Fractions in Ascending Order

Date: 03/22/99 at 21:39:20
From: Anonymous
Subject: Fractions in Ascending Order

My question is:

If a, b, c, and d are positive integers that satisfy a/b < c/d < 1,
arrange the 5 quantities b/a, d/c, bd/ac, b+d/a+c, 1 in ascending
order.

Date: 03/22/99 at 21:39:20
From: Doctor Mitteldorf
Subject: Re: Fractions in Ascending Order

We just have to compare these two-at-a-time until we have enough 
information to put them in order.  

Start with b/a > d/c because the reciprocals of these two numbers are 
in the opposite relation.

   b/a must be > 1, and d/c must be > 1, so that multiplying these two 
   together, you get a quantity bigger than either of them. So, 
   bd/ac > b/a, which is the larger of the two.

   So far, we have 1  <  d/c  <  b/a  <  bd/ac

Finally, where does (b+d)/(a+c) fit in? Let us use an example: if 
d/c = 5/4 and b/a = 3/2, then (3 + 5)/(2 + 4) = 8/6 = 4/3, which is 
smaller than d/c, but greater than 1.

Let us see if we can prove this: first, that it is greater than 1.  
That is easy: if b>a and d>c, then (b + d) has to be bigger than 
(a + c). 

Second, that it is less than d/c. Starting with d/c, to get to a 
denominator of (a + c) we must divide by (a + c)/c = 1 + a/c. 
To get the numerator from d to (b + d), we must multiply by 
(b + d)/d = 1 + b/d. The quantity we multiplied by is less than the 
quantity we divided by because a/c < b/d. (This is a re-writing of the 
original statement a/b < c/d, with each side multiplied by b/c.)  

So, we have shown that we can multiply d/c by a number < 1 and get 
(b + d)/(a + c). Therefore, the complete ordering is:

1  <  (b + d)/(a + c)  <  d/c  <  b/a  <  bd/ac

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