*Can a factor be a fraction?*

People sometimes agonise over whether a fraction such as $\frac{2}{3}$ can be called ‘a factor’ of another number, such as 6 (Foster, 2022). If factors are defined as “numbers that divide exactly into another number” (BBC Bitesize, n.d.), then, since $6÷\frac{2}{3}=9$, and $9$ is an integer, shouldn’t $\frac{2}{3}$ be regarded as a factor of $6$?

Perhaps we decide that we want to be able to say that the number $6$ has exactly $4$ factors ($1, 2, 3$ and $6$), and so we don’t allow numbers like $–2$ or $\frac{2}{3}$ to be factors. If so, we could use a tighter definition of factor, such as: *A factor is a positive integer that divides into another number a positive integer number of times* (or, equivalently, we could say ‘without any remainder’). However, in many instances we might want to think of factors more broadly than this. For example, when factorising $x^2±5x+6$, it might be helpful to think of the $6$ as having four possible *factor pairs* ($\{1, 6\}, \{2, 3\}, \{–1, –6\}$ and $\{–2, –3\}$). Similarly, when using the *factor theorem*, we typically treat negative numbers as ‘factors’. In other contexts, it might seem natural to regard a number like $\frac{2}{3}$ as being a factor of $\frac{4}{3}$, since it ‘goes into it’ twice, or even $x$ as a factor of $x^2$, since it goes into it $x$ times, regardless of whether or not $x$ might be an integer. We might even pull out an irrational number, such as $\pi$, from an expression like $2\pi r-\pi l$ to give $\pi(2r-l)$, and call this 'factorising', and refer to $\pi$ here as a ‘common factor’, although it is certainly not an integer, and is not even rational (Foster, Francome, Hewitt, & Shore, 2022).

‘Factor’ seems to be one of those words that is used differently in different contexts, even within school mathematics, and I think it isn’t really possible to settle on a fixed definition which will always apply (see Foster, Francome, Hewitt, & Shore, 2022, for a similar discussion about the word ‘fraction’). Perhaps the best approach to awkward issues like these is to acknowledge them and explore them. Turn the issue into a task: *What would happen if we allowed fractions to be factors?* Perhaps we call them ‘fraction factors’.

#### Exploring 'fraction factors'

Students often think of fractions as ‘numbers less than 1’, and they may initially think that *any* fraction would be a fraction factor of *any* integer, but of course this isn’t right. Although $\frac{2}{3}$ would be a fraction factor of $6$, it *wouldn’t* be a fraction factor of $5$, since $5÷\frac{2}{3}=\frac{15}{2}$, or $7.5$, which is not an integer. All *unit* fractions ($1/n$, where $n$ is an integer $\neq 0$) would be fraction factors of *every* integer, since they are by definition fraction factors of $1$, and $1$ is a factor of every integer. But when would a *non-unit* fraction be a fraction factor of an integer? Could we ask for *all* the fraction factors of $6$? Clearly not, because this list would include all of the unit fractions, and there are infinitely many of them.

There are many opportunities here for students to form conjectures and to find counterexamples – and, in each case, to try to find the *simplest* counterexample they can. It can also be helpful to look at the question the other way round, and ask what integers a fraction like $\frac{5}{12}$, say, would be a fraction factor of.

The conclusion is quite simple, but perhaps not that easy for students to arrive at without quite a bit of useful exploration. A fraction $\frac{p}{q}$, with $p,q \neq 0$, in its lowest terms, will divide an integer $m$ if and only if $m÷\frac{p}{q}$ is an integer. This is equivalent to saying that $\frac{mq}{p}$ must be an integer. Since $p$ and $q$ are co-prime, $\frac{mq}{p}$ will be an integer if and only if $p$ is a factor of $m$. So, the fraction factors of $6$ are fractions that, when simplified, have the (positive integer) factors of $6$ as their numerators:

$$\frac{1}{n}, \frac{2}{n}, \frac{3}{n}, \frac{6}{n}$$ for all integer $n \neq 0$.

Perhaps this seems obvious, but I think it can be quite unintuitive that, say, $\frac{2}{17}$ is a fraction factor of $6$, but $\frac{4}{3}$ isn’t.

A less algebraic – and perhaps clearer – way to appreciate why the numerator matters, but the denominator doesn’t, is to realise that $\frac{1}{q}$, as a unit fraction, will always be a fraction factor of *any* integer, regardless of what $q$ is, because $q$ of them will always fit into every $1$. For the same reason, $\frac{p}{q}$ will necessarily be a fraction factor of $p$, because $q$ of them will fit exactly into the integer $p$. So, $\frac{p}{q}$ will be a fraction factor of any integer of which $p$ is a factor.

### Things to reflect on

1. Do you agree that it isn't possible to have a single definition of 'factor' that applies across all of school mathematics? Why / why not?

2. If you agree, which other technical mathematical terms do you think may be problematic in this kind of fashion? (See Foster et al., (2022) for a discussion of 'fraction' in this regard.)

### References

BBC Bitesize (n.d.). What are factors? https://www.bbc.co.uk/bitesize/topics/zfq7hyc/articles/zp6wfcw

Foster, C. (2022, October 13). How open should a question be? [Blog post]. https://blog.foster77.co.uk/2022/10/how-open-should-question-be.html

Foster, C., Francome, T., Hewitt, D., & Shore, C. (2022). What is a fraction? *Mathematics in School, 51*(5), 25–27. https://www.foster77.co.uk/Foster%20et%20al.,%20Mathematics%20in%20School,%20What%20is%20a%20fraction.pdf

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