What is a Numeral in Base n?

When we talk about numerals in base n, we’re diving into a fascinating world of number systems beyond the familiar decimal (base 10) system. Let’s break it down step by step to understand what it means and how it works.

Understanding Number Systems

Decimal System (Base 10)

The decimal system is the most commonly used number system. It uses ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Each position in a number represents a power of 10. For example, in the number 345:

$345 = 3 times 10^2 + 4 times 10^1 + 5 times 10^0$

Binary System (Base 2)

The binary system is used in computers and digital systems. It uses only two digits: 0 and 1. Each position represents a power of 2. For example, in the binary number 1011:

$1011_2 = 1 times 2^3 + 0 times 2^2 + 1 times 2^1 + 1 times 2^0 = 8 + 0 + 2 + 1 = 11$

Other Bases

Numerals in base n follow the same principle but use n digits. For instance, in base 5, we use the digits 0, 1, 2, 3, and 4. Each position in a number represents a power of 5. For example, in the base 5 number 243:

$243_5 = 2 times 5^2 + 4 times 5^1 + 3 times 5^0 = 2 times 25 + 4 times 5 + 3 = 50 + 20 + 3 = 73$

Converting Between Bases

Converting from Base n to Decimal

To convert a number from base n to decimal, you multiply each digit by the corresponding power of n and sum the results. For example, let’s convert the base 7 number 321 to decimal:

$321_7 = 3 times 7^2 + 2 times 7^1 + 1 times 7^0 = 3 times 49 + 2 times 7 + 1 = 147 + 14 + 1 = 162$

Converting from Decimal to Base n

To convert a decimal number to base n, you repeatedly divide the number by n and record the remainders. These remainders, read in reverse, give you the number in base n. Let’s convert the decimal number 162 to base 7:

  1. $162 text{ divided by } 7 text{ gives quotient } 23 text{ and remainder } 1$
  2. $23 text{ divided by } 7 text{ gives quotient } 3 text{ and remainder } 2$
  3. $3 text{ divided by } 7 text{ gives quotient } 0 text{ and remainder } 3$

Reading the remainders from bottom to top, we get $321_7$

Practical Applications

Computing

The binary system (base 2) is fundamental in computing. Each bit in a computer’s memory is a binary digit, representing either 0 or 1. This simplicity allows for efficient data processing and storage.

Timekeeping

The hexadecimal system (base 16) is often used in computing for its compact representation of binary data. It uses sixteen digits: 0-9 and A-F. For example, the binary number 1111 1111 can be represented as FF in hexadecimal.

Everyday Use

While we mostly use the decimal system in everyday life, understanding other bases can be useful. For example, the sexagesimal system (base 60) is used in timekeeping (60 seconds in a minute, 60 minutes in an hour).

Examples and Exercises

Example 1: Converting Base 8 to Decimal

Convert the base 8 number 157 to decimal:

$157_8 = 1 times 8^2 + 5 times 8^1 + 7 times 8^0 = 1 times 64 + 5 times 8 + 7 = 64 + 40 + 7 = 111$

Example 2: Converting Decimal to Base 3

Convert the decimal number 45 to base 3:

  1. $45 text{ divided by } 3 text{ gives quotient } 15 text{ and remainder } 0$
  2. $15 text{ divided by } 3 text{ gives quotient } 5 text{ and remainder } 0$
  3. $5 text{ divided by } 3 text{ gives quotient } 1 text{ and remainder } 2$
  4. $1 text{ divided by } 3 text{ gives quotient } 0 text{ and remainder } 1$

Reading the remainders from bottom to top, we get $1200_3$

Exercise 1

Convert the base 6 number 254 to decimal.

Exercise 2

Convert the decimal number 78 to base 4.

Conclusion

Understanding numerals in base n opens up a world of different number systems, each with its unique applications and benefits. From computing to timekeeping, these systems play a crucial role in our daily lives and technological advancements.

3. Wikipedia – Positional Notation

Citations

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