expressions can be factored by using the following formulas. The ﬁrst three are simply Special Product Formulas written backward. FACTORING FORMULAS Formula 1. A2 B2 A B 21 1 2. A2 2AB B2 3. A2 2AB B2 A B 4. A3 B3 1 A B 5. A3 B3 A2 AB B2 A2 AB B2 2 2 E X AM P L E 6 \| Factoring Differences of Squares Name Difference of squares Perfect square Perfect square Difference of cubes Sum of cubes 2 z 2 Factor each polynomial: (a) 4x2 25 (b) x y 2 1 ▼ SO LUTI O N (a) Using the Difference of Squares Formula with A 2x and B 5, we have 4x2 25 2 52 2x 1 2 2x 5 1 2 1 2x 5 2 A2 B2 (A B) (A B) (b) We use the Difference of Squares Formula with A x y and B z. 2 z2 x y 2 1 ✎ Practice what you’ve learned: Do Exercises 21 and 57. x y z 1 x y z 21 2 E X AM P L E 7 \| Factoring Differences and Sums of Cubes (a) 27
x3 1 Factor each polynomial: (b) x 6 8 ▼ SO LUTI O N (a) Using the Difference of Cubes Formula with A 3x and B 1, we get 12 3 13 3x 1 2 1 1 9x2 3x 1 3x 1 3x 1 27x3 1 2 3x 3x b) Using the Sum of Cubes Formula with A x2 and B 2, we have x4 2x2 4 x2 2 ✎ Practice what you’ve learned: Do Exercises 23 and 25. x6 8 3 23 x2 2 2 1 1 1 2 A trinomial is a perfect square if it is of the form A2 2AB B2 or A2 2AB B2 ▲ ▲ 42 CHAPTER P \| Prerequisites 2AB or 2AB So we recognize a perfect square if the middle term 1 the product of the square roots of the outer two terms. 2 is plus or minus twice E X AM P L E 8 \| Recognizing Perfect Squares Factor each trinomial: (a) x 2 6x 9 (b) 4x 2 4xy y2 ▼ SO LUTI O N # (a) Here A x and B 3, so 2AB 2 x 3 6x. Since the middle term is 6x, the tri- # nomial is a perfect square. By the Perfect Square Formula, we have x2 6x 9 # # (b) Here A 2x and B y, so 2AB 2 2x y 4xy. Since the middle term is 4xy, the trinomial is a perfect square. By the Perfect Square Formula, we have x 3 1 2 2 ✎ Practice what you’ve learned: Do Exercises 53 and 55. ▲ 4x2 4xy y2 2x y 2 2 1 ■ Factoring an Expression Completely When we factor an expression, the result can sometimes be factored further. In general, we ﬁrst factor out common factors, then inspect the result to see whether it can be factored by any of the other methods of this section. We repeat this process until we have factored the expression completely. E X AM P L E 9 \| Factoring an Expression Completely Factor each expression completely. (a) 2x4 8x2 (b) x5y2 xy6 ▼ SO LUTI O N (a) We ﬁ
rst factor out the power of x with the smallest exponent. 2x4 8x2 2x2 2x2 1 1 x2 4 x 2 2 2 1 Common factor is 2x 2 x 2 2 Factor x2 4 as a difference of squares (b) We ﬁrst factor out the powers of x and y with the smallest exponents. 2 x5y2 xy6 xy2 xy2 xy2 1 x4 y4 x2 y2 x2 y2 1 x2 y2 x y 2 1 ✎ Practice what you’ve learned: Do Exercises 67 and 71 Common factor is xy2 Factor x 4 y 4 as a difference of squares Factor x 2 y 2 as a difference of squares ▲ In the next example we factor out variables with fractional exponents. This type of fac- toring occurs in calculus. E X AM P L E 10 \| Factoring Expressions with Fractional Exponents Factor each expression. (a) 3x3/2 9x1/2 6x1/2 (b) 2 x 2/3x 2 x 1/3 1 2 1 2 ▼ SO LUTI O N (a) Factor out the power of x with the smallest exponent, that is, x1 2. 3x3/2 9x1/2 6x1/2 3x1/2 3x1/2 x2 3x Factor out 3x–1/2 Factor the quadratic x2 3x +.7 \| Factoring 43 Check Your Answer To see that you have factored correctly, multiply using the Laws of Exponents. (a) (b) 3x1/2 x2 3x 2 3x3/2 9x1/2 6x1/2 2 1 2//3x /3 (b) Factor out the power of 2 x with the smallest exponent, that is, 2 x 2/3x 2 x 1/3 1 2 1 2 1 2 ✎ Practice what you’ve learned: Do Exercises 85 and 87. 1 2/3 1 2 1 2 2 2/3 2 x 2 4 2/ 2x 1 x 2 3. 2 x 2 1 Factor out (2 + x)– 2/3 Simplify Factor out 2 ■ Factoring by Grouping Terms Polynomials with at least four terms can sometimes be factored by grouping terms. The following example illustrates the
idea. E X AM P L E 11 \| Factoring by Grouping Factor each polynomial. (a) x 3 x 2 4x 4 (b) x 3 2x 2 3x 6 ▼ SO LUTI O N (a) x3 x2 4x 4 1 x2 1 x3 2x2 3x 6 (b) x3 x2 x2 4 2 1 x3 2x2 1 3 x 2 1 2 1 x 2 x2 3 2 1 x2 4x 4 2 x 1 2 2 3x 6 x 2 2 Group terms Factor out common factors Factor out x + 1 from each term Group terms 2 Factor out common factors Factor out x 2 from each term 1 ✎ Practice what you’ve learned: Do Exercise 29. 2 1 2 P. ▼ CONCE PTS 1. Consider the polynomial 2x 5 6x 4 4x 3. How many terms does this polynomial have? List the terms: What factor is common to each term? Factor the polynomial: 2x 5 6x 4 4x 3. 2. To factor the trinomial x 2 7x 10, we look for two integers whose product is and whose sum is. These integers are and, so the trinomial factors as. 3. The Special Factoring Formula for the “difference of squares” is A2 B 2. So 4x 2 25 factors as. 4. The Special Factoring Formula for a “perfect square” is A2 2AB B2. So x 2 10x 25 factors as. ✎ ✎ ✎ ✎ ✎ ✎ ✎ ✎ ✎ ▼ SKI LLS 5–12 ■ Factor out the common factor. 5. 5a 20 7. 2x3 16x 9. 2x2y 6xy2 3xy y 6 y 6 11. y 9 12. 1 2 1 2 6. 3b 12 8. 2x4 4x3 14x2 10. 7x4y2 14xy3 21xy4 13–20 ■ Factor the trinomial. 13. x2 2x 3 15. x2 2x 15 17. 3x2 16x 5 2 8 3x 2 2 2 5 a b 3x 2 a b 1 20. 2 19. 1 12 2 3 1 2 1 2 14. x2 6x 5 16. 2x2 5x 7 18. 5x2
