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Fractions in Ascending Order

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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.
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```
Date: 08/04/2008 at 21:39:20
From: Doctor Ian
Subject: Re: Fractions in Ascending Order

Hi David,

Here's the situation we're given:

a/b < c/d < 1           Integer a, b, c, d > 0

To keep from getting lost, let's make a table where we can keep track of
the comparisons we make.  We'll give each fraction a label (I through V),
and write the larger of the fractions at the intersection:

bd       b+d
b/a      d/c      --       ---        1
ac       a+c

(I)   b/a      -        -        -        -         -

(II)  d/c               -        -        -         -

bd
(III) --                         -        -         -
ac

b+d
(IV)  ---                                  -        -
a+c

(V)   1                                            -

We can multiply our original inequality by (bd)/(ac) to get

abd   cbd
--- < --- < bd
bac   dac

which simplifies to

d     b    bd
--- < --- < --
c     a    ac

II    I    III

Let's add that information to our table:

bd       b+d
b/a      d/c      --       ---        1
ac       a+c

(I)   b/a      -        -        -        -         -

(II)  d/c      I        -        -        -         -

bd
(III) --      III      III       -        -         -
ac

b+d
(IV)  ---                                  -        -
a+c

(V)   1                                            -

If a/b is less than 1, then b/a must be greater than 1.  Similarly for d/c.
And (bd)/(ac) is greater than both of those, so it must also be greater than 1:

bd       b+d
b/a      d/c      --       ---        1
ac       a+c

(I)   b/a      -        -        -        -         -

(II)  d/c      I        -        -        -         -

bd
(III) --      III      III       -        -         -
ac

b+d
(IV)  ---                                  -        -
a+c

(V)   1       I       II       III                 -

How can we compare b/a and (b+d)/(a+c)?  Well, we can look at the ratio,
and see how it compares to 1.  (If the ratio is less than 1, we know the
numerator is smaller; if greater than 1, the numerator is larger.)

b/a     b   a+c   ab + bc
------- = - * --- = -------
b+d     a   b+d   ab + ad
---
a+c

Now, we're golden if we know which is larger, bc or ad.  Let's go back to
our original condition:

a   c
- < - < 1
b   d

We can multiply by bd, to get

Since ad < bc, it must be true that

ab + bc
------- < 1

which means that b/a < (b+d)/(a+c).  We can use similar reasoning to show
that d/c < (b+d)/(a+c).  And since both b/a and d/c are greater than 1, we can
conclude that (b+d)/(a+c) is, too:

bd       b+d
b/a      d/c      --       ---        1
ac       a+c

(I)   b/a      -        -        -        -         -

(II)  d/c      I        -        -        -         -

bd
(III) --      III      III       -        -         -
ac

b+d
(IV)  ---     IV       IV                 -        -
a+c

(V)   1       I       II       III      IV        -

That leaves us with just one comparison to do.  If we try some examples, e.g.,

bd   3*2            b+d   3+2   5
1/3 < 1/2 < 1     -->     -- = --- = 6        --- = --- = -
ac   1*1            a+c   1+1   2

bd   3*5   15       b+d   3+5   8
2/3 < 4/5 < 1     -->     -- = --- = --       --- = --- = -
ac   2*4    8       a+c   2+4   6

it suggests that

bd   b+d
-- > ---
ac   a+c

but suggestion isn't proof.

A ratio is one way to compare two quantities.  Another is to subtract them,
and see if the difference is less than or greater than zero.  Let's try that,
and see if we can show that the difference must be positive:

bd   b+d
0 < -- - ---
ac   a+c

bd(a+c)   ac(b+d)
0 < ------- - -------
ac(a+c)   ac(a+c)

(abd + bcd) - (abc + acd)
0 < -------------------------
ac(a+c)

0 < (abd + bcd) - (abc + acd)

0 < abd + bcd - abc - acd

0 < ab(d - c) + cd(b - a)

But we know that c < d, so (d - c) is positive; and similarly, (b - a)
is positive.  So this is always true, and that means that

(bd)/(ac) > (b+d)/(a+c)

So that completes our table:

(I)      (II)   (III)     (IV)       (V)

bd       b+d
b/a      d/c      --       ---        1
ac       a+c

(I)   b/a      -        -        -        -         -

(II)  d/c      I        -        -        -         -

bd
(III) --      III      III       -        -         -
ac

b+d
(IV)  ---     IV       IV       III       -        -
a+c

(V)   1       I       II       III      IV        -

And now we have enough information to order everything:

I > II, V

II > V

III > I, II, IV, V

IV > I, II, V

V > nothing

So we have

III > IV > I > II > V

or

bd     b+d
--  >  ---  >  b/a  >  d/c  > 1
ac     a+c

Just to bolster our confidence, let's check that with an example again:

2/3 < 4/5 < 1,    a=2, b=3, c=4, d=5

That gives us

3*5     3+5     3     5
---  >  ---  >  -  >  -  >  1
2*4     2+4     2     4

15     8     3     5
--  >  -  >  -  >  -  >  1
8     6     2     4

15     8     3     5
--  >  -  >  -  >  -  >  1
8     6     2     4

1.875  >  1.333  >  1.5  >  1.25  >  1

And there ya go!

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Analysis

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