Subject: Roman Numerals
Date: Thu, 29 Apr 1999 01:55:26 -0400
Someone proposed this question to me and I do not know the answer, so I was wondering if you could help. How, using Roman Numeral, did people add, subtract, multiply, and divide with no zero or negative numbers?
Thanks in advance
The best reference that I know on arithmetic with Roman Numerals is
Arithmetic with Roman Numerals, James G. Kennedy, American Math. Monthly 88:1 (Jan. 1981) 29-32.
What follows is extracted from Math 101 Online which is a course taught at the University of Regina.
A Roman Experience
A nice example of a number system other than our base 10 system that involves different symbols and incorporates different base arithmetic is the Roman system. The symbols used are:
Further intricacies of the system are the way in which the Roman numbers are written using combinations of these symbols. The numbers 1, 2, and 3 are simply I, II, III but for 4, rather than use IIII, the symbol IV is typically used, the I preceding the V denoting subtraction. Similarly we use IX for 9, XC for 90, etc. while XI and CX represent 11 and 110 respectively. For example, 1999 may be written as MDCCCCLXXXXVIIII, or, in shorter form, as MCMXCIX.
M D CC L XXX V II D CCC XX IIII + CC L XXX V II MM D CCC L XXXX V IIIStarting from the right with the "I" column there are eight I's. Five I's make a V leaving three I's. In the "V" column there are 2 V's plus the one "carried" from the I column. Two V's make an X leaving one V. In the "X" column there are eight X's plus the one carried. Five X's make an L leaving three X's with an L carried, and so on.
In our base ten place value system:
1787 824 + 287 2898
Similarly for a simple multiplication:
CC L XXX V II X I CC L XXX V II MM D CCC L XX MMM C L V II
which corresponds, in base 10, to 287 11 = 3157, and a bit more complicated (and not so much fun),
CC L XXX V II XX III CC L XXX V II CC L XXX V II CC L XXX V II MM D CCC L XX MM D CCC L XX MMMMMM D C I
i.e. 287 23 = 6601.
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