basicmathematics_solutions.pdf -

Please enable JavaScript to view the full PDF

Basic Mathematics - S. Lang: Solutions Manual Hannes Thorsell Report any typographical or mathematical errors to: [email protected] December 2, 2020, version 1.0 1 2 Preface I wrote this solutions manual as part of my own reading of Basic Mathematics. It provides complete solutions to all exercises, complementing those found in the back of the book. For completeness even those exercises that do have solutions in Basic Mathematics also have solutions presented here, and it should be noted that they can, and sometimes do, differ from Lang’s provided solutions. This I believe is only a benefit as multiple valid, but different, solutions can then be perused. It also has the added benefit of providing corrections to incorrect solutions. I have tried to be as complete as possible in my solutions, while not being unnecessarily repetitious. If you find any errors (of any kind; be they mathe- matical, grammatical, etc.) please do send them to me at the e-mail address provided on the title page. Similarly do so if you find any solutions incomplete, or if a more elegant solution is present. I hope using this manual serves to aid you in your study of Basic Mathe- matics, and I wish you the best of luck. 3 Contents Preface 3 1 Numbers 7 1.1 The integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.2 Rules for addition . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.3 Rules for multiplication . . . . . . . . . . . . . . . . . . . . . . . 10 1.4 Even and odd integers; divisibility . . . . . . . . . . . . . . . . . 12 1.5 Rational numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.6 Multiplicative inverses . . . . . . . . . . . . . . . . . . . . . . . . 16 2 Linear Equations 19 2.1 Equations in two unknowns . . . . . . . . . . . . . . . . . . . . . 19 2.2 Equations in three unknowns . . . . . . . . . . . . . . . . . . . . 19 3 Real Numbers 21 3.1 Addition and multiplication . . . . . . . . . . . . . . . . . . . . . 21 3.2 Real numbers: positivity . . . . . . . . . . . . . . . . . . . . . . . 22 3.3 Powers and roots . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3.4 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 4 Quadratic Equations 28 Interlude 29 On reading books . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Sets and elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 5 Distance and Angles 30 5.1 Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 5.2 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 5.3 The Pythagoras theorem . . . . . . . . . . . . . . . . . . . . . . . 32 6 Isometries 37 6.1 Some standard mappings of the plane . . . . . . . . . . . . . . . 37 6.2 Isometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 6.3 Composition of isometries . . . . . . . . . . . . . . . . . . . . . . 45 6.4 Inverse of isometries . . . . . . . . . . . . . . . . . . . . . . . . . 58 6.5 Characterization of isometries . . . . . . . . . . . . . . . . . . . . 64 6.6 Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 7 Area and Applications 70 7.1 Area of a disc of radius r . . . . . . . . . . . . . . . . . . . . . . 70 7.2 Circumference of a circle of radius r . . . . . . . . . . . . . . . . 72 4 8 Coordinates and Geometry 73 8.1 Coordinate systems . . . . . . . . . . . . . . . . . . . . . . . . . . 73 8.2 Distance between points . . . . . . . . . . . . . . . . . . . . . . . 76 8.3 Equation of a circle . . . . . . . . . . . . . . . . . . . . . . . . . . 77 8.4 Rational points on a circle . . . . . . . . . . . . . . . . . . . . . . 78 9 Operations on Points 79 9.1 Dilations and reflections . . . . . . . . . . . . . . . . . . . . . . . 79 9.2 Addition, subtraction, and the parallelogram law . . . . . . . . . 79 10 Segments, Rays, and Lines 95 10.1 Segments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 10.2 Rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 10.3 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 10.4 Ordinary equation for a line . . . . . . . . . . . . . . . . . . . . . 99 11 Trigonometry 100 11.1 Radian measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 11.2 Sine and cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 11.3 The graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 11.4 The tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 11.5 Addition formulas . . . . . . . . . . . . . . . . . . . . . . . . . . 