7x 6 21–28 ■ Use a Factoring Formula to factor the expression. 21. 9a2 16 23. 27x3 y3 25. 8s3 125t 3 27. x2 12x 36 x 3 22. 2 1 24. a3 b6 26. 1 1000y3 28. 16z2 24z 9 2 4 4 2 1/2 1 x2 4 2 2x 1 5 4 1 3 2 2 1 2 x2 3 1 x 6 4/3 1 2 2 1 2 1 1/ 2x 3 3 2 1/2 2 2 2 1 1/2 2 3x 4 2 2 2 a2 b2 1 1 a b 2 1 2 a2 b2. 2 2 4a2b2. 1 5 3 x2 4 2 2x 1 1 x 6 1 31 x2 3 1 2 x1/2 1 2 2x 4 1 2 2 2 1 2/3 x 2 2 1 x 3 2 1 2x 3 1 1/3 2 3 x2 1 1/2 3 2 3x 4 96. 2 x1/2 2 1 ab 1 97. (a) Show that 2 1 a2 b2 (b) Show that (c) Show that a2 b2 c2 d2 2 1 1 2 1 (d) Factor completely: ac bd 2 2 a2 b2 c2 ad bc 2 1 2 2. 2 1 2 4a2c2 1 98. Verify Factoring Formulas 4 and 5 by expanding their right- hand sides. ▼ APPLICATIONS 99. Volume of Concrete A culvert is constructed out of large cylindrical shells cast in concrete, as shown in the ﬁgure. Using the formula for the volume of a cylinder given on the inside back cover of this book, explain why the volume of the cylindrical shell is 2 V pR2h pr2h Factor to show that V 2p average radius height # # thickness # Use the “unrolled” diagram to explain why this makes sense geometrically. R r h h 100. Mowing a Field A square ﬁeld in a certain state park is mowed around the edges every week. The rest of the ﬁeld is kept unmowed to serve as a habitat for birds and small animals (see the ﬁgure). The ﬁeld measures b feet by b feet, and the mowed strip is
x feet wide. (a) Explain why the area of the mowed portion is b2 b 2x 2. 2 1 (b) Factor the expression in (a) to show that the area of the mowed portion is also 4x b x. 2 1 44 CHAPTER P \| Prerequisites ✎ 29–34 ■ Factor the expression by grouping terms. 29. x 3 4x 2 x 4 31. 2x 3 x 2 6x 3 33. x 3 x 2 x 1 30. 3x 3 x 2 6x 2 32. 9x 3 3x 2 3x 1 34. x 5 x4 x 1 92. 93. 94. 95. 35–82 ■ Factor the expression completely. 35. 12x3 18x 37. 6y 4 15y 3 39. x 2 2x 8 41. y 2 8y 15 43. 2x 2 5x 3 45. 9x 2 36x 45 47. 6x 2 5x 6 49. x 2 36 51. 49 4y 2 53. t 2 6t 9 55. 4x 2 4xy y2 36. 30x 3 15x 4 38. 5ab 8abc 40. x 2 14 x 48 42. z2 6z 16 44. 2x 2 7x 4 46. 8x 2 10x 3 48. 6 5t 6t 2 50. 4x 2 25 52. 4t 2 9s 2 54. x 2 10x 25 56. r2 6rs 9s2 1 1 x b 58. a 2 ✎ 57 a2 1 2 1 b2 4 a2 1 60. 2 1 62. x 3 27 64. x 6 64 66. 27a 3 b 6 68. 3x 3 27x 70. x 3 3x 2 x 3 72. 18y 3x 2 2xy4 74. y 3 y 2 y 1 76. 3x 3 5x 2 6x 10 x 2 1 80 ✎ ✎ x2 1 2 2 1 9 x2 1 x2 59. 1 61. t 3 1 63. 8x 3 125 65. x 6 8y 3 67. x 3 2x 2 x 69. x 4 2x 3 3x 2 71. x 4y 3 x 2y 5 73. y 3 3y 2 4 y 12 75. 2x 3 4x 2 x 2 2 77 3x 2x2 x3 2 x 1 78. 79. 81. 82. 2 1 y 2
y4 1 a2 1 2 a2 2a 1 3 y5 a2 1 a2 2a 1 2 2 4 1 2 1 10 3 2 83–90 ■ Factor the expression completely. Begin by factoring out the lowest power of each common factor. 3x1/2 4x1/2 x3/2 x 1 x 1 7/2 1 2 1 3/2 2 ✎ ✎ 83. 85. 87. 88. 89. 90. 84. 86. 1/2 x5/2 x1/2 x3/2 2x1/2 x1/2 x2 1 1/2 2 1 x1/2 2 x 1 1 x 2 2x1/3 1 3x1/2 2 x2 1 1 2 x2 1 1 1/2 x1/2 1 2 2/3 5x4/3 1 5/4 x3/2 2 x 1 2 x 2 2 x2 1 1/2 1/3 1 91–96 ■ Factor the expression completely. (This type of expression arises in calculus in using the “product rule.”) 91. 3x2 1 4x 12 2 x3 2 2 1 2 1 4x 12 4 2 2 1 b x x x 1/4 2 b x ▼ DISCOVE RY • DISCUSSION • WRITI NG 101. The Power of Algebraic Formulas Use the Difference of Squares Formula to factor 17 2 162. Notice that it is easy to calculate the factored form in your head but not so easy to calculate the original form in this way. Evaluate each expression in your head: (a) 5282 5272 (b) 1222 1202 A B Now use the product formula 21 1 evaluate these products in your head: (d) 49 51 # (e) 998 1002 # (c) 10202 10102 A2 B2 to A B 2 102. Differences of Even Powers (a) Factor the expressions completely: A4 B 4 and A6 B 6. (b) Verify that 18,335 12 4 74 and that 2,868,335 126 76. SE CTIO N P.8 \| Rational Expressions 45 (c) Use the results of parts (a) and (b) to factor the integers 18,335 and 2,868,335. Show that in both of these factorizations, all the factors are prime numbers. 103. Factoring
An 1 Verify the factoring formulas in the list by expanding and simplifying the right-hand side in each case. A2 1 A3 1 A4 A2 A 1 A3 A2 A 1 2 1 2 1 On the basis of the pattern displayed in this list, how do you think A5 1 would factor? Verify your conjecture. Now generalize the pattern you have observed to obtain a factorization formula for An 1, where n is a positive integer. 