Roman arithmetic Guide, Meaning , Facts, Information and Description
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.
| Table of contents |
|
2 Compound operations |
| 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 | 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
Step 1 decodes the positional data in the terms and replaces it with primitive counts. In Step 2, like numerals are eliminated from both terms: a count of X and a count of I are each removed from each term, leaving a simplified problem of CV − XIII. Step 3 then expands the first term until it contains a common numeral (X) to the highest numeral in the second term. Step 2 is then repeated, followed by Step 3 until all of the numerals in the second term have been eliminated. Once all of the numerals of have been eliminated, the remaining numerals in the first term represent the solution as a primitive count. Step 5 reintroduces subtractive notation transforming the result back into a positional number.
Having defined the process where by addition and subtraction operations can be performed using only Roman numerals, the other two traditional operations of arithmetic, multiplication and division, can now be accomplished.Discussion
Compound operations
| Step | Description | Example |
|---|---|---|
| 1 | Remove subtractive notation | IV --> IIII |
| 2 | Add multiplicand to product | XIIII + “ “ --> XIIII |
| 3 | Subtract I from multiplicator | VII − I --> VI |
| 4 | Repeat Step 2 and 3 until multiplicator is empty | XIIII + XIIII --> XXVIII & VI - I --> V |
| 5 | Apply subtractive notation | LXXXX --> XC |
| 6 | Solution | XCVIII |
Solution: XIV × VII = XCVIII
| Step | Description | Example |
|---|---|---|
| 1 | Remove subtractive notation | none in this example |
| 2 | Subtract divisor from dividend | CXXI - V --> CXVI |
| 3 | Add I to quotient | I |
| 4 | Repeat Steps 2 and 3 until dividend is less than the divisor | |
| 5 | The count remaining in the dividend is the remainder | I |
| 6 | Apply subtractive notation to quotient | XXIIII --> XXIV |
| 7 | Solution | XXIV r I |
Solution: CXXI / V = XXIV remainder I