109 11.6 Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 12 Some Analytic Geometry 116 12.1 The straight line again . . . . . . . . . . . . . . . . . . . . . . . . 116 12.2 The parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 12.3 The ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 12.4 The hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 12.5 Rotation of hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . 150 13 Functions 157 13.1 Definition of a function . . . . . . . . . . . . . . . . . . . . . . . 157 13.2 Polynomial functions . . . . . . . . . . . . . . . . . . . . . . . . . 158 13.3 Graphs of functions . . . . . . . . . . . . . . . . . . . . . . . . . . 160 13.4 Exponential function . . . . . . . . . . . . . . . . . . . . . . . . . 173 13.5 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 14 Mappings 179 14.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 14.2 Formalism of mappings . . . . . . . . . . . . . . . . . . . . . . . 189 14.3 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 15 Complex Numbers 196 15.1 The complex plane . . . . . . . . . . . . . . . . . . . . . . . . . . 196 15.2 Polar form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 5 16 Induction and Summations 198 16.1 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 16.2 Summations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 16.3 Geometric series . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 17 Determinants 204 17.1 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 17.2 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 17.3 Properties of 2 × 2 determinants . . . . . . . . . . . . . . . . . . 206 17.4 Determinants of order 3 . . . . . . . . . . . . . . . . . . . . . . . 207 17.5 Properties of 3 × 3 determinants . . . . . . . . . . . . . . . . . . 208 17.6 Cramer’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 6 1 Numbers 1.1 The integers No exercises. 1.2 Rules for addition 1. (a + b) + (c + d) = (a + b) + (d + c) = (commutativity) a + (b + d) + c = (associativity) a + (d + b) + c = (commutativity) (a + d) + (b + c) (associativity) 2. (a + b) + (c + d) = a + (b + c) + d = (associativity) a + (c + b) + c = (commutativity) (a + c) + (b + d) (associativity) 3. (a − b) + (c − d) = a + (−b + c) − d = (associativity) a + (c − b) − d = (commutativity) (a + c) + (−b − d) (associativity) 4. (a − b) + (c − d) = (a + c) + (−b − d) = (from exercise 3) (a + c) − (b + d) (N5) 5. (a − b) + (c − d) = (a − b) + (−d + c) = (commutativity) a + (−b − d) + c = (associativity) a + (−d − b) + c = (commutativity) (a − d) + (−b + c) = (associativity) (a − d) + (c − b) (commutativity) 7 6. (a − b) + (c − d) = (a + c) − (b + d) = (from exercise 4) −(b + d) + (a + c) (commutativity) 7. (a − b) + (c − d) = −(b + d) + (a + c) = (from exercise 5) −(b + d) + (−(−a) − (−c)) = (N4) −(b + d) − (−a − c) (N5) 8. ((x + y) + z) + w = (x + (y + z)) + w = (associativity) (x + (z + y)) + w = (commutativity) ((x + z) + y) + w = (associativity) (x + z) + (y + w) (associativity) 9. (x − y) − (z − w) = (x − y) + (−z − (−w)) = (N5) (x − y) + (−z + w) = (N4) (x − y) + (w − z) = (commutativity) x + (−y + w) − z = (associativity) x + (w − y) − z = (commutativity) (x + w) − y − z (associativity) 10. (x − y) − (z − w) = (x + w) − y − z = (from exercise 9) (x + w) + (−z − y) = (commutativity) x + (w − z) − y = (associativity) x + (−z + w) − y = (commutativity) (x − z) + (w − y) (associativity) 8 11. −(a + b + c) = −((a + b) + c) (associativity) = −(a + b) + (−c) (N5) = −a + (−b) + (−c) (N5) 12. −(a − b − c) = −a + (−(−b)) + (−(−c)) (from exercise 11) = −a + b + c) (N4) 13. −(a − b) = −a + (−(−b)) (N5) = −a + b (N4) =b−a (commutativity) 14. 17. −2 + x = 4 ⇐⇒ −x + 4 = −1 ⇐⇒ 2−2+x=2+4 ⇐⇒ x−x+4=x−1 ⇐⇒ 0+x=6 ⇐⇒ 0+4+1=x−1+1 ⇐⇒ x=6 5=x+0 ⇐⇒ x=5 15. 18. 2−x=5 ⇐⇒ 4−x=8 ⇐⇒ 2−x+x=5+x ⇐⇒ 4−x+x=8+x ⇐⇒ 2+0−5=5+x−5 ⇐⇒ 4+0−8=8+x−8 ⇐⇒ −3 = 0 + x ⇐⇒ −4 = x + 0 ⇐⇒ x = −3 x = −4 19. 16. −5 − x = −2 ⇐⇒ x−3=7 ⇐⇒ −5 − x + x = −2 + x ⇐⇒ x−3+3=7+3 ⇐⇒ −5 + 0 + 2 = −2 + x + 2 ⇐⇒ x + 0 = 10 ⇐⇒ −3 = x + 0 ⇐⇒ x=7 x = −3 9 20. 