Addition and Multiplication with Roman NumeralsDate: Sun, 27 Aug 1995 08:04:26 From: "Fred Holtby" Subject: roman numerals Dear Dr. Math, I have a question about roman numerals. How did those old Romans use them? Let's start with addition. How would they add what we call 436 and 27 and 248? I've been asking math teachers about this for decades now, to no avail. Please help. Best, Fred Date: 8/28/95 at 15:21:42 From: Doctor Heather Subject: Re: roman numerals Hi Fred, The 'old Romans' used letters to symbolize the concept of quantity just as we use numbers. I played around with these things a lot as a child, so maybe that's why it doesn't seem so strange to me. So, let's do the addition of 436, 27, and 248. First , let's switch over to Roman numerals. Just as a reminder, I'll list the Roman numerals. M-1000 D-500 C-100 L-50 X-10 V-5 I-1 O.k., so we have CDXXXVI + XXVII + CCXLVIII I see where a problem may arise: The convention of saying CD = 400, or XL = 40. But if we think of it like those expanded notation problems, you know, instead of 248 + 35, it is 100 + 100 + 10 + 10 + 10 + 10 + 8 + 10 + 10 + 10 + 5 it will be easier to understand. So, let's start with adding the hundreds. CD + CC = DC Oh! I just see the easy way to handle the CD = 400 notation. Think of the C before the D as a -C (negative C). That's what it means anyway (i.e., 500 - 100) Now add the tens: XXX + XX + XL = LXXXX, or XC Now add the ones VI + VII + VIII = XXI So, put all of the letter together, and we have DCCXI, or in our terms, 711. Now we'll do multiplication, though againI'm not guaranteeing my methods are going to be exactly those of the Greeks. It'll be just an idea of how they did it. O.k., let's try XLVIII x XCIII (or, in our terms, 48 x 93) We're going to use basically the same methods we use to multiply today, but we have to remember to carefully transfer our thinking to symbolic numbers. Now, when we multiply, what is the first thing we do? We take the number in the ones place in the second number and multiply it by the whole first number. In this case, the number in the ones place in the second number is III. So, our first problem is XLVIII x III Let's do this a really simplistic way. We are going to just add together three of each letter that appears in the first number: -XXX + LLL + VVV + III + III + III = -XXX + CL + XV + IX = CXX + XXIV = CXLIV Now, from the second expression to the third, I subtracted the -XXX from CL to get CXX, and I subtracted the I in IX from the V in XV to get XXIV. We can see that so far we are doing well 48 x 3 = 144 = CXLIV Now, we need to do XLVIII x XC, because we remember that when we do multiplication, that is what we really do, multiply by 90 instead of 9. XLVIII x XC This is a little tougher to just write out 90 of every letter that appears in the number. We have two choices: we can multiply the number by 90 now, or multiply it by 9 and then multiply it by 10 at the end. Let's just multiply it by 90 from the beginning, and we'll see how it goes. -X x XC + L x XC + V x XC + I x XC + I x XC + I x XC = -CM + MMMMD + CDL + XC + XC + XC = -(-C) + -M + MMMMD + -C + DL + -XXX + CCC = C + MMMD + DXX + CC = MMMDDCCCXX = MMMMCCCXX We're still o.k. because 90 x 48 = 4320 Now, all we need to do is add MMMMCCCXX and CXLIV MMMMCCCXX + CXLIV = MMMMCCCCLXIV = MMMMCDLXIV And we got the right answer! 48 x 93 = 4464 = MMMMCDLXIV I'm sure they skipped a lot of steps, rather than how I wrote out every gory detail, but at least this gives you an idea. We can multiply or add with anything, really. Think about having different number systems, like ! = 1, @ = 5, * = 10, etc. And maybe there are number systems that AREN't based on 10. You can add and multiply in those systems, too. It's just a matter of training your mind. Dr. Heather, The Geometry Forum |
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