# How to resolve the new definition of subtraction and division seen in college algebra?

Here's the foundational thing that irritates me the most when teaching college algebra.

Up through the secondary level, I think that instructors and students are trained to understand subtraction and division in terms of the inverse operation. Focusing on division here, if one asked "Why is $6/2 = 3$?", then one would most likely say it's because:

$$3 \times 2 = 6$$

But in every college-level algebra book I've seen, a different definition is given (and this goes for any texts in remedial elementary algebra, intermediate algebra, college algebra, etc.). Specifically, such books begin with the accepted "properties of real numbers", which are basically a restatement of the axioms for a field. In particular, one of the basic axioms is the existence of inverses: e.g., for multiplication, for any $a \ne 0$, there exists a value $1/a$ such that $a \times 1/a = 1$. (This is already problematic because students at this level are not yet familiar with statements involving existential quantifiers.) Thereafter, division is defined this way: $a/b$ means $a \times 1/b$. Of course, that's exactly what we see for a definition in most abstract algebra texts. But then technically this commits us to justifying "Why is $6/2 = 3$?" with something like the following chain of reasoning from the axioms:

$$6/2 = 6 \times 1/2 = (3 \times 2) \times 1/2 = 3 \times (2 \times 1/2) = 3 \times 1 = 3$$

Which I'm pretty sure no one actually ever does. Rather, they continue to use the secondary-school justification, even though this is technically out-of-synch (although, obviously, provably consistent with) our starting textbook axiom-properties.

Furthermore: When radicals are defined in the college algebra text, then the definition will once again look like the understanding of inverses from secondary-school subtraction and division (so it is additionally irritating to have these definitions and justifications out-of-synch with each other).

In summary: Advantages of the secondary-school definition: (1) it's what students are familiar with, (2) it provides shorter justifications, (3) it better lays the groundwork for the definition of radicals. Advantages of the standard college-algebra definition: (1) it complies with any standard textbook, and (2) it synchronizes with standard abstract algebra definitions.

So I go back-and-forth about this proud nail every semester. It seems like there would be more advantages to redefining subtraction and division as per the customary secondary-school rules, and thus smooth the way for student entry and understanding of the course; but the labor of going off-book and rewriting everything always deters me.

What is the best resolution to this problem?

• Does $6/3$ denote a rational number, or an expression involving a division "operation"? Which textbook does the course employ? – Bill Dubuque May 29 '18 at 15:22
• One further question: does your division apply to reals or only to integers? If the latter, are you essentially trying to show that $\Bbb R$ (or any field containing $\Bbb Z)$ also contains $\Bbb Q$ (up to ring isomorphism)? This follows immediately from the universal properties of fraction fields but the idea is so simple that its essence can be explained to a bright high-school student. If this is of interest let me know and I will elaborate in an answer. – Bill Dubuque May 30 '18 at 3:03
• @Number: As the question says, this is about the definition of the division (and subtraction) operation. This could be in the context of a half-dozen or more college-level algebra texts that I've seen. I don't think the other question is relevant. – Daniel R. Collins May 30 '18 at 18:51
• So the scope of your question includes division in $\Bbb R$ (or any field), not simply integer division? Are you essentially asking if there are pedagogical (dis)advantages of using alternative field (or group) axioms that replace inversion by division (or negation by subtraction in the additive case)? And the motivation for such is to clarify the relationship with fraction fields (or their additive analog = difference groups)? – Bill Dubuque May 30 '18 at 20:03
• Normal definition makes more sense to me too. But does it really even come up though? How much time are you spending on arithmetic? Can't you gloss over it and move on in the course? Maybe it just bugs you. – guest May 31 '18 at 7:07

But then technically this commits us to justifying "Why is $6/2 = 3$?" with something like the following chain of reasoning from the axioms:

$$6/2 = 6 \times 1/2 = (3 \times 2) \times 1/2 = 3 \times (2 \times 1/2) = 3 \times 1 = 3$$

Which I'm pretty sure no one actually ever does.

I am reminded of Principia Mathematica famously taking hundreds of pages to prove that $1+1=2$. I'm pretty sure no-one actually ever does that either. (Actually, I doubt that anyone at secondary-school level, much level college level, bothers to justify $6/2=3$ at all).

