Roman arithmetic

From Wikipedia, the free encyclopedia

In Rome, merchants used Roman numerals to perform basic arithmetic operations. In modern education, the Roman arithmetic used by the Romans is seldom taught. The preferred method is to convert the Roman numeral into an Hindu-Arabic numeral and solve the equation using a modern positional notation system. While this is 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 notation system. In the algorithms that follow, the first step is to remove the subtractive notation from the numerals before any arithmetic operations. The subtractive notation 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.

Contents 1 Basic operations 1.1 Addition 1.1.1 Example 1.1.2 Discussion 1.2 Subtraction 1.2.1 Example 1.2.2 Discussion 2 Compound operations 2.1 Multiplication 2.1.1 Example 2.1.2 Discussion 2.2 Division 2.2.1 Example 2.2.2 Discussion

[edit] Basic operations

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

[edit] Addition

[edit] Example

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; VV → X; CXXXVIIIII → CXXXX
5	Apply subtractive notation	XXXX → XL
6	Solution	CXL

Solution: CXVI + XXIV = CXL

[edit] 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.

[edit] Subtraction

[edit] Example

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

[edit] Discussion

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.

[edit] Compound operations

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.

[edit] Multiplication

multiplicand × multiplier = product

[edit] Example

XIV × VII = ?

Step	Description	Example
1	Remove subtractive notation	IV → IIII
2	Add multiplicand to product	XIIII + " " → XIIII
3	Subtract I from multiplier	VII − I → VI
4	Repeat Step 2 and 3 until multiplier is empty	XIIII + XIIII → XXVIII & VI − I → V
5	Apply subtractive notation	LXXXX → XC
6	Solution	XCVIII

Solution: XIV × VII = XCVIII

[edit] Discussion

Step 1 decodes the positional data in the terms and replaces it with primitive counts. Step 2 adds the multiplicand to product. Since subtractive notation has been removed in Step 1 and is later encoded in Step 5, there is no longer a requirement to perform the same processes when performing addition or subtraction within the multiplication operation. Step 3 reduces the number if iterations remaining for the addition operation in Step 2 by decreasing the value of the multiplier. Step 5 reintroduces subtractive notation transforming the result back into a positional number.

[edit] Division

dividend / divisor = quotient

[edit] Example

CXXI / V = ?

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 rem I

Solution: CXXI / V = XXIV remainder I

[edit] Discussion

Step 1 decodes the positional data in the terms and replaces it with primitive counts. Step 2 subtracts the divisor from the dividend. Since subtractive notation has been removed in Step 1 and is later encoded in Step 5, there is no longer a requirement to perform the same processes when performing addition or subtraction within the division operation. Step 3 increases the counter used for the quotient if remaining count of the dividend is greater than the divisor. Step 5 reintroduces subtractive notation transforming the result back into a positional number.