In grade school when students are taught subtraction, they eventually get to ‘stacked’ arithmetic to utilize our place value system.

For example, \(483 – 37\) is written as

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

Then students are taught to ‘borrow’ (for subtraction) and ‘carry’ (for addition). Thus, in the above example, the subtraction would look like

$$\require{cancel}\begin{array}{rrrr}
&4&\cancel{8}^{7}&\cancel{3}^{13}\\
-&&3&7\\
&\hline\\
&4&4&6\\
\end{array}$$

Somewhere in all this is a discussion of place value and the meaning of ‘borrowing’. Specifically, in the example above, working out the place-value arithmetic right to left, we reason that since \(3 – 7\) would yield a negative value, we need to ‘borrow’ from the tens place. Since there are \(8\) tens, we can ‘break up’ one of those tens into ten, ones and add it to the three ones that we already have. Hence, the thirteen in the “ones” place and the seven in the tens place.

When we arrive at a problem like
$$\begin{array}{rr}
&4008\\
-&39\\
&\hline
\end{array}$$

things can get confusing. First, since \(8-9\) yields a negative value, we have to borrow from the tens place. But there are zero tens, hence we need to borrow from the hundreds place so we can borrow from the tens place. But there are zero hundreds. So we need to borrow from the thousands place so that we can borrow from the hundreds place so that we can borrow from the tens place to make up the deficit in the ones place. This leads to a bit of tangled mess.

$$\begin{array}{rrrrr}
&\overset{3}{\cancel{4}}&\overset{9}{\cancel{\overset{10}{\cancel{0}}}}&\overset{9}{\cancel{\overset{10}{\cancel{0}}}}&\overset{18}{\cancel{8}}\\
-&&&3&9\\
&\hline\\
&3&9&6&9\\
\end{array}$$

And, I have no idea if I did anything correctly.

Now, the general messiness of this method has been discussed plenty of times and I am most certainly not the first person to critique it.

The advantages, however, of performing the subtraction in the ‘traditional’ way are that work stays fairly compact in terms of the writing that’s needed, is systematic (double-edged sword in that once the system is known we can go through the motions blindly, but that’s a user issue not an issue with the method), and efficient in terms of total computation required (leveraging the place value system keeps the numbers ‘tame’).

I actually don’t have any problems with teaching and performing the subtraction in the way described above. However, given the messiness that can ensue and the potential of loss of meaning in the method, not all students seem to get a handle on this. An alternative, “modern” approach (sometimes described as the Common Core approach for (positive / negative) propaganda purposes) would look something like this.

$$\begin{align}
4008-39 & = (4000 – 30) + (8 – 9)\\
& = 3970 – 1\\
& = 3969
\end{align}$$

This approach leverages the ‘simplicity’ of working with multiples of tens and single digit subtraction. Students do have to know how to work with negative numbers and (re-)grouping. But that’s not a major hindrance. The advantages of subtracting this way is that there is an emphasis on ‘value’ at the expense of ‘place’. Whereas the traditional stacked arithmetic approach puts more emphasis on ‘place’ than on ‘value’. Regardless, I have no qualms with teaching students to subtract with (re-)grouping.

A Third Approach

When working with adult students who have been out of formal pencil-and-paper arithmetic for decades, I’ve found that the following third approach works reasonably well, even if it looks a bit cryptic. And it only looks cryptic because when I explain this method to adults there is a lot of talking and leading the student through the method. When reading mathematics, it is often best to re-read it since there is no dynamic dialogue available in a text-only medium.

Here’s how it works with explanations of each step following:
$$\begin{array}{rrrrrl}
&4&0&0&8&\\
-&&&3&9&\\
&\hline\\
&4&0&-3&-1&\mbox{[a]}\\
&4&0&-4&9&\mbox{[b]}\\
&4&-1&6&9&\mbox{[c]}\\
&3&9&6&9&\mbox{[d]}\\
\end{array}$$

• [a] place by place subtraction
• [b] add ten to the ones place by subtracting one from the tens place
• [c] add ten to the tens place by subtracting one from the hundreds place
• [d] add ten to the hundreds place by subtracting one from the thousands place

For whatever reason, many adults I’ve shown this to see this as fairly logical and express amazement that they didn’t learn it this way from the get go. It’s not anything revolutionary, but I like it mostly because I feel that it puts equal emphasis on place and value.

