What is the intersection of two intervals?

In mathematics, intervals are a way to describe a range of numbers. For example, the interval [1, 5] includes all numbers between 1 and 5, inclusive. When we talk about the intersection of two intervals, we are looking for the range of numbers that are common to both intervals. This concept is particularly useful in various fields, such as algebra, calculus, and even in computer science.

Understanding Intervals

Types of Intervals

There are several types of intervals, and it’s essential to understand them before diving into their intersections:

  1. Closed Interval [a, b]: Includes all numbers between a and b, including a and b.
  2. Open Interval (a, b): Includes all numbers between a and b but not a and b themselves.
  3. Half-Open Interval [a, b): Includes all numbers between a and b, including a but not b.
  4. Half-Open Interval (a, b]: Includes all numbers between a and b, including b but not a.

Examples

  • Closed Interval: [2, 6] includes 2, 3, 4, 5, and 6.
  • Open Interval: (2, 6) includes 3, 4, and 5 but not 2 and 6.
  • Half-Open Interval: [2, 6) includes 2, 3, 4, and 5 but not 6.
  • Half-Open Interval: (2, 6] includes 3, 4, 5, and 6 but not 2.

Finding the Intersection

To find the intersection of two intervals, we need to determine the range of numbers that are common to both intervals. Here’s a step-by-step guide to help you understand this process:

  1. Identify the Intervals

    Let’s consider two intervals, A and B.

    • Interval A: [a1, a2]
    • Interval B: [b1, b2]

  1. Compare the Intervals

    To find the intersection, we need to compare the starting and ending points of both intervals. The intersection will start from the maximum of the two starting points and end at the minimum of the two ending points.

  1. Determine the Intersection

    The intersection of two intervals A and B can be represented as:

    $text{Intersection} = [max(a1, b1), min(a2, b2)]$

  1. Check for Validity

    For the intersection to be valid, the starting point must be less than or equal to the ending point. If this condition is not met, the intervals do not intersect.

Examples

Example 1: Closed Intervals

  • Interval A: [2, 6]
  • Interval B: [4, 8]

The intersection is:

$text{Intersection} = [max(2, 4), min(6, 8)] = [4, 6]$

Example 2: Open Intervals

  • Interval A: (1, 5)
  • Interval B: (3, 7)

The intersection is:

$text{Intersection} = (max(1, 3), min(5, 7)) = (3, 5)$

Example 3: Half-Open Intervals

  • Interval A: [1, 4)
  • Interval B: (2, 6]

The intersection is:

$text{Intersection} = [max(1, 2), min(4, 6)] = [2, 4)$

Special Cases

No Intersection

If the maximum starting point is greater than the minimum ending point, the intervals do not intersect. For example:

  • Interval A: [1, 3]
  • Interval B: [4, 6]

There is no intersection because:

$max(1, 4) = 4$

$min(3, 6) = 3$

Since 4 > 3, there is no intersection.

Equal Intervals

If the intervals are identical, the intersection is the interval itself. For example:

  • Interval A: [2, 5]
  • Interval B: [2, 5]

The intersection is:

$text{Intersection} = [2, 5]$

Practical Applications

Algebra and Calculus

In algebra and calculus, finding the intersection of intervals can help solve inequalities and understand the domain and range of functions.

Computer Science

In computer science, interval intersections are used in algorithms for scheduling, memory allocation, and more.

Real-Life Examples

  • Scheduling: Finding common available time slots for meetings.
  • Geography: Overlapping regions on a map.

Conclusion

Understanding the intersection of two intervals is a fundamental concept in mathematics with various practical applications. By comparing the starting and ending points of the intervals, you can easily determine the range of numbers common to both. Whether you’re solving algebraic inequalities or scheduling meetings, this concept is a valuable tool in your mathematical toolkit.

Citations

  1. 1. Khan Academy – Intervals
  2. 2. Purplemath – Interval Notation
  3. 3. MathIsFun – Inequalities

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