2 P.8 Rational Expressions LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find the domain of an algebraic expression ■ Simplify rational expressions ■ Add and subtract rational expressions ■ Multiply and divide rational expressions ■ Simplify compound fractions ■ Rationalize a denominator or numerator ■ Avoid common errors A quotient of two algebraic expressions is called a fractional expression. Here are some examples: 2x x 1 1x 3 x 1 y 2 y2 4 A rational expression is a fractional expression in which both the numerator and denominator are polynomials. For example, the following are rational expressions: 2x x 1 x x2 1 x3 x x2 5x 6 Expression Domain In this section we learn how to perform algebraic operations on rational expressions. 1 x 1x 1 1x ■ The Domain of an Algebraic Expression In general, an algebraic expression might not be deﬁned for all values of the variable. The domain of an algebraic expression is the set of real numbers that the variable is permitted to have. The table in the margin gives some basic expressions and their domains. 46 CHAPTER P \| Prerequisites E X AM P L E 1 \| Finding the Domain of an Expression Consider the expression 2x 4 x 3 (a) Find the value of the expression for x 2. (b) Find the domain of the expression. ▼ SO LUTI O N (a) We ﬁnd the value by substituting 2 for x in the expression: 2 4 2 2 1 2 3 8 Substitute x = 2 (b) The denominator is zero when x 3. Since division by zero is not deﬁned, we must have x 3. Thus, the domain is all real numbers except 3. We can write this in set notation as ✎ Practice what you’ve learned: Do Exercise 9. x 5 0 x 3 6 E X AM P L E 2 \| Finding the Domain of
an Expression Find the domains of the following expressions. (a) 2x2 3x 1 (b) x x2 5x 6 (c) 1x x 5 ▼ SO LUTI O N (a) This polynomial is deﬁned for every x. Thus, the domain is the set of real numbers. (b) We ﬁrst factor the denominator. x x2 5x Denominator would be 0 if x 2 or x 3. Since the denominator is zero when x 2 or 3, the expression is not deﬁned for these numbers. The domain is x 2 (c) For the numerator to be deﬁned, we must have x 0. Also, we cannot divide by x 3 and x 6 5. 0 zero, so x 5. Must have x 0 to take square root. 1x x 5 Denominator would be 0 if x 5. Thus, the domain is x x 0 and x 5 0 ✎ Practice what you’ve learned: Do Exercise 13. 5. 6 ▲ ▲ ■ Simplifying Rational Expressions To simplify rational expressions, we factor both numerator and denominator and use the following property of fractions: AC BC A B This allows us to cancel common factors from the numerator and denominator. E X AM P L E 3 \| Simplifying Rational Expressions by Cancellation SE CTIO N P.8 \| Rational Expressions 47 Simplify: x2 1 x2 x 2 ▼ SO LUTI O N We can’t cancel the x2’s in x2 1 x2 x 2 because x2 is not a factor. x2 1 x2 Factor Cancel common factors ✎ Practice what you’ve learned: Do Exercise 23. ▲ ■ Multiplying and Dividing Rational Expressions To multiply rational expressions, we use the following property of fractions: A B # C D AC BD This says that to multiply two fractions, we multiply their numerators and multiply their denominators. E X AM P L E 4 \| Multiplying Rational Expressions Perform the indicated multiplication and simplify: x2 2x 3 x2 8x 16 # 3x 12 x 1 ▼ SO LUTI O N We ﬁrst factor. x2 2x 3 x2 8x 16 # 3x 12 Factor Cancel common factors 2 Property of fractions ✎ Practice what you’ve learned:
Do Exercise 31. ▲ To divide rational expressions, we use the following property of fractions This says that to divide a fraction by another fraction we invert the divisor and multiply. E X AM P L E 5 \| Dividing Rational Expressions x 4 x2 4 Perform the indicated division and simplify: x2 3x 4 x2 5x 6 48 CHAPTER P \| Prerequisites ▼ SO LUTI O N x 4 x2 4 x2 3x 4 x2 5x 6 x 4 x2 4 # x2 5x 6 x2 3x Invert divisor and multiply Factor 2 Cancel common factors ▲ 1 ✎ Practice what you’ve learned: Do Exercise 37. 2 1 2 Avoid making the following error: A B C A B A C For instance, if we let A 2, B 1, and C 1, then we see the error Wrong! ■ Adding and Subtracting Rational Expressions To add or subtract rational expressions, we ﬁrst ﬁnd a common denominator and then use the following property of fractions: A C B C A B C Although any common denominator will work, it is best to use the least common denominator (LCD) as explained in Section P.2. The LCD is found by factoring each denominator and taking the product of the distinct factors, using the highest power that appears in any of the factors. E X AM P L E 6 \| Adding and Subtracting Rational Expressions Perform the indicated operations and simplify. (a) 3 x 1 x x 2 (b) 1 x2 1 2 x 1 1 2 2 ▼ SO LUTI O N (a) Here the LCD is simply the product 21 b) The LCD of x2 1 1 x2 and 1 x 1 2 is 3x 6 x2 x x 2 x 1 2 2 1 x2 2x 21 2x ✎ Practice what you’ve learned: Do Exercise 49. 2 1 2 Write fractions using LCD 2 Add fractions Combine terms in numerator x 1 2. 2 21 Factor Combine fractions using LCD Distributive Property Combine terms in numerator ▲ Diophantus lived in Alexandria about 250 A.D. His book Arithmetica is considered the ﬁrst book on algebra. In it he gives methods for ﬁnding integer solutions of algebraic equations. Arithmetica was read and studied for more than a thousand years.
