Babylonian Mathematics and Numeration Systems

Definition of Hindu-Arabic Numeration System: This is the formal name for our standard base 10 number system. It is important to highlight and recognize this terminology.
Expanded Notation: This method represents numbers as a sum of their digits multiplied by their corresponding place values. * Example: The number $285$ in expanded notation is $2 \times 100 + 8 \times 10 + 5 \times 1$ .
Exponent Representation: In base 10, each place value is a power of 10. The number $285$ can be expressed as: * $2 \times 10^2 + 8 \times 10^1 + 5 \times 10^0$
Place Value Chart: * Going left from the decimal: $10^0$ (ones), $10^1$ (tens), $10^2$ (hundreds), $10^3$ (thousands), $10^4$ (ten-thousands), and so on. * Going right from the decimal (fractions/decimals): $10^{-1}$ ( $0.1$ ), $10^{-2}$ ( $0.01$ ), $10^{-3}$ ( $0.001$ ), and so on. * Example: $5436.45$ fits into the chart with 5 in the $10^3$ place, 4 in the $10^2$ place, 3 in the $10^1$ place, 6 in the $10^0$ place, 4 in the $10^{-1}$ place, and 5 in the $10^{-2}$ place.

Base 60 System: Unlike our base 10 system, Babylonian mathematics uses a base 60 (sexagesimal) system.
Cuneiform Symbols: Instead of hieroglyphics, Babylonian writing is called cuneiform. They utilized only two primary symbols to represent all numbers from 1 to 59: * Unit (1): Represented by an upside-down triangle (pointing downward). * Ten (10): Represented by a cluster or chevron symbol (sideways triangle).
Spacing as a Key Element: Spacing is critical in Babylonian numeration to distinguish between place values. * Example of 12: A single "10" symbol clustered with two "1" symbols represents $12$. * Example of 602: A "10" symbol, followed by a distinct space, followed by two "1" symbols. This indicates $(10 \times 60^1) + (2 \times 60^0) = 600 + 2 = 602$ .
The Babylonian Zero: The textbook uses a symbol consisting of two stacked triangles or chevrons (similar to $\\ll$ ) to represent zero. This symbol is essential for identifying empty place values.

To convert, identify clusters of symbols separated by spaces. Each cluster corresponds to a power of 60, starting with $60^0$ on the far right.
Case Study: Land Units in a Large Valley: * Cluster 1: One "10" and two "1"s = $12$. * Cluster 2: The zero symbol = $0$. * Cluster 3: Four "10"s and one "1" = $41$. * Chart Calculation: * $60^2 \text{ place}: 12$ * $60^1 \text{ place}: 0$ * $60^0 \text{ place}: 41$ * Expanded Formula: $12 \times 60^2 + 0 \times 60^1 + 41 \times 60^0$ * Result: $43,241$
Testing and Homework Requirements: When performing these conversions, students must show either the place value chart ( $60^2, 60^1, 60^0$ ) or the expanded multiplication formula to receive full credit.

Positional List Notation: To avoid drawing numerous cuneiform symbols, the system can be written as a list of numbers in parentheses with a subscript 60 (e.g., $(1, 1, 5)_{60}$ ). * Example $1$: $(1, 1, 5)_{60}$ is $1 \times 60^2 + 1 \times 60^1 + 5 \times 60^0 = 3600 + 60 + 5 = 3665$ . * Example $2$: $(13, 50)_{60}$ is $13 \times 60^1 + 50 \times 60^0 = 780 + 50 = 830$ .
The Semicolon (";"): In Babylonian notation, the semicolon represents the decimal place. Numbers to the left are powers of 60 ( $60^0, 60^1$ ), while numbers to the right are negative powers ( $60^{-1}, 60^{-2}$ ). * Example $3$: $(1, 1; 5)_{60}$ * Calculation: $1 \times 60^1 + 1 \times 60^0 + 5 \times 60^{-1}$ * Expanded: $60 + 1 + \frac{5}{60} = 61.0833...$ * Example $4$: $(3, 12, 30; 45, 6)_{60}$ * Place Values: $60^2, 60^1, 60^0, 60^{-1}, 60^{-2}$ * Calculation: $3 \times 60^2 + 12 \times 60^1 + 30 \times 60^0 + 45 \times 60^{-1} + 6 \times 60^{-2}$ * Result: $11,550.75166...$ (rounded to $11,550.752$ )

To convert from our system to Babylonian, use long division by the base 60.
Example: 82 * $82 \div 60 = 1$ with a remainder of $22$. * Written as: $(1, 22)_{60}$ . * Cuneiform: One triangle, space, two chevrons and two triangles.
Example: 139 * $139 \div 60 = 2$ with a remainder of $19$. * Written as: $(2, 19)_{60}$ . * Cuneiform: Two triangles, space, one chevron and nine triangles.
Note on Drawing: It is acceptable to stack the triangles in cuneiform clusters to save space.

Babylonians developed an extremely accurate estimation for square roots long before calculators or computers.
The Formula: * Define the number as $n = s^2 + e$ , where $s^2$ is the largest perfect square that does not exceed $n$ , and $e$ is the remaining difference. * Estimation formula: $\sqrt{n} \approx s + \frac{e}{2s} - \frac{e^2}{8s^3}$
Demonstration: Finding $\sqrt{150}$ * Step 1: Find the closest perfect square. $12^2 = 144$ is the largest square without exceeding $150$ (since $13^2 = 169$ , which is too high). * Step 2: Define variables. $s = 12$ and $e = 6$ ( $150 - 144 = 6$ ). * Step 3: Plug into formula: $12 + \frac{6}{2(12)} - \frac{6^2}{8(12^3)}$ * Step 4: Execute calculation: $12 + \frac{6}{24} - \frac{36}{8(1728)}$ * Step 5: Compare results: * Calculator actual value: $12.24744871$ * Babylonian estimate: $12.24739583$
Significance: The estimate is correct up to three decimal places. This represents high mathematical sophistication contemporary with Egyptian hieroglyphics.

Babylonians performed division using reciprocals and specific tables (refer to page 29 in the textbook).
Procedure for Division (e.g., $60 \div 5$ ): 1. Separate into multiplication: Re-write the problem as $60 \times \frac{1}{5}$ . 2. Convert fraction to Babylonian form: Using the table, $\frac{1}{5}$ is written as zero semicolon twelve ( $0; 12$ ). 3. Multiply: Take the integer and multiply it by the digits from the fraction. $60 \times 12 = 720$ . 4. Convert back to Babylonian: Use long division to find the places. $720 \div 60 = 12$ remainder $0$. In comma notation, this is $(12, 0)_{60}$ . 5. Finalize Notation: Change the comma to a semicolon to represent the Babylonian division result: $12; 0$ .
Example: $23 \div 12$ * Re-write: $23 \times \frac{1}{12}$ . * Fraction conversion: From the table, $\frac{1}{12} = 0; 5$ . * Multiply: $23 \times 5 = 115$ . * Long division: $115 \div 60 = 1$ remainder $55$. * Final Answer: $1; 55$

Student Question: How should we write these answers on a test?
Instructor Response: I need to see the work. Either write out the place value chart ( $60^0, 60^1, 60^2$ ) or show the full multiplication and addition steps. If you can do it in your calculator, that's great, but I am looking for the work to accompany the answer.
Note on the semicolon/comma: The book might write values with commas for whole number positional notation, but use the semicolon specifically for the division/decimal boundary.