21. −7 + x = −10 ⇐⇒ −3 + x = 4 ⇐⇒ 7 − 7 + x = −10 + 7 ⇐⇒ 3−3+x=4+3 ⇐⇒ 0 + x = −3 ⇐⇒ 0+x=7 ⇐⇒ x = −3 x=7 22. a+b=a+c ⇐⇒ (−a) + a + b = (−a) + a + c ⇐⇒ 0+b=0+c ⇐⇒ b=c 23. By exercise 22 we have that if (a + b) = (a + 0) then b = 0. 1.3 Rules for multiplication 1. (a) 28 33 a7 b4 (d) 26 33 a8 b8 (b) 25 3a6 b10 (e) 24 35 a6 b8 (c) 210 33 a6 b10 (f) 29 34 a11 b19 2. (a+b)3 = (a+b)(a2 +2ab+b2 ) = a3 +2a2 b+ab2 +ba2 +2ab2 +b3 = a3 +3a2 b+3ab2 +b3 (a−b)3 = (a−b)(a2 −2ab+b2 ) = a3 −2a2 b+ab2 −a2 b+2ab2 +b3 = a3 −3a2 b+3ab2 −b3 3. (a + b)4 = (a + b)(a3 + 3a2 b + 3ab2 + b3 ) = a4 + 3a3 b + 3a2 b2 + ab3 + a3 b + 3a2 b2 + 3ab3 + b4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 (a − b)4 = (a − b)(a3 − 3a2 b + 3ab2 − b3 ) = a4 − 3a3 b + 3a2 b2 − ab3 − a3 b + 3a2 b2 − 3ab3 + b4 = a4 − 4a3 b + 6a2 b2 − 4ab3 + b4 4. 4 − 16x + 16x2 6. 4x2 + 20x + 25 5. 1 − 4x + 4x2 7. x2 − 2x + 1 10 8. x2 − 1 19. 4 − 9x2 9. 2x2 + 11x + 5 20. 2x3 + 3x2 − 9x − 10 10. x4 − 1 21. −6x3 − x2 + 5x + 2 11. 1 − x6 22. 6x3 + 25x2 + 3x − 4 12. x4 + 2x2 + 1 23. 2x3 − 3x2 − 3x + 2 13. x4 − 2x2 + 1 24. 4x3 + 3x2 − 25x + 6 14. x4 + 4x2 + 4 25. x3 − 2x2 − x + 2 15. x4 − 4x2 + 4 26. −x3 + 5x2 − 7x + 3 16. x6 − 8x3 + 16 27. −x3 + 4x2 − 5x + 2 17. x6 − 16 28. 16x3 − 4x2 − 8x + 3 18. 4x4 − 1 29. −12x3 − 4x2 + 5x + 2 30. (a) The population doubles six times, so in total it is 26 · 50, 000 = 3, 200, 000 (b) The population doubles eight times, so in total it is 28 · 50, 000 = 12, 800, 000 (c) The population doubles nine times, so in total it is 29 · 50, 000 = 25, 600, 000 31. (a) The population doubles two times, so in total it is 22 · 100, 000 = 400, 000 (b) The population doubles four times, so in total it is 24 · 100, 000 = 1, 600, 000 (c) The population doubles six times, so in total it is 26 · 100, 000 = 6, 400, 000 32. (a) The population triples six times, so in total it is 36 · 200, 000 = 145, 800, 000 (b) The population triples five times, so in total it is 35 · 200, 000 = 48, 600, 000 33. (a) The population triples three times, so in total it is 33 · 25, 000 = 675, 000 (b) The population triples four times, so in total it is 34 · 25, 000 = 2, 025, 000 11 1.4 Even and odd integers; divisibility 1. For the first one, suppose a is even and b is even. Then a = 2m and b = 2n for integers m and n. Hence a + b = 2m + 2n = 2(m + n), whence a + b is even. The third one follows from the second with the roles of a and b reversed. For the fourth one, suppose a is odd and b is odd. Then a = 2m + 1 and b = 2n+1 for integers m and n. Hence a+b = 2m+1+2n+1 = 2(m+n+1), whence a + b is even.  2. Suppose a is even and b is any positive integer. Then a = 2m for some integer n. Hence ab = 2mb, whence ab is even.  3. Suppose a is even. Then a = 2m for some integer m. Hence a3 = (2m)3 = 2(4m3 ), whence a3 is even.  4. Suppose a is odd. Then a = 2m + 1 for some integer m. Hence a3 = (2m + 1)3 = 8m3 + 12m2 + 6m + 1 = 2(4m3 + 6m2 + 3m) + 1, whence a3 is odd.  5. Suppose n is even. Then n = 2k for some integer k. Hence (−1)n = (−1)2k = ((−1)2 )k = 1k = 1.  6. Suppose n is odd. Then n = 2k + 1 for some integer k. Hence (−1)n = (−1)2k+1 = (−1)2k · (−1) = 1 · (−1) = −1.  7. Suppose m and n are both odd. Then m = 2a+1 and n = 2b+1 for integers a and b. Hence ab = (2a+1)(2b+1) = 4ab+2a+2b+1 = 2(2ab+a+b)+1, whence ab is odd.  8. 4 16. 1 9. 3 17. 3 10. 5 18. 2 11. 2 19. 2 12. 1 20. 1 13. 6 21. 0 14. 2 22. 1 15. 2 23. 2 24. Suppose a ≡ b (mod 5) and x ≡ y (mod 5). Then a − b = 5m for some integer m and x − y = 5n for some integer n. Hence (a − b) + (x − y) = (a + x) − (b + y) = 5m + 5n = 5(m + n), and so a + x ≡ b + y (mod 5). Furthermore, since a = 5m + b, we have ax = 5mx + bx. Similarly since 12 y = 5n+x we have by = 5bn+bx. All in all, ax−by = 5mx+bx−5bn−bx = 5(mx − bn), so that ax ≡ by (mod 5).  25. Suppose a ≡ b (mod d) and x ≡ y (mod d). Then a − b = dm for some integer m and x − y = dn for some integer n. Hence (a − b) + (x − y) = (a + x) − (b + y) = dm + dn = d(m + n), and so a + x ≡ b + y (mod d). Furthermore, since a = dm + b, we have ax = dmx + bx. Similarly since y = dn+x we have by = dbn+bx. All in all, ax−by = dmx+bx−dbn−bx = d(mx − bn), so that ax ≡ by (mod d).  