Fermat (see page 159) made some of his most important discoveries while studying this book. Diophantus’ major contribution is the use of symbols to stand for the unknowns in a problem. Although his symbolism is not as simple as what we use today, it was a major advance over writing everything in words. In Diophantus’ notation the equation x5 7x2 8x 5 24 is written K©å h ©zM°´iskd c Our modern algebraic notation did not come into common use until the 17th century. SE CTIO N P.8 \| Rational Expressions 49 ■ Compound Fractions A compound fraction is a fraction in which the numerator, the denominator, or both are themselves fractional expressions. E X AM P L E 7 \| Simplifying a Compound Fraction Simplify: x y 1 1 y x ▼ SO LUTI O N 1 We combine the terms in the numerator into a single fraction. We do the same in the denominator. Then we invert and multiply ▼ SO LUTI O N 2 We ﬁnd the LCD of all the fractions in the expression, then multiply the numerator and denominator by it. In this example the LCD of all the fractions is xy. Thus # xy xy Multiply numerator and denominator by xy x2 xy xy y2 Simplify x 1 y 1 ✎ Practice what you’ve learned: Do Exercises 67 and 69. Factor x y x y 2 2 ▲ The next two examples show situations in calculus that require the ability to work with fractional expressions. E X AM P L E 8 \| Simplifying a Compound Fraction Simplify: 1 a 1 a h h ▼ SO LUTI O N We begin by combining the fractions in the numerator using a common denominator. 50 CHAPTER P \| Prerequisites Combine fractions in the numerator Invert divisor and multiply Distributive Property Simplify Cancel common factors a ✎ Practice what you’ve learned: Do Exercise 77. 1 2 E X AM P L E 9 \| Simplifying a Compound Fraction Simplify: 1 1 x2 2 1 x2 1/2 x2 1 1 x2 1/2 2 1 x2 ▼ SO LUTI O N 1 Factor 1 2 1 2 from the numerator. ▲ x2 2 Factor out the power of 1 x2 with
the smallest exponent, in this case 1 x2 1 12. 2 1 x2 2 1 1 x2 1/2 x2 1 1 x2 1/2 2 1 1 1 x2 2 1 x2 2 1 x2 1 x2 1/2 1 1 x2 1/2 1 1 x2 1 3/2 2 ▼ SO LUTI O N 2 Since 1 fractions by multiplying numerator and denominator by 1/2 1/ 1 x2 1 x2 1/2 1 2 1 x2 2 1 1 x2 1/2 x2 1 1 x2 1/2 2 1 1 x2 2 2 1 2. is a fraction, we can clear all 1 x2 1 1/2 x2 1 1 x2 2 1 x2 1 x2 2 1 x2 1/2 1/2 1/2 2 # 1 1 2 1 1 x2 2 1 x2 x2 3/2 1 2 1 1 x2 1 3/2 2 ✎ Practice what you’ve learned: Do Exercise 85. ▲ ■ Rationalizing the Denominator or the Numerator we may rationalize the denominaIf a fraction has a denominator of the form tor by multiplying numerator and denominator by the conjugate radical This is effective because, by Product Formula 1 in Section P.6, the product of the denominator and its conjugate radical does not contain a radical: A B2C, A B2C. A B2C 1 2 1 A B2C 2 A2 B2C E X AM P L E 10 \| Rationalizing the Denominator Rationalize the denominator: 1 1 22 SE CTIO N P.8 \| Rational Expressions 51 ▼ SO LUTI O N We multiply both the numerator and the denominator by the conjugate radical of 1 12 1 12, which is. 1 1 12 # 1 12 1 12 Multiply numerator and denominator by the conjugate radical 2 Product Formula 1: (a + b)(a – b) = a2 – b2 1 1 12 1 12 12 12 1 1 12 1 2 2 1 12 1 ✎ Practice what you’ve learned: Do Exercise 89. 12 1 E X AM P L E 11 \| Rationalizing the Numerator Rationalize the numerator: 24 h 2 h ▼ SO LUTI O N We multiply numerator and denominator by the conjug
ate radical 14 h 2. 14 h 2 h # 14 h 2 14 h 2 14 h 2 h 2 22 14 h 1 2 14 h 2 h 4 h 4 14 h 2 h 2 1 2 1 h 1 14 h 2 14 h 2 ✎ Practice what you’ve learned: Do Exercise 95. h 2 1 Multiply numerator and denominator by the conjugate radical Product Formula 1: (a + b)(a – b) = a2 – b2 Cancel common factors ▲ ▲ ■ Avoiding Common Errors Don’t make the mistake of applying properties of multiplication to the operation of addition. Many of the common errors in algebra involve doing just that. The following table states several properties of multiplication and illustrates the error in applying them to addition. 2 a, b 0 Correct multiplication property 2 a2 # b2 a # b 1 1a # b 1a1b 2a2 # b2 ab b a a1 # b1 1 a Common error with addition a b 2 a2 b2 2 1 2a b 1a 1b 2a2 b2 a1 b1 a b 1 1 2 52 CHAPTER P \| Prerequisites To verify that the equations in the right-hand column are wrong, simply substitute numbers for a and b and calculate each side. For example, if we take a 2 and b 2 in the fourth error, we ﬁnd that the left-hand side is whereas the right-hand side is Since ror in each of the other equations. (See Exercise 113.) the stated equation is wrong. You should similarly convince yourself of the er- P. ▼ CONCE PTS 1. Which of the following are rational expressions? (a) 3x x2 1 (b) 1x 1 2x 3 (c) x 1 x2 1 x 3 2 2. To simplify a rational expression, we cancel factors that are common to the. True or false? and. So the expression simpliﬁes to. (a) x2 3 x2 5 simpliﬁes to 3 5. (b) 3x2 5x2 simpliﬁes to 3 5. 4. (a) To multiply two rational expressions, we multiply their together and multiply their together. So 2 x 1 # x x 3 is the same as. (b) To divide two rational expressions, we the divisor, then multiply. So 3 x 5 x x 2 is the
same as. 5. Consider the expression 2 1 x (a) How many terms does this expression have? (b) Find the least common denominator of all the terms. (c) Perform the addition and simplify. 2x 1 x 4, x 7 11. 1x 3, x 6 ✎ 13. x2 1 x2 x 2, x 2 10. 12. 14. 2t2 5 3t 6, t 1, x 5 1 1x 1 22x x 1, x 8 15–28 ■ Simplify the rational expression. 15. 17. 19. 21. ✎ 23. 25. 27. 12x 6x2 x 1 2 5y2 10y y2 x2 4 x2 6x 8 x2 5x 4 y2 y y2 1 2x3 x2 6x 2x2 7x 6 2 16. 18. 20. 22. 24. 26. 28. 81x3 18x 14t2 t 7t 2 x2 1 4 1 x 2 2 x 1 12 1 2 1 x2 x 2 x2 1 x2 x 12 x2 5x 6 y2 3y 18 2y2 5y 3 1 x2 x3 1 29–44 ■ Perform the multiplication or division and simplify. 6. True or false? 1 2 (a) 1 x 1 x (b) 1 2 is the same as is the same as 1 2 x x 2 2x.. 29. 4x x2 4 # x 2 16x ✎ 31. x2 2x 15 x2 9 # x 3 x 5 ▼ SKI LLS 7–14 ■ An expression is given. (a) Evaluate it at the given value. (b) Find its domain. 7. 4x2 10x 3, 8. x4 x3 9x, x 1 x 5 33. 35. 36. # t 3 t 3 t2 9 t2 9 x2 7x 12 x2 3x 2 # x2 5x 6 x2 6x 9 x2 2xy y2 x2 y2 # 2x2 xy y2 x2 xy 2y2 30. 32. 34. x2 25 x2 16 # x 4 x 5 x2 2x 3 x2 2x 3 # x2 x 6 x2 2x # 3 x 3 x x3 x2 x2 2x 3 ✎ 37. x 3 4x2 9 x2 7
