What number is between 1/2 and 8/9?

My answer to a question on Quora: What number is between 1/2 and 8/9?

John K Williamsson gave a good answer: for what he called the “dirty sum” of \(\frac{1}{2}\) and \( \frac{8}{9}\) (the mathematical term for that is mediant):

\(\displaystyle{\frac{1}{2} < \frac{1+8}{2+9} < \frac{8}{9}}\)

He suggests to use algebra to prove the mediant inequality: if \(a,b,c,d\) are positive numbers and

\(\displaystyle{\frac{a}{b} < \frac{c}{d}}\)

then

\(\displaystyle{\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}}.\)

I wish to add that, at the primary school level, the mediant inequality does not need an algebraic proof, it is sufficiently self-evident.

Indeed consider fractions \(\frac{1}{2}\) and \(\frac{8}{9}\) as descriptions of real-life situations:

\(\frac{1}{2}\) : \(2\) children have \(1\) bag of fruits.

\(\frac{8}{9}\) : \(9\) children have \(8\) bags of fruits.

They come together and share equally: \(1 + 8\) bags of fruit between \(2+9 = 11\) kids, that is, they form the mediant:

\(\displaystyle{\frac{1+8}{2+9}}\)

In this sharing, which group of kids looses and and which one gains? Of course \(2\) children with \(1\) bag gain: they have \(\frac{1}{2}\) bags per head, the other group comes with bigger share per head: \(\frac{8}{9}\). For the same reason, kids in the second group lose.

I use an example with kids and bags of sweets in my lectures; here I replaced sweets by more politically correct fruits — perhaps I have to go further and use green vegetables in place of fruits. The original idea belonged to the great Israel Gelfand, and was stated in a more colorful language:

You can explain mathematics to everyone, even to drunkards. If you ask some people drinking vodka on a park bench, what is is bigger, \(\frac{2}{3}\) or \(\frac{3}{4}\), they will respond with expletives. But if you ask them, what is better, \(2\) bottles of vodka for \(3 \) people or \(3\) bottles of vodka for \(4 \) people, they will instantly give you the right answer: of course, \(3\) bottles for \(4\) people.

And this instant conclusion comes from an argument which is the reversal of the informal proof of the mediant inequality: how to get from the situation “\(2\) bottles for \(3\) people” to the situation “\(3\) bottles for \(4\) people”? Of course, it means that a fourth man comes and brings with him a whole bottle — can you imagine, a whole bottle of vodka! In the mediant inequality,

\(\displaystyle{\frac{2}{3} < \frac{2+1}{3+1} < \frac{1}{1}},\)

or

\(\displaystyle{ \frac{2}{3} < \frac{3}{4} < 1.}\)

I have seen some papers which confirm that this is a typical pattern of arithmetical thinking, as done by “normal” people in real life situations (for example, I have seen a claim that it is used by hospital nurses for comparing doses of medication, which one is bigger and which one is smaller).