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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - 0.8 Number 1 – 32

Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 Section 0.8

CHAPTER 0 PRELIMINARIES

SECTION 0.8 Chapter Review


Calculus 9th Purcell Chapter 0 - 0.8

Problem Set 0.8, Number 1 – 32
Respond with true or false to each of the following assertions. Be prepared to justify your answer. Normally, this means that you should supply a reason if you answer true and provide a counter- example if you answer false.
  1. Any number that can be written as a fraction p/q is rational.
    2. 💬 SOLUTION:
     
    False:
    p and q must be integers.

  2. The difference of any two rational numbers is rational.
    3. 💬 SOLUTION:
     
    True:
    (p1)/(q1) – (p2)/(q2) = (p1q2 – p2q1)/(q1q2); since p1, q1, p2 and q2 are integers, so are p1q2 – p2q1 and q1q2.

  3. The difference of any two irrational numbers is irrational.
    4. 💬 SOLUTION:
     
    False:
    If the numbers are opposites (–π and π) then the sum is 0, which is rational.

  4. Between two distinct irrational numbers, there is always another irrational number.
    5. 💬 SOLUTION:
     
    True:
    Between any two distinct real numbers there are both a rational and an irrational number.

  5. 0.999... (repeating 9s) is less than 1.
    6. 💬 SOLUTION:
     
    False:
    0.999... is equal to 1.

  6. The operation of exponentiation is commutative; that is, (am)n = (an)m.
    7. 💬 SOLUTION:
     
    True:
    (am)n = (an)m = amn

  7. The operation * defined by m * n = mn is associative.
    8. 💬 SOLUTION:
     
    False:
    (a * b) * c = abc; a * (b * c) = abc

  8. The inequalities x ≤ y, y ≤ z and z ≤ x together imply that x = y = z.
    9. 💬 SOLUTION:
     
    True:
    Since x ≤ y ≤ z and x ≥ z, x = y = z

  9. If |x| < ε for every positive number ε, then x = 0.
    10. 💬 SOLUTION:
     
    True:
    If x was not 0, then ε = |x|/2 would be a positive number less than |x|.

  10. If x and y are real numbers, then (x – y)(y – x) ≤ 0.
    11. 💬 SOLUTION:
     
    True:
    y – x = –(x – y) so
    (x – y)(y – x) = (x – y)(–1)(x – y)
    = (–1)(x – y)2.
    (x – y)2 ≥ 0 for all x and y, so
    –(x – y)2 ≤ 0.

    Semoga Bermanfaat 😁

  11. If a < b < 0, then 1/a > 1/b.
    12. 💬 SOLUTION:
     
    True:
    a < b < 0; a < b; (a/b) > 1; (1/b) < (1/a)

  12. It is possible for two closed intervals to have exactly one point in common.
    13. 💬 SOLUTION:
     
    True:
    [a, b] and [b, c] share point b in common.

  13. If two open intervals have a point in common, then they have infinitely many points in common.
    14. 💬 SOLUTION:
     
    True:
    If (a, b) and (c, d) share a point then c < b so they share the infinitely many points between b and c.

  14. If x < 0 then √(x2) = –x
    15. 💬 SOLUTION:
     
    True:
    √(x2) = |x| = –x if x < 0.

  15. If x is a real number, then |–x| = x.
    16. 💬 SOLUTION:
     
    False:
    For example, if x = –3, then |–x| = |–(–3)| = |3| = 3 which does not equal x.

  16. If |x| < |y|, then x < y.
    17. 💬 SOLUTION:
     
    False:
    For example, take x = 1 and y = –2.

  17. If |x| < |y| then x4 < y4.
    18. 💬 SOLUTION:
     
    True:
    |x| < |y| ⇔ |x|4 < |y|4
    |x|4 = x4 and y4 = y4, so x4 < y4

  18. If x and y are both negative, then |x + y| = |x| + |y|.
    19. 💬 SOLUTION:
     
    True:
    |x + y| = –(x + y)
    |x + y| = –x + (–y) = |x| + |y|

  19. If |r| < 1 then 1/(1 + |r|) ≤ 1/(1 – r) ≤ 1/(1 – |r|).
    20. 💬 SOLUTION:
     
    True:
    If r = 0, then
    1/(1 + |r|) = 1/(1 – r) = 1/(1 – |r|) = 1.
    For any r, 1 + |r| ≥ 1 – |r|. Since
    |r| < 1, 1 – |r| > 0 so 1/(1 + |r|) ≤ 1/(1 – |r|);
    also, –1 < r < 1.
    If –1 < r < 0, then |r| = –r and
    1 – r = 1 + |r|, so
    1/(1 + |r|) = 1/(1 – r) ≤ 1/(1 – |r|).
    If 0 < r < 1, then |r| = r and
    1 – r = 1 – |r|, so
    1/(1 + |r|) ≤ 1/(1 – r) = 1/(1 – |r|).

