Thinking about the Mediant of Two Fractions

Date: 12/10/2009 at 23:24:08
From: Teresa
Subject: new fraction formula

Students were trying to find a fraction between two given fractions 
with 1 as the numerators (such as 1/2 and 1/3).  After working on 
several problems, one person noticed that when given 1/a and 1/b, the 
fraction 2/(a+b) always seems to be in between 1/a and 1/b.  Is this 
a new fraction rule, or has this been discovered and proved before?

I created the following "proof" but would like it to be checked, 
since it has been many years since I've tried to prove a new idea:

If a and b are positive integers and a > b, then would 

         1/a           <     2/(a+b)     <       1/b?

    [b(a+b)]/[ab(a+b)] < 2ab / [ab(a+b)] < [a(a+b)]/[ab(a+b)],
                b(a+b) <      2ab        < a(a+b)
                b(a+b) < 2ab   and   2ab < a(a+b)
                   a+b < 2a    and    2b < a+b
                     b < a     and     b < a,

so it is true?

Date: 12/11/2009 at 10:46:42
From: Doctor Peterson
Subject: Re: new fraction formula

Hi, Teresa.

Yes, this is true.  In fact, more generally, given ANY two fractions 
a/b and c/d (with a, b, c, and d positive numbers), the fraction 
(a+c)/(b+d), sometimes called the mediant, is between a/b and c/d.

Your proof contains all the right elements, but just needs to be 
rearranged or explained more fully.  What you've done is to show that 
IF 2/(a+b) is between 1/a and 1/b, THEN b < a.  You want to show the 
converse of this.  Since all the statements are equivalent to one 
another, you could essentially just reverse the order of your 
statements and it constitutes the proof you want.  It would read a 
little oddly that way, though, so it needs to be explained carefully.

Let's see if I can write a cleaner version of your proof for the 
general case.

Given positive integers a, b, c, and d, such that a/b < c/d, we want 
to prove that a/b < (a+c)/(b+d) < c/d. 

The given condition, that a/b < c/d, is equivalent to

  ad   bc
  -- < --, and therefore to ad < bc
  bd   bd

What we want to show, writing all the fractions with a common 
denominator, is

  ad(b+d)   bd(a+c)   bc(b+d)
  ------- < ------- < -------
  bd(b+d)   bd(b+d)   bd(b+d)

This is equivalent to showing that the numerators fall in the same 
order:

  ad(b+d) < bd(a+c) < bc(b+d)

or

  abd + ad^2 < abd + bcd < b^2c + bcd

We will take this in two parts:

  abd + ad^2 < abd + bcd, and abd + bcd < b^2c + bcd

The first desired inequality is true because

  ad < bc
  ad^2 < bcd
  abd + ad^2 < abd + bcd

The second is true because

  ad < bc
  abd < b^2c
  abd + bcd < b^2c + bcd

We're done.  (This could be written without working backward from the 
goal, but it would be very unclear why we did what we did.)

There is also an interesting visual "proof" of this.  Interpret the 
fractions as slopes, and consider any two points A (b,a) and D (d,c) 
in the first quadrant:

     ^
     |
  a+c|                  ____---C
     |          ____----    . /
    c|      B---         .   /
     |     /          .     /
     |    /        .       /
     |   /      .         /
     |  /    .           /
    a| /  .      ____---A
     |/. ____----
     O----------------------------->
            d           b     b+d

Here the slope of OA, a/b, is less than the slope of OB, c/d. When 
we add the two vectors A and B (that is, make a parallelogram OACB 
so that AC and OB have the same slope and length), C has coordinates 
(b+d, a+c), so that the slope of the diagonal OC is (a+c)/(b+d). 
Clearly this diagonal goes through the middle of the parallelogram, 
so that its slope is between those of OA and OB.

For more on this, including a more subtle proof, see:

  Wikipedia: Mediant (Mathematics)
    http://en.wikipedia.org/wiki/Mediant_(mathematics) 

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 12/11/2009 at 15:37:03
From: Teresa
Subject: Thank you (new fraction formula)

Thank you so much!  I am going to share your answer with the person 
who noticed the rule (who usually doesn't feel very good at math) and 
tell him that he can name it.  That way we can always call it by his 
name in our class!  Thanks again!
Teresa

Date: 12/11/2009 at 20:58:54
From: Doctor Peterson
Subject: Re: Thank you (new fraction formula)

Hi, Teresa.

I just realized that, although I focused on a more general rule that I
find interesting, your student's rule can be explained much more 
simply, which may even be what the student had in mind.

We're looking at two unit fractions (fractions with numerator 1), and
trying to find a fraction that is between them.  We know that if a >
b, then 1/a < 1/b.  If we pick any number c BETWEEN a and b, then

  1/a < 1/c < 1/b

In particular, if c is the average (mean) of a and b, this will be
true.  For example, if a is 5 and b is 3, the average, c, is 4, and we
know that

  1/5 < 1/4 < 1/3

But in the case you were looking at, 1/2 and 1/3, there is no WHOLE
number between 2 and 3.  Their average is (2+3)/2 = 5/2. If we ignore
the fact that this is not whole and just take its reciprocal (1/c), we
get the fraction 2/5. The reasoning that said that 1/c is between the
others is still valid, so we do know that

  1/3 < 2/5 < 1/2

as desired.

The general rule is that we can take c = (a+b)/2 and use its
reciprocal, 2/(a+b):

  1/a < 2/(a+b) < 1/b

So this is a very reasonable choice to have made, and its truth
follows from two elementary facts, namely that the average is between
the two numbers, and that increasing the denominator decreases the
fraction.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/