Understanding Absolute Value

In mathematics, the absolute value of a number is its distance from zero on the number line, regardless of its direction. It’s always a non-negative value, meaning it’s either zero or positive. The absolute value of a number is denoted by two vertical bars surrounding the number. For example, the absolute value of 5 is written as |5| and the absolute value of -5 is written as |-5|.

Visualizing Absolute Value

Imagine a number line, with zero at the center. Positive numbers are located to the right of zero, and negative numbers are located to the left. The absolute value of a number is simply its distance from zero, measured in units along the number line.

For instance, the number 5 is 5 units to the right of zero, so its absolute value is 5. Similarly, the number -5 is 5 units to the left of zero, so its absolute value is also 5.

Key Properties of Absolute Value

  1. Non-negativity: The absolute value of any number is always greater than or equal to zero. This means that |x| ≥ 0 for any number x.

  2. Symmetry: The absolute value of a number is equal to the absolute value of its negative. This means that |x| = |-x| for any number x.

  3. Triangle Inequality: The absolute value of the sum of two numbers is less than or equal to the sum of their absolute values. This means that |x + y| ≤ |x| + |y| for any numbers x and y.

Applications of Absolute Value

Absolute value has numerous applications in various fields, including:

  • Distance: In geometry, absolute value is used to calculate the distance between two points on a number line. For example, the distance between the points 3 and -2 is |3 – (-2)| = |5| = 5 units.

  • Error Analysis: In engineering and science, absolute value is used to represent the magnitude of errors or deviations from expected values. For example, if a measurement is expected to be 10 units and the actual measurement is 9.8 units, the absolute error is |10 – 9.8| = |0.2| = 0.2 units.

  • Programming: In computer programming, absolute value is used in various algorithms, such as sorting and searching. For example, the absolute value of a difference between two numbers can be used to determine their proximity to each other.

  • Finance: In finance, absolute value is used to calculate the difference between a stock’s current price and its previous price. For example, if a stock’s price increases from $100 to $105, the absolute price change is |105 – 100| = |5| = $5.

Examples of Absolute Value

Here are some examples of calculating absolute values:

  • |7| = 7
  • |-3| = 3
  • |0| = 0
  • |-10.5| = 10.5
  • |√2| = √2

Solving Equations with Absolute Value

Equations involving absolute values can be solved by considering two cases:

  1. Case 1: The expression inside the absolute value is positive or zero. In this case, we simply remove the absolute value bars and solve the equation normally.

  2. Case 2: The expression inside the absolute value is negative. In this case, we multiply both sides of the equation by -1, remove the absolute value bars, and solve the equation normally.

Example:

Solve the equation |x – 2| = 5.

Case 1: x – 2 ≥ 0

We remove the absolute value bars and solve the equation:

x – 2 = 5
x = 7

Case 2: x – 2 < 0

We multiply both sides of the equation by -1, remove the absolute value bars, and solve the equation:

-(x – 2) = 5
-x + 2 = 5
-x = 3
x = -3

Therefore, the solutions to the equation |x – 2| = 5 are x = 7 and x = -3.

Conclusion

The concept of absolute value is fundamental in mathematics and has wide-ranging applications in various fields. Understanding its properties and how to solve equations involving absolute values is essential for a solid mathematical foundation. By visualizing absolute value as distance on the number line and applying the appropriate rules, we can confidently work with this important mathematical concept.

Citations

  1. 1. Math is Fun – Absolute Value
  2. 2. Khan Academy – Absolute Value
  3. 3. Purplemath – Absolute Value

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