x 12 2x2 7x 15 6x2 x 2 x 3 x2 6x 5 2x2 7x 3 2y2 y 3 y2 5y 6 38. 39. 40. 41. 2x 1 2x2 x 15 2x2 3x 1 x2 2x 15 4y2 9 2y2 9y 18 x3 x 1 x x2 2x 1 43. x/y z 42. 2x2 3x 2 x2 1 2x2 5x 2 x2 x 2 44. x y/z 45–64 ■ Perform the addition or subtraction and simplify. 46. 48. 50. 52. 54. 56. 58. 60. 2x 2x 3 3 2 a2 ab 1 1 x2 x 3 2x 3 2 2 1 4 b2 1 x3 1 x 2 x x2 4 x x2 x 2 2 x2 5x 4 45. 47. ✎ 49 x2 1 x2 x 51. 53. 55. 57. 59. 61. 62. 63. 64. 1 x2 7x 12 1 x2 x2 x x 1 x x2 x 6 1 x2 3x 2 1 x 2 2 x 3 1 x2 2x 3 3 x2 65–76 ■ Simplify the compound fractional expression 65. ✎ 67. ✎ 69. 66. 68. 70. 1 1 x2 x 1 x2 SE CTIO N P.8 \| Rational Expressions 53 71. 73. 75. y x x y 1 1 y2 x2 x2 y2 x1 y1 1 1 1 1 x 72. 74. 76. y x y x x y x1 y1 77–82 ■ Simplify the fractional expression. (Expressions like these arise in calculus.) 1 1x 1 1x h h 78. ✎ 77. 79 x2 80. 1 2 1 x h h 2 x3 7x 1 2 81. 1 B 2 x 21 x2 b a 82. 1 B a 2 x3 1 4x3 b 83–88 ■ Simplify the expression. (This type of expression arises in calculus when using the “quotient rule.”) 3 1 x 2 x 3 2 2 1 83/2 2 1 x2 1/2 2 1 2 4 x2 x 6 1 1/2 x 1 x 1 1/2 x2 1 1 x2 2x 1 84. x 6
✎ 85. 1 x 2 1 2 1 x2 2 86. 1 1 x 3 1 87. 7 3x 88. 1 2 2 1/3 x 1 x 2 1 1/2 3 2x 1 7 3x 1 x 1 2/3 2/3 2 7 3x 1/2 2 89–94 ■ Rationalize the denominator. ✎ 89. 91. 93. 1 2 13 2 12 17 y 13 1y 90. 92. 94. 2 3 15 1 1x 1 x y 2 2 1x 1y 1 95–100 ■ Rationalize the numerator. ✎ 95. 97. 1 15 3 1r 12 5 96. 98. 13 15 2 1x 1x h h 1x1x h 99. 2x2 1 x 100. 2x 1 2x 54 C HAPTER P \| Prerequisites 101–108 ■ State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) ▼ DISCOVE RY • DISCUSSION • WRITI NG 111. Limiting Behavior of a Rational Expression The rational 101. 103. 105. 107. 16 a 16 2 4 x 1 a 16 102. 104. 106. 108 2a b b 2b 1 x x2 x 1 x 1 x ▼ APPLICATIONS 109. Electrical Resistance If two electrical resistors with resistances R1 and R2 are connected in parallel (see the ﬁgure), then the total resistance R is given by R 1 1 R2 1 R1 (a) Simplify the expression for R. 10 ohms and R2 (b) If R1 20 ohms, what is the total resistance R? R⁄ R2 110. Average Cost A clothing manufacturer ﬁnds that the cost of producing x shirts is 500 6x 0.01x2 dollars. (a) Explain why the average cost per shirt is given by the rational expression A 500 6x 0.01x2 x (b) Complete the table by calculating the average cost per shirt for the given values of x. x 10 20 50 100 200 500 1000 Average cost CHAPTER P \| REVIEW expression x 2 9 x 3 is not deﬁned for x 3. Complete the tables and determine what value the expression approaches as x gets closer and closer to 3. Why is this reasonable? Factor the numerator of the expression and simplify to see why. x
2 9 x 3 x 2.80 2.90 2.95 2.99 2.999 x2 9 x 3 x 3.20 3.10 3.05 3.01 3.001 112. Is This Rationalization? In the expression 2/ 1x we would eliminate the radical if we were to square both numerator and denominator. Is this the same thing as rationalizing the denominator? 113. Algebraic Errors The left-hand column in the table lists some common algebraic errors. In each case, give an example using numbers that shows that the formula is not valid. An example of this type, which shows that a statement is false, is called a counterexample. Algebraic error Counterexample a2 b2 2 1 2a2 b2 a b a b a b a3 b3 1/3 a b 2 1 am/an am/n a1/n 1 an b 0 0 0 0 ab RTI LAS Properties of Real Numbers (p. 9) Commutative: a b b a ab ba a b ab 2 c a bc c a 2 1 2 Associative if a 0 if a 0 Distributive: b c a 1 2 ab ac Absolute Value (p. 16) Distance between a and b is b a 0 0 d a, b 1 2 Exponents (p. 21) aman amn am an am ab amn n amn n anbn n an bn a b b 2 2 1 1 a If n is even, then 2n an a 0 0 am/n 2n am Special Product Formulas (p. 35) Sum and difference of same terms: A2 B2 A B 21 1 A B 2 Square of a sum or difference A2 2AB B2 2 A2 2AB B2 Cube of a sum or difference: Radicals (p. 27) 2n a b means bn a 2n ab 2n a2n b 2n a 2n b 2n a 2a If n is odd, then n a B b 2m 2n an a mn A B 2 1 A B 2 1 3 A3 3A2B 3AB2 B3 3 A3 3A2B 3AB2 B3 Factoring Formulas (p. 41) Difference of squares: A2 B2 Perfect squares: A B 21 1 A B 2 A2 2AB B2 A2 2AB B2 CHAPTER P \| Review 55
Sum or difference of cubes: A3 B3 A3 B3 A B 21 1 A B 21 1 Rational Expressions (p. 46) A2 AB B2 2 A2 AB B2 2 We can cancel common factors: AC BC A B To multiply two fractions, we multiply their numerators together and their denominators together: A B C D AC BD To divide fractions, we invert the divisor and multiply: A B C D A B D C To add fractions, we ﬁnd a common denominator: A C B C A B C ▼ CO N C E P T S U M MARY Section P.1 ■ Make and use algebra models Section P.2 ■ Classify real numbers ■ Use properties of real numbers ■ Add, subtract, multiply, and divide fractions Section P.3 ■ Graph numbers on the real line ■ Use the order symbols,,, ■ Work with set and interval notation ■ Work with absolute values ■ Find distances on the real line Section P.4 ■ Use exponential notation ■ Simplify expressions using the Laws of Exponents ■ Write numbers in scientiﬁc notation Section P.5 ■ Simplify expressions involving radicals ■ Simplify expressions involving rational exponents ■ Express radicals using rational exponents ■ Rationalize a denominator Section P.6 ■ Add, subtract, and multiply polynomials ■ Multiply algebraic expressions ■ Use the Special Product Formulas Review Exercises 1–2 Review Exercises 3–4 5–8 9–12 Review Exercises 13–16 13–20 13–24 25–28 37–38 Review Exercises 29–32, 39–46 39–46 61–64 Review Exercises 33–36, 40, 93 29–32, 59, 77 47–50 57–58 Review Exercises 87–92 87–95 88, 89, 92, 94, 95 56 CHAPTER P \| Prerequisites Section P.7 ■ Factor out common factors ■ Factor trinomials by trial and error ■ Use the Special Factoring Formulas ■ Factor algebraic expressions completely ■ Factor by grouping terms Section P.8 ■ Find the domain of an algebraic expression ■ Simplify rational expressions ■ Add, subtract, multiply, and divide rational expressions ■ Simplify compound fractions ■ Rationalize a numerator or a denominator ■ Avoid common algebraic errors ▼ E X E RC I S E S 1–2 ■ Make and use an algebra model to solve the problem
