Chapter 1

Exercise 1.1

Q1

(Fernando Li)

  1. \[ A \cap B = \lbrace 2,3,5,7 \rbrace \]
  2. \[ A \cap C = \lbrace 2,4,6,8,10 \rbrace \]
  3. \[ B \cap C = \lbrace 2 \rbrace \]
  4. \[ A \cup B = \lbrace 1,2,3,4,5,6,7,8,9,10,11,13,17,19 \rbrace \]
  5. \[ A \cup C = \lbrace 1,2,3,4,5,6,7,8,9,10,12,14,16,18 \rbrace \]
  6. \[ B \cup C = \lbrace 2,3,4,5,6,7,8,10,11,12,13,14,16,17,18,19 \rbrace \]
  7. \[ A \cup B \cup C = \lbrace 1,2,3,4,5,6,7,8,9,10,11,12,13,14,16,17,18,19 \rbrace \]
  8. \[ A \cap B \cap C = \lbrace 2 \rbrace \]
  9. \[ (A \cup B) \cap C = \lbrace 2,4,6,8,10 \rbrace \]
  10. \[ (A \cap B) \cup C = \lbrace 2,3,4,5,6,7,8,10,12,14,16,18 \rbrace \]
  11. \[ A \cap B' = \lbrace 1,4,6,8,9,10 \rbrace \]
  12. \[ 'A \cap B' = \lbrace 12,14,15,16,18 \rbrace \]

Q2

(David Rosa)

(a)false, a counterexample is as follows \[ A = \lbrace 1,2,3,4 \rbrace \\ B = \lbrace 3,4,5,6 \rbrace \\ U = \lbrace 1,2,3,4,5,6,7,8 \rbrace \\\] Then \[ A - B = \lbrace 1,2 \rbrace \\ A' = \lbrace 5,6,7,8 \rbrace \\ A' \cap B = \lbrace 5,6 \rbrace \\\] Thus \[ A - B = A' \cap B \] is false.

(b)false, a counterexample is as follows \[ A = \lbrace 1,2 \rbrace \\ B = \lbrace 2,3 \rbrace \\ C = \lbrace 3,4 \rbrace \\ U = \lbrace 1,2,3,4,5 \rbrace \\\] Then \[ A \cup B = \lbrace 1,2,3 \rbrace \\ (A \cup B) \cap C = \lbrace 3 \rbrace \\ B \cap C = \lbrace 3 \rbrace \\ A \cup ( B \cap C) = \lbrace 1,2,3 \rbrace \\ \] Thus \[ (A \cup B) \cap C = A \cup ( B \cap C) \] is false.

(c)true As \[ B' \cap B = \emptyset \] \[ (A' \cup B') \cap B = (A' \cap B) \cup (B' \cap B) \\ = (A' \cap B) \cup \emptyset \\ = A' \cap B \\ = \lbrace x \in U : x \in B \ and \ x \in A' \rbrace\\ = \lbrace x \in U : x \in B \ and \ x \notin A \rbrace\\ = B -A \]

Exercise 1.2

Q1

  1. \[ \lbrace x:\pm \sqrt2 \rbrace \]
  2. \[ \lbrace x: \sqrt2 \rbrace \] (c)Q means rational numbers, so \[\lbrace x \in Q: x^2 = 2 \rbrace = \emptyset \]

Q2

(Jiin Kim)

(a)\[ A \cap B = [3,5] \] (b)\[ A \cup B = [1,9) \] (c)\[ A - C = (1,5) \] (d)\[ B \cap C = \lbrace 5 \rbrace \] (e)\[ C - B = \lbrace 1 \rbrace \] (f)\[ B - C = [3,5) \cup (5,9) \] (g)\[ B - (B - C) = \lbrace 5 \rbrace \] (h)\[ A \cup D = [1,\infty) \] (i)\[ C \cap D = \lbrace 5 \rbrace \]

Exercise 1.3

Q1

(Maria Kahlert)

(a)\[ 2x-3 \geq 4+7x \\ 2x-7x \geq 4+3 \\ -5x \geq 7 \\ x \leq -\frac{7}{5}\]

