SOLUTION: what is one possible value of x for which x < 2 < 1/x ?

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Question 309760: what is one possible value of x for which x < 2 < 1/x ?
Found 2 solutions by Edwin McCravy, palanisamy:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
what is one possible value of x for which x < 2 < 1/x ?

1%2F3 is a possible value for x because

1%2F3+%3C++2++%3C++1%2F%281%2F3%29

Inverting and multiplying that last expression,

1%2F3+%3C++2++%3C++1%2A%283%2F1%29

or

1%2F3+%3C++2++%3C++3

and that is true!

Other possible values of x would be 1%2F4,1%2F5,1%2F6, etc. 

How did I get that?

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

We try Case 1:  x+%3E+0  (which means x is a positive number)

x%3E0AND+x+%3C++2++%3C++1%2Fx

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

x%3E0AND+x%5E2+%3C++2x++%3C++1

x%3E0ANDx%5E2+%3C+2x+%3C+1

x%3E0ANDx%5E2%3C2xAND2x%3C1

x%3E0ANDx%28x-2%29%3C0ANDx%3C1%2F2

Since x is a positive number in this case, and the product
x%28x-2%29 is negative, and since a positive number must 
be multiplied by a negative number in order to get a negative 
number, then x-2%29 must be a negative number, so

x%3E0ANDx-2%3C0ANDx%3C1%2F2

x%3E0ANDx%3C2ANDx%3C1%2F2

In order for a positive number to be both less than 2
and also less than 1%2F2, it needs to be less
than 1%2F2

So therefore:

0%3Cx%3C1%2F2

We try Case 2:  x+%3C+0  (which means x is a negative number)

x%3C0AND+x+%3C++2++%3C++1%2Fx

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


x%3C0ANDx%5E2+%3E+2x+%3E+1

x%3C0ANDx%5E2%3E2xAND2x%3C1

x%3C0ANDx%5E2-2x%3C0ANDx%3C1%2F2

x%3C0ANDx%28x-2%29%3C0ANDx%3C1%2F2

Since x is a negative number in this case, and the product
x%28x-2%29 is negative, and since a negative number must 
be multiplied by a positive number in order to get a negative 
number, then x-2%29 must be a positive number. But then
x-2%3E0 implies x%3E2 with contradicts x%3C0.

So Case 2 is impossible.

We try Case 3:  x = 0

This is impossible since 1%2Fx would not be defined.

Therefore only Case 1 is possible, so the solution is

0%3Cx%3C1%2F2

and the solution set is %22%28%220%22%2C%221%2F2%22%29%22

So x can be any positive value between 0 and 1%2F2, exclusive of
1%2F2.  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

Answer by palanisamy(496) About Me  (Show Source):
You can put this solution on YOUR website!
x< 2 < 1/x
Put x = 1/4
Then,
1/4< 2 < 4
The given inequalities are satisfied.