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Addition and Multiplication with Roman Numerals

Date: 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.


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


So, put all of the letter together, and we have DCCXI, or in our terms,

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

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 


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:


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.


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 =


We're still o.k. because 90 x 48 = 4320

Now, all we need to do is add MMMMCCCXX and CXLIV


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
Associated Topics:
Middle School Puzzles

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