. 1. Elena regularly takes a multivitamin and mineral supplement. She purchases a bottle of 250 tablets and takes two tablets every day. (a) Find a formula for the number of tablets T that are left in the bottle after she has been taking the tablets for x days. (b) How many tablets are left after 30 days? (c) How many days will it take for her to run out of tablets? 2. Alonzo's Delivery is having a sale on calzones. Each calzone costs $2, and there is a $3 delivery charge for phone-in orders. (a) Find a formula for the total cost C of ordering x calzones for delivery. (b) How much would it cost to have 4 calzones delivered? (c) If you have $15, how many calzones can you order? 3–4 ■ Determine whether each number is rational or irrational. If it is rational, determine whether it is a natural number, an integer, or neither. 3. (a) 16 8 3 (e) 4. (a) 5 (e) 24 16 (f) (b) 16 8 2 25 6 (f) 1020 (b) (c) 116 (d) 12 (c) 125 (d) 3p 5–8 ■ State the property of real numbers being used. 5. 3 2x 2x 3 a b 1 21 a b 2 6. a b a b 2 2 1 1 Ax Ay x y 7 21 8–12 ■ Evaluate each expression. Express your answer as a fraction in lowest terms. 2 9. (a) 5 6 10. (a) 2 3 11 15 7 10 (b) (b) 5 6 7 10 2 3 11 15 Review Exercises 65, 66, 74, 77, 78, 81, 83 67–72, 77, 86 72–75, 78, 80, 85, 123, 124 65–86 76, 82, 84 Review Exercises 111–114 98–101 99–108 107–108 57–58, 110 115–122 11. (a) 12. (a) 15 8 30 7 # 12 5 12 35 (b) (b) 15 8 30 7 12 5 # 12 35 13–16 ■ Express the interval in terms of inequalities, and then graph the interval. 13. [2, 6 15. 2 q, 4’ 1 14. 0, 10’ 1 16.
[2, q 2 17–20 ■ Express the inequality in interval notation, and then graph the corresponding interval. 17. x 5 19. 1 x 5 18. x 3 20. 0 x 1 2 21–24 ■ The sets A, B, C, and D are deﬁned as follows: A C 1, 0, 1, 2, 4 2 5 1, 1 1 4 6 Find each of the following sets. 21. (a) A B 22. (a) C D 23. (a) A C 24. (a) A D (b) A B (b) C D (b) B D (b) B C 25–36 ■ Evaluate the expression. 25. 7 10 27. 3 9 29. 23 32 31. 2161/3 3 2 5 26. 28. 1 1 1 30. 21/281/2 32. 642/3 1242 12 23 125 33. 35. 34. 24 4 24 324 36. 12 150 CHAPTER P \| Review 57 3 x 2 92. 2x 1 1 2 3 95. x2 1 2 x x 2 1 x2 2x 3 2x2 5x 3 x2 2x 3 x2 8x 16 x 2 2 2 # 3x 12 x 1 97. 99. 90. 91. 93. 94. x 3 3 x 21 1 x 3 2 2 1x 1 x 6 1 2 2 1 x 1 21 x 1 1 1x 2 x 2 x 2 21 1x 1 21 x3 2x2 3x x 96. 98. 100. 101. 102. 3 1 t 2 1 t x 1 x3/ 2 1 x3 1 x2/ x2 2x 15 x2 6x 5 x 1 1 2 x 1 x2 x 12 x2 1 103. 105. 1 x 1 1 x 1 x x2 1 2 x2 1 104 106. 107 x2 4 2 x2 x 2 108 109. 110 3x2 5x 2 1x h 1x h (rationalize the numerator) 111–114 ■ Find the domain of the algebraic expression. 111. 113. x 5 x 10 1x x2 3x 4 112. 114. 2x x2 9 1x 3 x2 4x 4 115–122 ■ State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator 0.) 115. 117. x y 2
1 12 y y 3 x 3 y 3 12 y 1 119. 2a2 a 116. 1 1a 1 a 1 1 1a 118. 23 a b 23 a 23 b 120. 1 x 4 1 x 1 4 121. x 3 y 3 x2 1 x2 2x 1 122. x y 21 1 1 2x 1 x 2 xy y2 2 37–38 ■ Express the distance between the given numbers on the real line using an absolute value. Then evaluate this distance. 37. (a) 3 and 5 38. (a) 0 and 4 (b) 3 and 5 (b) 4 and 4 39–46 ■ Write the expression as a power of x. 39. 1 x2 m 41. x2xm x3 1 2 43. xaxbxc x2c1 45. xc1 1 2 2 40. x1x 42. 44. 46. 11 xm n 2 2 2 c b 2 2 nx5 xa 11 x2 1 2 xn 47–50 ■ Express the radical as a power with a rational exponent. 47. (a) 23 7 (b) 25 74 49. (a) 26 x5 (b) 1x 9 B A 51–60 ■ Simplify the expression. 48. (a) 23 57 50. (a) 2y3 (b) (b) A A 14 5 B 28 y 3 2 B 51. 53. 3x1y2 1 2 2 2x 3y 2 1 x4 3x 1 x3 2 55. 23 2y4 x3y 1 2 57. 59. x 2 1x 8r1/2s3 2r2s4 2 52. 54. 56. 58. 60. 4 b3 1 2 2 2 6 a2 1 3 a3b 2 1 r 2s4/3 r 1/3s b a 2x2y4 1x 1 1x 1 ab2c 3 2a2b4 b a 2 61. Write the number 78,250,000,000 in scientiﬁc notation. 62. Write the number 2.08 108 in decimal notation. 63. If a 0.00000293, b 1.582 1014, and c 2.8064 1012, use a calculator to approximate the number ab/c. 64. If your heart beats 80 times per minute and you live to be 90 years old, estimate the number of times your
heart beats during your lifetime. State your answer in scientiﬁc notation. 65–86 ■ Factor the expression. 65. 2x2y 6xy2 67. x2 9x 18 69. 3x2 2x 1 71. 4t2 13t 12 73. 25 16t2 75. x6 1 77. x1/2 2x1/2 x3/2 79. 4x3 8x2 3x 6 66. 12x2y4 3xy5 9x3y2 68. x2 3x 10 70. 6x2 x 12 72. x4 2x2 1 74. 2y6 32y2 76. y3 2y2 y 2 78. a4b2 ab5 80. 8x3 y6 2 1 1 x 2 2 5/2 2x x 2 2 81. 2 82. 3x 3 2 x 2 18x 12 84. ax 2 bx 15 86. 2 2 3/2 x 22x 2 2 83. a2y b2y x 1 1 85. 2 2 2 87–110 ■ Perform the indicated operations. 5 87. 4 x 1 1 2 2x 1 1 2y 7 1 21 3x 2 21 2y 7 2 2 88. 89. 2a2 b 1 2 2 x 1 1 2 1 123. If m n 0 and a 2mn, b m 2 n 2, c m 2 n 2, show that a 2 b2 c2. 124. If t 1 2 a x 3 1 x3 b and x 0, show that 21 t 2 1 2 a x3 1 x3 b ■ CHAPTER P \| TEST 1. A pizzeria charges $9 for a medium plain cheese pizza, plus $1.50 for each extra topping. (a) Find a formula that models the cost C of a medium pizza with x toppings. (b) Use your model from part (a) to ﬁnd the cost of a medium pizza with the following extra toppings: anchovies, ham, sausage, and pineapple. 2. Determine whether each number is rational or irrational. If it is rational, determine whether it is a natural number, an integer, or neither. (a) 5 3. Let A 5 (a) A B 2, 0, 1, 3, 5 (b) 15 and B 6, 1, 5, 7 1 0, 2 (b) A B 5 (c)
