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Home > Roman arithmetic


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In Rome, merchants used Roman numerals to perform basic arithmetic operations. In modern education, the Roman arithmetic used by the Romans is seldom taught, preferring to instruct students to convert the Roman numeral into an Arabic numeral and solve the equation using a modern positional notation system. While that's more practical, it is not really learning how to add, subtract, multiply and divide Roman numerals, it is only making the student practice converting from Roman to Arabic and back again. Except for historical purposes, none of this is particularly useful to the grade student unless it is used to demonstrate the existence of different numeral systems and their impact on Arithmetic and to do that, the student needs to learn how to perform arithmetic operations in the native numeral system.

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.

1 Basic operations

All arithmetic operations can be broken down to combinations of addition and subtraction.

1.1 Addition

1.1.1 Example

CXVI + XXIV = ?


Step Description Example
1Remove subtractive notationIV --> IIII
2 Concatenate termsCXVI + XXIIII --> CXVIXXIIII
3Sequence numerals high to lowCXVIXXIIII --> CXXXVIIIII
4Simplify result by summation of internal numeralsIIIII --> V then VV --> X

CXXVIIIII --> CXXXX

5Apply subtractive notationXXXX --> XL
6SolutionCXL


Solution: CXVI + XXIV = CXL

1.1.2 Discussion

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.

1.2 Subtraction

1.2.1 Example

CXVI − XXIV = ?


Step Description Example
1Remove subtractive notationIV --> IIII
2Eliminate common numerals between termsCXVI − XXIIII --> CV − XIII
3Expand numerals in first term until common denominator in second term is produced. CV − XIII --> LLIIIII − XIII --> LXXXXXIIIII − XIII
4Repeat steps 2 and 3 until second term is empty LXXXXXIIIII − XIII --> LXXXXII
5Apply subtractive notation LXXXXII --> XCII
6SolutionXCII


Solution: CXVI − XXIV = XCII



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