How to Convert Base n to Base Ten?

Understanding how to convert numbers from base n (where n can be any integer greater than 1) to base ten (our familiar decimal system) is a fundamental skill in mathematics. This process involves breaking down the base n number into its individual digits and then calculating its value in base ten.

Understanding Number Bases

What is a Number Base?

A number base or radix is the number of unique digits, including zero, used to represent numbers in a positional numeral system. The most common base is ten (decimal), but there are many others, such as binary (base 2), octal (base 8), and hexadecimal (base 16).

Why Convert Between Bases?

Different bases are used for various applications. For example, computers use binary (base 2) because they operate on binary logic. Hexadecimal (base 16) is often used in programming because it is more compact than binary.

Steps to Convert Base n to Base Ten

Here’s a step-by-step guide to convert a number from base n to base ten.

  1. Understand the Place Values
    Each digit in a number has a place value, which depends on its position and the base of the number. In base n, the rightmost digit has a place value of $n^0$, the next digit to the left has a place value of $n^1$, and so on.

  1. Write Down the Digits and Their Place Values
    Let’s say we have a number in base n, represented as $d_k d_{k-1} … d_1 d_0$. Each $d_i$ is a digit in the number, and its place value is $n^i$

  1. Multiply Each Digit by Its Place Value
    Multiply each digit by its corresponding place value. This will give you the value of each digit in base ten.

  1. Sum the Results
    Add up all the values obtained in the previous step. The sum is the base ten equivalent of the base n number.

Example: Converting Base 5 to Base Ten

Let’s convert the base 5 number 243 to base ten.

  1. Write Down the Digits and Their Place Values
    In base 5, the number 243 can be written as $2 times 5^2 + 4 times 5^1 + 3 times 5^0$

  1. Calculate the Place Values
    • $5^2 = 25$
    • $5^1 = 5$
    • $5^0 = 1$

  1. Multiply Each Digit by Its Place Value
    • $2 times 25 = 50$
    • $4 times 5 = 20$
    • $3 times 1 = 3$

  1. Sum the Results
    $50 + 20 + 3 = 73$

    So, the base ten equivalent of the base 5 number 243 is 73.

Example: Converting Base 2 to Base Ten

Let’s convert the binary number 1101 to base ten.

  1. Write Down the Digits and Their Place Values
    In binary (base 2), the number 1101 can be written as $1 times 2^3 + 1 times 2^2 + 0 times 2^1 + 1 times 2^0$

  1. Calculate the Place Values
    • $2^3 = 8$
    • $2^2 = 4$
    • $2^1 = 2$
    • $2^0 = 1$

  1. Multiply Each Digit by Its Place Value
    • $1 times 8 = 8$
    • $1 times 4 = 4$
    • $0 times 2 = 0$
    • $1 times 1 = 1$

  1. Sum the Results
    $8 + 4 + 0 + 1 = 13$

    So, the base ten equivalent of the binary number 1101 is 13.

Example: Converting Base 16 to Base Ten

Let’s convert the hexadecimal number 1A3 to base ten.

  1. Write Down the Digits and Their Place Values
    In hexadecimal (base 16), the number 1A3 can be written as $1 times 16^2 + A times 16^1 + 3 times 16^0$. Note that A in hexadecimal represents the decimal number 10.

  1. Calculate the Place Values
    • $16^2 = 256$
    • $16^1 = 16$
    • $16^0 = 1$

  1. Multiply Each Digit by Its Place Value
    • $1 times 256 = 256$
    • $10 times 16 = 160$
    • $3 times 1 = 3$

  1. Sum the Results
    $256 + 160 + 3 = 419$

    So, the base ten equivalent of the hexadecimal number 1A3 is 419.

Conclusion

Converting numbers from base n to base ten involves understanding place values, multiplying each digit by its place value, and summing the results. This method can be applied to any base, whether it’s binary, octal, hexadecimal, or any other base. With practice, this process becomes straightforward and essential for various applications in mathematics and computer science.

3. Wikipedia – Positional Notation

Citations

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

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