Babylonian Algebra

Calculate the length

Babylonian Solution | Modern Solution |
---|---|

7 4 = 28 28 – 10 = 18 18 1/3 = 6 (the length) 10 - 6 = 4 (the width) |
4L + W = 28 L + W = 10 3 L = 18L = 6W = 10 – 6 = 4 |

**Example 2**

**Solve X^{2} + 6X = 16. **

**Firstly write the equation in the form: **

*X*(*X* + 6) = 16

*X* + 6 = *Y*

**Then our problem is to find numbers X and Y such that Y – X = 6 and XY =16.**

**Now let Y = a + 3 and X = a – 3. Then (a + 3)(a – 3) = 16, so a^{2} – 9 = 16, giving a = 5 and hence X = 2.**

From these examples it can be seen that the Babylonians thought very much as we do today. However, they did not use abstract symbols or letters to express unknown numbers. The Babylonians thus rarely demonstrated a general method; instead they gave many numerical examples.

**As well as the tables of reciprocals, the Babylonians constructed tables of squares, cubes and exponents for the integers 1 to 30. A tablet containing the combination n^{2} + n^{3} for this range of integers has also been found. Such a tablet would have been used to solve cubic equations of the form X^{3} + X^{2} = b. **

Two interesting (finite) series have also been discovered on a tablet dated around 300 B.C.:

**1 + 2 + 2 ^{2} + ... + 2^{9} = 2^{9} + 2^{9} – 1,**

**and**

**1 ^{2} + 2^{2} + 3^{2} + ... + 10^{2} = [1(^{1}**

**This raises the question of whether the Babylonians were familiar with the formulas
**

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