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

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

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

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

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

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

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?

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

The answer is $\text{VIII}$ 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.

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

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

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.