Exploring Sets A And B A Detailed Solution

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In the fascinating realm of mathematics, set theory holds a pivotal role, serving as the bedrock for numerous advanced concepts. This article delves into the intricacies of two distinct sets, A and B, each defined by unique conditions. Our journey will involve a meticulous exploration of these sets, aiming to understand their individual natures and their interplay. We will begin by demonstrating that set A can be expressed as a closed interval, a fundamental concept in real analysis. Subsequently, we will navigate the complexities of set B, which is defined over integers and involves an intriguing rational expression. The climax of our exploration will be the calculation of the intersection of sets A and B, a process that will require a synthesis of our understanding of both sets. This endeavor is not merely an exercise in mathematical manipulation; it is an opportunity to hone our analytical skills and appreciate the elegance of set theory.

Decoding Set A: The Interval [-2, 3]

The first step in our exploration is to decipher the nature of set A. Defined as A = {x ∈ ℝ | |2x − 1| ≤ 5}, this set comprises all real numbers x that satisfy the inequality |2x − 1| ≤ 5. To unveil the structure of A, we must解开 the absolute value inequality. This involves considering two separate cases:

  1. When 2x − 1 ≥ 0, the absolute value |2x − 1| simplifies to 2x − 1. Thus, we have the inequality 2x − 1 ≤ 5. Adding 1 to both sides yields 2x ≤ 6, and dividing by 2 gives x ≤ 3. Additionally, from the condition 2x − 1 ≥ 0, we get x ≥ 1/2. Combining these inequalities, we have 1/2 ≤ x ≤ 3.
  2. When 2x − 1 < 0, the absolute value |2x − 1| becomes −(2x − 1). Hence, the inequality transforms to −(2x − 1) ≤ 5. Distributing the negative sign, we get -2x + 1 ≤ 5. Subtracting 1 from both sides gives -2x ≤ 4, and dividing by -2 (and flipping the inequality sign) yields x ≥ -2. From the condition 2x − 1 < 0, we have x < 1/2. Thus, in this case, we have -2 ≤ x < 1/2.

Combining both cases, we find that set A includes all real numbers x such that -2 ≤ x ≤ 3. This precisely defines the closed interval [-2, 3]. Therefore, we have successfully demonstrated that A = [-2, 3]. This interval notation provides a concise and intuitive representation of set A, allowing us to visualize its extent on the real number line. The endpoints -2 and 3 are included in the set, signifying the closed nature of the interval. This understanding of A is crucial for our subsequent task of determining its intersection with set B.

Unraveling Set B: Integer Solutions to a Rational Expression

Having characterized set A, we now turn our attention to the enigmatic set B. Defined as B = {x ∈ ℤ | (2x-3)/(x+3) ∈ ℤ}, set B comprises all integers x for which the rational expression (2x-3)/(x+3) evaluates to an integer. This definition presents a unique challenge, as we must identify integer values of x that result in an integer quotient. To tackle this, we employ a clever algebraic manipulation. We rewrite the rational expression using polynomial long division or by adding and subtracting a constant in the numerator:

(2x-3)/(x+3) = (2(x+3) - 9)/(x+3) = 2 - 9/(x+3)

This transformation reveals a crucial insight: the expression (2x-3)/(x+3) will be an integer if and only if 9/(x+3) is an integer. In other words, (x+3) must be a divisor of 9. The divisors of 9 are ±1, ±3, and ±9. Therefore, we have the following possibilities:

  1. x + 3 = 1 => x = -2
  2. x + 3 = -1 => x = -4
  3. x + 3 = 3 => x = 0
  4. x + 3 = -3 => x = -6
  5. x + 3 = 9 => x = 6
  6. x + 3 = -9 => x = -12

Thus, set B consists of the integers {-12, -6, -4, -2, 0, 6}. This discrete set of integer values stands in stark contrast to the continuous interval that defines set A. Understanding the elements of B is paramount for our final step: determining the intersection of A and B. This intersection will reveal the integers that lie within the interval [-2, 3], providing a concrete link between the two sets.

The Grand Finale: Calculating A ∩ B

With the individual natures of sets A and B now elucidated, we arrive at the culminating step of our exploration: calculating their intersection, denoted as A ∩ B. The intersection of two sets is the set of elements that are common to both. In our case, A ∩ B will consist of all integers that are both members of the interval [-2, 3] and elements of the set {-12, -6, -4, -2, 0, 6}.

Set A, as we established, is the interval [-2, 3], encompassing all real numbers between -2 and 3, inclusive. Set B, on the other hand, is the discrete set {-12, -6, -4, -2, 0, 6}. To find their intersection, we simply identify the elements of B that also fall within the interval [-2, 3].

Examining the elements of B, we observe that:

  • -12 is not in the interval [-2, 3].
  • -6 is not in the interval [-2, 3].
  • -4 is not in the interval [-2, 3].
  • -2 is in the interval [-2, 3].
  • 0 is in the interval [-2, 3].
  • 6 is not in the interval [-2, 3].

Therefore, the elements common to both A and B are -2 and 0. Consequently, the intersection of A and B is the set {-2, 0}. This result elegantly encapsulates the interplay between the continuous set A and the discrete set B. The intersection, a small set of integers, represents the points where these two distinct mathematical entities converge. This final calculation underscores the power of set theory in dissecting and understanding mathematical structures.

In conclusion, our journey through sets A and B has been a testament to the beauty and precision of mathematics. We successfully demonstrated that A = [-2, 3], unraveled the intricacies of B to find its elements, and culminated our exploration by calculating A ∩ B = {-2, 0}. This exercise not only reinforces our understanding of set theory but also highlights the importance of careful analysis and algebraic manipulation in solving mathematical problems.