# Divide and Conquer: Division

Of the four basic arithmetic operations (addition, subtraction, multiplication, and division), division is probably the hardest for students new to integer arithmetic. It’s a little weird. With addition, subtraction, and multiplication everything works out “cleanly” in the sense that we start with whole numbers and end with whole numbers. With division, we have the problem of the remainder and the problem of the inverse.

First the problem of the inverse. Consider this problem $$37 \div 8$$

Students are taught to know their multiplication facts [a point of heated math debate], but not their division facts. Division facts, in some sense, don’t make too much sense. For example, what’s $$37 \div 8$$? Well, that’s $$4 \mbox{ r } 5$$ and that doesn’t make sense to have that as something to memorize (or if some people object to the remainder notation, then $$4\frac{5}{8}$$, which is ugly in a different way, but whatever — six of one, half dozen of another and a different topic). Where some students get stuck is that they can’t recall “$$8$$ times what equals $$37$$”. That makes sense that the student would be stuck since there is no integer that satisfies this division. And hence multiplication and division don’t feel like inverse operations when working without restriction in the integers.

As a consequence of perceived “non-invertibility” and perhaps an unintended side-effect of the language of “pre-division” (ie division in which the remainder is zero), the notion of a remainder becomes a little harder to grasp. The language of “pre-division” is things like: “$$8$$ can’t go into $$37$$” or “you can’t divide $$37$$ by $$8$$” (slightly better is “$$8$$ doesn’t go into $$37$$ evenly”, though I don’t prefer “evenly” (even though I find myself saying it from time to time out of laziness)). By using this type of restrictive language, the (unintended) side-effect is that it wrongly conditions the student to believe that something is or is not possible. In a technical sense, it is true that “$$8$$ doesn’t go into $$37$$” provided that we qualify this with “in the integers” or “with zero remainder”. As an anecdote, a buddy of mine once told me that he resented being told as a kid that “you couldn’t divide thirty-seven by eight” only to find out years later that you could indeed divide thirty-seven by eight.

A final confusion about the remainder is a result of “non-invertibiility”. In the case of $$37 \div 8$$, if the student accepts that a remainder exists, then it is a matter of finding the greatest integer $$x$$ such that $$8 \times x \leq 37$$. For the student new to this, that means a little bit of fumbling with multiplication facts. The process goes something like this: $$8 \times 4 = 32$$ which is less than $$37$$, so let’s try $$5$$ which gives $$8 \times 5 = 40$$. Thus, the whole part is $$4$$ and the remainder? Oh wait, is it $$37 – 32$$ or $$40 – 37$$? Of course, the remainder is $$37 – 32$$, but when the idea of a remainder is confusing, then from where the difference should be consider is also confusing and vice versa.

For these reasons, there are a few different approaches a teacher can take to teaching integer division. First, let’s modify the traditional approach. Rather than starting explicitly with “zero remainder” type division problems, I would suggest starting explicitly with “non-zero remainder” type division problems — $$37 \div 8$$ for example. This immediately gets the student used to the fact that integer division has some extras.

Now, let’s consider $$37 \div 8$$. The long division approach typically taught would have the student reason like so:

Q: How many whole times does $$8$$ go into $$3$$?
A: Zero times.
Q: Then $$8 \times 0 = 0$$ and $$3 – 0 = 3$$ and then we “bring down” the $$7$$ to get $$37$$. So how many whole times does $$8$$ go into $$37$$
A: Four times.
Final: Then $$8 \times 4 = 32$$ and $$37 – 32 = 5$$ which is the remainder. Thus, the answer is four remainder five.

That’s a clean and straightforward algorithm, but one thing it misses is what the heck we’re doing. More specifically, the notion of place and value are hidden or simply forgotten in this discussion. I think it’s important to have a thought process rooted in place and value so that the student understands why we would consider $$3$$ divided by $$8$$ when the number is actually $$37$$.

There are tactile methods for doing this involving physical counters representing ones, tens, hundreds, thousands, etc. One of the pedagogical problems with using such methods is that the translation back to a pencil-and-paper method is sometimes sacrificed and thus reliance on these props becomes the only way to tackle such questions.

