Addition and Subtraction: Complement and Willetts's method

For early addition and subtraction, consider starting with ten's complement methods. United States Common Core K.OA.4 requires knowing the complements {(1,9),(2,8),(3,7),(4,6),(5,5)} and 1.OA.6 requires fluency for addition and subtraction within 10. Classrooms also tend to flashcard addition and subtraction up to 20 for K.OA.4 because of regrouping methods (borrowing, Austrian); but regrouping methods are extremely broken, borrowing in particular is known to significantly increase error rate and decrease comprehension of subtraction.

I provide both algorithms here because the addition and subtraction algorithm share a common form. This unifies both and reduces the number of different rules a student needs to remember: the subtraction rule is basically the addition rule and it even uses the exact same carrying logic.

For addition and subtraction I generally recommend strong foundations first and then the algorithms. Students need to understand math, not simply blindly apply a mechanical rule. The explanations for commutativity, the carry mechanism, "taking back," equal subtractions, and equal additions all establish properties of addition and subtraction and the underlying mechanisms of these rules, rather than making them mystic spells to chant at the paper in hopes you conjure addition.

Always use long addition and subtraction: line up column by column, not horizontally.

Note that while this is first-grade addition and subtraction, these algorithms start at 2 digit numbers but generalize well enough that taking them out to 4-6 digits is relatively trivial for first graders. They're designed so that you don't get lost bookkeeping, which is a real problem for seven year olds; recovery from working memory failure is an important part of learning addition.

Addition: Completing the Ten

For addition, we first need to understand the commutative property of addition (Common Core 1.OA.3) and the carry mechanism. Both of these are demonstrated visually.

For the commutativity property of addition, use a visual demonstration with objects. For example, 3 objects on the left, 5 objects on the right, together we can count 8 objects. Then move the same objects around to put the 5 objects on the left, and the 3 on the right—well these are the same objects, so we also have 8 objects. So addition is commutative: 3+5 and 5+3 are the same thing. Note that this doesn't work with subtraction, if you have 5 and you take away 3 you have 2 left, but if you have 3 you don't have enough to take away 5 (this is the difference between "have" and "owe").

For the carry mechanism, consider adding two numbers. If we take one of those numbers away from 10, then if the other number is as large as or larger than the result, we will get a number greater than 10. This would mean our result would add 1 to the column to the left, which is carrying.

The Addition Algorithm

Once we have established the above justifications, we have the following algorithm, starting right to left, per digit.

First, by commutativity, begin with the larger digit. Take the tens complement of that digit (from our 5 pairs above). If the smaller digit is less than this, add the larger and smaller digit and use the result as the sum.

If the smaller digit is not less than the complement, then we must do something else. That something else is very mechanical, and therefor we are going to not explain what to do but rather first explain why to do it.

First, consider that if we add the complement to the larger number, it will give us 10. For example if we are using 5+8, then adding 8+2 gives us 10, which leaves us a very easy problem: 10+5 is just 0+5 with a 1 carried to the left. But wait, we added 2 already to get this 10, so we must take back the 2 before we can continue. To do this, we subtract 2 from the 5. That leaves us with 5-2=3, therefor our result is 3 carry 1.

The mechanical explanation, once we've gone through and explained that we are taking back the complement, is that if the two numbers add up to greater than ten—if the smaller number is at least as big as the complement of the larger number—we subtract the complement of the larger number from the smaller number and carry.

Notation and Cascades

It's worth noting that whenever you are adding two numbers, you will never have a carry above 1: 9+9=18, and if you had a carry in you will get 19, which is still 9 carry 1. You will never carry a 2 when adding two numbers.

As such, when you carry, you only need to cross out the digit in the next column left of the top row. Because it's crossed out, you know you carried a 1.

In the case of a 9, this carry will cause a carry itself, so continue crossing out 9s until you reach something that isn't a 9, then cross that out too. That's a 9 cascade (it's a carry-skip, same as in computer architecture).