Nevertheless, since this troubles your conscience, the best resolution to an apparent mis-match between an old, familiar, system and a new system is surely to prove their equivalence and then to work in whichever system is most convenient for the task at hand. So if you prove the lemma $a / b = c \iff a = b \times c$ you can then apply your secondary-school justification without any twinges of conscience. And if you explicitly teach the principle of proving equivalence and then working in the more convenient system you're doing your students a big favour. I think it can be argued that that principle is as foundational to mathematical thinking as axiomatisation, and probably more so.

• For the students I have in these classes, it is definitely necessary to justify that $6/2 = 3$ numerous times each semester... if only as a warm-up to knowing how to check polynomial division, factoring, radicals, why division by zero is undefined, etc. Likewise, the principle of proving bidirectional equivalence would be beyond them. – Daniel R. Collins May 28 '18 at 18:53

I'm very much with you concerning the problem.

Striving for conceptual understanding and not merely exercising procedure following, I actually do bother with the question to justify why 6/2=3. But maybe one has to clarify more clearly, what 'justify' in school context actually means. To me, it doesn't mean to give a mathematically valid proof as in the Principia Mathematica, but merely an explanation that allows students to connect to other concepts and to model meaning; loosely speaking...

In the case you mention, I see the problem mainly in a change of concept: In middle school - students develop an understanding of inverse operation: They learn to see division as inverse operation of multiplication. In high school or later in college, students then should develop an understanding of inverse element: they learn that there is a neutral element with respect to an operation and the meaning of an inverse element is that its action on (i.e. operation with) the element itself results in the neutral element.

This is the way that I read your line of reasoning. The first equal sign reads as interpretation what 'division' means: multiplication with the inverse element.

In my experience, students often have a lot of trouble with this abstraction: To see as object what they used to know as an operation. Frankly I don't know what the best way is to approach this problem. That might much depend on the local context you encounter. But I do have some strategies as an approach. Said in advance, it anyway just takes time, and patience, and care...

I explicitly discuss different models for the objects and the operations. For example, taking the number line as model for the numbers (objects), what are models for the operations? addition and subtraction might be straight forward, but if you model muliplication as concatenation of "arrows", that lets you explain what $5 \times \frac{2}{3}$ is, but does not work for a calculation like $\frac{8}{7}\times \frac{2}{3}$. That works if you model multiplication as scaling (e.g. I use the following Geogebra-Applet as part of the many ways to explain why the product of two negative numbers yields a positive number as a result). You can model division in a similar way.

Using explicit models I'm usually pretty successful in talking about the distinction of inverse operation and inverse element.

I'm sure there are other ideas and better ways to address this problem with students, I would love to hear more about how what people in this community think about this question.

The problem stems from effectively defining the division operator twice. You should begin by defining division on $R$ as usual:

For all $x, y, z \in R$ where $y\neq 0$, we have $x/y=z$ iff $x=y\cdot z$

So, $6/2=3$ iff $6 = 2\times 3$ by the definition of division on $R$.

From the field axioms, we know that

For all $x\in R$ where $x\neq 0$, there exists $y\in R$ such that $x\cdot y=1$

Note that I do not make use of the division operator here.

Now, we can prove:

For all $a\in R$ and $a\neq 0$, we have $1/a\in R$ and $a\cdot (1/a)=1$

Suppose $a\in R$ and $a\neq 0$.

Then, from the field axioms, there must exist $b\in R$ such that $a\cdot b =1$

Applying the definition of division, we have $b=1/a$

Substituting, we have $a\cdot (1/a)=1$

We conclude as required that:

For all $a\in R$ and $a\neq 0$, we have $1/a\in R$ and $a\cdot (1/a)=1$

• So you think that it's profitable to be off-book definitions in that way? – Daniel R. Collins Jun 2 '18 at 15:36
• @DanielR.Collins What else can you do when the book definition leads to confusion? – Dan Christensen Jun 2 '18 at 15:41
• @DanielR.Collins See for example web.stanford.edu/~jchw/2015Math110Material/… – Dan Christensen Jun 2 '18 at 15:46
• @DanielR.Collins If you are stuck with the book definitions, it might help to use the $x^{-1}$ notation for the multiplicative inverse of $x$ and define $x/y=x\cdot y^{-1}$. Then derive the "definition" of division that I give above to establish a link to their high-school definition. – Dan Christensen Jun 2 '18 at 19:55