Why do we need fractions such as $\dfrac{3}{-5}$? I need a convincing answer suitable for 8th grade students. Here's what I've already thought of (which don't fully satisfy me!):

  • Solving the equation $-5x+3=6$, we get to $x=\dfrac{3}{-5}$.
  • In the formula relating distances of image and object from the mirror and the its focal length, sometimes we need to put negative numbers for $f$ or $q$: $$\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{1}{f}$$

do you have any other suggestion?

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    $\begingroup$ Since 3/-5 is -3/5, I don't see how we really need them at all. It is ultimately a matter of convention if you want to allow or forbid fractions to be written with a negative sign in the denominator. Having said that, there is no good reason to forbid it and is clearly more flexible to allow it. $\endgroup$ – John Coleman Sep 2 '16 at 18:56
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    $\begingroup$ @JohnColeman: with the same argument, one never really need $3/6$ nor $\sqrt{4}$. Rejecting non-normalized expressions is deeply flawed: we have to work with them, and teach to simplify (or normalize) them. The first given argument in the question seems very compelling to me. $\endgroup$ – Benoît Kloeckner Sep 5 '16 at 19:34
  • $\begingroup$ @BenoîtKloeckner Of course you are correct -- but I was addressing the part of the question when OP asked why we needed expressions with a negative denominator. Of course we don't need them (or things like 3/6 for that matter), but notation should be geared to convenience rather than strict necessity. $\endgroup$ – John Coleman Sep 5 '16 at 23:44
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    $\begingroup$ @JohnColeman: well, if we are lead to write these expressions at some point, then we do need them at that point. I don't quite get how it could be otherwise, even if we decide that the ultimate result of a computation should avoid them. $\endgroup$ – Benoît Kloeckner Sep 6 '16 at 15:19

I would say you're doing your student a disservice if you were to seriously disallow a negative denominator. A fraction is simply a ratio of two integers (where the denominator is not allowed to be zero). I disagree with @yoniLavi that we never need such fractions. Since division by negative numbers makes sense, such a fraction with a negative denominator also does. (Though without any further context, I would generally say it's bad form to leave the negative in the denominator; however, in certain real world applications, it might be more desirable to leave it there in the denominator.)

Such fractions come up all the time in algebra and calculus. For example, in the definition of slope. We just say given $(x_1,y_1)$ and $(x_2,y_2)$, then

$$ m= \frac{y_2 - y_1}{x_2 - x_1} $$

we don't define it as

$$ m= \frac{y_2 - y_1}{x_2 - x_1} \text{ }\text{ if $x_2-x_1 > 0$} $$ and $$ m= \frac{y_1 - y_2}{x_1 - x_2} \text{ }\text{ if $x_2-x_1 < 0$} $$

Moreover, in application problems, a negative number can easily arise in the denominator.

As for how to convince an 8th grader?

Well, I would start by considering the student's background, interests, and (mathematical) proclivities. If the student has a high interest in the mathematical motivation for allowing a negative denominator, I would start with what I said in the first paragraph above.

Regardless of whether the student had a strong interest in mathematics, I would also give a concrete example that the student could relate to. I would remind the student that fractions can be used not only to express ratios as in the comparison on lengths (like the relative lengths of the sides of a triangle--they'll see this as soon as they're in high school geometry), but that fractions can also be used to measure rates of change: how one quantity varies with respect to the variation of another quantity. This is where knowing something about the student's interests and background are relevant. Come up with something the student is interested in where you're comparing the changes of two quantities and where the second quantity can decrease (ie., it's negative). I would think that that should be enough.

Having worked with the tail-end of this age group quite extensively the past 8 years, I would be sure to feed their curiosity, congratulate them on the interesting question, and perhaps encourage them to come up with a real life application which makes sense to them where a negative denominator makes sense.

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    $\begingroup$ The example of the slope calculation perfectly illustrates why fractions with a negative denominator are needed: they naturally arise in the middle of calculations, even if at the end you can rewrite the result so denominators are not negative. And actually you can't always be sure about the denominator's sign. In algebra we may want to consider the function $1/(x^2-1)$, or $1/(x^2-a^2)$ for all $x$, and sometimes the denominator is negative. Or in trigonometry consider $\tan x = (\sin x)/(\cos x)$. For many $x$-intervals the denominator is negative. How awkward to try to avoid that! $\endgroup$ – KCd Sep 4 '16 at 16:39

There are numbers that one (or many) may never use, is this a reason to eliminate them. Is there a reason to have 1 to any non-negative integer power? When responding to questions of this type "Why do we need fractions such as...? " It is not a question of popularity of use but is a question of "Does this numeral name a well define number"? Then depending upon the the response you may want to elaborate on expected or agreed form. Do we need a VII or are we forced to use '7' or like the Maya :| Give the student the definition of a fraction and ask "does this example follow the definition.

