# Why was I taught to convert "improper fractions" into mixed numbers?

Thinking back, a significant part of my middle school math education was spent converting "improper fractions" such as 9/8 into "mixed numbers such as 1 1/8. This went beyond understanding that a fraction is just like any other number: we were told that you should never write fractions with a greater numerator than denominator. Since then, I have not used a "mixed number" once in any sort of serious math. I am around 2 years in to a math heavy college degree and every single time some number of the form a/b where a>b showed up, it was written as a/b or as a decimal. Maybe it has some use in higher mathematics, but I am wondering what the Chicago Math people were thinking when they spent so much tie on this weird way to think about fractions. What use is there for teaching "improper fractions" as a core part of a middle school math curriculum?

• Short answer: It's easier than determining decimal representation, but it still helps you estimate (e.g. $149/150$ is a tiny bit harder to process than $2\,\frac{49}{50}$). It also provides a motivation for learning about prime numbers. Finally, you need mixed numbers for baking. ;) Commented Apr 11, 2017 at 23:42
• (I'm guessing @apnorton meant $149/50$, not $149/150$) in his/her comment, providing even more evidence for the utility of mixed numbers.) Commented Apr 12, 2017 at 0:03
• I did indeed, @JasonDeVito. Whoops. Commented Apr 12, 2017 at 0:05

It's useful in everyday life. Most people will only ever come across fractions when dividing objects between groups, and then it is useful. For example, if you have to split 16 things between 5 people you do $\frac{16}{5} = 3\frac{1}{5}$ and you know you'll have three each plus one left over. This is much easier (and practically more useful) than actually doing the division to find the decimal expansion of $\frac{16}{5} = 3.2$.
Think of it as the first step in finding the continued fraction of your number. If your beginning number is rational, this leads to finding the gcd of numerator and denominator. Also, if $\gcd(p,q) = 1,$ it leads to solving $px-qy=1$
This is $\sqrt 2,$ for which the continued fraction is infinite. $$\small \begin{array}{cccccccccccccccccccccccccccccc} & & 1 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{1}{1} & & \frac{3}{2} & & \frac{7}{5} & & \frac{17}{12} & & \frac{41}{29} & & \frac{99}{70} & & \frac{239}{169} & & \frac{577}{408} & & \frac{1393}{985} & & \frac{3363}{2378} & & \frac{8119}{5741} \\ \\ & 1 & & -1 & & 1 & & -1 & & 1 & & -1 & & 1 & & -1 & & 1 & & -1 & & 1 & & -1 \end{array}$$