26. Suppose n is an integer not divisible by 3. Then n = 3k + 1 or n = 3k + 2. In the first case n2 = (3k + 1)2 = 9k 2 + 6k + 1 = 3(3k 2 + 2k) + 1, which is not divisible by 3. In the other case n2 = (3k + 2)2 = 9k 2 + 12k + 4 = 3(3k 2 + 4k + 1) + 1, which also is not divisible by 3. This shows that if the square of an integer is divisible by 3, then that integer must be divisible by 3 also, because if it were not, then its square would not be divisible by 3.  1.5 Rational numbers 1. (a) a = 83 3 (c) a = − 20 (b) a = − 35 3 5 2. (a) x = 3 (c) x = − 27 5 (b) x = 2 2 3. (a) 5 (e) 5 1 31 (b) 3 (f) 2 6 1 (c) 5 (g) 2 10 2 (d) 3 (h) 5 n mn 4. Let b = m. Then ab = ba = mn = 1.  1 5. (a) x = 14 (f) x = 3 11 (b) x = 3 3 (g) x = 5 9 (c) x = 32 8 (d) x = 1 (h) x = 39 3 5 31 (e) x = 4 (i) x = 14 50 69 6. (a) x = 63 (d) x = 8 (b) x = − 20 31 (e) x = 11 26 13 16 (c) x = 41 (f) x = 29 13 7. (a) 5! = 120, 6! = 720, 7! = 5040, 8! = 40, 320. (b) 30 = 1, 31 = 3, 32 = 3, 33 = 1, 40 = 1, 41 = 4, 42 = 6, 43 = 4,







4 5 5 5 5 5 5






4 = 1, 0 = 1, 1 = 5, 2 = 10, 3 = 10, 4 = 1, 0 = 1. (c)     m m! m! m = = = n n!(m − n!) (m − (m − n))!(m − n)! m−n  (d)     m m m! m! + = + n n−1 n!(m − n)! (n − 1)!(m − n + 1)! m!(m − n + 1) m!n = + n!(m − n + 1)! n!(m − n + 1)! m!(m − n + 1 + n) = n!(m − n + 1)! (m + 1)! = n!((m + 1) − n)!   m+1 = n  8. Suppose there is a rational number a = pq where p and q are integers, and q 6= 0, such that a3 = 2. Suppose furthermore that a is written in lowest 3 form. Then pq3 = 2 and so p3 = 2q 3 . Thus p3 is even, and so p is even also. Hence p = 2k for some integer k. This gives us (2k)3 = 2q 3 , i.e. 2(2k 3 ) = q 3 and so q 3 is even, and so q is even also. But this contradicts our assumption that a was written in lowest form.  9. Suppose there was a rational number a such that a4 = 2. Then a2 would be a rational number which would equal the square root of 2, but we know such a number doesn’t exist.  10. Suppose there is a rational number a = pq where p and q are integers, and q 6= 0, such that a2 = 3. Suppose furthermore that a is written in lowest form. As such p2 = 3q 2 . We know (from an earlier exercise) that if p2 is divisible by 3, then so too must p be divisible by 3. Thus p = 3k for some integer k. Hence (3k)2 = 3q 2 and so q 2 = 3k 2 . Thus q 2 is divisible by 3, and so too is q. But then both p and q are divisible by 3, which contradicts the fact that a was in lowest form.  11. 14 (a) 1.414 (b) 1.4142 12. (a) 1.73 (b) 1.732 13. (a) 2.23 (b) 2.236 14. (a) 1.25 (b) 1.259 15. (a) 1.44 (b) 1.442 1 16. (a) It has halved once, so there are 6 · 2 = 3 grams remaining. (b) It has halved nine times, so there are 6 · ( 12 )9 = 6 512 = 3 256 grams remaining. (c) It has halved twelve times, so there are 6 · ( 21 )12 = 6 4096 = 3 2048 grams remaining. 17. If one third decomposes, then two thirds remain. (a) It has been reduced twice, so there are 15·( 32 )2 = 20 3 grams remaining. (b) It has been reduced three times, so there are 15 · ( 23 )3 = 40 9 grams remaining. (c) It has been reduced eleven times, so there are 15 · ( 23 )11 = 10240 59049 grams remaining. 18. If one fourth dissolves away, then three fourths remain undissolved. 3 75 (a) It has been reduced once, so there are 25 · 4 = 4 grams remaining. (b) It has been reduced three times, so there are 25 · ( 34 )3 = 675 64 grams remaining. (c) It has been reduced five times, so there are 25 · ( 34 )5 = 6075 1024 grams remaining. 9 19. (a) The bacteria population has been reduced once, so there are 106 · 10 = 5 9 · 10 bacteria remaining. (b) The bacteria population has been reduced three times, so there are 9 3 106 · ( 10 ) = 93 · 103 bacteria remaining. (c) The bacteria population has been reduced five times, so there are 9 5 106 · ( 10 ) = 95 · 10 bacteria remaining. (d) Between the 60 and 70 minute mark. (e) Between the 110 and 120 minute mark. (f) Between the 150 and 160 minute mark. 2 20. (a) The fish population has been reduced once, so there are 50, 000 · 3 ≈ 33, 300 fish remaining. 15 (b) The fish population has been reduced twice, so there are 50, 000·( 32 )2 ≈ 22, 200 fish remaining. (c) The fish population has been reduced four times, so there are 50, 000 · ( 23 )4 ≈ 9, 900 fish remaining. (d) The fish population has been reduced six times, so there are 50, 000 · ( 23 )6 ≈ 4, 400 fish remaining. (e) The second month, as can be seen in (a) and (b). (f) Between the third and fourth month, that is, during the third month. 