Question 309760
what is one possible value of x for which  x <  2  <  1/x  ?
<pre><font size = 3><b>
{{{1/3}}} is a possible value for x because

{{{1/3 <  2  <  1/(1/3)}}}

Inverting and multiplying that last expression,

{{{1/3 <  2  <  1*(3/1)}}}

or

{{{1/3 <  2  <  3}}}

and that is true!

Other possible values of x would be {{{1/4}}},{{{1/5}}},{{{1/6}}}, etc. 

How did I get that?

Greater than zero means "positive"
Less than zero means "negative"

We try Case 1:  {{{x > 0}}}  (which means x is a positive number)

{{{x>0}}}{{{AND}}}{{{ x <  2  <  1/x}}}

We multiply through the right inequality by positive number x
and the inequality sign does not reverse:

{{{x>0}}}{{{AND}}}{{{ x^2 <  2x  <  1}}}

{{{x>0}}}{{{AND}}}{{{x^2 < 2x < 1}}}

{{{x>0}}}{{{AND}}}{{{x^2<2x}}}{{{AND}}}{{{2x<1}}}

{{{x>0}}}{{{AND}}}{{{x(x-2)<0}}}{{{AND}}}{{{x<1/2}}}

Since x is a positive number in this case, and the product
{{{x(x-2)}}} is negative, and since a positive number must 
be multiplied by a negative number in order to get a negative 
number, then {{{x-2)}}} must be a negative number, so

{{{x>0}}}{{{AND}}}{{{x-2<0}}}{{{AND}}}{{{x<1/2}}}

{{{x>0}}}{{{AND}}}{{{x<2}}}{{{AND}}}{{{x<1/2}}}

In order for a positive number to be both less than 2
and also less than {{{1/2}}}, it needs to be less
than {{{1/2}}}

So therefore:

{{{0<x<1/2}}}

We try Case 2:  {{{x < 0}}}  (which means x is a negative number)

{{{x<0}}}{{{AND}}}{{{ x <  2  <  1/x}}}

If we multiply the second inequality by x, which in this case is 
negative, we reverse the inequality symbols:


{{{x<0}}}{{{AND}}}{{{x^2 > 2x > 1}}}

{{{x<0}}}{{{AND}}}{{{x^2>2x}}}{{{AND}}}{{{2x<1}}}

{{{x<0}}}{{{AND}}}{{{x^2-2x<0}}}{{{AND}}}{{{x<1/2}}}

{{{x<0}}}{{{AND}}}{{{x(x-2)<0}}}{{{AND}}}{{{x<1/2}}}

Since x is a negative number in this case, and the product
{{{x(x-2)}}} is negative, and since a negative number must 
be multiplied by a positive number in order to get a negative 
number, then {{{x-2)}}} must be a positive number. But then
{{{x-2>0}}} implies {{{x>2}}} with contradicts {{{x<0}}}.

So Case 2 is impossible.

We try Case 3:  x = 0

This is impossible since {{{1/x}}} would not be defined.

Therefore only Case 1 is possible, so the solution is

{{{0<x<1/2}}}

and the solution set is {{{"("}}}{{{0}}}{{{","}}}{{{1/2}}}{{{")"}}}

So x can be any positive value between 0 and {{{1/2}}}, exclusive of
{{{1/2}}}.  x could be any fraction whose numerator is less than half 
its denominator.  Or x could be any decimal whose tenths digit is less \
than 5, such as .437 or .33339, etc.

Edwin</pre>