How to add and subtract Roman numerals the right way, like the Romans calculated: steps and explanations.

• Let's learn by an example:

• XC + MCMLIII + CCXIX - MCCXLVIII - XC = ?

• 90 + 1.953 + 219 - 1.248 - 90 = ? the Romans did not have the Arabic numbers

1. Cancel equal but different signs numerals.

• Matching numerals / symbols are colored the same.
• XC + MCMLIII + CCXIX - MCCXLVIII - XC = MCMLIII + CCXIX - MCCXLVIII;

2. Simplify the operation right away by crossing out common symbols in the numerals of different signs.

• Matching numerals / symbols are colored the same.
• MCMLIII + CCXIX - MCCXLVIII = CML + XIX - XLV;

When crossing out, do not mix subtractive and additive groups symbols.

• For example, the numerals XI and IV:
• XI is an additive group - largest symbol on the left, down to the smallest on the right - to calculate the value add up the symbols: XI = X + I = 10 + 1 = 11;
• IV is a subtractive group - a smaller symbol precedes a larger one - to calculate the value subtract the first symbol from the second: IV = V - I = 5 - 1 = 4.
• When subtracting the two numerals, XI - IV = (uncompact the subtractive) XI - IIII = X - III = VIIIII - III = VII (= 7), CORRECT. But if you cross out the common symbol regardless of the fact that it is part of a subtractive group: XI - IV = X - V = VV - V = V (= 5), WRONG.

3. Substitute for any subtractives in the roman numerals; that is, "uncompact" them using only the additive notation.

• Subtractive notation = a group of two numerals, one of a lower value preceding another larger one, the only allowed ones are: IV, IX, XL, XC, CD, CM, M(V), M(X), (X)(L), (X)(C), (C)(D), (C)(M) - to calculate the value subtract the first symbol from the second.
• Additive notation = a group of two or more numerals, equal or sorted in descending order from high to low, largest symbol on the left, down to the smallest on the right - to calculate the value add up the symbols.
• CML: CM = M - C = D CCCCC - C = D CCCC = DCCCC; CML = DCCCCL;
• XIX: IX = X - I = V IIIII - I = V IIII = VIIII; XIX = XVIIII;
• XLV: XL = L - X = XXXXX - X = XXXX; XLV = XXXXV;

4. Put the Roman numerals together, catenate them.

• Catenate all the positive numerals together.
• DCCCCL + XVIIII = DCCCCLXVIIII;

5. Simply "cross out" (remove) any symbols occurring in both values.

• DCCCCLXVIIII - XXXXV = DCCCCLIIII - XXX;

6. Using only the additive notation, convert larger value symbols to some smaller ones, a necessary step in order to cross out all the common symbols.

• Replacements used, in additive notation: L = XXXXX;
• DCCCCLIIII - XXX = DCCCCXXXXXIIII - XXX = DCCCCXXIIII;

7. Rewrite the symbols written in excessive additive notation by using the substractive notation.

• The numeral C should not repeat itself more than 3 times in a row, rewrite:
• DCCCCXXIIII = CMXXIIII;
• The numeral I should not repeat itself more than 3 times in a row, rewrite:
• CMXXIIII = CMXXIV

Final answer:

• XC + MCMLIII + CCXIX - MCCXLVIII - XC = CMXXIV;

• 90 + 1,953 + 219 - 1,248 - 90 = 924 the Romans did not have the Arabic numbers