9 3 (d) 1,000,000. Find each of the following sets. 6 4. (a) Graph the intervals 4, 2 3 2 and 0, 3 3 4 on a real line. (b) Find the intersection and the union of the intervals in part (a), and graph each of them on a real line. (c) Use an absolute value to express the distance between 4 and 2 on the real line, and then evaluate this distance. 5. Evaluate each expression: (a) 26 (e) 2 3 2 b a (b) (f) 2 6 2 1 15 32 116 (c) 26 (g) 4 38 B 216 6. Write each number in scientiﬁc notation. (a) 186,000,000,000 (b) 0.0000003965 (d) 710 712 (h) 813/4 7. Simplify each expression. Write your ﬁnal answer without negative exponents. (a) 1200 132 (b) 3a3b3 1 2 4ab2 2 2 (c) 3 125 x9 B 21 3x3/2y3 x2y1/2 b 8. Perform the indicated operations and simplify. 2x 5 1 (b) (d) a 2 4x 5 x 6 (a) 3 2 1 2x 3 1 (d 21 3 2 (e) 2 (c) 1 (f) x2 1a 1b x 3 1 2 1 x 3 21 2 1a 1b 2 9. Factor each expression completely. (a) 4x2 25 (d) x4 27x (b) 2x2 5x 12 (e) 3x3/2 9x1/2 6x1/2 (c) x3 3x2 4x 12 (f) x3y 4xy 10. Simplify the rational expression. (a) x2 3x 2 x2 x 2 (c) x2 x2 4 x 1 x 2 # x 3 2x 1 (b) (d) 2x2 x 1 x2 9 x y 1 x y x 1 y 11. Rationalize the denominator and simplify. 16 2 13 (b) (a) 6 23 4 58 Image not available due to copyright restrictions George Polya (1887–1985) is famous among mathematicians for his ideas on problem solving. His lectures on problem solving at Stanford University attracted overﬂ
ow crowds whom he held on the edges of their seats, leading them to discover solutions for themselves. He was able to do this because of his deep insight into the psychology of problem solving. His well-known book How To Solve It has been translated into 15 languages. He said that Euler (see page 100) was unique among great mathematicians because he explained how he found his results. Polya often said to his students and colleagues, “Yes, I see that your proof is correct, but how did you discover it?” In the preface to How To Solve It, Polya writes, “A great discovery solves a great problem but there is a grain of discovery in the solution of any problem. Your problem may be modest; but if it challenges your curiosity and brings into play your inventive faculties, and if you solve it by your own means, you may experience the tension and enjoy the triumph of discovery.” GENERAL PRINCIPLES There are no hard and fast rules that will ensure success in solving problems. However, it is possible to outline some general steps in the problem-solving process and to give principles that are useful in solving certain problems. These steps and principles are just common sense made explicit. They have been adapted from George Polya’s insightful book How To Solve It. 1. Understand the Problem The ﬁrst step is to read the problem and make sure that you understand it. Ask yourself the following questions: What is the unknown? What are the given quantities? What are the given conditions? For many problems it is useful to draw a diagram and identify the given and required quantities on the diagram. Usually, it is necessary to introduce suitable notation In choosing symbols for the unknown quantities, we often use letters such as a, b, c, m, n, x, and y, but in some cases it helps to use initials as suggestive symbols, for instance, V for volume or t for time. 2. Think of a Plan Find a connection between the given information and the unknown that enables you to calculate the unknown. It often helps to ask yourself explicitly: “How can I relate the given to the unknown?” If you don’t see a connection immediately, the following ideas may be helpful in devising a plan. ■ TRY TO RECOGNIZE SOMETHING FAMILIAR Relate the given situation to previous knowledge. Look at the unknown and try to recall a more familiar problem that has a
similar unknown. ■ TRY TO RECOGNIZE PATTERNS Certain problems are solved by recognizing that some kind of pattern is occurring. The pattern could be geometric, numerical, or algebraic. If you can see regularity or repetition in a problem, then you might be able to guess what the pattern is and then prove it. ■ USE ANALOGY Try to think of an analogous problem, that is, a similar or related problem but one that is easier than the original. If you can solve the similar, simpler problem, then it might give you the clues you need to solve the original, more difﬁcult one. For instance, if a problem involves very large numbers, you could ﬁrst try a similar problem with smaller numbers. Or if the problem is in three-dimensional geometry, you could look for something similar in two-dimensional geometry. Or if the problem you start with is a general one, you could ﬁrst try a special case. ■ INTRODUCE SOMETHING EXTRA You might sometimes need to introduce something new—an auxiliary aid—to make the connection between the given and the unknown. For instance, in a problem for which 59 60 Focus on Problem Solving a diagram is useful, the auxiliary aid could be a new line drawn in the diagram. In a more algebraic problem the aid could be a new unknown that relates to the original unknown. ■ TAKE CASES You might sometimes have to split a problem into several cases and give a different argument for each case. For instance, we often have to use this strategy in dealing with absolute value. ■ WORK BACKWARD Sometimes it is useful to imagine that your problem is solved and work backward, step by step, until you arrive at the given data. Then you might be able to reverse your steps and thereby construct a solution to the original problem. This procedure is commonly used in solving equations. For instance, in solving the equation 3x 5 7, we suppose that x is a number that satisﬁes 3x 5 7 and work backward. We add 5 to each side of the equation and then divide each side by 3 to get x 4. Since each of these steps can be reversed, we have solved the problem. ■ ESTABLISH SUBGOALS In a complex problem it is often useful to set subgoals (in which the desired situation is only partially fulﬁlled). If you can attain or accomplish these subgoals