(b)\[ 8(x+1)-2 < 5(x-6)+7 \\ 8x+8-2 < 5x-30+7 \\ 3x < -29 \\ x < -\frac{29}{3}\]

(c)As \[ x^2+1 > 0 \] \[ \frac{x^2-3x+7}{x^2+1} < 1 \\ x^2-3x+7 < x^2+1 \\ -3x < -6 \\ x > 2\] (d)\[ (2x+7)(5-11x) \leq 0 \\ (x+\frac{7}{2})(x - \frac{5}{11}) \geq 0 \\ x \geq \frac{5}{11} \ or\ x \leq -\frac{7}{2}\]

(e)\[ (x^2-2x-3) <0 \\ (x-3)(x+1) <0 \\ -1<x<3\] (f)\[ 2x^2-3x > 4 \\ 2x^2-3x-4 > 0 \] When \[ 2x^2-3x-4 = 0 \] \[ x = \frac{-(-3) \pm \sqrt{[(-3)^2-4\times2\times(-4)]} }{2\times2}\\ = \frac{3 \pm \sqrt{41} }{4}\] Thus \[ 2x^2-3x-4 > 0 \\ x < \frac{3 - \sqrt{41} }{4} \ or \ x > \frac{3 + \sqrt{41} }{4}\]

(g)\[ 2x^2-3x < -4 \\ 2x^2-3x+4 < 0 \] When \[ 2x^2-3x+4 = 0 \] \[ (-3)^2-4\times2\times4 = -23 <0 \] Thus the equation has no solution. Thus \[ 2x^2-3x+4 < 0 \\ x \in \emptyset \] (h)\[ \frac{2x+3}{x-4} \geq 0 \\ (2x+3)(x-4) \geq 0 \ and\ x \neq 4 \\ (x+\frac{3}{2})(x-4) \geq 0 \ and\ x \neq 4 \\ x > 4 \ or \ x \leq -\frac{3}{2}\] (i)\[ \frac{2x+3}{x-4} < 1 \\ \frac{2x+3}{x-4} -1 <0 \\ \frac{2x+3}{x-4} - \frac{x-4}{x-4} <0\\ \frac{x+7}{x-4} < 0 \\ (x+7)(x-4) < 0 \\ -7<x<4\]

Q2

(a)\[ 2x^3+7x^2-15x = x(2x^2+7x-15) \\ = x(x+5)(2x-3)\] (b)\[ 2x^3+3x^2-2x-3 = (2x^3-2x)+(3x^2-3) \\ = 2x(x^2-1) + 3(x^2-1) \\ = (2x+3)(x^2-1) \\ = (2x+3)(x+1)(x-1)\] (c)\[ x^3-x^2-x-2 = (x^3-1)-(x^2+x+1) \\ = (x-1)(x^2+x+1)-(x^2+x1) \\ = (x-1-1)(x^2+x+1) \\ = (x-2)(x^2+x+1)\] (d)\[ x^4-3x^3-13x^2+15x = x(x^3-3x^2-13x+15) \\ = x[(x^3-x)+(-3x^2-12x+15)] \\ = x[x(x^2-1)-3(x^2+4x-5)] \\ = x[x(x+1)(x-1)-3(x-1)(x+5)] \\ = x(x-1)[x(x+1)-3(x+5)] \\ = x(x-1)(x^2-2x-15)\\ = x(x-1)(x+3)(x-5)\] (e)\[ x^4-3x^3+x^2+3x-2 = (x^4+x^2-2)+(-3x^3+3x) \\ = (x^2+2)(x^2-1)-3x(x^2-1) \\ = (x^2+2-3x)(x^2-1) \\ = [(x-2)(x-1)][(x+1)(x-1)] \\ = (x-2)(x+1)(x-1)^2\]

(f)\[ x^4-x^3+x^2-3x+2 = (x^4-x^3)+(x^2-3x+2) \\ = x^3(x-1)+(x-1)(x-2) \\ = (x-1)(x^3+x-2) \\ = (x-1)[(x^3-1)+(x-1)] \\ = (x-1)[(x-1)(x^2+x+1)+(x-1)] \\ = (x-1)^2(x^2+x+2)\]