  20. If |r| > 1 then 1/(1 – |r|) ≤ 1/(1 – r) ≤ 1/(1 + |r|).
    21. 💬 SOLUTION:
     
    True:
    If |r| > 1, then 1 – |r| < 0. Thus,
    since 1 + |r| ≥ 1 – |r|, 1/(1 – |r|) ≤ 1/(1 + |r|).
    If r > 1, |r| = r, and 1 – r = 1 – |r|, so
    1/(1 – |r|) = 1/(1 – r) ≤ 1/(1 + |r|).
    If r < –1, |r| = –r, and 1 – r = 1 + |r|, so
    1/(1 – |r|) ≤ 1/(1 – r) = 1/(1 + |r|).

    Semoga Bermanfaat 😁

  21. It is always true that ||x| – |y|| ≤ |x + y|.
    22. 💬 SOLUTION:
     
    True:
    If x and y are the same sign, then ||x| – |y|| = |x – y|.
    |x – y| ≤ |x + y| when x and y are the same sign,
    so ||x| – |y|| ≤ |x + y|.
    If x and y have opposite signs then either
    ||x| – |y|| = |x – (–y)| = |x + y|
    (x > 0, y < 0) or
    ||x| – |y|| = |–x – y| = |x + y|
    (x < 0, y > 0).
    In either case ||x| – |y|| = |x + y|.
    If either x = 0 or y = 0, the inequality is easily seen to be true.

  22. For every positive real number y, there exists a real number x such that x2 = y.
    23. 💬 SOLUTION:
     
    True:
    If y is positive, then x = √y satisfies x2 = (√y)2 = y.

  23. For every real number y, there exists a real number x such that x3 = y.
    24. 💬 SOLUTION:
     
    True:
    For every real number y, whether it is positive, zero, or negative, the cube root x = ∛y satisfies x3 = (∛y)3 = y.

  24. It is possible to have an inequality whose solution set consists of exactly one number.
    25. 💬 SOLUTION:
     
    True:
    For example x2 ≤ 0 has solution [0].

  25. The equation x2 + y2 + ax + y = 0 represents a circle for every real number a.
    26. 💬 SOLUTION:
     
    True:
    x2 + ax + y2 + y = 0
    x2 + ax + ((a2)/4) + y2 + y + (1/4) = ((a2)/4) + (1/4)
    (x + (a/2))2 + (y + (1/2))2 = ((a2 + 1)/4)
    Is a circle for all values of a.

  26. The equation x2 + y2 + ax + by = c represents a circle for all real numbers a, b, c.
    27. 💬 SOLUTION:
     
    False:
    If a = b = 0 and c < 0, the equation does not represent a circle.

  27. If (a, b) is on a line with slope (3/4), then (a + 4, b + 3) is also on that line.
    28. 💬 SOLUTION:
     
    True:
    y – b = (3/4)(x – a)
    y = (3/4)x – (3a/4) + b;
    If x = a + 4:
    y = (3/4)(a + 4) – (3a/4) + b
    = (3a/4) + 3 – (3a/4) + b
    = b + 3

  28. If (a, b), (c, d) and (e, f) are on the same line, then (a – c)/(b – d) = (a – e)/(b – f) = (e – c)/(f – d) provided all three points are different.
    29. 💬 SOLUTION:
     
    True:
    If the points are on the same line, they have equal slope. Then the reciprocals of the slopes are also equal.

  29. If ab > 0, then (a, b) lies in either the first or third quadrant.
    30. 💬 SOLUTION:
     
    True:
    If ab > 0, a and b have the same sign, so (a, b) is in either the first or third quadrant.

  30. For every ε > 0 there exists a positive number x such that x < ε.
    31. 💬 SOLUTION:
     
    True:
    If x = ε/2. If ε > 0, then x > 0 and x < ε.

  31. If ab = 0, then (a, b) lies on either the x-axis or the y-axis.
    32. 💬 SOLUTION:
     
    True:
    If ab = 0, a or b is 0, so (a, b) lies on the x-axis or the y-axis. If a = b = 0, (a, b) is the origin.

  32. If √[(x2 – x1)2 + (y2 – y1)2] = |x2 – x1|, then (x1, y1) and (x2, y2) lie on the same horizontal line.
    33. 💬 SOLUTION:
     
    True:
    y1 = y2, so (x1, y1) and (x2, y2) are on the same horizontal line.

Semoga Bermanfaat 😁

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