In [a] we can see that we have four thousands, zero hundreds, a short fall of three tens, and a shortfall of one one. This is equivalent to \(4000 – 30 – 1\). Now, resolving values in the various places is fairly straightforward. We just borrow from the next place on the left even if there is already a shortfall. All we’re doing is just accruing the shortfall.

Hence, in [b] we subtract one ten and add ten ones giving \(4, 0, -4, 9\) and since we had a shortfall of three tens to start with, we now have a shortfall of four tens. This sequence is equivalent to \(4000 – 40 + 9\)

We proceed similarly in [c] by subtracting one hundred and adding ten tens to obtain \(4,-1,6,9\) which is equivalent to \(4000 – 100 + 60 + 9\). And in [d] we resolve the deficit in the hundreds place by adding ten hundreds and subtracting one thousand giving \(3, 9, 6, 9\) which is \(3000 + 900 + 60 + 9 = 3969\) as desired.

What about negative results?

What if our problem, instead, were
$$\begin{array}{rrrrrl}
&&&3&9&\\
-&4&0&0&8&\\
&\hline\\
\end{array}$$

Clearly, we can work out the arithmetic in all of the approaches as \(-(4000 – 39)\). But if we just wanted to play around with this third approach, the arithmetic would look like this

$$\begin{array}{rrrrrl}
&&&3&9&\\
-&4&0&0&8&\\
&\hline\\
&-4&0&3&1&\\
\end{array}$$

And this is as far as we can go. The “answer” is \(-4000 + 30 + 1\), which is, as expected \(-3969\). We can’t proceed any further since we can’t resolve the \(-4000\). (If we wanted to go nuts, we could borrow from the ten thousands place to get \(-1, 6, 0, 3, 1\) which would be \(-10000 + 6000 + 30 + 1\). If we wanted to establish convention, we could disallow this.)

Here’s another example.

$$\begin{array}{rrrrrl}
&1&5&3&9&\\
-&4&6&2&8&\\
&\hline\\
&-3&-1&1&1&\mbox{[a]}\\
&-4&9&1&1&\mbox{[b]}
\end{array}$$

Here [b] gives \(-4000 + 900 + 10 + 1\) which is \(-3089\). Alternatively, notice that

$$\begin{array}{rrrrrl}
&4&6&2&8&\\
-&1&5&3&9&\\
&\hline\\
&3&1&-1&-1&\\
&3&1&-2&9&\\
&3&0&8&9&
\end{array}$$

Thus, taking the negative, gives the desired result of \(-3089\) to the original subtraction.

Thoughts

This approach is just a combination of the stacked arithmetic that’s traditionally taught and the re-grouping that’s now offered as an alternative approach, except that this approach writes out the steps more explicitly. It also requires that the student has a grasp of negative values. As I said, my adult students tend to be receptive to this after a few runthroughs. If you have a student (adult or not) struggling with place value subtraction, you can try this as another approach. If you do, I highly recommend a lot of speaking to explain every single step and the rationale attached to each step. Students should be doing this regardless of the approach, but for this method that I give it is even more important since there are not standard reference points (or other people to go to).

Props can help as well so that students can associate what a shortfall of ten but a surplus of one would mean.

If you want to make this consistent with addition, then look at how it would work with this example:
$$\begin{array}{rrrrrl}
&4&6&2&8&\\
+&1&5&3&9&\\
&\hline\\
&5&11&5&17&\mbox{[a]}\\
&5&11&6&7&\mbox{[b]}\\
&6&1&6&7&\mbox{[c]}
\end{array}$$

Notice that in [a] all I did was direct place value addition without worrying about borrowing. Then in [b] we have in the last two places \(50 + 17\) which is “better” written as \(60 + 7\) and hence the “carry”. Then in [c] we have \(5000 + 1100\) which is \(6000 + 100\) giving the modification to the first two columns.

Again, this is no different than what students currently do in either of the two standard approaches, except here in this third approach we just write out the steps more explicitly bringing to attention the place and value at each step.

Let me know your thoughts.