, then you might be able to build on them to reach your ﬁnal goal. ■ INDIRECT REASONING Sometimes it is appropriate to attack a problem indirectly. In using proof by contradiction to prove that P implies Q, we assume that P is true and Q is false and try to see why this cannot happen. Somehow we have to use this information and arrive at a contradiction to what we absolutely know is true. ■ MATHEMATICAL INDUCTION In proving statements that involve a positive integer n, it is frequently helpful to use the Principle of Mathematical Induction, which is discussed in Section 9.4. 3. Carry Out the Plan In Step 2, a plan was devised. In carrying out that plan, you must check each stage of the plan and write the details that prove that each stage is correct. 4. Look Back Having completed your solution, it is wise to look back over it, partly to see whether any errors have been made and partly to see whether you can discover an easier way to solve the problem. Looking back also familiarizes you with the method of solution, which may be useful for solving a future problem. Descartes said, “Every problem that I solved became a rule which served afterwards to solve other problems.” We illustrate some of these principles of problem solving with an example. P RO B L E M \| Average Speed A driver sets out on a journey. For the ﬁrst half of the distance, she drives at the leisurely pace of 30 mi/h; during the second half she drives 60 mi/h. What is her average speed on this trip? 01A-W4525.qxd 1/8/08 11:05 AM Page 61 General Principles 61 Thinking About the Problem It is tempting to take the average of the speeds and say that the average speed for the entire trip is 30 60 2 45 mi/h Try a special case But is this simple-minded approach really correct? Let’s look at an easily calculated special case. Suppose that the total distance traveled is 120 mi. Since the ﬁrst 60 mi is traveled at 30 mi/h, it takes 2 h. The second 60 mi is traveled at 60 mi/h, so it takes one hour. Thus, the total time is 2 1 3 hours and the average speed is 120 3 40 mi/h So our guess of 45 mi/h was wrong. Understand the problem ▼ SO LUTI O N We
need to look more carefully at the meaning of average speed. It is deﬁned as average speed distance traveled time elapsed Introduce notation State what is given Let d be the distance traveled on each half of the trip. Let t1 and t2 be the times taken for the ﬁrst and second halves of the trip. Now we can write down the information we have been given. For the ﬁrst half of the trip we have 30 d t1 and for the second half we have 60 d t2 Identify the unknown Now we identify the quantity that we are asked to ﬁnd: average speed for entire trip total distance total time 2d t1 t2 Connect the given with the unknown To calculate this quantity, we need to know t1 and t2, so we solve the above equations for these times: Now we have the ingredients needed to calculate the desired quantity: t1 d 30 t2 d 60 average speed 2d t1 t1 2d d 60 d 30 60 1 60 d a 30 120d 2d d Multiply numerator and denominator by 60 2d 2 d 60 b 120d 3d 40 So the average speed for the entire trip is 40 mi/h. ▲ 62 Focus on Problem Solving © Don’t feel bad if you can’t solve these problems right away. Problems 2 and 6 were sent to Albert Einstein by his friend Wertheimer. Einstein (and his friend Bucky) enjoyed the problems and wrote back to Wertheimer. Here is part of his reply: Your letter gave us a lot of amusement. The ﬁrst intelligence test fooled both of us (Bucky and me). Only on working it out did I notice that no time is available for the downhill run! Mr. Bucky was also taken in by the second example, but I was not. Such drolleries show us how stupid we are! (See Mathematical Intelligencer, Spring 1990, page 41.) Problems 1. Distance, Time, and Speed A man drives from home to work at a speed of 50 mi/h. The return trip from work to home is traveled at the more leisurely pace of 30 mi/h. What is the man’s average speed for the round-trip? 2. Distance, Time, and Speed An old car has to travel a 2-mile route, uphill and down. Be- cause it is so old, the car can climb the �
��rst mile—the ascent—no faster than an average speed of 15 mi/h. How fast does the car have to travel the second mile—on the descent it can go faster, of course—to achieve an average speed of 30 mi/h for the trip? 3. A Speeding Fly A car and a van are parked 120 mi apart on a straight road. The drivers start driving toward each other at noon, each at a speed of 40 mi/h. A ﬂy starts from the front bumper of the van at noon and ﬂies to the bumper of the car, then immediately back to the bumper of the van, back to the car, and so on, until the car and the van meet. If the ﬂy ﬂies at a speed of 100 mi/h, what is the total distance it travels? 4. Comparing Discounts Which price is better for the buyer, a 40% discount or two succes- sive discounts of 20%? 5. Cutting up a Wire A piece of wire is bent as shown in the ﬁgure. You can see that one cut through the wire produces four pieces and two parallel cuts produce seven pieces. How many pieces will be produced by 142 parallel cuts? Write a formula for the number of pieces produced by n parallel cuts. 6. Amoeba Propagation An amoeba propagates by simple division; each split takes 3 minutes to complete. When such an amoeba is put into a glass container with a nutrient ﬂuid, the container is full of amoebas in one hour. How long would it take for the container to be ﬁlled if we start with not one amoeba, but two? 7. Running Laps Two runners start running laps at the same time, from the same starting position. George runs a lap in 50 s; Sue runs a lap in 30 s. When will the runners next be side by side? 8. Batting Averages Player A has a higher batting average than player B for the ﬁrst half of the baseball season. Player A also has a higher batting average than player B for the second half of the season. Is it necessarily true that player A has a higher batting average than player B for the entire season? 9. Coffee and Cream A spoonful of cream is taken from a pitcher of cream and put into a cup of coffee. The coffee is stirred

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