
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 (mathematics) 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. == Basic operations == All arithmetic operations can be broken down to combinations of addition and subtraction. === Addition === ====Example==== CXVI + XXIV = ? {| border=1 cellspacing=0 cellpadding=2 !Step !Description !Example |- |1||Remove subtractive notation||IV → IIII |- |2||concatenation 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 ====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. === Subtraction === ====Example==== CXVI − XXIV = ? {| border=1 cellspacing=0 cellpadding=2 !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 ====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. == 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. === Multiplication === ''multiplicand × multiplicator = product'' ====Example==== XIV × VII = ? {| border=1 cellspacing=0 cellpadding=2 !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 ====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 multiplicator. Step 5 reintroduces subtractive notation transforming the result back into a positional number. === Division === ''dividend / divisor = quotient'' ====Example==== CXXI / V = ? {| border=1 cellspacing=0 cellpadding=2 !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 ====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. Ancient Rome Elementary arithmetic la:Numeri Romani
== CPU Arithmetic == Shibboleth is apparently correct about multiplying and dividing in modern machine language (I haven't tried it much since about 1980), although I wish he had explained it a little to Mr. Hardy. The mul, div, imul, idiv, fmul and fdiv commands at x86 instruction listings are apparently intended for multiplying and dividing. User:Art LaPella 02:38, Sep 4, 2004 (UTC) :''The CPU in modern computers generally is only able to directly perform these two simple operations.'' was deleted from the article. :The machine language may have the instructions, but that does not mean the CPU actually multiplied or divided as a single step. In the ALU/FPU of many CPUs, the actual multiplication is a series of shifts or additions performed by the processor. The same holds true for division. The orginal text did not state that CPUs lacked instructions to perform more than addition or subtraction (which was true with the Z80's instruction set). I wish Shibboleth had taken the time to discussed it here first, rather than just making the deletion from the article. --User:Denorris 04:31, Sep 4, 2004 (UTC)
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Words begining with Roman_arithmetic:
Roman_arithmetic
Roman_arithmetic
Roman arithmetic
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 (mathematics) 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. == Basic operations == All arithmetic operations can be broken down to combinations of addition and subtraction. === Addition === ====Example==== CXVI + XXIV = ? {| border=1 cellspacing=0 cellpadding=2 !Step !Description !Example |- |1||Remove subtractive notation||IV → IIII |- |2||concatenation 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 ====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. === Subtraction === ====Example==== CXVI − XXIV = ? {| border=1 cellspacing=0 cellpadding=2 !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 ====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. == 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. === Multiplication === ''multiplicand × multiplicator = product'' ====Example==== XIV × VII = ? {| border=1 cellspacing=0 cellpadding=2 !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 ====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 multiplicator. Step 5 reintroduces subtractive notation transforming the result back into a positional number. === Division === ''dividend / divisor = quotient'' ====Example==== CXXI / V = ? {| border=1 cellspacing=0 cellpadding=2 !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 ====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. Ancient Rome Elementary arithmetic la:Numeri Romani
Roman arithmetic
== CPU Arithmetic == Shibboleth is apparently correct about multiplying and dividing in modern machine language (I haven't tried it much since about 1980), although I wish he had explained it a little to Mr. Hardy. The mul, div, imul, idiv, fmul and fdiv commands at x86 instruction listings are apparently intended for multiplying and dividing. User:Art LaPella 02:38, Sep 4, 2004 (UTC) :''The CPU in modern computers generally is only able to directly perform these two simple operations.'' was deleted from the article. :The machine language may have the instructions, but that does not mean the CPU actually multiplied or divided as a single step. In the ALU/FPU of many CPUs, the actual multiplication is a series of shifts or additions performed by the processor. The same holds true for division. The orginal text did not state that CPUs lacked instructions to perform more than addition or subtraction (which was true with the Z80's instruction set). I wish Shibboleth had taken the time to discussed it here first, rather than just making the deletion from the article. --User:Denorris 04:31, Sep 4, 2004 (UTC)
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