# 1999 = MIM?

I have a basic understanding of Roman numerals, but why can’t MIM = 1999? To me, that looks like 1,000, plus 1 off of 1,000. I’m guessing the right way is MXMIX.

http://boards.straightdope.com/sdmb/showthread.php?threadid=44801 contains a link to Cecil’s answer.

I haven’t read the link, but I think I know the answer – the “subtract” method only works on the next-largest and the next-next largest. You can subtract I from V or X, but not L (50) or C (100). Similarly, you can subtract X from L or C, but not D (500) or M (1000).

CalMeachem - Not QUITE the right way to state it. You have to add that you can only subtract powers of ten, not the 5 times a power of ten symbols. So you can’t do VL = 45 either, even though L is the next-next largest after V. 45 has to be XLV. If you DO read Cecil’s link you find that people historically wrote all sorts of variants using the subtraction rule all sorts of ways, and a lot of Romans would probably have just written 45 as XXXXV, with no implied subtraction at all.

I’d express the rule for the usually recognized “canonical” form as:

If the current tens digit you are representing is a 9, represent it as the symbol for the next greatest power of 10, preceded by the current power of 10 (CM, XC, IX). If it is a 4, use the symbol for 5 times the current power of 10, prefixed by the current power of 10 (CD, XL, IV). Otherwise use the 5 times current power symbol if needed, and as many of the current power of ten symbols as required. Then proceed to the next tens digit of your number.

So, 1999 yields 1 - M, 9 - CM, 9 - XC, 9 - IX - MCMXCIX, the usual “accepted” form.

CalMeacham, with an a. Sorry.