The Subtraction Algorithm

The subtraction algorithm is the inverse of addition and, interestingly, only requires single-digit addition and subtraction. We need an equal subtraction and an equal addition demonstration to show the important properties of the difference. This should be the first subtraction algorithm taught (borrowing mangles place value by suggesting you can have a 12 in the ones place). This rule is Willetts's rule.

Again emphasize that subtraction is not commutative.

Equal Subtraction

The first demonstration is equal subtraction. Set up 12 objects on the left and 8 on the right, in rows of 4. This gives 12-8=?.

Now move all 8 objects on the right down, and 8 of the objects on the left down. This will create an equal setup at the bottom: each side is 2 rows of 4 objects, and the right side has none left. We can count the remaining objects, giving 4, and rewrite 4-0=4.

This is the equal subtraction property: If we subtract the same amount from the minuend and subtrahend, the difference does not change, the gap is still just as wide.

Equal Addition

For the second demonstration, again set up our 12-8 problem. Then, add two more objects to each side—let's say we start with 12 and 8 orange objects, and we add 2 blue or turquoise or cyan objects to each side. This gives 14-10=?.

First, move down the two new objects. We now have two equal sets of objects moved away, and the result looks like our 12-8 problem and the action looks like our equal subtraction, which we know doesn't affect the difference.

Then, move the remaining 8 down, leaving us again with 4-0=4.

This is the equal additions property: if we add the same amount to the minuend and subtrahend, the difference does not change, the gap is still just as wide.

Full Justification

The full justification involves converting the problem to a minus-zero problem.

For each column, take the digit of the minuend (top) and subtrahend (bottom). If the minuend's digit is greater than or equal to the subtrahend's, then subtract the subtrahend's digit from both the minuend and subtrahend, leaving the subtrahend's digit as zero.

If the minuend's digit is not greater than or equal to the subtrahend's, then take the complement of the subtrahend's digit. Add that complement to the subtrahend—causing a carry—and to the minuend.

Neither of these actions will ever cause a carry or borrow or anything else in the minuend.

Let's say we have 235 - 163.

First, we see 5>3, so begin with 5-3=2. Now we have 232-160.

Next, we see 3<6, so by (6,4) we will add 4 to the tens place. We have 272-200, we had 160+40 so we had to carry from the tens place.

We now have 2=2, so we will now subtract 2 from the hundreds place, 72-0=72.

Mechanical Explanation

With all of that, the simple mechanical explanation is as follows:

If the minuend's digit is greater than or equal to the subtrahend's, then the digit for the difference is the minuend's digit minus the subtrahend's.

Otherwise, carry in the subtrahend (all the same implications and notation as addition) and use the difference as the complement of the subtrahend's digit plus the minuend's digit.

Moser's Method of Addition

Moser's method of addition is a slight addition optimization to mess with carrying.

First, drill 5-10 examples of subtraction failing commutativity digit by digit. For example, show 235-162 is not the same thing as 265-132. Emphasize this. Strongly.

Next, recall that addition is commutative, and that we commute each digit in the addition algorithm anyway to operate based on the larger one because it's easier. Well, carry-skip over nines cascades is also easier, therefor, 23965 + 19498 = 29998+13465. 8+5 gives us a carry and a complement of 2, 5-2=3 so that's the digit for our sum, and we crash directly into a 9 cascade so do the carry-skip and we're now working with 30,000+13463, which quickly gets us 43463.

Moser's method also sets up for things like rounding (29998 is close to 30,000, add 2 and subtract 2). It's a simple swap so it implicitly adds and subtracts the difference, rather than for example converting all the bottom digits to zero or, when not possible, reducing them by enough to make the top digit 9, which actually requires more work.

On Carry Look-Ahead Subtraction

A carry look-ahead approach is also viable, although it should not be the first taught. CLA takes advantage of Willetts's rule allowing for very fast carry evaluation, thus very fast checking. This can be done with a modus notation, but I recommend just doing the carry phase and working from the crutches in minus notation instead of recopying into modus notation.