21. (a) ( 54 )3 < 2 while ( 54 )4 > 2 so the population doubles roughly every 40 years. (b) ( 45 )4 < 3 while ( 45 )5 > 3 so the population triples roughly every 50 years. 1.6 Multiplicative inverses 1. (a) x = − 15 19 (f) x = − 22 27 5 (b) x = 10 (g) x = 6 25 31 (c) x = 2 31 (h) x = (d) x = − 37 11 5 (e) x = − 137 (i) x = − 65 19 1 1 (x−y)−(x+y) −2y 2. (a) x+y + x−y = (x+y)(x−y) = x2 −y 2 . x3 −1 (x−1)(1+x+x2 ) (b) x−1 = x−1 = 1 + x + x2 . x4 −1 (x−1)(1+x+x2 +x3 ) (c) x−1 = x−1 = 1 + x + x2 + x3 . xn −1 (x−1)(1+x+x2 +···+xn−1 ) (d) x−1 = x−1 = 1 + x + x2 + · · · + xn−1 . 1 1 (2x−y)+(2x+y) 4x 3. (a) 2x+y + 2x−y = (2x+y)(2x−y) = 4x2 −y 2 . 2x 3x+1 (2x)(2x+1)−(x+5)(3x+1) x2 −14x−5 (b) x+5 − 2x+1 = (x+5)(2x+1) = 2x2 +11x+5 . 1 1 (x−3y)+(x+3y) 2x (c) x+3y + x−3y = (x+3y)(x−3y) = x2 −9y 2 . 1 x (x+y)+x(3x−2y) x+y+3x2 −2xy x+y+3x2 −2xy (d) 3x−2y + x+y = (3x−2y)(x+y) = 3x2 +3xy−2xy−2y 2 = 3x2 +xy−2y 2 . x3 −y 3 (x−y)(x2 +xy+y 2 ) 4. (a) x−y = x−y = x2 + xy + y 2 . x4 −y 4 (x−y)(x3 +x2 y+xy 2 +y 3 ) (b) x−y = x−y = x3 + x2 y + xy 2 + y 3 . 2 1−2t2 −t4 +4t2 1+2t2 +t4 (1+t2 )2 (c) x2 + y 2 = ( 1−t 2 2t 2 1+t2 ) + ( 1+t2 ) = (1+t2 )2 = (1+t2 )2 = (1+t2 )2 = 1. x3 +1 (x+1)(x2 −x+1) 5. (a) x+1 = x+1 = x2 − x + 1. 16 x5 +1 (x+1)(x4 −x3 +x2 −x+1) (b) x+1 = x+1 = x4 − x3 + x2 − x + 1. xn +1 (x+1)(xn−1 −xn−2 +···−x+1) (c) x+1 = x+1 = xn−1 − xn−2 + · · · − x + 1. d 6. The time taken t is equal to s where d is distance and s is speed. Thus it took t = 5/4 25 2/5 = 8 seconds. 3/10 7. (a) d = m v = 9 2/3 = 20 lb/in . 3 (c) v = m d = 15 2/3 = 45 2 in3 . m 6 9 3 (b) d = v = 4/3 = 2 lb/in . 340 8. (a) C = 0 (d) C = 9 (b) C = 10 (c) C = 335 9 (e) C = −40 9. From exercise 8 we can rearrange the formula to get F = 59 C + 32. Then we can proceed to plug in C: 493 (a) F = 32 (d) F = 5 (b) F = 14 (e) F = 104 (c) F = −40 (f) F = 212 E 10 E 220 10. (a) R = I = 3 . (b) R = I = 10 = 22. 11. (a) Let x be the amount of water added. We want to find x such that 65 80 12 · 100 + x = (12 + x) 100 . Solving for x yields x = 9. (b) Similarly, x = 30. (c) Similarly, x = 72. 12. Let x be the number of hours the wind was favorable. Then the time in unfavorable wind is 4 − x. The distance travelled was 3,000 mi, and so 900x + (4 − x)500 = 3000. Solving for x gives x = 25 hours. 13. Let x be the number of $5.00 tickets sold. Then the number of $2.00 tickets sold is 1300 − x. The total amount collected was $4,100, and so 5x + (1300 − x)2 = 4100. Solving for x yields x = 500. Thus there were 500 $5.00 tickets sold, and 800 $2.00 tickets sold. 14. (a) We have 8 g of salt in the solution. We add x pure water. Thus the 8 8 solution consists of (80+x) parts salt. For 4% salt we want (80+x) = 4 100 . Solving for x yields x = 120 g. 160 (b) Similarly, x = 3 g. (c) Similarly, x = 20 g. 17 15. (a) Let x be the number of quarts of water added. Then the mixture is 6 6 25 6+x parts alcohol. For a 25% mixture we require 6+x = 100 . Solving for x yields x = 18 qt. (b) Similarly, x = 24 qt. (c) Similarly, x = 34 qt. 16. Let x be the number of hours the boat was along the first river. Then it was along the other river for 50 − x hours. It travelled a total distance of 500 mi, thus 20x + (50 − x)8 = 500. Solving for x yields x = 253 = 83 1 hours. 17. (a) Let x be the amount of water that evaporates. Thus the solution will 25 end up weighing 2 − x lb. We initially have 100 · 2 lbs of salt. Thus 25 100 ·2 the mixture will contain 2−x parts salt. For the mixture to contain 25 100 ·2 40 3 40% salt we require 2−x = 100 . Solving for x yields x = 4 lb. 3 (b) Similarly, x = 5 lb. 18. (a) Let x be the amount of water added. Then the total amount of liquid 60 in the tank is 5 + x. The tank initially has 100 · 5 gal of antifreeze. 60 100 ·5 Thus the tank will contain 5+x parts antifreeze. For a mixture of 60 40 100 ·5 40% antifreeze we require 5+x = 100 . Solving for x yields x = 52 gal. The total amount of liquid is then 5 + 25 which is less than 10, so it fits in the tank. (b) Similarly, x = 25 gal, which of course will not fit in the tank. 18 2 Linear Equations 2.1 Equations in two unknowns 5 1. x = 3 5. x = − 21 1 y= 3 y = − 23 2. x = 30 6. x = 0 y = 17 y=0 1 3. x = −2 7. x = 2 3 y=1 y= 2 47 4. x = −3 8. x = 10 y = −5 y = − 17 5 d−2b −2d−3b 9. (a) x = ad−bc (c) x = ad−bc 2a−c 3a+2c y= ad−bc . y= ad−bc . 