Fractions are Hard!

4.4 Means and Mediants

The average of two numbers \(x\) and \(y\), also called the mean of the two numbers, is \(\dfrac{x+y}{2}\) and is a value sure to lie between \(x\) and \(y\). We can see this in a (two-dimensional) sandbox.


Make two piles of sand, each a unit wide, one \(x\) units high and the other \(y\) units high. Then level out the sand. The result is a pile of some height between the two original heights. Since the total area of sand does not change, the final height is \(\dfrac{x+y}{2}\).



Challenge: Show that this result is still true if one of \(x\) or \(y\) or both is negative. (Or zero?)

Challenge: Draw (or just imagine) a picture to show why the average of ten numbers is sure to lie between the highest and lowest of those numbers.




Incorrectly adding two fractions this way gives what is called the mediant of the two fractions.


\(\dfrac{a}{b}+\dfrac{c}{d} \longrightarrow  \dfrac{a+c}{b+d}\).


The sandbox model shows that the value of the mediant is also sure to lie between the two original numbers.



Challenge: Does this result hold true if one or both fractions are negative or zero?

Challenge: Draw a picture to show that if \(0<\dfrac{a_{1}}{b_{1}}<\dfrac{a_{2}}{b_{2}}<\cdots\dfrac{a_{n}}{b_{n}}\),


\(\dfrac{a_{1}}{b_{1}}<\dfrac{a_{1}+a_{2}+\cdots +a_{n}}{b_{1}+b_{2}+\cdots +b_{n}}<\dfrac{a_{n}}{b_{n}}\).


(Is the condition that the fractions be positive necessary?)


There is a useful interpretation of the mediant of two fractions: If a team of \(a\) girls has \(b\) pizzas to share (thus each girl is currently entitled to amount \(\dfrac{a}{b}\) of pizza) meets up with a team of \(c\) boys with \(d\) pizzas to share (thus each boy is currently entitled to amount \(\dfrac{c}{d}\) of pizza), then, as a big group, each person is entitled to amount \(\dfrac{a+c}{b+d}\) of pizza. (And think about what this means physically: this number can’t be worse than the smaller of \(\dfrac{a}{b}\) and \(\dfrac{c}{d}\), nor can it be better than the biggest of \(\dfrac{a}{b}\) and \(\dfrac{c}{d}\).


CHALLENGE: Suppose \(w_{1}\) and \(w_{2}\) are two positive numbers that add to \(1\). Then \(w_{1}x+w_{2}y\) is a weighted average of \(x\) and \(y\). (For \(w_{1}=w_{2}=\frac{1}{2}\) we get the usual average.) Is there a sandbox picture that proves that this weighted average of \(x\) and \(y\) is sure to lie between the two numbers?

Is the weighted mediant \(\dfrac{w_{1}a+w_{2}c}{w_{1}b+w_{2}d}\) of \(\dfrac{a}{b}\) and \(\dfrac{c}{d}\) sure to lie between the two original fractions? (If so, a picture to prove it?)

Do any results, and picture proofs, you obtain here extend to three or more terms?


Mediants do appear in solutions to problems.


It is said that swans mate for life. (Is this true?)

In a flock of swans two-thirds of the males are bonded to five-eighths of the females. What fraction of the entire flock is bonded to a partner? 


Represent the fraction of bonded males and bonded females each pictorially.


Actually any multiple of these diagrams depicts the same proportion of bonded males and bonded females.


But since each male is matched with one female, and vice versa, a diagram that shows this matching must have an equal number of green dots on each line. This suggests we should think \(\dfrac{10}{15}\) of the males and \(\dfrac{10}{16}\) of the females.



It is now clear, for the entire adult population, 20 out of every 31 swans are bonded. And notice that \(\dfrac{20}{31}\) is the mediant of \(\dfrac{10}{15}\) and \(\dfrac{10}{16}\).


EXERCISE: More generally, suppose \(\dfrac{a}{b}\) of the male swans are bonded to \(\dfrac{c}{d}\) of the female swans. Show that the fraction of bonded swans is found by writing these fractions with a common numerator and then taking the mediant of those two fractions.  (Where else in arithmetic is it helpful to find a common numerator?)


For an application of mediants to the study of FORD CIRCLES see my April 2012 COOL MATH ESSAY.

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