  • $\begingroup$ The point is, if we never ever encounter a weird fraction with negative denominator, why should we even think about them? $\endgroup$ – Behzad Sep 6 '16 at 4:18
  • $\begingroup$ One reason is because the student asked about it. If the student had asked why do we need 18/12 we never use it. You would respond, yes we do. It has meaning but is most often seen as 3/2 or 1.5. The same type of response should be given to the question re. 3/-5. Sometimes we write it as -3/5 It also asks the question, "What is 3 divided by -5? $\endgroup$ – Sid Hollander Sep 7 '16 at 15:24
  • $\begingroup$ As a matter of fact, there's a straightforward answer to the question " why do we need $\dfrac{18}{12}$?": we need it to be able to calculate $\dfrac{3}{2}+\dfrac{1}{12}$. $\endgroup$ – Behzad Sep 7 '16 at 15:38
  • $\begingroup$ Some might say 3/2 + 1/12 = 1.5 + 0.08.3333333333333. $\endgroup$ – Sid Hollander Sep 8 '16 at 20:42
  • $\begingroup$ @Behzad There's nothing to calculate with $\frac{3}{2}+\frac{1}{12}$; it's already a well-defined and exact number. Your point is well taken, questions about negative denominators are much like questions about unreduced fractions. These things exist and we should know how to work with them and perhaps even understand their value. Reduced isn't always better. Slice a pizza into 16 pieces, there's definitely a sense in which it's much easier to manage to eat $\frac{8}{16}$ of that pizza than $\frac{1}{2}$ of a pizza only sliced in half. Same quantity of food, different approach to eating. $\endgroup$ – A.Ellett Sep 30 '16 at 1:33

I'm actually on the student's side here and would say that we never need fractions such as $\dfrac{3}{-5}$.

Like @JohnColeman said, it is more flexible to allow examples such as your first one, but I don't remember seeing anyone use these beyond middle school, except to demonstrate a specific point in a calculation.

And I think your second example is misleading, since it's confusing two issues. When we have a rational algebraic expression, the denominator can surely turn out negative, but when you plug in the numbers and write it as a numeric fraction, you wouldn't usually write it with the minus in the denominator, just like you wouldn't usually write the fraction in a non-reduced form.

So, to sum up, in my opinion, there is never a need for these fractions except to direct the readers' gaze to something meta-mathematical. I think a similar example is with writing a decimal with trailing zeros in the sciences (e.g. 13.20mL), where the trailing zero has no mathematical meaning, but is used to demonstrate the precision of the measurement.

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    $\begingroup$ As an applied mathematician who's taught at the high school and college level, and tutored students from 7th grade through graduate school, the point of a trailing zero is noteworthy (and of mathematical interest). If we write $13.2mL$ this could be the result of rounding $13.24mL$ to one decimal place. But, $13.20mL$ could never be the result of such rounding. And, in high school and earlier, this is an important matter for the students to be able to comprehend. I have asked on quizzes for my students to explain why $0.5$ (inexact answer) and $1/2$ (exact answer) may not be equal. $\endgroup$ – A.Ellett Sep 2 '16 at 22:16
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    $\begingroup$ @A.Ellett, I think that your comment is actually saying exactly what I meant but with a disagreement of semantics. I think we both agree that 0.5 and 1/2 are the same exact number, but the measurements they represent might be different - you say that the difference is of "mathematical interest", while I consider it to be only of meta-mathematical interest. $\endgroup$ – yoniLavi Sep 3 '16 at 20:53
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    $\begingroup$ (Part 1) Not to quibble, but I disagree that $0.5$ and $1/2$ are the same exact number. Firstly, a decimal is never an exact numerical representation for the simple reason that we do not know how we got it, whether it was rounded or computed. A fraction is an exact number; its value is unquestionable. Philosophically I'm a bit of a constructivist. While that may seem extremely ivory tower, it's quite relevant in the context of students who think decimals and the numbers their calculators spit out are "exact values", which they aren't. $\endgroup$ – A.Ellett Sep 4 '16 at 4:29
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    $\begingroup$ (Part 2) The "mathematical interest" is in the context of high school where students should be acquiring an understanding of the significance of the different ways we represent numerical quantities. Students struggle to understand that a letter used to represent a given quantity is not therefore a variable, but a constant. Students often fail to grasp that, if $k$ is a fixed value, then $k^2+2k+1$ is not an expression of unknown value which still must be determined, rather it's a well-defined number. $\endgroup$ – A.Ellett Sep 4 '16 at 4:49

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