Sets A & B: Intervals And Operations (A ∩ B, A \ B)

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Hey guys! Let's dive into some set theory and tackle a problem involving sets A and B, defined using inequalities. We'll express these sets as intervals or unions of intervals and then perform some common set operations like intersection (A ∩ B), set difference (A \ B and B \ A), and intersection with the set of natural numbers (B ∩ N). Get ready to brush up on your algebra and set theory skills!

Problem Breakdown

We're given two cases, each with its own definitions for sets A and B. For each case, our mission is to:

  1. Express sets A and B as intervals or unions of intervals. This means solving the inequalities that define each set.
  2. Calculate the following set operations:
    • A ∩ B (the intersection of A and B): elements present in both A and B.
    • A \ B (the set difference of A and B): elements present in A but not in B.
    • B \ A (the set difference of B and A): elements present in B but not in A.
    • B ∩ N (the intersection of B and the set of natural numbers): natural numbers present in B.

Let's get started with the first case!

Case a) A = {x ∈ R ||-x + 1| < 3}; B = {x ∈ R |(5x+4)/3 ∈ (-2;7 2/3)}

Defining Set A

The set A is defined by the inequality |-x + 1| < 3. This involves an absolute value, so we need to consider two scenarios:

  1. -x + 1 < 3
    • Subtracting 1 from both sides: -x < 2
    • Multiplying both sides by -1 (and flipping the inequality sign): x > -2
  2. -(-x + 1) < 3 which simplifies to x - 1 < 3
    • Adding 1 to both sides: x < 4

Combining these two inequalities, we get -2 < x < 4. Therefore, set A can be expressed as the open interval A = (-2, 4). It's important to remember that the parentheses indicate that the endpoints -2 and 4 are not included in the set. This is because the original inequality was a strict inequality (<), not (≤).

Defining Set B

Set B is defined by the condition (5x+4)/3 ∈ (-2;7 2/3). Let's break this down. First, 7 2/3 is a mixed number, which we can convert to an improper fraction: 7 2/3 = (7 * 3 + 2) / 3 = 23/3. So, the condition for set B can be rewritten as:

(5x + 4) / 3 ∈ (-2, 23/3)

This means that the expression (5x + 4) / 3 must lie between -2 and 23/3. We can write this as a compound inequality:

-2 < (5x + 4) / 3 < 23/3

Now, let's solve for x:

  1. Multiply all parts of the inequality by 3: -6 < 5x + 4 < 23
  2. Subtract 4 from all parts: -10 < 5x < 19
  3. Divide all parts by 5: -2 < x < 19/5

So, set B can be expressed as the open interval B = (-2, 19/5). Since 19/5 = 3.8, this means B includes all real numbers between -2 and 3.8, excluding the endpoints.

Performing Set Operations for Case a)

Now that we have A = (-2, 4) and B = (-2, 19/5), we can perform the required set operations.

  • A ∩ B (Intersection): This is the set of elements common to both A and B. Since B is contained within A (B's upper bound 19/5 is less than A's upper bound of 4), the intersection is simply set B itself. Therefore, A ∩ B = (-2, 19/5). Think of it like this: if all of B is inside A, then where they overlap is just B. This is a fundamental concept in set theory.
  • A \ B (Set Difference): This is the set of elements in A but not in B. Visually, this is the part of the interval A that lies beyond the interval B. Since both intervals start at -2, the difference will be the interval from 19/5 (the upper bound of B) to 4 (the upper bound of A). Therefore, A \ B = [19/5, 4). Notice the square bracket on 19/5, which means 19/5 is included in the result because it's not in B (B is open at 19/5).
  • B \ A (Set Difference): This is the set of elements in B but not in A. Since B is entirely contained within A, there are no elements in B that are not also in A. Therefore, B \ A = ∅ (the empty set). This makes sense because if B is a subset of A, there's nothing left in B once you remove A.
  • B ∩ N (Intersection with Natural Numbers): This is the set of natural numbers that are also in B. Remember that natural numbers are positive integers (1, 2, 3, ...). B = (-2, 19/5), which is the same as (-2, 3.8). The natural numbers within this interval are 1, 2, and 3. Therefore, B ∩ N = {1, 2, 3}. Understanding the different types of numbers (natural, integers, real) is crucial for set operations.

Case b) A = {x ∈ R |2|x| ≤ 30-7|x|}; B = {x ∈ R | x^2 - 4x - 21 ≤ 0}

Let's move on to the second case and tackle these sets!