Q3

(Ashley a-d; Byoungchurl Lee e-h)

(a)\[ (x-4)(9-5x)(2x+3) < 0 \\ (x-4)(x-\frac{9}{5})(x+\frac{3}{2}) >0 \\ x \in ( -\frac{3}{2},\frac{9}{5}) \cup (4, \infty)\] (b)\[ (x-3)(2x+1)^2 \leq 0 \\ (x-3)(x+\frac{1}{2})^2 \leq 0 \\ x \in (-\infty, 3] \] (c)\[ x^3-2x^2-5x+6 < 0 \\ (x^3-x)+(-2x^2-4x+6) < 0\\ x(x^2-1)-2(x^2+2x-3) < 0\\ x(x-1)(x+1)-2(x+3)(x-1) < 0\\ (x-1)[x(x+1)-2(x+3)] < 0 \\ (x-1)(x^2-x-6) < 0 \\ (x-1)(x-3)(x+2) <0 \\ x \in (-\infty,-2) \cup (1,3) \]

(d)\[ -2x^3+x^2+15x-18 \leq 0 \\ (-2x^3+8x)+(x^2+7x-18) \leq 0 \\ -2x(x^2-4)+(x-2)(x+9) \leq 0 \\ -2x(x+2)(x-2)+(x-2)(x+9) \leq 0 \\ (x-2)[-2x(x+2)+(x+9)] \leq 0 \\ (x-2)(-2x^2-3x+9) \leq 0 \\ (x-2)(-2x+3)(x+3) \leq 0 \\ (x-2)(x-\frac{3}{2})(x+3) \geq 0 \\ x \in [-3,\frac{3}{2}] \cup [2,\infty)\] (e)\[ x^3-x^2-5x-3 > 0 \\ (x^3-x)+(-x^2-4x-3) > 0 \\ x(x^2-1)-(x^2+4x+3) > 0 \\ x(x-1)(x+1)-(x+3)(x+1) > 0 \\ (x+1)[x(x-1)-(x+3)] > 0\\ (x+1)(x^2-2x-3) > 0 \\ (x+1)(x-3)(x+1) > 0 \\ (x+1)^2(x-3) > 0 \\ x \in (3,\infty)\]

(f)\[ x^3+3x^2+5x+3 \leq 0 \\ (x^3-x)+(3x^2+6x+3) \leq 0 \\ x(x^2-1)+3(x^2+2x+1) \leq 0 \\ x(x-1)(x+1)+3(x+1)^2 \leq 0 \\ (x+1)[x(x-1)+3(x+1)] \leq 0 \\ (x+1)(x^2+2x+3) \leq 0 \\ \] As \[ x^2+2x+3 = (x^2+2x+1)+2 = (x+1)^2+2 > 0 \] Thus \[ x^3+3x^2+5x+3=(x+1)(x^2+2x+3) \leq 0 \\ x \in (-\infty,-1] \]

(g)\[ x^4+2x^3-13x^2-14x+24 > 0 \\ (x^4+2x^3-3x^2)+(-10x^2-14x+24) > 0 \\ x^2(x^2+2x-3)-2(5x^2+7x-12) > 0 \\ x^2(x+3)(x-1)-2(x-1)(5x+12) > 0 \\ (x-1)[x^2(x+3)-2(5x+12)] > 0 \\ (x-1)(x^3+3x^2-10x-24) > 0 \\ (x-1)[(x^3+3x^2+2x)+(-12x-24) >0\\ (x-1)[x(x^2+3x+2)-12(x+2)]>0\\ (x-1)[x(x+1)(x+2)-12(x+2)]>0\\ (x-1)(x+2)[x(x+1)-12]>0\\ (x-1)(x+2)(x^2+x-12) > 0\\ (x-1)(x+2)(x+4)(x-3) > 0\\ x \in (-\infty,-4) \cup (-2,1) \cup (3,\infty)\] (h)\[ 6x^4+x^3-15x^2 \leq 0 \\ x^2(6x^2+x-15) \leq 0 \\ x^2(2x-3)(3x+5) \leq 0 \\ x \in [-\frac{5}{3},\frac{3}{2}]\]