3d+4b 5d−7b (b) x = ad−bc (d) x = ad−bc −3c−4a 7a−5c y= ad−bc . y= ad−bc . 10. It is clear that x = 0 and y = 0 satisfy the system. By multiplying the first equation by d and the second equation by b and subtracting the second from the first, we get (ad − bc)x = 0. Since ad − bc 6= 0 then x = 0. We plug this back into the two equations which yields by = 0 and dy = 0. But note that b and d can not both be 0, since then ad − bc = 0. WLOG suppose b 6= 0. Then, analogously to above, y = 0.  ud−bv 11. x = ad−bc va−uc y= ad−bc . 2.2 Equations in three unknowns 1. x = 1 z=0 y = 12 z = − 12 5. x = 21 9 7 y = − 16 2. x = 16 13 z = 16 1 y= 16 3 z= 16 15 6. x = 13 6 3. x = 51 y = − 13 67 6 y= 29 z = 13 67 25 z= 67 7. x = 17 4 4. x = 0 y = − 43 y=0 z = − 72 19 8. x = 12 z= 7 3 y=0 z = − 12 11. x = 16 15 38 y = − 75 9. x = 205 33 43 z = − 25 y = − 130 43 z = − 94 43 32 12. x = 131 8 315 10. x = 3 y = − 131 8 95 y= 3 z = − 131 20 3 Real Numbers 3.1 Addition and multiplication 1. (a) Associativity: ((E + E) + E) = E = (E + (E + E)) ((E + E) + I) = I = (E + (E + I)) ((E + I) + E) = I = (E + (I + E)) ((E + I) + I) = E = (E + (I + I)) ((I + E) + E) = I = (I + (E + E)) ((I + E) + I) = E = (I + (E + I)) ((I + I) + E) = E = (I + (I + E)) ((I + I) + I) = I = (I + (I + I)) Commutativity: E+E =E =E+E I +I =E =I +I I +E =I =E+I That E acts as zero for addition can be seen above. E added to either E or I results in E or I, respectively. The additive inverse of E is E, since E + E = E. The additive inverse of I is I, since I + I = E. (b) Associativity: (E(EE)) = E = ((EE)E) (E(EI)) = E = ((EE)I) (E(IE)) = E = ((EI)E) (E(II)) = E = ((EI)I) (I(EE)) = E = ((IE)E) (I(EI)) = E = ((IE)I) (I(IE)) = E = ((II)E) (I(II)) = I = ((II)I) Commutativity: EE = E = EE II = I = II EI = E = IE That I acts as 1 may be a bit unclear, but note that it does in fact satisfy the fact that I · a = a. Also, E acts as 0 since Ea = E for any a. 21 Finally, for distributivity (commutativity means we only have to show distributivity from one side): E(E + E) = E = EE + EE E(E + I) = E = EE + EI E(I + E) = E = EI + EE E(I + I) = E = EI + EI I(E + E) = E = IE + IE I(E + I) = I = IE + II I(I + E) = I = II + IE I(I + I) = E = II + II 3.2 Real numbers: positivity 1. (a) There is a slight error in the question. We need in fact to assume also a 6= 0. Suppose a 6= 0 is a real number. If a is positive then by POS 1 a · a = a2 is positive. If a is not positive then −a is positive. But of course (−a)(−a) = a2 , and by POS 1 again, a2 is positive.  (b) Suppose a is positive and b is negative. Then −b is positive. Thus a(−b) = −ab is positive. Hence ab is negative.  (c) Suppose a and b are both negative. Then −a and −b are positive, and hence (−a)(−b) = ab is positive.  2. Suppose a is positive, but a−1 was negative. Then aa−1 = 1 would be negative, which is impossible.  3. Suppose a is negative, but a−1 is positive. Then aa−1 = 1 would be negative, which is impossible.  4. √  √ 2 r s a a a √ = √ = b b b  5. √ 1 1+ 2 √ √ = √ √ = −(1 + 2) 1− 2 (1 − 2)(1 + 2)  1√ 6. Since the multiplicative inverse is 2+ 3 , we have √ 1 2− 3 √ √ = √ √ =2− 3 2+ 3 (2 − 3)(2 + 3) so c = 2 and d = −1. 22 1√ 7. Since the multiplicative inverse is 3+ 5 , we have √ 1 3− 5 3 1√ √ = √ √ = − 5 3+ 5 (3 − 5)(3 + 5) 4 4 3 so c = 4 and d = − 14 . 1√ 8. Since the multiplicative inverse is a+b 2 , we have √ 1 a−b 2 a b √ √ = √ √ = 2 2 − 2 2 2 a+b 2 (a − b 2)(a + b 2) a − 2b a − 2b a b so c = a2 −2b2 and d = − a2 −2b 2. 1√ 9. Since the multiplicative inverse is a+b 3 , we have √ 1 a−b 3 a b √ √ = √ √ = 2 2 − 2 2 3 a+b 3 (a − b 3)(a + b 3) a − 3b a − 3b a b so c = a2 −3b2 and d = − a2 −3b 2. 10. √ √ √ √ √ (x+y 5)(z+w 5) = xz+xw 5+yz 5+5yw = (xz+5yw)+(xw+yz) 5 Where c = xz + 5yw and d = xw +yz. These are both rational since x,y,z, w are all rational. 11. √ √ √ √ √ (x+y a)(z+w a) = xz+xw a+yz a+ayw = (xz+ayw)+(xw+yz) a Where c = xz + ayw and d = xw + yz. These are both rational since x, y, z, w, a are all rational. 12. (a) √ x+1 (d) √ 2x−h √ 2( 2x+3−1) h( x−h− x) (b) √x−8 2x+h √ (e) − h(√x+h− 2( 1+x+3) x) 1 √ 2 √ (c) − √x−h+ x (f) √ x+2h+ x 13. (a) x = −1 and x = 3 (d) x = −7 and x = 5 (b) x = −5 and x = 5 (e) x = −7 and x = −1 (c) x = −1 and x = 7 (f) x = 1 and x = 3 14. 