Solution manual for precalculus 3rd edition by axler

Instructor’s Solutions Manual, Section 0.1 Problem Solutions to Problems, Section 0.1 The problems in this section may be harder than typical problems found in the rest of this book √ Show that 67 + is an irrational number √ solution Suppose 67 + is a rational number Because √ = ( 67 + √ 2) − 76 , √ this implies that is the difference of two rational numbers, which implies√that √ is a rational number, which is not true Thus our assumption that 67 + is √ a rational number must be false In other words, 67 + is an irrational number Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.1 Show that − Problem √ is an irrational number √ solution Suppose − is a rational number Because √ = − (5 − √ 2), √ this √ implies that is the difference of two rational numbers, which implies√that is a rational number, which is not true Thus our √ assumption that − is a rational number must be false In other words, − is an irrational number Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.1 Problem √ Show that is an irrational number √ solution Suppose is a rational number Because √ √ 2= , √ this implies that is the quotient of two rational numbers, which implies √ √ that is a rational number, which is not true Thus √ our assumption that is a rational number must be false In other words, is an irrational number Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.1 Show that √ Problem is an irrational number solution Suppose √ is a rational number Because √ √ 2= · , √ this implies that is the product of two rational numbers, which implies that √ √ 2 is a rational number, which is not true Thus our assumption that is a rational number must be false In other words, √ is an irrational number Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.1 Problem √ Show that + is an irrational number √ solution Suppose + is a rational number Because √ √ = (4 + 2) − 4, √ this implies that is the difference of two rational numbers, which implies √ that is a rational number Because √ √ 2= , √ this √ implies that is the quotient of two rational numbers, which implies that √ is a rational number, which is not true Thus our assumption that + √ is a rational number must be false In other words, + is an irrational number Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.1 Problem 6 Explain why the sum of a rational number and an irrational number is an irrational number solution We have already seen the pattern for this solution in Problems and We can repeat that pattern, using arbitrary numbers instead of specific numbers Suppose r is a rational number and x is an irrational number We need to show that r + x is an irrational number Suppose r + x is a rational number Because x = (r + x ) − r, this implies that x is the difference of two rational numbers, which implies that x is a rational number, which is not true Thus our assumption that r + x is a rational number must be false In other words, r + x is an irrational number Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.1 Problem 7 Explain why the product of a nonzero rational number and an irrational number is an irrational number solution We have already seen the pattern for this solution in Problems and We can repeat that pattern, using arbitrary numbers instead of specific numbers Suppose r is a nonzero rational number and x is an irrational number We need to show that rx is an irrational number Suppose rx is a rational number Because x= rx , r this implies that x is the quotient of two rational numbers, which implies that x is a rational number, which is not true Thus our assumption that rx is a rational number must be false In other words, rx is an irrational number Note that the hypothesis that r is nonzero is needed because otherwise we would be dividing by in the equation above Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.1 Problem 8 Suppose t is an irrational number Explain why t is also an irrational number solution Suppose 1t is a rational number Then there exist integers m and n, with n = 0, such that m = t n Note that m = 0, because t cannot equal The equation above implies that t= n , m which implies that t is a rational number, which is not true Thus our assumption that 1t is a rational number must be false In other words, 1t is an irrational number Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.1 