Understanding Number Systems

In mathematics, a number system is a way of representing numbers using symbols and rules. The most common number system we use is the base-10 system, also known as the decimal system. This system uses ten distinct digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) and the concept of place value to represent numbers. Each digit’s position in a number determines its value.

For example, in the number 325, the digit ‘3’ represents 3 hundreds (3 x 10²), ‘2’ represents 2 tens (2 x 10¹), and ‘5’ represents 5 ones (5 x 10⁰). This is how we understand the value of the number 325.

Base-4 Number System

The base-4 system is a number system that uses only four distinct digits: 0, 1, 2, and 3. Each digit’s position in a number represents a power of 4, just like in base-10, where each position represents a power of 10.

Converting from Base-4 to Base-10

To convert a base-4 number to its equivalent in base-10, we follow these steps:

  1. Identify the place values: In a base-4 number, each digit’s position represents a power of 4, starting from the rightmost digit as 4⁰ (which is 1), then 4¹, 4², 4³, and so on.
  2. Multiply each digit by its corresponding place value: Multiply each digit in the base-4 number by its corresponding power of 4.
  3. Sum the results: Add up the products obtained in step 2. The result is the base-10 equivalent of the given base-4 number.

Example

Let’s convert the base-4 number 3210 to base-10.

  1. Identify the place values:
    t* The rightmost digit ‘0’ is in the 4⁰ place (1).
    t* The digit ‘1’ is in the 4¹ place (4).
    t* The digit ‘2’ is in the 4² place (16).
    t* The digit ‘3’ is in the 4³ place (64).

  2. Multiply each digit by its place value:
    t* 3 x 4³ = 3 x 64 = 192
    t* 2 x 4² = 2 x 16 = 32
    t* 1 x 4¹ = 1 x 4 = 4
    t* 0 x 4⁰ = 0 x 1 = 0

  3. Sum the results:
    t* 192 + 32 + 4 + 0 = 228

Therefore, the base-4 number 3210 is equivalent to 228 in base-10.

Another Example

Let’s convert the base-4 number 132 to base-10.

  1. Identify the place values:
    t* The digit ‘2’ is in the 4⁰ place (1).
    t* The digit ‘3’ is in the 4¹ place (4).
    t* The digit ‘1’ is in the 4² place (16).

  2. Multiply each digit by its place value:
    t* 1 x 4² = 1 x 16 = 16
    t* 3 x 4¹ = 3 x 4 = 12
    t* 2 x 4⁰ = 2 x 1 = 2

  3. Sum the results:
    t* 16 + 12 + 2 = 30

Therefore, the base-4 number 132 is equivalent to 30 in base-10.

Conclusion

Converting between different number systems is a fundamental concept in computer science and mathematics. Understanding how to convert from base-4 to base-10 helps us appreciate the flexibility and power of different number systems and their applications in various fields.

3. Tutorials Point – Number Systems

Citations

  1. 1. Khan Academy – Number Systems
  2. 2. Math is Fun – Number Systems

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