23 2 (a) x = −1 and x = 2 (d) x = 0 and x = 3 (b) x = −1 and x = 13 (c) x = − 52 and x = − 32 (e) x = − 14 and x = 11 4 15. (a) √ x−y √ (d) √ 3 √ ( x− y)2 ( x−3− x)2 (b) √ x √ (e) √ x+y−1 √ ( x+y+ y)2 (3+ x+y)( x+y+1) 2 (c) √ 2√ (f) √ x+y−x√ ( x+1− x−1)2 ( x+y−x) x+y √ √ √ √ ( x+ y)2 ( x−3+ x)2 16. (a) x−y (d) 3 √ √ √ √ ( x+y− y)2 ( x+y−1)(3− x+y) (b) (e) 9−x−y x √ √ √ √ x+1+ x−1 ( x+y+x) x+y (c) 2 (f) x+y 17. √ √ x−1=3+ x only if √ x−1=9+6 x+x only if √ −10 = 6 x only if √ 5 x=− 3 which is impossible. 18. √ √ x−1=3− x only if √ x−1=9−6 x+x only if √ −10 = −6 x only if √ 5 x= only if 3 25 x= 9 19. The answers are all acquired as in exercise 17 and 18. 13 √4 (a) No solution exists. those curious: x = 3 − 3 (b) There is an error in the book. (c) x = 1 Lang writes that this has no so- lution, however, it does. Unfor- (d) No solution exists. tunately it requires being able (e) No solution exists. to solve quadratic equations. We’ll state the solution here for (f) x = 169 36 24 20. |a − b| = | − (b − a)| = |b − a|  3.3 Powers and roots 1. (a) 22 30 a1 b−4 (c) 2−1 31 a−2 b−2 (b) 22 3−6 a−1 b5 (d) 27 3−2 a−10 b−9 2. 3 3. 8 √ √ 4. ( 2)5 = 4 5. This is not an integer, however, this needs to be proven. Since all integers are also rational, we postpone this proof to after the next exercise. √ √ √ 5. ( 2)−5 = 4√ 1 2 , and ( 2)5 = 4 2 as before. We need to prove two things. First, that if x is irrational, then so is ax for any rational number a. Second, that if x is irrational, then x−1 is irrational. So firstly suppose a is rational and x is irrational, but ax is rational. Then ax = pq for integers p and q. Furthermore a = rs for integers r and s. Thus x = ps qr , but that would mean x is rational, which we know it’s not. Secondly, suppose x is irrational but x−1 is rational. Then x−1 = x1 = pq for integers p and q. Then x = pq , but that would mean that x is rational, which we know it’s not. From this we see that the answer is no to all questions in exercises 4 and 5. 6. (a) 2 (e) 16 (b) 2 (f) 16 (c) 27 (d) 1 (g) 125 7. (a) 0.3 (c) 0.25 (b) 0.3 (d) 1.1 4 125 8. (a) 9 (c) 64 2 343 (b) 3 (d) 8 √ 3 9. (a) x = 2 + 5 (d) x = 0 2 (b) x = −5 or x = −1 (e) x = 3 √ √ 14 16 −5 √8+1 −5 √8−1 (c) x = 3 or x = 3 (f) x = 3 8 or x = 3 8 25 3.4 Inequalities 1. Suppose a > b and c < 0. Then a − b > 0 and −c > 0. Hence −c(a − b) = bc − ac > 0 so that ac < bc.  2. Suppose 0 < a < b and 0 < c < d. Since a < b and c > 0 we have ac < bc. Similarly since c < d and b > 0 we have bc < bd. Thus ac < bc < bd.  3. Suppose a < b < 0 and c < d < 0. Since a < b and c < 0 we have ac > bc. Similarly since c < d and b < 0 we have bc > bd. Thus ac > bc > bd.  4. (a) Suppose 0 < x < y. Then x−1 and y −1 are both positive. Thus since x < y we get my multiplying both sides by x−1 y −1 that y1 < x1 .  a (b) By multiplying both sides by bd we see that b < dc implies ad < bc. a If ad < bc we divide both sides by bd whence b < dc .  5. Suppose a < b and c is any real number. Then b − a = (b + c) − (a + c) > 0. So a + c < b + c. To show a < b implies a − d < b − d just let d = −c.  6. If a < b then immediately by exercise 2 a2 < b2 . Now assuming a2 < b2 , and a < b we can apply exercise 2 again, yielding a3 < b3 . In fact, we can apply exercise 2 repeatedly, so that in general an < bn . 7. (This is a proof by contraposition. In the future no note will be given when this technique is used.) Let a and b both be positive. Suppose a1/n ≥ b1/n . If they are equal then clearly a = b. Otherwise by exercise 6 we have a > b.  8. (a) We start by showing the left inequality. ad < bc ⇐⇒ ab + ad < bc + ab ⇐⇒ a(b + d) < b(a + c) ⇐⇒ a a+c < b b+d For the right inequality ad < bc ⇐⇒ cd + ad < bc + cd ⇐⇒ d(a + c) < c(b + d) ⇐⇒ a+c c < b+d d 26 (b) We start by showing the left inequality. ad < bc ⇐⇒ rad < rbc ⇐⇒ ab + rad < rbc + ab ⇐⇒ a(b + rd) < b(a + rc) ⇐⇒ a a + rc < b b + rd For the right inequality ad < bc ⇐⇒ cdr + ad < bc + cdr ⇐⇒ d(a + rc) < c(b + rd) ⇐⇒ a + rc c < b + rd d (c) Suppose r < s. Then r(bc − ad) < s(bc − ad). Expanding into rbc − rad < sbc − sad we add like terms on each side as such ab + rbc + sad + rscd < ab + sbc + rad + rscd. Factoring yields (b + sd)(a + rc) < (a + sc)(b + rd), whence a+rc a+sc b+rd < b+sd .  