# Mediant Fractions

The mediant fraction of two given fractions a1/b1 and a2/b2 is defined as (a1 + a2)/(b1 + b2),which should not be confused with the addition of fractions. Mediant fractions have appeared in the construction of the Farey series and the Stern-Brocot tree where they were proven to possess a couple of characteristic number-theoretic properties. Mediant fractions are known to lie between their progenitors, viz., for a1/b1 < a2/b2,

 (1) a1/b1 < (a1 + a2)/(b1 + b2) < a2/b2

(For a visual proof check one of Kevin Brown's pages.) Use (1) as a clue to solve the following

### Problem

Let m and n be two coprime (positive) integers. Consider the set A of all the fractions i/m, i = 1, ..., m-1, and j/n, j = 1, ..., n-1 in the unit interval [0,1]. Fractions k/(n+m), k = 0, 1, ..., (n+m) divide the interval [0,1] into (n+m) smaller subintervals. The question is about the distribution of points from the set A among those intervals. Since 1/m > 1/(m+n) and 1/n > 1/(m+n), the first and the last of the subintervals contains no point from A. Prove that all other intervals contain exactly 1 point each.

Solution (1) is not a very strong property. E.g., the average of two fractions, besides lying between the two fractions, possesses the property of monotonicity:

If A1 < A2 and B1 < B2, then (A1 + B1)/2 < (A2 + B2)/2

That mediants lack that property leads to some unexpected results. The following is known as Simpson's paradox although it contains no logical contradiction. Envisage three tables with a pair of red and blue pots on top of each. On the first table, the red pot contains 5 black and 6 white balls. The blue pot contains 3 black and 4 white balls, respectively. The distribution of balls on the other tables is presented in the diagram.

You are given one chance to extract a black ball from the pots on the first table. Which pot will you choose? Well, the probability of picking the black ball from the red pot is 5/11 while that for the blue pot is 3/7. Since 5/11 > 3/7 you, of course, choose the red pot.

Next, approach the second table with the same task. You again will choose the red pot because 6/9 > 9/14. Now to the interesting part. Note that, on the third table, the pots contain the total of balls collected from the pots of the same color from the first two tables. For instance, the red pot contains 11 black balls (5 came from the red pot on the first and 6 from the red pot on the second table.) The task is the same. Which pot will you choose? Surprisingly, selecting from the blue pot gives you a better chance of picking up the black ball. The direction of the inequality is now reversed: 11/20 < 12/21, as you can easily verify.

In statistics, unless used judiciously, Simpson's paradox may wreak havoc. Let, for example, give a different interpretation of the above example. Two studies (= the two left tables) have been performed to evaluate first year student enrollment in engineering (red color) and liberal arts (blue color) colleges depending on whether they took (first number, in thousands) or did not take (the second number, in thousands) AP classes in school.

The two studies tell us that the proportion of students who took AP classes is greater in engineering than in liberal arts colleges. However, if we combine the results of the two studies, the conclusion is different: the proportion of the AP takers is greater in liberal arts than in engineering colleges.

Here's another example, now from consumer economics. Two different, say A and B, items could be purchased at two stores, about which it is known that the average cost of one item to the consumer in the first store is greater than that in the second store. Which store would you go shopping at? Let's take up some numerics.

Over a period of time, the first store sells on average, 20 items A at \$3 per item and 80 items B at \$5 per item. In the second store, the numbers are: 80 items A @ \$4 and 20 items B @ 6\$. The average item price at the first store is (\$3·20 + \$5·80)/100 = \$4.6. In the second store, the number is (\$4·80 + \$6·80)/100 = \$4.4. However, whichever the item one is looking for, it would be stupid to go shopping at the second store, as the per item prices are higher there for both items A and B.

## References

1. R. Falk, Understanding Probability and Statistics, A K Peters, 1993
2. M. Gardner, Time Travel And Other Mathematical Bewilderments, W.H.Freeman and Co, 1988
3. J. A. Paulos, A Mathematician Read the Newspaper, Anchor Books, 1995 (You can find a more comprehensive discussion of this problem in the Proofs section of the site.)

First of all, note that since m and n are assumed to be coprime, all the points at hand are distinct. So we should not worry about points from A lying at the end points of the given subintervals.

Because of the inequalities 1/m > 1/(m+n) and 1/n > 1/(m+n) no two points from the set A that have the same denominator can lie in a single interval of length 1/(n+m). Assume then that r/m and s/n do. Then, for some t,

t/(n + m) < r/m < (t +1)/(n + m) and t/(n + m) < s/n < (t +1)/(n + m)

But then from (1) the same two inequalities hold for the mediant fraction (r + s)/(m + n) which is an obvious contradiction. • 