Question 697295
<pre>
Here is the correct solution:

First we must prove another theorem:

The reciprocals of two given unequal positive numbers are
positive numbers unequal in the opposite order from the 
two given positive numbers.

[Maybe you've already proved this, but in case you haven't,
here is the proof of that]:

Proof:

Suppose we are given:

0 < x < y  and we are to prove that 0 < {{{1/y}}} < {{{1/x}}}

Starting with

0 < x < y

divide all three sides by y:

{{{0/y}}} < {{{x/y}}} < {{{y/y}}}

Simplify:

0 < {{{x/y}}} < 1

Divide all three sides by x:

{{{0/x}}} < {{{expr(x/y)}}}÷{{{x}}} < {{{1/x}}}
 
0 < {{{expr(x/y)}}}÷{{{x/1}}} < {{{1/x}}} 

0 < {{{x/y}}}·{{{1/x}}} < {{{1/x}}}

0 < {{{cross(x)/y}}}·{{{1/cross(x)}}} < {{{1/x}}}

0 < {{{1/y}}} < {{{1/x}}}

----------------------------------

Now that we have proved that, we can turn to your
problem:

4 < a < 7 < b < 9

We are given 

      4 < a  

Divide both sides by 9

(1)   {{{4/9}}} < {{{a/9}}} 

We are given 

      b < 9  

By the theorem above:

     {{{1/9}}} < {{{1/b}}}

Multiply both sides by "a":

(2)   {{{a/9}}} < {{{a/b}}}

 
We are given 

      7 < b  

By the theorem above:

     {{{1/b}}} < {{{1/7}}}

Multiply both sides by "a":

(3)   {{{a/b}}} < {{{a/7}}}

We are given 

      a < 7  

Divide both sides by 7

(4)   {{{a/7}}} < {{{7/7}}} 

      {{{a/7}}} < 1

Putting (1), (2), (3) and (4) together:

      {{{4/9}}} < {{{a/b}}} < {{{a/7}}} < 1

So omit {{{a/7}}} and we have

      {{{4/9}}} < {{{a/b}}} < 1

as the only one of the inequalities given that 
we can be sure of.

Edwin</pre>