9. Suppose 3x − 1 > 0. Then 3x > 1, i.e. x > 31 .  10. Suppose 4x + 5 < 0. Then 4x < −5, i.e. x < − 54 .  23 11. x > −1 22. 5 <x<5 12. x > − 32 23. 1 3 <x< 1 2 13. x < − 13 24. x < 3 or x > 19 3 14. x < − 73 25. −1 < x < 1 √ √ 15. x > −6 26. − 2 < x < 2 √ √ 16. x > − 43 27. − 3 < x < 3 17. x < − 18 28. −2 < x < 2 18. x < − 35 29. x < −1 or x > 1 √ √ 19. − 21 < x < 2 30. x < − 2 or x > 2 2 √ √ 20. x < −3 or x > 3 31. x < − 3 or x > 3 21. − 21 < x < 0 32. x < −2 or x > 2 27 4 Quadratic Equations √ 1. x = −3± 17 2 9. x = − 27 and x = 1 √ 3± 17 3 2. x = 2 10. x = 4 and x = −2 √ √ √ 4± −4 2± −2 3. x = 2 , i.e. no solution in 11. x = 2 , i.e. no solutions in the real numbers. the real numbers. √ 4. x = −1 and x = 5 12. x = −1± √ 3 2 1 5. x = −1 and x = 3 √ √ 13. x = −2 − 2 and x = −1 + 2 1 6. x = 3 and x = 1 √ √ 5+ 5 1− 5 √ 19 14. x = 2 and 2 1 −2± 3 7. x = 2 √ √ 5− 5 1+ 5 √ 15. x = and 2 2 5± −31 8. x = −4 , i.e. no solutions in p √ the real numbers. 16. x = 1 ± 1 + 3 28 Interlude On Logic and Mathematical Expres- sions On reading books No exercises. Logic No exercises. Sets and elements No exercises. Notation No exercises. 29 5 Distance and Angles 5.1 Distance No exercises. 5.2 Angles 1. (a) 10 in2 (b) 20 in2 (c) 45 2 in2 (d) 80 3 in2 30 (e) 35 in2 (f) 75 2 in2 (g) 35 in2 (h) 45 in2 31 θ2 −θ1 2 2. (a) 360 πr (b) πr22 − πr12 = π(r22 − r12 ) θ2 −θ1 2 θ2 −θ1 2 θ2 −θ1 2 (c) 360 πr2 − 360 πr1 = 360 π(r2 − r12 ) 4π π 3. (a) 9 (c) 3 π 4π (b) 2 (d) 3 4. (a) 21π (c) 32π (b) 7π (d) 24π 16π 4π 5. (a) 9 (c) 3 16π (b) 2π (d) 3 7π 77π 6. (a) 36 (c) 72 28π 49π (b) 9 (d) 36 5.3 The Pythagoras theorem √ √ 1. (a) 2 2 (d) 5 2 √ (b) 3 2 √ √ (c) 4 2 (e) r 2 √ √ 2. (a) 5 (d) 5 5r √ √ (b) 3 34 (e) r 34r √ √ (c) 4 65 (f) r 65r √ √ 3. (a) 3 (d) 4 3 √ (b) 2 3 √ √ (c) 3 3 (e) r 3 4. 32 √ √ (a) 5 2 (d) 26 √ (b) 21 √ √ (c) 38 (e) 35 √ √ 5. a2 + b2 + c2 and r a2 + b2 + c2 , respectively. √ 6. 100 26 ft. √ √ 7. (a) 51 ft. (c) 133 ft. √ (b) 2 42 ft. √ √ 8. (a) 20 10 ft. (c) 10 105 ft. (b) 40 ft. 9. Suppose d(P, O) = d(Q, O). Then d(P, M )2 = d(P, O)2 + d(O, M )2 , and d(P, Q)2 = d(Q, O)2 +d(O, M )2 = d(P, O)2 +d(O, M )2 , whence d(P, M ) = d(Q, M ).  10. In the first case we have where from tright triangles we know m(A) + m(D) = 90 and m(B) + m(C) = 90 Thus, m(A) + m(D) + m(C) + m(B) = 180 In the second case we have 33 where from right triangles we know m(D) + m(E) = 90 and m(D) + m(A) + m(B) = 90 Furthermore, m(E) + m(C) = 180. Thus, m(B) + m(C) + m(A) = m(B) + m(C) + m(A) + m(D) − m(D) = 90 + m(C) − m(D) = 90 + m(C) − m(D) + m(E) − m(E) = 90 + 180 − 90 = 180 11. In the first case we have where the total area is the sum of the two right triangles, i.e. area = ah bh (a+b)h 2 + 2 = 2 . In the second case we have where the total area is the difference of the smaller right triangle from the larger right triangle, i.e. area = (a+b)h 2 − ah bh 2 = 2 .  12. (a) Consider a right triangle with legs of √ √ b, with a ≥ b, distance a and and length of hypotenuse c. Then c = a2 + b2 ≥ 2a > a > b.  34 (b) By Pythagoras, d(P, M )2 = d(P, Q)2 + d(Q, M )2 , which is smallest when d(Q, M ) = 0, i.e. when Q and M are the same point. 13. Consider (a) We know m(C) + m(A) = 90 and m(C) + m(B) = 90. So m(A) = m(B). (b) See Theorem 1 § 3. (c) By part (a) m(A) = m(B) and by part (b) m(B) = m(B 0 ), so that m(A) = m(B 0 ).  14. Consider 35 Since A bisects P M we have d(P, O) = d(M, O), by the corollary to Pythagoras’ theorem. Similarly d(P, O) = d(Q, O), i.e. all perpendicular bisectors meet in a point.  36 6 Isometries 6.1 Some standard mappings of the plane 1. (a) All points of the plane are fixed. (b) Only the point O is fixed. (c) Only points on the line are fixed. (d) Only the point O is fixed. (e) No point is fixed. (f) Only the point O is fixed, if r 6= 1. If r = 1 all points are fixed. 2. (a) (−1)360 + 330 (g) (0)360 + 120 (b) (−1)360 + 270 (h) (1)360 + 180 (c) (−1)360 + 180 (i) (−2)360 + 320 (d) (−1)360 + 90 (j) (1)360 + 240 (e) (−1)360 + 315 (k) (2)360 + 0 (f) (−1)360 + 135 (l) (1)360 + 90 3. (a) . (b) . (c) . 37 (d) . (e) . (f) . (g) . (h) . 38 (i) . (j) . (k) . (l) . 39 6.2 Isometries 1. (a) . (b) . (c) With respect to the origin. 40