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The two most basic operations of arithmetic are addition and subtraction. Multiplication is specialized form of addition where you quickly add identical numbers and division is a specialized form of subtraction where you quickly remove identical numbers.
The use of subtractive notation with Roman numerals increased the complexity of performing basic arithmetic operations without conveying the benefits of a full positional notion system. In the algorithms that follow, the first step is to remove the subtractive notation from the numerals before any arithmetic operations. The subtractive notion is then reapplied to the solution as the end of the operation.
The Roman abacus was a hand-held tool for assisting in the computations using Roman numerals.
All arithmetic operations can be broken down to combinations of addition and subtraction.
CXVI + XXIV = ?
| Step | Description | Example |
|---|---|---|
| 1 | Remove subtractive notation | IV --> IIII |
| 2 | Concatenate terms | CXVI + XXIIII --> CXVIXXIIII |
| 3 | Sequence numerals high to low | CXVIXXIIII --> CXXXVIIIII |
| 4 | Simplify result by summation of internal numerals | IIIII --> V then VV --> X
CXXVIIIII --> CXXXX |
| 5 | Apply subtractive notation | XXXX --> XL |
| 6 | Solution | CXL |
Solution:
CXVI + XXIV = CXL
Step 1 decodes the positional data in the terms and replaces it with primitive counts. Now represented as a pure counting system, the concatenation of the terms in Step 2 gives the correct solution to the problem: CXVIXXIIII represents the same number as CXL - both terms convert to 140 in Arabic numerals. Steps 3 & 4 now reduce the result to the simplest expression possible and Step 5 reintroduces subtractive notation transforming the result back into a positional number.
CXVI − XXIV = ?
| Step | Description | Example |
|---|---|---|
| 1 | Remove subtractive notation | IV --> IIII |
| 2 | Eliminate common numerals between terms | CXVI − XXIIII --> CV − XIII |
| 3 | Expand numerals in first term until common denominator in second term is produced. | CV − XIII --> LLIIIII − XIII --> LXXXXXIIIII − XIII |
| 4 | Repeat steps 2 and 3 until second term is empty | LXXXXXIIIII − XIII --> LXXXXII |
| 5 | Apply subtractive notation | LXXXXII --> XCII |
| 6 | Solution | XCII |
Solution:
CXVI − XXIV = XCII