# Simple But Evil #5 — Roman Numeral Arithmetic

This edition of Simple But Evil, really is simple and evil. The prerequisite is that students should know at the least, multi-digit stacked addition and should have seen how roman numerals work. For example, they should be able to do this problem

$$\begin{array}{rr} &483\\ +&37\\ &\hline \end{array}$$

and they should be able to convert, say $$\mbox{XLVI}$$ to $$46$$

Here’s the challenge, suppose we never discovered our current place-value arithmetic, how would we do this addition?

$$\begin{array}{rr} &XLVI\\ +&LIX\\ &\hline \end{array}$$

One simple but evil rule is that we can’t convert this into $$46 + 59$$ and solve from here using using our base-10 arithmetic. All the work must be done using Roman numerals.

What’s the educational value of this?

Consider,
$$\begin{array}{rr} &XLVI\\ +&LIX\\ &\hline \end{array}$$

With near certainty, students will try to do this by inventing stacked arithmetic rules for the Roman numeral system. This won’t really work out and the eye-opener for us teaching is that most of our students have been following an algorithm.

In all the times I have given this problem, regardless of age group, only one student showed a method that was consistent, understandable, extendable, and rooted in mathematical thinking.

Of course, we can ask questions about subtraction. How would we do this subtraction without our place-value system?

$$\begin{array}{rr} &XI\\ -&III\\ &\hline \end{array}$$

The answer is $$\mbox{VIII}$$ defying the usual expectations of place-value arithmetic.

What I am looking for is reasoning within the Roman numeral system. This is difficult for students because they are conditioned to see the above problem as $$11-3$$ and with a little bit of memorization or finger counting, they’ll have that the answer is $$8$$. Now, in the Roman numeral system, what’s a good, consistent, understandable, and mathematical sound way of handling this?

There are probably many ways, but I like this one the best:
$$XI = V + V + I = V + IIIII + I$$
Thus,
$$XI – III = V + IIIIII – III = V + III = VIII$$

This is the basic counting that students do when they first learn numbers. It’s also similar to methods of regrouping — i.e., $$11-3 = (8 + 3) – 3 = 8$$. But in the Roman numeral system we can see the negation with the “one sticks”.

Here’s another example with commentary.

$$\begin{array}{rr} &LIX\\ -&XLVI\\ &\hline \end{array}$$

First, we can establish a convention that we want our numbers written in a “counting form”.
That is, we note that the biggest unit is $$L$$ and we put $$XL$$ in counting form as $$XXXX$$. Next, we put $$IX$$ in counting form $$VIIII$$. This gives that $$LIX = XXXXXVIIII$$ and $$XLVI = XXXXVI$$

And now, our Roman numeral subtraction is $$LIX – XLVI = XXXXXVIIII – XXXXVI = XIII$$ as expected. The nice thing (maybe?) is that this does become a “blind” algorithm for Roman numeral arithmetic — we are matching “like” numerals and performing the subtraction.

Going back to our original addition problem
$$\begin{array}{rr} &LIX\\ +&XLVI\\ &\hline \end{array}$$

We can do the same thing by utilizing Roman numeral facts: $$LIX + XLVI = LVIIII + XXXXVI = LXXXXVVIIIII = LXXXXXV = LLV = CV$$ and indeed $$59 + 46 = 105$$. Notice that this Roman numeral addition is similar to $$(50 + 9) + (40 + 6) = 90 + 15 = 90 + 10 + 5 = 100 + 5 = 105$$.

Let me know if you try this. I’ve found that I can have lots of conversations about arithmetic and can get students to see the standard arithmetic more clearly by stepping out into Roman numeral land.

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