The carry phase

Willetts's rule produces carry in the subtrahend. This allows us to do the following mechanically:

If the current column's subtrahend digit, including carry (if it's crossed out, then consider its value +1, with a crossed out 9 being 0), is greater than the column's minuend digit, emit a carry.
If carrying into a 9, cascade all the way to the first non-9 digit, being sure to cross out the first non-nine digit as well as all the 9s

So let's consider a really big subtraction, 533241536781 - 325243499942.

2>1, so carry. Cross out the 4.
(4+1)<8, no carry.
9>7, carry. Cross out the 9 (the next one to the left, not this 9)
Carry-skip: cross out all the 9s and the 4. Skip past all the crossed out 9s.
(4+1)=5, no carry.
3>1, carry. Cross out the 4.
(4+1)>4, carry. Cross out the 2.
(2+1)>2, carry. Cross out the 5.
(5+1)>3, carry. Cross out the 2.
(2+1)=3, no carry.
3<5, no carry.

We now have all of our carries taken care of. The student can now return to the subtraction problem digit by digit, without worrying about the carries.

CLA without Modus

Modas notation, described next, might provide confusion. CLA without the modas notation still uses the first and third pass, but skips the second: the student will have to reason that the subtrahend's digit +1 is the digit to work with if it's struck, which is slightly more cognitive load but it's also harder to get lost.

During the regular Willetts's method, the student must consider on each column:

Is the subtraheand's digit a struck nine?
If so, skip to the first column that's not a struck nine.
Is this digit receiving a carry? (struck)
If so, is the digit plus one (with rollover to zero) bigger than the digit it's being subtracted from?
If not, do the subtraction while remembering to address the carry, not the number written on the page.
If it is bigger, strike the next column's subtrahend digit and use the number plus its carry to find the correct tens complement (skip through nines to save time).
Use that tens complement to perform addition on the minuend.

When using CLA without modus, the first phase is:

Is this digit plus carry (is it struck?) bigger than the minuend's?
If so, strike the next column of the subtrahend (skip through nines to save time, and skip all of those carry checks because you no longer need to read that column at all).

The second phase then becomes:

Is the subtraheand's digit a struck nine?
If so, skip to the first column that's not a struck nine.
Is the next digit left in the subtrahend struck?
If so, find the tens complement of the subtrahend plus carry (+1 if struck, +0 if not struck) (discard knowledge of the subtrahend), then add that to the minuend (discard knowledge of the minuend) to find the difference.
If not, subtract the subtrahend's digit plus carry from the minuend to find the difference.

This at least skips the "Am I struck? Then do the comparison to decide whether to even use the complement using subtrahend+1" decision in favor of "is the next column struck? If so, I am using the complement; is the subtrahend struck and thus +1?" The one-digit comparison is still easier, so this isn't theoretically as easy as using modus notation, but it also doesn't risk overwhelming a seven year old with two types of subtraction.

Modus versus Minus

We could write this as 533241536781∸336353500952 (modus, i.e. mod-10 minus; yes this is also the monas operator, but it's unlikely anyone will ever encounter that outside specialized grad-school math) but for now that notation would be confusing. Don't do this. However, for illustrative purposes, I'll compare the carry look-ahead in full with the regular approach.

With CLA using modas notation, there are three phases. First,

Is this digit plus carry (is it struck?) bigger than the minuend's?
If so, strike the next column of the subtrahend (skip through nines to save time, and skip all of those carry checks because you no longer need to read that column at all).

Second, when recopying the problem, bring down the minuend, then copy the subtrahend as follows:

Is this digit struck?
If so, copy it down as itself +1 with rollover (9 becomes 0).
If not, just copy it down.

Finally,

Is the subtrahend's digit bigger than the minuend's?
If so, find the tens complement of the subtrahend (discard knowledge of the subtrahend), then add that to the minuend (discard knowledge of the minuend) to find the difference.
If not, subtract the subtrahend's digit from the minuend to find the difference.

Note in this last step we're doing the modus operation: address the minuend by adding or subtracting to get the difference, but don't operate on the subtrahend (no carries). That's the minus with a dot above it, only act on the upper row.

It's very likely best to stick with minus notation and just do the crutches before the subtraction pass.