Problem 9 Give an example of two irrational numbers whose sum is an irrational number √ √ solution Problem implies that 2 and are irrational numbers Because √ √ √ + 2 = 2, we have an example of two irrational numbers whose sum is an irrational number Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.1 Problem 10 10 Give an example of two irrational numbers whose sum is a rational number solution Note that √ + (5 − √ Thus we have two irrational numbers (5 − sum equals a rational number 2) = √ is irrational by Problem 2) whose Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 6 ( x + y − r )(z + w − t) solution ( x + y − r )(z + w − t) = x (z + w − t) + y(z + w − t) − r (z + w − t) = xz + xw − xt + yz + yw − yt − rz − rw + rt Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 7 (2x + 3)2 solution (2x + 3)2 = (2x )2 + · (2x ) · + 32 = 4x2 + 12x + Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 8 (3b + 5)2 solution (3b + 5)2 = (3b)2 + · (3b) · + 52 = 9b2 + 30b + 25 Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 9 (2c − 7)2 solution (2c − 7)2 = (2c)2 − · (2c) · + 72 = 4c2 − 28c + 49 Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 10 10 (4a − 5)2 solution (4a − 5)2 = (4a)2 − · (4a) · + 52 = 16a2 − 40a + 25 Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 11 11 ( x + y + z)2 solution ( x + y + z )2 = ( x + y + z)( x + y + z) = x ( x + y + z) + y( x + y + z) + z( x + y + z) = x2 + xy + xz + yx + y2 + yz + zx + zy + z2 = x2 + y2 + z2 + 2xy + 2xz + 2yz Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 12 12 ( x − 5y − 3z)2 solution ( x − 5y − 3z)2 = ( x − 5y − 3z)( x − 5y − 3z) = x ( x − 5y − 3z) − 5y( x − 5y − 3z) − 3z( x − 5y − 3z) = x2 − 5xy − 3xz − 5yx + 25y2 + 15yz − 3zx + 15zy + 9z2 = x2 + 25y2 + 9z2 − 10xy − 6xz + 30yz Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 13 13 ( x + 1)( x − 2)( x + 3) solution ( x + 1)( x − 2)( x + 3) = ( x + 1)( x − 2) ( x + 3) = ( x2 − 2x + x − 2)( x + 3) = ( x2 − x − 2)( x + 3) = x3 + 3x2 − x2 − 3x − 2x − = x3 + 2x2 − 5x − Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 14 14 (y − 2)(y − 3)(y + 5) solution (y − 2)(y − 3)(y + 5) = (y − 2)(y − 3) (y + 5) = (y2 − 3y − 2y + 6)(y + 5) = (y2 − 5y + 6)(y + 5) = y3 + 5y2 − 5y2 − 25y + 6y + 30 = y3 − 19y + 30 Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 15 15 ( a + 2)( a − 2)( a2 + 4) solution ( a + 2)( a − 2)( a2 + 4) = ( a + 2)( a − 2) ( a2 + 4) = ( a2 − 4)( a2 + 4) = a4 − 16 Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 16 16 (b − 3)(b + 3)(b2 + 9) solution (b − 3)(b + 3)(b2 + 9) = (b − 3)(b + 3) (b2 + 9) = (b2 − 9)(b2 + 9) = b4 − 81 Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 17 xy( x + y) x − Exercise 17 y solution xy( x + y) 1 − x y = xy( x + y) y x − xy xy = xy( x + y) y−x xy = ( x + y)(y − x ) = y2 − x Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 18 a2 z(z − a) z + Exercise 18 a solution a2 z ( z − a ) 1 + z a = a2 z ( z − a ) a z + az az = a2 z ( z − a ) a+z az = a(z − a)( a + z) = a ( z2 − a2 ) = az2 − a3 Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 19 19 (t − 2)(t2 + 2t + 4) solution (t − 2)(t2 + 2t + 4) = t(t2 + 2t + 4) − 2(t2 + 2t + 4) = t3 + 2t2 + 4t − 2t2 − 4t − = t3 − Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler Instructor’s Solutions Manual, Section 0.2 Exercise 20 20 (m − 2)(m4 + 2m3 + 4m2 + 8m + 16) solution (m − 2)(m4 + 2m3 + 4m2 + 8m + 16) = m(m4 + 2m3 + 4m2 + 8m + 16) − 2(m4 + 2m3 + 4m2 + 8m + 16) = m5 + 2m4 + 4m3 + 8m2 + 16m − 2m4 − 4m3 − 8m2 − 16m − 32 = m5 − 32 Direct.eu/Solution-Manual-for-Precalculus-3rd-Edition-by-Axler ... irrational number Direct.eu /Solution- Manual- for- Precalculus- 3rd- Edition- by- Axler Instructor’s Solutions Manual, Section 0.2 Exercise Solutions to Exercises, Section 0.2 For Exercises 1–4, determine... Direct.eu /Solution- Manual- for- Precalculus- 3rd- Edition- by- Axler Instructor’s Solutions Manual, Section 0.2 Exercise 7 (2x + 3)2 solution (2x + 3)2 = (2x )2 + · (2x ) · + 32 = 4x2 + 12x + Direct.eu /Solution- Manual- for- Precalculus- 3rd- Edition- by- Axler. .. Direct.eu /Solution- Manual- for- Precalculus- 3rd- Edition- by- Axler Instructor’s Solutions Manual, Section 0.2 Exercise 8 (3b + 5)2 solution (3b + 5)2 = (3b)2 + · (3b) · + 52 = 9b2 + 30b + 25 Direct.eu /Solution- Manual- for- Precalculus- 3rd- Edition- by- Axler