Though I don’t particularly like the language of “groups of” to discuss multiplication, in the context of division we can get away with it a bit more easily, especially when we consider place and value. A parenthetical: (not to distract hugely from this article: consider the infamous $$3 \times 5 = 15$$ problem where there is endless debate about what $$3 \times 5$$ means and how it is different from $$5 \times 3$$. My response is that any time we invoke the language of “group” we are invoking units. Thus, if we say that three times five is to mean “three groups of five” we’ve imposed units. “Three times five” never was a statement of “three groups of five” nor was it a statement of “five groups of three”. In fact, if we wanted to consider multiplication this way, we should have always written $$3 \mbox{ [groups] }\times 5 \mbox{ [units per group]} = 15 \mbox{ units}$$. Note, however, that this is no less different than the previous statement: $$5 \mbox{ [ units per group] }\times 3 \mbox{ [groups]} = 15 \mbox{ units}$$. This, then, is different from $$5 \mbox{ [groups] }\times 3 \mbox{ [units per group]} = 15 \mbox{ units}$$ since the physical grouping is different. But where was the physical grouping ever imposed on the original question of $$3 \times 5$$? That’s a question in the abstract, to which the answer is $$15$$ and to which $$3 \times 5 = 5 \times 3 = 15$$ — a statement of equality (and we can debate equality vs equivalence once we can resolve where the units are in the multiplication (and where are the units?)).).

With that said, where are the groups in division if they never were in multiplication? There aren’t any, but there is a place value and so we can have abstract place-value groups (but ye be warned: units matter and if we translate back to discussing about multiplication, it’s best to solidify units; otherwise consider talking about ‘scaling’ to discuss multiplication and division, which is still a different discussion). In any case, using groups isn’t inaccurate, it just may be incomplete.

Let’s continue to consider $$37 \div 8$$ and break a few things down a bit.

First, $$37 = 30 + 7$$. Next, let’s understand $$30 + 7$$ as $$3 \mbox{ tens } + 7 \mbox{ ones }$$. Now, when we consider $$37 \div 8$$ we can reason as follows:

Q: If I have $$3$$ tens, then how many whole groups of $$8$$ do I have?
A: I have $$0$$ whole groups.
Q: So what should I do with my $$3$$ sets of tens?
A: Break them into $$30$$ ones. And since I already have $$7$$ ones, I now have $$37$$ ones.
Q: So, how many whole groups of $$8$$ do I have from $$37$$ ones?
A: I have $$4$$ whole groups.
Q: How many do I have left over?
A: I have $$5$$ left over since $$8 \mbox{ [ones per group] } \times 4 \mbox{ [groups] } = 32 \mbox{ [ones]}$$. Thus, $$37 \mbox{ [ones] } – 32 \mbox{ [ones] } = 5 \mbox{ [ones (left over)]}$$

And hopefully, huzzah!

Let’s try this with a larger problem, and I will drop the Q & A format. Consider $$428 \div 5$$.

We have $$4$$ hundreds, $$2$$ tens, and $$8$$ ones. Since $$4 \mbox{ hundreds } \div 5 \mbox{ hundreds per group }$$ gives me $$0$$ whole groups, then the remainder is $$4 \mbox{ hundreds } = 40 \mbox{ tens}$$. Adding this to the already present $$2$$ tens, we have $$42$$ tens. Thus, $$42 \mbox{ [tens] } \div 5 \mbox{ [tens per group] }$$ gives $$8$$ groups (of tens) with a remainder of $$2$$ tens. The remaining $$2$$ tens now get broken down into ones to give a total of $$28$$ ones. And dividing by $$5$$ similarly, we have $$5$$ groups (of ones) and $$3$$ ones left over. So what do we have in the end? $$8 \mbox{ [groups of tens] } + 5 \mbox{ [groups of ones]}$$ and $$3$$ ones left over (remainder). Therefore, $$428 \div 5 = 85 \mbox{ remainder } 3$$.

Now, this is a really wordy write-up of the the standard division algorithm. But this is a way to explain what is going on! Try this long-winded approach with your students. See what happens. Do they have a clearer understanding of the process? If you are using (or have to use) tactile methods (physical counters, eg), try to get your students to translate from physical space to the abstract space as well as the written space.

As a final, aside: before introducing division, once students understand multiplication reasonably well, have them work through problems like $$xy + z$$. For example $$8\times 4 + 5 = 37$$. There’s no need to mention the connection to division just yet, but then rather than go from $$8\times 4 = 32$$ to $$32 \div 8$$ to $$37 \div 8$$, I think it’s worth trying $$8 \times 4$$ to $$8 \times 4 + 5$$ to $$37 \div 8$$ since zero remainder division problems are a subset of general integer division. Once students have been able to resolve that $$37 \div 8 = 4 \mbox{ r } 5$$, it may be enlightening to show them how it relates to $$8\times 4 + 5$$.

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