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.