Defining Set A

Set A is defined by the inequality 2|x| ≤ 30 - 7|x|. Again, we have an absolute value, but this time we can simplify the inequality first. Remember, when working with absolute values, isolating the absolute value term is often a good first step. Let's see how it works here.

  1. Add 7|x| to both sides: 9|x| ≤ 30
  2. Divide both sides by 9: |x| ≤ 30/9
  3. Simplify the fraction: |x| ≤ 10/3

Now we can deal with the absolute value. This inequality means that x must be within 10/3 units of 0. This translates to two inequalities:

  1. x ≤ 10/3
  2. -x ≤ 10/3, which is equivalent to x ≥ -10/3

Combining these, we get -10/3 ≤ x ≤ 10/3. So, set A can be expressed as the closed interval A = [-10/3, 10/3]. The square brackets indicate that the endpoints -10/3 and 10/3 are included in the set. This is because the original inequality was a non-strict inequality (≤).

Defining Set B

Set B is defined by the inequality x^2 - 4x - 21 ≤ 0. This is a quadratic inequality. To solve it, we need to first find the roots of the corresponding quadratic equation: x^2 - 4x - 21 = 0.

We can factor the quadratic as follows: (x - 7)(x + 3) = 0. Factoring quadratics is a useful skill for solving inequalities and equations. Practice makes perfect!

This gives us two roots: x = 7 and x = -3. These roots divide the number line into three intervals: (-∞, -3], [-3, 7], and [7, ∞). To determine where the inequality x^2 - 4x - 21 ≤ 0 holds true, we can test a value from each interval:

  1. Interval (-∞, -3): Let's test x = -4. (-4)^2 - 4(-4) - 21 = 16 + 16 - 21 = 11 > 0. The inequality does not hold in this interval.
  2. Interval [-3, 7]: Let's test x = 0. (0)^2 - 4(0) - 21 = -21 ≤ 0. The inequality does hold in this interval.
  3. Interval [7, ∞): Let's test x = 8. (8)^2 - 4(8) - 21 = 64 - 32 - 21 = 11 > 0. The inequality does not hold in this interval.

Therefore, set B can be expressed as the closed interval B = [-3, 7]. Remember, because the inequality was non-strict (≤), the endpoints are included in the solution, hence the square brackets. Testing intervals is a standard technique for solving inequalities.

Performing Set Operations for Case b)

Now that we have A = [-10/3, 10/3] and B = [-3, 7], let's perform the set operations. Remember that -10/3 ≈ -3.33 and 10/3 ≈ 3.33.

  • A ∩ B (Intersection): This is the set of elements common to both A and B. A spans from approximately -3.33 to 3.33, and B spans from -3 to 7. The overlap is the interval from -3 to 3.33 (10/3). Therefore, A ∩ B = [-3, 10/3]. Visualizing these intervals on a number line can help you understand the intersection. This is a critical step in understanding set operations.
  • A \ B (Set Difference): This is the set of elements in A but not in B. A extends from -3.33 to 3.33, and B starts at -3. The part of A that's not in B is the interval from -3.33 (-10/3) up to, but not including, -3. Therefore, A \ B = [-10/3, -3). Notice the parenthesis on -3, indicating it's not included because it's part of B.
  • B \ A (Set Difference): This is the set of elements in B but not in A. B extends from -3 to 7, and A ends at 3.33 (10/3). The part of B that's not in A is the interval from 3.33 (10/3) up to 7. Therefore, B \ A = (10/3, 7]. Again, the parenthesis on 10/3 indicates it's not included because it's part of A.
  • B ∩ N (Intersection with Natural Numbers): This is the set of natural numbers that are also in B. B = [-3, 7]. The natural numbers within this interval are 1, 2, 3, 4, 5, 6, and 7. Therefore, B ∩ N = {1, 2, 3, 4, 5, 6, 7}. Remember, natural numbers are positive integers, so we only consider whole numbers greater than zero.

Wrapping Up

So, guys, we've successfully navigated through two cases of set operations! We've expressed sets as intervals, found intersections and set differences, and even intersected with the set of natural numbers. This exercise highlights the importance of understanding inequalities, absolute values, quadratic equations, and the definitions of set operations. By mastering these concepts, you'll be well-equipped to tackle more complex problems in mathematics and computer science. Keep practicing, and you'll become a set theory pro in no time! 🚀